How can I create a zip archive of a directory structure in Python?
The easiest way is to use shutil.make_archive. It supports both zip and tar formats.
import shutil
shutil.make_archive(output_filename, 'zip', dir_name)
If you need to do something more complicated than zipping the whole directory (such as skipping certain files), then you'll need to dig into the zipfile module as others have suggested.
As others have pointed out, you should use zipfile. The documentation tells you what functions are available, but doesn't really explain how you can use them to zip an entire directory. I think it's easiest to explain with some example code:
import os
import zipfile
def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
for file in files:
ziph.write(os.path.join(root, file),
os.path.relpath(os.path.join(root, file),
os.path.join(path, '..')))
with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
zipdir('tmp/', zipf)
To add the contents of mydirectory to a new zip file, including all files and subdirectories:
import os
import zipfile
zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
zf.write(dirname)
for filename in files:
zf.write(os.path.join(dirname, filename))
zf.close()
How can I create a zip archive of a directory structure in Python?
In a Python script
In Python 2.7+, shutil has a make_archive function.
from shutil import make_archive
make_archive(
'zipfile_name',
'zip', # the archive format - or tar, bztar, gztar
root_dir=None, # root for archive - current working dir if None
base_dir=None) # start archiving from here - cwd if None too
Here the zipped archive will be named zipfile_name.zip. If base_dir is farther down from root_dir it will exclude files not in the base_dir, but still archive the files in the parent dirs up to the root_dir.
I did have an issue testing this on Cygwin with 2.7 - it wants a root_dir argument, for cwd:
make_archive('zipfile_name', 'zip', root_dir='.')
Using Python from the shell
You can do this with Python from the shell also using the zipfile module:
$ python -m zipfile -c zipname sourcedir
Where zipname is the name of the destination file you want (add .zip if you want it, it won't do it automatically) and sourcedir is the path to the directory.
Zipping up Python (or just don't want parent dir):
If you're trying to zip up a python package with a __init__.py and __main__.py, and you don't want the parent dir, it's
$ python -m zipfile -c zipname sourcedir/*
And
$ python zipname
would run the package. (Note that you can't run subpackages as the entry point from a zipped archive.)
Zipping a Python app:
If you have python3.5+, and specifically want to zip up a Python package, use zipapp:
$ python -m zipapp myapp
$ python myapp.pyz
This function will recursively zip up a directory tree, compressing the files, and recording the correct relative filenames in the archive. The archive entries are the same as those generated by zip -r output.zip source_dir.
import os
import zipfile
def make_zipfile(output_filename, source_dir):
relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
for root, dirs, files in os.walk(source_dir):
# add directory (needed for empty dirs)
zip.write(root, os.path.relpath(root, relroot))
for file in files:
filename = os.path.join(root, file)
if os.path.isfile(filename): # regular files only
arcname = os.path.join(os.path.relpath(root, relroot), file)
zip.write(filename, arcname)
Use shutil, which is part of python standard library set.
Using shutil is so simple(see code below):
1st arg: Filename of resultant zip/tar file,
2nd arg: zip/tar,
3rd arg: dir_name
Code:
import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
Modern Python (3.6+) using the pathlib module for concise OOP-like handling of paths, and pathlib.Path.rglob() for recursive globbing. As far as I can tell, this is equivalent to George V. Reilly's answer: zips with compression, the topmost element is a directory, keeps empty dirs, uses relative paths.
from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile
from os import PathLike
from typing import Union
def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
src_path = Path(source_dir).expanduser().resolve(strict=True)
with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
for file in src_path.rglob('*'):
zf.write(file, file.relative_to(src_path.parent))
Note: as optional type hints indicate, zip_name can't be a Path object (would be fixed in 3.6.2+).
With python 3.9, pathlib & zipfile module you can create a zip files from anywhere in the system.
def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
"""Zip the provided directory without navigating to that directory using `pathlib` module"""
# Convert to Path object
dir = Path(dir)
with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
for entry in dir.rglob("*"):
zip_file.write(entry, entry.relative_to(dir))
It is neat, typed, and has less code.
For adding compression to the resulting zip file, check out this link.
You need to change:
zip = zipfile.ZipFile('Python.zip', 'w')
to
zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
I've made some changes to code given by Mark Byers. Below function will also adds empty directories if you have them. Examples should make it more clear what is the path added to the zip.
#!/usr/bin/env python
import os
import zipfile
def addDirToZip(zipHandle, path, basePath=""):
"""
Adding directory given by \a path to opened zip file \a zipHandle
#param basePath path that will be removed from \a path when adding to archive
Examples:
# add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
zipHandle.close()
# add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir', 'dir')
zipHandle.close()
# add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
zipHandle.close()
# add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir')
zipHandle.close()
# add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir')
zipHandle.close()
# add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
addDirToZip(zipHandle, 'otherDir')
zipHandle.close()
"""
basePath = basePath.rstrip("\\/") + ""
basePath = basePath.rstrip("\\/")
for root, dirs, files in os.walk(path):
# add dir itself (needed for empty dirs
zipHandle.write(os.path.join(root, "."))
# add files
for file in files:
filePath = os.path.join(root, file)
inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
#print filePath + " , " + inZipPath
zipHandle.write(filePath, inZipPath)
Above is a simple function that should work for simple cases. You can find more elegant class in my Gist:
https://gist.github.com/Eccenux/17526123107ca0ac28e6
I have another code example that may help, using python3, pathlib and zipfile.
It should work in any OS.
from pathlib import Path
import zipfile
from datetime import datetime
DATE_FORMAT = '%y%m%d'
def date_str():
"""returns the today string year, month, day"""
return '{}'.format(datetime.now().strftime(DATE_FORMAT))
def zip_name(path):
"""returns the zip filename as string"""
cur_dir = Path(path).resolve()
parent_dir = cur_dir.parents[0]
zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
p_zip = Path(zip_filename)
n = 1
while p_zip.exists():
zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
date_str(), n))
p_zip = Path(zip_filename)
n += 1
return zip_filename
def all_files(path):
"""iterator returns all files and folders from path as absolute path string
"""
for child in Path(path).iterdir():
yield str(child)
if child.is_dir():
for grand_child in all_files(str(child)):
yield str(Path(grand_child))
def zip_dir(path):
"""generate a zip"""
zip_filename = zip_name(path)
zip_file = zipfile.ZipFile(zip_filename, 'w')
print('create:', zip_filename)
for file in all_files(path):
print('adding... ', file)
zip_file.write(file)
zip_file.close()
if __name__ == '__main__':
zip_dir('.')
print('end!')
To retain the folder hierarchy under the parent directory to be archived:
import glob
import os
import zipfile
with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
for fp in glob(os.path.join(parent, "**/*")):
base = os.path.commonpath([parent, fp])
zipf.write(fp, arcname=fp.replace(base, ""))
If you want, you could change this to use pathlib for file globbing.
If you want a functionality like the compress folder of any common graphical file manager you can use the following code, it uses the zipfile module. Using this code you will have the zip file with the path as its root folder.
import os
import zipfile
def zipdir(path, ziph):
# Iterate all the directories and files
for root, dirs, files in os.walk(path):
# Create a prefix variable with the folder structure inside the path folder.
# So if a file is at the path directory will be at the root directory of the zip file
# so the prefix will be empty. If the file belongs to a containing folder of path folder
# then the prefix will be that folder.
if root.replace(path,'') == '':
prefix = ''
else:
# Keep the folder structure after the path folder, append a '/' at the end
# and remome the first character, if it is a '/' in order to have a path like
# folder1/folder2/file.txt
prefix = root.replace(path, '') + '/'
if (prefix[0] == '/'):
prefix = prefix[1:]
for filename in files:
actual_file_path = root + '/' + filename
zipped_file_path = prefix + filename
zipf.write( actual_file_path, zipped_file_path)
zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()
So many answers here, and I hope I might contribute with my own version, which is based on the original answer (by the way), but with a more graphical perspective, also using context for each zipfile setup and sorting os.walk(), in order to have a ordered output.
Having these folders and them files (among other folders), I wanted to create a .zip for each cap_ folder:
$ tree -d
.
├── cap_01
| ├── 0101000001.json
| ├── 0101000002.json
| ├── 0101000003.json
|
├── cap_02
| ├── 0201000001.json
| ├── 0201000002.json
| ├── 0201001003.json
|
├── cap_03
| ├── 0301000001.json
| ├── 0301000002.json
| ├── 0301000003.json
|
├── docs
| ├── map.txt
| ├── main_data.xml
|
├── core_files
├── core_master
├── core_slave
Here's what I applied, with comments for better understanding of the process.
$ cat zip_cap_dirs.py
""" Zip 'cap_*' directories. """
import os
import zipfile as zf
for root, dirs, files in sorted(os.walk('.')):
if 'cap_' in root:
print(f"Compressing: {root}")
# Defining .zip name, according to Capítulo.
cap_dir_zip = '{}.zip'.format(root)
# Opening zipfile context for current root dir.
with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:
# Iterating over os.walk list of files for the current root dir.
for f in files:
# Defining relative path to files from current root dir.
f_path = os.path.join(root, f)
# Writing the file on the .zip file of the context
new_zip.write(f_path)
Basically, for each iteration over os.walk(path), I'm opening a context for zipfile setup and afterwards, iterating iterating over files, which is a list of files from root directory, forming the relative path for each file based on the current root directory, appending to the zipfile context which is running.
And the output is presented like this:
$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03
To see the contents of each .zip directory, you can use less command:
$ less cap_01.zip
Archive: cap_01.zip
Length Method Size Cmpr Date Time CRC-32 Name
-------- ------ ------- ---- ---------- ----- -------- ----
22017 Defl:N 2471 89% 2019-09-05 08:05 7a3b5ec6 cap_01/0101000001.json
21998 Defl:N 2471 89% 2019-09-05 08:05 155bece7 cap_01/0101000002.json
23236 Defl:N 2573 89% 2019-09-05 08:05 55fced20 cap_01/0101000003.json
-------- ------- --- -------
67251 7515 89% 3 files
You probably want to look at the zipfile module; there's documentation at http://docs.python.org/library/zipfile.html.
You may also want os.walk() to index the directory structure.
To give more flexibility, e.g. select directory/file by name use:
import os
import zipfile
def zipall(ob, path, rel=""):
basename = os.path.basename(path)
if os.path.isdir(path):
if rel == "":
rel = basename
ob.write(path, os.path.join(rel))
for root, dirs, files in os.walk(path):
for d in dirs:
zipall(ob, os.path.join(root, d), os.path.join(rel, d))
for f in files:
ob.write(os.path.join(root, f), os.path.join(rel, f))
break
elif os.path.isfile(path):
ob.write(path, os.path.join(rel, basename))
else:
pass
For a file tree:
.
├── dir
│ ├── dir2
│ │ └── file2.txt
│ ├── dir3
│ │ └── file3.txt
│ └── file.txt
├── dir4
│ ├── dir5
│ └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip
You can e.g. select only dir4 and root.txt:
cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]
with zipfile.ZipFile("selective.zip", "w" ) as myzip:
for f in files:
zipall(myzip, f)
Or just listdir in script invocation directory and add everything from there:
with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
for f in os.listdir():
if f == "listdir.zip":
# Creating a listdir.zip in the same directory
# will include listdir.zip inside itself, beware of this
continue
zipall(myzip, f)
Here is a variation on the answer given by Nux that works for me:
def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
basePath = os.path.split( srcPath )[ 0 ]
for root, dirs, files in os.walk( srcPath ):
p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
# add dir
zipHandle.write( root, p, zipOperation )
# add files
for f in files:
filePath = os.path.join( root, f )
fileInZipPath = os.path.join( p, f )
zipHandle.write( filePath, fileInZipPath, zipOperation )
Try the below one .it worked for me.
import zipfile, os
zipf = "compress.zip"
def main():
directory = r"Filepath"
toZip(directory)
def toZip(directory):
zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )
list = os.listdir(directory)
for file_list in list:
file_name = os.path.join(directory,file_list)
if os.path.isfile(file_name):
print file_name
zippedHelp.write(file_name)
else:
addFolderToZip(zippedHelp,file_list,directory)
print "---------------Directory Found-----------------------"
zippedHelp.close()
def addFolderToZip(zippedHelp,folder,directory):
path=os.path.join(directory,folder)
print path
file_list=os.listdir(path)
for file_name in file_list:
file_path=os.path.join(path,file_name)
if os.path.isfile(file_path):
zippedHelp.write(file_path)
elif os.path.isdir(file_name):
print "------------------sub directory found--------------------"
addFolderToZip(zippedHelp,file_name,path)
if __name__=="__main__":
main()
Say you want to Zip all the folders(sub directories) in the current directory.
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
Zip a file or a tree (a directory and its sub-directories).
from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED
def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
paths = [Path(tree_path)]
while paths:
p = paths.pop()
if p.is_dir():
paths.extend(p.iterdir())
if skip_empty_dir:
continue
zf.write(p)
To append to an existing archive, pass mode='a', to create a fresh archive mode='w' (the default in the above). So let's say you want to bundle 3 different directory trees under the same archive.
make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
A solution using pathlib.Path, which is independent of the OS used:
import zipfile
from pathlib import Path
def zip_dir(path: Path, zip_file_path: Path):
"""Zip all contents of path to zip_file"""
files_to_zip = [
file for file in path.glob('*') if file.is_file()]
with zipfile.ZipFile(
zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
for file in files_to_zip:
print(file.name)
zip_f.write(file, file.name)
current_dir = Path.cwd()
zip_dir = current_dir / "test"
tools.zip_dir(
zip_dir, current_dir / 'Zipped_dir.zip')
The obvious way to go would be to go with shutil, Like the second top answer says so, But if you still wish to go with ZipFile for some reason, And if you are getting some trouble doing that (Like ERR 13 in Windows etc), You can use this fix:
import os
import zipfile
def retrieve_file_paths(dirName):
filePaths = []
for root, directories, files in os.walk(dirName):
for filename in files:
filePath = os.path.join(root, filename)
filePaths.append(filePath)
return filePaths
def main(dir_name, output_filename):
filePaths = retrieve_file_paths(dir_name)
zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
main("my_dir", "my_dir_archived")
This one recursively iterates through every sub-folder/file in your given folder, And writes them to a zip file instead of attempting to directly zip a folder.
Here's a modern approach, using pathlib, and a context manager. Puts the files directly in the zip, rather than in a subfolder.
def zip_dir(filename: str, dir_to_zip: pathlib.Path):
with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
# Use glob instead of iterdir(), to cover all subdirectories.
for directory in dir_to_zip.glob('**'):
for file in directory.iterdir():
if not file.is_file():
continue
# Strip the first component, so we don't create an uneeded subdirectory
# containing everything.
zip_path = pathlib.Path(*file.parts[1:])
# Use a string, since zipfile doesn't support pathlib directly.
zipf.write(str(file), str(zip_path))
I prepared a function by consolidating Mark Byers' solution with Reimund and Morten Zilmer's comments (relative path and including empty directories). As a best practice, with is used in ZipFile's file construction.
The function also prepares a default zip file name with the zipped directory name and '.zip' extension. Therefore, it works with only one argument: the source directory to be zipped.
import os
import zipfile
def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
path_file_zip = os.path.join(
os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
for root, dirs, files in os.walk(path_dir):
for file_or_dir in files + dirs:
zip_file.write(
os.path.join(root, file_or_dir),
os.path.relpath(os.path.join(root, file_or_dir),
os.path.join(path_dir, os.path.pardir)))
# import required python modules
# You have to install zipfile package using pip install
import os,zipfile
# Change the directory where you want your new zip file to be
os.chdir('Type your destination')
# Create a new zipfile ( I called it myfile )
zf = zipfile.ZipFile('myfile.zip','w')
# os.walk gives a directory tree. Access the files using a for loop
for dirnames,folders,files in os.walk('Type your directory'):
zf.write('Type your Directory')
for file in files:
zf.write(os.path.join('Type your directory',file))
Well, after reading the suggestions I came up with a very similar way that works with 2.7.x without creating "funny" directory names (absolute-like names), and will only create the specified folder inside the zip.
Or just in case you needed your zip to contain a folder inside with the contents of the selected directory.
def zipDir( path, ziph ) :
"""
Inserts directory (path) into zipfile instance (ziph)
"""
for root, dirs, files in os.walk( path ) :
for file in files :
ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )
def makeZip( pathToFolder ) :
"""
Creates a zip file with the specified folder
"""
zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
zipDir( pathToFolder, zipf )
zipf.close()
print( "Zip file saved to: " + pathToFolder)
makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
Function to create zip file.
def CREATEZIPFILE(zipname, path):
#function to create a zip file
#Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file
zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
zipf.setpassword(b"password") #if you want to set password to zipfile
#checks if the path is file or directory
if os.path.isdir(path):
for files in os.listdir(path):
zipf.write(os.path.join(path, files), files)
elif os.path.isfile(path):
zipf.write(os.path.join(path), path)
zipf.close()
One thing completely missed by previous answers is the fact that using os.path.join() can easily return POSIX-incompatible paths when you run the code on Windows. The resulting archive will contain files with literal backslashes in their names when processing it with any common archive software on Linux, which is not what you want. Use path.as_posix() for the arcname parameter instead!
import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
for path in Path("include_all_of_this_folder").rglob("*"):
zf.write(path, path.as_posix())
Related
How can I create a zip archive of a directory structure in Python?
The easiest way is to use shutil.make_archive. It supports both zip and tar formats.
import shutil
shutil.make_archive(output_filename, 'zip', dir_name)
If you need to do something more complicated than zipping the whole directory (such as skipping certain files), then you'll need to dig into the zipfile module as others have suggested.
As others have pointed out, you should use zipfile. The documentation tells you what functions are available, but doesn't really explain how you can use them to zip an entire directory. I think it's easiest to explain with some example code:
import os
import zipfile
def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
for file in files:
ziph.write(os.path.join(root, file),
os.path.relpath(os.path.join(root, file),
os.path.join(path, '..')))
with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
zipdir('tmp/', zipf)
To add the contents of mydirectory to a new zip file, including all files and subdirectories:
import os
import zipfile
zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
zf.write(dirname)
for filename in files:
zf.write(os.path.join(dirname, filename))
zf.close()
How can I create a zip archive of a directory structure in Python?
In a Python script
In Python 2.7+, shutil has a make_archive function.
from shutil import make_archive
make_archive(
'zipfile_name',
'zip', # the archive format - or tar, bztar, gztar
root_dir=None, # root for archive - current working dir if None
base_dir=None) # start archiving from here - cwd if None too
Here the zipped archive will be named zipfile_name.zip. If base_dir is farther down from root_dir it will exclude files not in the base_dir, but still archive the files in the parent dirs up to the root_dir.
I did have an issue testing this on Cygwin with 2.7 - it wants a root_dir argument, for cwd:
make_archive('zipfile_name', 'zip', root_dir='.')
Using Python from the shell
You can do this with Python from the shell also using the zipfile module:
$ python -m zipfile -c zipname sourcedir
Where zipname is the name of the destination file you want (add .zip if you want it, it won't do it automatically) and sourcedir is the path to the directory.
Zipping up Python (or just don't want parent dir):
If you're trying to zip up a python package with a __init__.py and __main__.py, and you don't want the parent dir, it's
$ python -m zipfile -c zipname sourcedir/*
And
$ python zipname
would run the package. (Note that you can't run subpackages as the entry point from a zipped archive.)
Zipping a Python app:
If you have python3.5+, and specifically want to zip up a Python package, use zipapp:
$ python -m zipapp myapp
$ python myapp.pyz
This function will recursively zip up a directory tree, compressing the files, and recording the correct relative filenames in the archive. The archive entries are the same as those generated by zip -r output.zip source_dir.
import os
import zipfile
def make_zipfile(output_filename, source_dir):
relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
for root, dirs, files in os.walk(source_dir):
# add directory (needed for empty dirs)
zip.write(root, os.path.relpath(root, relroot))
for file in files:
filename = os.path.join(root, file)
if os.path.isfile(filename): # regular files only
arcname = os.path.join(os.path.relpath(root, relroot), file)
zip.write(filename, arcname)
Use shutil, which is part of python standard library set.
Using shutil is so simple(see code below):
1st arg: Filename of resultant zip/tar file,
2nd arg: zip/tar,
3rd arg: dir_name
Code:
import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
Modern Python (3.6+) using the pathlib module for concise OOP-like handling of paths, and pathlib.Path.rglob() for recursive globbing. As far as I can tell, this is equivalent to George V. Reilly's answer: zips with compression, the topmost element is a directory, keeps empty dirs, uses relative paths.
from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile
from os import PathLike
from typing import Union
def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
src_path = Path(source_dir).expanduser().resolve(strict=True)
with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
for file in src_path.rglob('*'):
zf.write(file, file.relative_to(src_path.parent))
Note: as optional type hints indicate, zip_name can't be a Path object (would be fixed in 3.6.2+).
With python 3.9, pathlib & zipfile module you can create a zip files from anywhere in the system.
def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
"""Zip the provided directory without navigating to that directory using `pathlib` module"""
# Convert to Path object
dir = Path(dir)
with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
for entry in dir.rglob("*"):
zip_file.write(entry, entry.relative_to(dir))
It is neat, typed, and has less code.
For adding compression to the resulting zip file, check out this link.
You need to change:
zip = zipfile.ZipFile('Python.zip', 'w')
to
zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
I've made some changes to code given by Mark Byers. Below function will also adds empty directories if you have them. Examples should make it more clear what is the path added to the zip.
#!/usr/bin/env python
import os
import zipfile
def addDirToZip(zipHandle, path, basePath=""):
"""
Adding directory given by \a path to opened zip file \a zipHandle
#param basePath path that will be removed from \a path when adding to archive
Examples:
# add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
zipHandle.close()
# add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir', 'dir')
zipHandle.close()
# add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
zipHandle.close()
# add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir')
zipHandle.close()
# add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir')
zipHandle.close()
# add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
addDirToZip(zipHandle, 'otherDir')
zipHandle.close()
"""
basePath = basePath.rstrip("\\/") + ""
basePath = basePath.rstrip("\\/")
for root, dirs, files in os.walk(path):
# add dir itself (needed for empty dirs
zipHandle.write(os.path.join(root, "."))
# add files
for file in files:
filePath = os.path.join(root, file)
inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
#print filePath + " , " + inZipPath
zipHandle.write(filePath, inZipPath)
Above is a simple function that should work for simple cases. You can find more elegant class in my Gist:
https://gist.github.com/Eccenux/17526123107ca0ac28e6
I have another code example that may help, using python3, pathlib and zipfile.
It should work in any OS.
from pathlib import Path
import zipfile
from datetime import datetime
DATE_FORMAT = '%y%m%d'
def date_str():
"""returns the today string year, month, day"""
return '{}'.format(datetime.now().strftime(DATE_FORMAT))
def zip_name(path):
"""returns the zip filename as string"""
cur_dir = Path(path).resolve()
parent_dir = cur_dir.parents[0]
zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
p_zip = Path(zip_filename)
n = 1
while p_zip.exists():
zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
date_str(), n))
p_zip = Path(zip_filename)
n += 1
return zip_filename
def all_files(path):
"""iterator returns all files and folders from path as absolute path string
"""
for child in Path(path).iterdir():
yield str(child)
if child.is_dir():
for grand_child in all_files(str(child)):
yield str(Path(grand_child))
def zip_dir(path):
"""generate a zip"""
zip_filename = zip_name(path)
zip_file = zipfile.ZipFile(zip_filename, 'w')
print('create:', zip_filename)
for file in all_files(path):
print('adding... ', file)
zip_file.write(file)
zip_file.close()
if __name__ == '__main__':
zip_dir('.')
print('end!')
To retain the folder hierarchy under the parent directory to be archived:
import glob
import os
import zipfile
with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
for fp in glob(os.path.join(parent, "**/*")):
base = os.path.commonpath([parent, fp])
zipf.write(fp, arcname=fp.replace(base, ""))
If you want, you could change this to use pathlib for file globbing.
If you want a functionality like the compress folder of any common graphical file manager you can use the following code, it uses the zipfile module. Using this code you will have the zip file with the path as its root folder.
import os
import zipfile
def zipdir(path, ziph):
# Iterate all the directories and files
for root, dirs, files in os.walk(path):
# Create a prefix variable with the folder structure inside the path folder.
# So if a file is at the path directory will be at the root directory of the zip file
# so the prefix will be empty. If the file belongs to a containing folder of path folder
# then the prefix will be that folder.
if root.replace(path,'') == '':
prefix = ''
else:
# Keep the folder structure after the path folder, append a '/' at the end
# and remome the first character, if it is a '/' in order to have a path like
# folder1/folder2/file.txt
prefix = root.replace(path, '') + '/'
if (prefix[0] == '/'):
prefix = prefix[1:]
for filename in files:
actual_file_path = root + '/' + filename
zipped_file_path = prefix + filename
zipf.write( actual_file_path, zipped_file_path)
zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()
So many answers here, and I hope I might contribute with my own version, which is based on the original answer (by the way), but with a more graphical perspective, also using context for each zipfile setup and sorting os.walk(), in order to have a ordered output.
Having these folders and them files (among other folders), I wanted to create a .zip for each cap_ folder:
$ tree -d
.
├── cap_01
| ├── 0101000001.json
| ├── 0101000002.json
| ├── 0101000003.json
|
├── cap_02
| ├── 0201000001.json
| ├── 0201000002.json
| ├── 0201001003.json
|
├── cap_03
| ├── 0301000001.json
| ├── 0301000002.json
| ├── 0301000003.json
|
├── docs
| ├── map.txt
| ├── main_data.xml
|
├── core_files
├── core_master
├── core_slave
Here's what I applied, with comments for better understanding of the process.
$ cat zip_cap_dirs.py
""" Zip 'cap_*' directories. """
import os
import zipfile as zf
for root, dirs, files in sorted(os.walk('.')):
if 'cap_' in root:
print(f"Compressing: {root}")
# Defining .zip name, according to Capítulo.
cap_dir_zip = '{}.zip'.format(root)
# Opening zipfile context for current root dir.
with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:
# Iterating over os.walk list of files for the current root dir.
for f in files:
# Defining relative path to files from current root dir.
f_path = os.path.join(root, f)
# Writing the file on the .zip file of the context
new_zip.write(f_path)
Basically, for each iteration over os.walk(path), I'm opening a context for zipfile setup and afterwards, iterating iterating over files, which is a list of files from root directory, forming the relative path for each file based on the current root directory, appending to the zipfile context which is running.
And the output is presented like this:
$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03
To see the contents of each .zip directory, you can use less command:
$ less cap_01.zip
Archive: cap_01.zip
Length Method Size Cmpr Date Time CRC-32 Name
-------- ------ ------- ---- ---------- ----- -------- ----
22017 Defl:N 2471 89% 2019-09-05 08:05 7a3b5ec6 cap_01/0101000001.json
21998 Defl:N 2471 89% 2019-09-05 08:05 155bece7 cap_01/0101000002.json
23236 Defl:N 2573 89% 2019-09-05 08:05 55fced20 cap_01/0101000003.json
-------- ------- --- -------
67251 7515 89% 3 files
You probably want to look at the zipfile module; there's documentation at http://docs.python.org/library/zipfile.html.
You may also want os.walk() to index the directory structure.
To give more flexibility, e.g. select directory/file by name use:
import os
import zipfile
def zipall(ob, path, rel=""):
basename = os.path.basename(path)
if os.path.isdir(path):
if rel == "":
rel = basename
ob.write(path, os.path.join(rel))
for root, dirs, files in os.walk(path):
for d in dirs:
zipall(ob, os.path.join(root, d), os.path.join(rel, d))
for f in files:
ob.write(os.path.join(root, f), os.path.join(rel, f))
break
elif os.path.isfile(path):
ob.write(path, os.path.join(rel, basename))
else:
pass
For a file tree:
.
├── dir
│ ├── dir2
│ │ └── file2.txt
│ ├── dir3
│ │ └── file3.txt
│ └── file.txt
├── dir4
│ ├── dir5
│ └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip
You can e.g. select only dir4 and root.txt:
cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]
with zipfile.ZipFile("selective.zip", "w" ) as myzip:
for f in files:
zipall(myzip, f)
Or just listdir in script invocation directory and add everything from there:
with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
for f in os.listdir():
if f == "listdir.zip":
# Creating a listdir.zip in the same directory
# will include listdir.zip inside itself, beware of this
continue
zipall(myzip, f)
Here is a variation on the answer given by Nux that works for me:
def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
basePath = os.path.split( srcPath )[ 0 ]
for root, dirs, files in os.walk( srcPath ):
p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
# add dir
zipHandle.write( root, p, zipOperation )
# add files
for f in files:
filePath = os.path.join( root, f )
fileInZipPath = os.path.join( p, f )
zipHandle.write( filePath, fileInZipPath, zipOperation )
Try the below one .it worked for me.
import zipfile, os
zipf = "compress.zip"
def main():
directory = r"Filepath"
toZip(directory)
def toZip(directory):
zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )
list = os.listdir(directory)
for file_list in list:
file_name = os.path.join(directory,file_list)
if os.path.isfile(file_name):
print file_name
zippedHelp.write(file_name)
else:
addFolderToZip(zippedHelp,file_list,directory)
print "---------------Directory Found-----------------------"
zippedHelp.close()
def addFolderToZip(zippedHelp,folder,directory):
path=os.path.join(directory,folder)
print path
file_list=os.listdir(path)
for file_name in file_list:
file_path=os.path.join(path,file_name)
if os.path.isfile(file_path):
zippedHelp.write(file_path)
elif os.path.isdir(file_name):
print "------------------sub directory found--------------------"
addFolderToZip(zippedHelp,file_name,path)
if __name__=="__main__":
main()
Say you want to Zip all the folders(sub directories) in the current directory.
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
Zip a file or a tree (a directory and its sub-directories).
from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED
def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
paths = [Path(tree_path)]
while paths:
p = paths.pop()
if p.is_dir():
paths.extend(p.iterdir())
if skip_empty_dir:
continue
zf.write(p)
To append to an existing archive, pass mode='a', to create a fresh archive mode='w' (the default in the above). So let's say you want to bundle 3 different directory trees under the same archive.
make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
A solution using pathlib.Path, which is independent of the OS used:
import zipfile
from pathlib import Path
def zip_dir(path: Path, zip_file_path: Path):
"""Zip all contents of path to zip_file"""
files_to_zip = [
file for file in path.glob('*') if file.is_file()]
with zipfile.ZipFile(
zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
for file in files_to_zip:
print(file.name)
zip_f.write(file, file.name)
current_dir = Path.cwd()
zip_dir = current_dir / "test"
tools.zip_dir(
zip_dir, current_dir / 'Zipped_dir.zip')
The obvious way to go would be to go with shutil, Like the second top answer says so, But if you still wish to go with ZipFile for some reason, And if you are getting some trouble doing that (Like ERR 13 in Windows etc), You can use this fix:
import os
import zipfile
def retrieve_file_paths(dirName):
filePaths = []
for root, directories, files in os.walk(dirName):
for filename in files:
filePath = os.path.join(root, filename)
filePaths.append(filePath)
return filePaths
def main(dir_name, output_filename):
filePaths = retrieve_file_paths(dir_name)
zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
main("my_dir", "my_dir_archived")
This one recursively iterates through every sub-folder/file in your given folder, And writes them to a zip file instead of attempting to directly zip a folder.
Here's a modern approach, using pathlib, and a context manager. Puts the files directly in the zip, rather than in a subfolder.
def zip_dir(filename: str, dir_to_zip: pathlib.Path):
with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
# Use glob instead of iterdir(), to cover all subdirectories.
for directory in dir_to_zip.glob('**'):
for file in directory.iterdir():
if not file.is_file():
continue
# Strip the first component, so we don't create an uneeded subdirectory
# containing everything.
zip_path = pathlib.Path(*file.parts[1:])
# Use a string, since zipfile doesn't support pathlib directly.
zipf.write(str(file), str(zip_path))
I prepared a function by consolidating Mark Byers' solution with Reimund and Morten Zilmer's comments (relative path and including empty directories). As a best practice, with is used in ZipFile's file construction.
The function also prepares a default zip file name with the zipped directory name and '.zip' extension. Therefore, it works with only one argument: the source directory to be zipped.
import os
import zipfile
def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
path_file_zip = os.path.join(
os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
for root, dirs, files in os.walk(path_dir):
for file_or_dir in files + dirs:
zip_file.write(
os.path.join(root, file_or_dir),
os.path.relpath(os.path.join(root, file_or_dir),
os.path.join(path_dir, os.path.pardir)))
# import required python modules
# You have to install zipfile package using pip install
import os,zipfile
# Change the directory where you want your new zip file to be
os.chdir('Type your destination')
# Create a new zipfile ( I called it myfile )
zf = zipfile.ZipFile('myfile.zip','w')
# os.walk gives a directory tree. Access the files using a for loop
for dirnames,folders,files in os.walk('Type your directory'):
zf.write('Type your Directory')
for file in files:
zf.write(os.path.join('Type your directory',file))
Well, after reading the suggestions I came up with a very similar way that works with 2.7.x without creating "funny" directory names (absolute-like names), and will only create the specified folder inside the zip.
Or just in case you needed your zip to contain a folder inside with the contents of the selected directory.
def zipDir( path, ziph ) :
"""
Inserts directory (path) into zipfile instance (ziph)
"""
for root, dirs, files in os.walk( path ) :
for file in files :
ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )
def makeZip( pathToFolder ) :
"""
Creates a zip file with the specified folder
"""
zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
zipDir( pathToFolder, zipf )
zipf.close()
print( "Zip file saved to: " + pathToFolder)
makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
Function to create zip file.
def CREATEZIPFILE(zipname, path):
#function to create a zip file
#Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file
zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
zipf.setpassword(b"password") #if you want to set password to zipfile
#checks if the path is file or directory
if os.path.isdir(path):
for files in os.listdir(path):
zipf.write(os.path.join(path, files), files)
elif os.path.isfile(path):
zipf.write(os.path.join(path), path)
zipf.close()
One thing completely missed by previous answers is the fact that using os.path.join() can easily return POSIX-incompatible paths when you run the code on Windows. The resulting archive will contain files with literal backslashes in their names when processing it with any common archive software on Linux, which is not what you want. Use path.as_posix() for the arcname parameter instead!
import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
for path in Path("include_all_of_this_folder").rglob("*"):
zf.write(path, path.as_posix())
How can I create a zip archive of a directory structure in Python?
The easiest way is to use shutil.make_archive. It supports both zip and tar formats.
import shutil
shutil.make_archive(output_filename, 'zip', dir_name)
If you need to do something more complicated than zipping the whole directory (such as skipping certain files), then you'll need to dig into the zipfile module as others have suggested.
As others have pointed out, you should use zipfile. The documentation tells you what functions are available, but doesn't really explain how you can use them to zip an entire directory. I think it's easiest to explain with some example code:
import os
import zipfile
def zipdir(path, ziph):
# ziph is zipfile handle
for root, dirs, files in os.walk(path):
for file in files:
ziph.write(os.path.join(root, file),
os.path.relpath(os.path.join(root, file),
os.path.join(path, '..')))
with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
zipdir('tmp/', zipf)
To add the contents of mydirectory to a new zip file, including all files and subdirectories:
import os
import zipfile
zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
zf.write(dirname)
for filename in files:
zf.write(os.path.join(dirname, filename))
zf.close()
How can I create a zip archive of a directory structure in Python?
In a Python script
In Python 2.7+, shutil has a make_archive function.
from shutil import make_archive
make_archive(
'zipfile_name',
'zip', # the archive format - or tar, bztar, gztar
root_dir=None, # root for archive - current working dir if None
base_dir=None) # start archiving from here - cwd if None too
Here the zipped archive will be named zipfile_name.zip. If base_dir is farther down from root_dir it will exclude files not in the base_dir, but still archive the files in the parent dirs up to the root_dir.
I did have an issue testing this on Cygwin with 2.7 - it wants a root_dir argument, for cwd:
make_archive('zipfile_name', 'zip', root_dir='.')
Using Python from the shell
You can do this with Python from the shell also using the zipfile module:
$ python -m zipfile -c zipname sourcedir
Where zipname is the name of the destination file you want (add .zip if you want it, it won't do it automatically) and sourcedir is the path to the directory.
Zipping up Python (or just don't want parent dir):
If you're trying to zip up a python package with a __init__.py and __main__.py, and you don't want the parent dir, it's
$ python -m zipfile -c zipname sourcedir/*
And
$ python zipname
would run the package. (Note that you can't run subpackages as the entry point from a zipped archive.)
Zipping a Python app:
If you have python3.5+, and specifically want to zip up a Python package, use zipapp:
$ python -m zipapp myapp
$ python myapp.pyz
This function will recursively zip up a directory tree, compressing the files, and recording the correct relative filenames in the archive. The archive entries are the same as those generated by zip -r output.zip source_dir.
import os
import zipfile
def make_zipfile(output_filename, source_dir):
relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
for root, dirs, files in os.walk(source_dir):
# add directory (needed for empty dirs)
zip.write(root, os.path.relpath(root, relroot))
for file in files:
filename = os.path.join(root, file)
if os.path.isfile(filename): # regular files only
arcname = os.path.join(os.path.relpath(root, relroot), file)
zip.write(filename, arcname)
Use shutil, which is part of python standard library set.
Using shutil is so simple(see code below):
1st arg: Filename of resultant zip/tar file,
2nd arg: zip/tar,
3rd arg: dir_name
Code:
import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
Modern Python (3.6+) using the pathlib module for concise OOP-like handling of paths, and pathlib.Path.rglob() for recursive globbing. As far as I can tell, this is equivalent to George V. Reilly's answer: zips with compression, the topmost element is a directory, keeps empty dirs, uses relative paths.
from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile
from os import PathLike
from typing import Union
def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
src_path = Path(source_dir).expanduser().resolve(strict=True)
with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
for file in src_path.rglob('*'):
zf.write(file, file.relative_to(src_path.parent))
Note: as optional type hints indicate, zip_name can't be a Path object (would be fixed in 3.6.2+).
With python 3.9, pathlib & zipfile module you can create a zip files from anywhere in the system.
def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
"""Zip the provided directory without navigating to that directory using `pathlib` module"""
# Convert to Path object
dir = Path(dir)
with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
for entry in dir.rglob("*"):
zip_file.write(entry, entry.relative_to(dir))
It is neat, typed, and has less code.
For adding compression to the resulting zip file, check out this link.
You need to change:
zip = zipfile.ZipFile('Python.zip', 'w')
to
zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
I've made some changes to code given by Mark Byers. Below function will also adds empty directories if you have them. Examples should make it more clear what is the path added to the zip.
#!/usr/bin/env python
import os
import zipfile
def addDirToZip(zipHandle, path, basePath=""):
"""
Adding directory given by \a path to opened zip file \a zipHandle
#param basePath path that will be removed from \a path when adding to archive
Examples:
# add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
zipHandle.close()
# add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir', 'dir')
zipHandle.close()
# add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
zipHandle.close()
# add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir', 'dir')
zipHandle.close()
# add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir/subdir')
zipHandle.close()
# add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
zipHandle = zipfile.ZipFile('test.zip', 'w')
addDirToZip(zipHandle, 'dir')
addDirToZip(zipHandle, 'otherDir')
zipHandle.close()
"""
basePath = basePath.rstrip("\\/") + ""
basePath = basePath.rstrip("\\/")
for root, dirs, files in os.walk(path):
# add dir itself (needed for empty dirs
zipHandle.write(os.path.join(root, "."))
# add files
for file in files:
filePath = os.path.join(root, file)
inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
#print filePath + " , " + inZipPath
zipHandle.write(filePath, inZipPath)
Above is a simple function that should work for simple cases. You can find more elegant class in my Gist:
https://gist.github.com/Eccenux/17526123107ca0ac28e6
I have another code example that may help, using python3, pathlib and zipfile.
It should work in any OS.
from pathlib import Path
import zipfile
from datetime import datetime
DATE_FORMAT = '%y%m%d'
def date_str():
"""returns the today string year, month, day"""
return '{}'.format(datetime.now().strftime(DATE_FORMAT))
def zip_name(path):
"""returns the zip filename as string"""
cur_dir = Path(path).resolve()
parent_dir = cur_dir.parents[0]
zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
p_zip = Path(zip_filename)
n = 1
while p_zip.exists():
zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
date_str(), n))
p_zip = Path(zip_filename)
n += 1
return zip_filename
def all_files(path):
"""iterator returns all files and folders from path as absolute path string
"""
for child in Path(path).iterdir():
yield str(child)
if child.is_dir():
for grand_child in all_files(str(child)):
yield str(Path(grand_child))
def zip_dir(path):
"""generate a zip"""
zip_filename = zip_name(path)
zip_file = zipfile.ZipFile(zip_filename, 'w')
print('create:', zip_filename)
for file in all_files(path):
print('adding... ', file)
zip_file.write(file)
zip_file.close()
if __name__ == '__main__':
zip_dir('.')
print('end!')
To retain the folder hierarchy under the parent directory to be archived:
import glob
import os
import zipfile
with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
for fp in glob(os.path.join(parent, "**/*")):
base = os.path.commonpath([parent, fp])
zipf.write(fp, arcname=fp.replace(base, ""))
If you want, you could change this to use pathlib for file globbing.
If you want a functionality like the compress folder of any common graphical file manager you can use the following code, it uses the zipfile module. Using this code you will have the zip file with the path as its root folder.
import os
import zipfile
def zipdir(path, ziph):
# Iterate all the directories and files
for root, dirs, files in os.walk(path):
# Create a prefix variable with the folder structure inside the path folder.
# So if a file is at the path directory will be at the root directory of the zip file
# so the prefix will be empty. If the file belongs to a containing folder of path folder
# then the prefix will be that folder.
if root.replace(path,'') == '':
prefix = ''
else:
# Keep the folder structure after the path folder, append a '/' at the end
# and remome the first character, if it is a '/' in order to have a path like
# folder1/folder2/file.txt
prefix = root.replace(path, '') + '/'
if (prefix[0] == '/'):
prefix = prefix[1:]
for filename in files:
actual_file_path = root + '/' + filename
zipped_file_path = prefix + filename
zipf.write( actual_file_path, zipped_file_path)
zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()
So many answers here, and I hope I might contribute with my own version, which is based on the original answer (by the way), but with a more graphical perspective, also using context for each zipfile setup and sorting os.walk(), in order to have a ordered output.
Having these folders and them files (among other folders), I wanted to create a .zip for each cap_ folder:
$ tree -d
.
├── cap_01
| ├── 0101000001.json
| ├── 0101000002.json
| ├── 0101000003.json
|
├── cap_02
| ├── 0201000001.json
| ├── 0201000002.json
| ├── 0201001003.json
|
├── cap_03
| ├── 0301000001.json
| ├── 0301000002.json
| ├── 0301000003.json
|
├── docs
| ├── map.txt
| ├── main_data.xml
|
├── core_files
├── core_master
├── core_slave
Here's what I applied, with comments for better understanding of the process.
$ cat zip_cap_dirs.py
""" Zip 'cap_*' directories. """
import os
import zipfile as zf
for root, dirs, files in sorted(os.walk('.')):
if 'cap_' in root:
print(f"Compressing: {root}")
# Defining .zip name, according to Capítulo.
cap_dir_zip = '{}.zip'.format(root)
# Opening zipfile context for current root dir.
with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:
# Iterating over os.walk list of files for the current root dir.
for f in files:
# Defining relative path to files from current root dir.
f_path = os.path.join(root, f)
# Writing the file on the .zip file of the context
new_zip.write(f_path)
Basically, for each iteration over os.walk(path), I'm opening a context for zipfile setup and afterwards, iterating iterating over files, which is a list of files from root directory, forming the relative path for each file based on the current root directory, appending to the zipfile context which is running.
And the output is presented like this:
$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03
To see the contents of each .zip directory, you can use less command:
$ less cap_01.zip
Archive: cap_01.zip
Length Method Size Cmpr Date Time CRC-32 Name
-------- ------ ------- ---- ---------- ----- -------- ----
22017 Defl:N 2471 89% 2019-09-05 08:05 7a3b5ec6 cap_01/0101000001.json
21998 Defl:N 2471 89% 2019-09-05 08:05 155bece7 cap_01/0101000002.json
23236 Defl:N 2573 89% 2019-09-05 08:05 55fced20 cap_01/0101000003.json
-------- ------- --- -------
67251 7515 89% 3 files
You probably want to look at the zipfile module; there's documentation at http://docs.python.org/library/zipfile.html.
You may also want os.walk() to index the directory structure.
To give more flexibility, e.g. select directory/file by name use:
import os
import zipfile
def zipall(ob, path, rel=""):
basename = os.path.basename(path)
if os.path.isdir(path):
if rel == "":
rel = basename
ob.write(path, os.path.join(rel))
for root, dirs, files in os.walk(path):
for d in dirs:
zipall(ob, os.path.join(root, d), os.path.join(rel, d))
for f in files:
ob.write(os.path.join(root, f), os.path.join(rel, f))
break
elif os.path.isfile(path):
ob.write(path, os.path.join(rel, basename))
else:
pass
For a file tree:
.
├── dir
│ ├── dir2
│ │ └── file2.txt
│ ├── dir3
│ │ └── file3.txt
│ └── file.txt
├── dir4
│ ├── dir5
│ └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip
You can e.g. select only dir4 and root.txt:
cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]
with zipfile.ZipFile("selective.zip", "w" ) as myzip:
for f in files:
zipall(myzip, f)
Or just listdir in script invocation directory and add everything from there:
with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
for f in os.listdir():
if f == "listdir.zip":
# Creating a listdir.zip in the same directory
# will include listdir.zip inside itself, beware of this
continue
zipall(myzip, f)
Here is a variation on the answer given by Nux that works for me:
def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
basePath = os.path.split( srcPath )[ 0 ]
for root, dirs, files in os.walk( srcPath ):
p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
# add dir
zipHandle.write( root, p, zipOperation )
# add files
for f in files:
filePath = os.path.join( root, f )
fileInZipPath = os.path.join( p, f )
zipHandle.write( filePath, fileInZipPath, zipOperation )
Try the below one .it worked for me.
import zipfile, os
zipf = "compress.zip"
def main():
directory = r"Filepath"
toZip(directory)
def toZip(directory):
zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )
list = os.listdir(directory)
for file_list in list:
file_name = os.path.join(directory,file_list)
if os.path.isfile(file_name):
print file_name
zippedHelp.write(file_name)
else:
addFolderToZip(zippedHelp,file_list,directory)
print "---------------Directory Found-----------------------"
zippedHelp.close()
def addFolderToZip(zippedHelp,folder,directory):
path=os.path.join(directory,folder)
print path
file_list=os.listdir(path)
for file_name in file_list:
file_path=os.path.join(path,file_name)
if os.path.isfile(file_path):
zippedHelp.write(file_path)
elif os.path.isdir(file_name):
print "------------------sub directory found--------------------"
addFolderToZip(zippedHelp,file_name,path)
if __name__=="__main__":
main()
Say you want to Zip all the folders(sub directories) in the current directory.
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
Zip a file or a tree (a directory and its sub-directories).
from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED
def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
paths = [Path(tree_path)]
while paths:
p = paths.pop()
if p.is_dir():
paths.extend(p.iterdir())
if skip_empty_dir:
continue
zf.write(p)
To append to an existing archive, pass mode='a', to create a fresh archive mode='w' (the default in the above). So let's say you want to bundle 3 different directory trees under the same archive.
make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
A solution using pathlib.Path, which is independent of the OS used:
import zipfile
from pathlib import Path
def zip_dir(path: Path, zip_file_path: Path):
"""Zip all contents of path to zip_file"""
files_to_zip = [
file for file in path.glob('*') if file.is_file()]
with zipfile.ZipFile(
zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
for file in files_to_zip:
print(file.name)
zip_f.write(file, file.name)
current_dir = Path.cwd()
zip_dir = current_dir / "test"
tools.zip_dir(
zip_dir, current_dir / 'Zipped_dir.zip')
The obvious way to go would be to go with shutil, Like the second top answer says so, But if you still wish to go with ZipFile for some reason, And if you are getting some trouble doing that (Like ERR 13 in Windows etc), You can use this fix:
import os
import zipfile
def retrieve_file_paths(dirName):
filePaths = []
for root, directories, files in os.walk(dirName):
for filename in files:
filePath = os.path.join(root, filename)
filePaths.append(filePath)
return filePaths
def main(dir_name, output_filename):
filePaths = retrieve_file_paths(dir_name)
zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
main("my_dir", "my_dir_archived")
This one recursively iterates through every sub-folder/file in your given folder, And writes them to a zip file instead of attempting to directly zip a folder.
Here's a modern approach, using pathlib, and a context manager. Puts the files directly in the zip, rather than in a subfolder.
def zip_dir(filename: str, dir_to_zip: pathlib.Path):
with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
# Use glob instead of iterdir(), to cover all subdirectories.
for directory in dir_to_zip.glob('**'):
for file in directory.iterdir():
if not file.is_file():
continue
# Strip the first component, so we don't create an uneeded subdirectory
# containing everything.
zip_path = pathlib.Path(*file.parts[1:])
# Use a string, since zipfile doesn't support pathlib directly.
zipf.write(str(file), str(zip_path))
I prepared a function by consolidating Mark Byers' solution with Reimund and Morten Zilmer's comments (relative path and including empty directories). As a best practice, with is used in ZipFile's file construction.
The function also prepares a default zip file name with the zipped directory name and '.zip' extension. Therefore, it works with only one argument: the source directory to be zipped.
import os
import zipfile
def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
path_file_zip = os.path.join(
os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
for root, dirs, files in os.walk(path_dir):
for file_or_dir in files + dirs:
zip_file.write(
os.path.join(root, file_or_dir),
os.path.relpath(os.path.join(root, file_or_dir),
os.path.join(path_dir, os.path.pardir)))
# import required python modules
# You have to install zipfile package using pip install
import os,zipfile
# Change the directory where you want your new zip file to be
os.chdir('Type your destination')
# Create a new zipfile ( I called it myfile )
zf = zipfile.ZipFile('myfile.zip','w')
# os.walk gives a directory tree. Access the files using a for loop
for dirnames,folders,files in os.walk('Type your directory'):
zf.write('Type your Directory')
for file in files:
zf.write(os.path.join('Type your directory',file))
Well, after reading the suggestions I came up with a very similar way that works with 2.7.x without creating "funny" directory names (absolute-like names), and will only create the specified folder inside the zip.
Or just in case you needed your zip to contain a folder inside with the contents of the selected directory.
def zipDir( path, ziph ) :
"""
Inserts directory (path) into zipfile instance (ziph)
"""
for root, dirs, files in os.walk( path ) :
for file in files :
ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )
def makeZip( pathToFolder ) :
"""
Creates a zip file with the specified folder
"""
zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
zipDir( pathToFolder, zipf )
zipf.close()
print( "Zip file saved to: " + pathToFolder)
makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
Function to create zip file.
def CREATEZIPFILE(zipname, path):
#function to create a zip file
#Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file
zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
zipf.setpassword(b"password") #if you want to set password to zipfile
#checks if the path is file or directory
if os.path.isdir(path):
for files in os.listdir(path):
zipf.write(os.path.join(path, files), files)
elif os.path.isfile(path):
zipf.write(os.path.join(path), path)
zipf.close()
One thing completely missed by previous answers is the fact that using os.path.join() can easily return POSIX-incompatible paths when you run the code on Windows. The resulting archive will contain files with literal backslashes in their names when processing it with any common archive software on Linux, which is not what you want. Use path.as_posix() for the arcname parameter instead!
import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
for path in Path("include_all_of_this_folder").rglob("*"):
zf.write(path, path.as_posix())
I am trying to create a backup script in Python 3.5 that will zip all the contents of a specified folder. Here is what I am doing now:
import shutil
import datetime
import os
import os.path
import dropbox
INPUT_FOLDER = '~/Documents/db-backup-test/'
ARCHIVE_PATH = '~/Documents/tmp/'
ARCHIVE_FILE_NAME = "db-backup_iOS__"
def compress_folder(origem, destino):
try:
arq = shutil.make_archive(destino, 'zip', root_dir=origem, base_dir=origem)
except Exception as e:
print('\n\nAn error was found during compression:')
print(e)
timestamp = str(datetime.datetime.now())
input_path = os.path.expanduser(INPUT_FOLDER)
archive_path = os.path.expanduser(ARCHIVE_PATH + ARCHIVE_FILE_NAME + timestamp)
compress_folder(input_path, archive_path)
I goes well, except for one thing: the resulting zip archive extracts its contents with a reproduction of the full original path. What I would like is to get is a simple copy of the folder, with its original name and its contents with relative paths, when extracting the archive.
I was looking into a simple and short solution that would not require to iterate through the folder contents.
Is this possible with shutil.make_archive(), or will I need to dig into other modules, like zipfile?
Use the arcname parameter to control the name/path in the zip file.
#!/usr/bin/env python2.7
import os
import zipfile
def zip(src, dst):
zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
abs_src = os.path.abspath(src)
for dirname, subdirs, files in os.walk(src):
for filename in files:
absname = os.path.abspath(os.path.join(dirname, filename))
arcname = absname[len(abs_src) + 1:]
print 'zipping %s as %s' % (os.path.join(dirname, filename), arcname)
zf.write(absname, arcname)
zf.close()
zip("src", "dst")
With a directory structure like this:
src
└── a
├── b
│ └── bar
└── foo
I've been trying to use the zipfile and shutil.make_archive modules to recursively create a zip file of a directory. Both modules work great--except empty directories do not get added to the archive. Empty directories containing other empty directories are also silently skipped.
I can use 7Zip to create an archive of the same path and empty directories are preserved. Therefore I know this is possible within the file format itself. I just don't know how to do it within Python. Any ideas? Thanks!
There is a example using zipfile:
import os, zipfile
from os.path import join
def zipfolder(foldername, filename, includeEmptyDIr=True):
empty_dirs = []
zip = zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED)
for root, dirs, files in os.walk(foldername):
empty_dirs.extend([dir for dir in dirs if os.listdir(join(root, dir)) == []])
for name in files:
zip.write(join(root ,name))
if includeEmptyDIr:
for dir in empty_dirs:
zif = zipfile.ZipInfo(join(root, dir) + "/")
zip.writestr(zif, "")
empty_dirs = []
zip.close()
if __name__ == "__main__":
zipfolder('test1/noname/', 'zip.zip')
You'll need to register a new archive format to do that, since the default ZIP archiver does not support that. Take a look at the meat of the existing ZIP archiver. Make your own archiver that creates directories using that currently-unused dirpath variable. I looked for how to create an empty directory and found this:
zip.writestr(zipfile.ZipInfo('empty/'), '')
With that, you should be able to write the necessary code to make it archive empty directories.
This is lifted from Adding folders to a zip file using python but is the only function I have tried that works. The one listed as the answer does not work under Python 2.7.3 (doesn't copy empty directories and is inefficient). The following is tried and tested:
#!/usr/bin/python
import os
import zipfile
def zipdir(dirPath=None, zipFilePath=None, includeDirInZip=True):
if not zipFilePath:
zipFilePath = dirPath + ".zip"
if not os.path.isdir(dirPath):
raise OSError("dirPath argument must point to a directory. "
"'%s' does not." % dirPath)
parentDir, dirToZip = os.path.split(dirPath)
#Little nested function to prepare the proper archive path
def trimPath(path):
archivePath = path.replace(parentDir, "", 1)
if parentDir:
archivePath = archivePath.replace(os.path.sep, "", 1)
if not includeDirInZip:
archivePath = archivePath.replace(dirToZip + os.path.sep, "", 1)
return os.path.normcase(archivePath)
outFile = zipfile.ZipFile(zipFilePath, "w",
compression=zipfile.ZIP_DEFLATED)
for (archiveDirPath, dirNames, fileNames) in os.walk(dirPath):
for fileName in fileNames:
filePath = os.path.join(archiveDirPath, fileName)
outFile.write(filePath, trimPath(filePath))
#Make sure we get empty directories as well
if not fileNames and not dirNames:
zipInfo = zipfile.ZipInfo(trimPath(archiveDirPath) + "/")
#some web sites suggest doing
#zipInfo.external_attr = 16
#or
#zipInfo.external_attr = 48
#Here to allow for inserting an empty directory. Still TBD/TODO.
outFile.writestr(zipInfo, "")
outFile.close()
def zip_dir(src_dir, dst_zip, *, skip_suffixes=None, dry=False):
import logging
from pathlib import Path
from os import walk
from tempfile import TemporaryDirectory
from zipfile import ZipFile, ZipInfo
_log = logging.getLogger(zip_dir.__name__)
_log.addHandler(logging.NullHandler())
_sep = 50 * "-"
skip_suffixes = skip_suffixes or []
src_dir, dst_zip = Path(src_dir), Path(dst_zip)
_log.info("zipping dir: '%s' to: '%s", str(src_dir), str(dst_zip))
if not src_dir.exists():
raise FileNotFoundError(str(src_dir))
if not src_dir.is_dir():
raise NotADirectoryError(str(src_dir))
if dst_zip.exists():
raise FileExistsError(str(dst_zip))
with TemporaryDirectory() as tmp_dir:
tmp_zip_path = Path(tmp_dir).joinpath(dst_zip.name)
with ZipFile(str(tmp_zip_path), mode="w") as zip_out:
for root, dirs, files in walk(src_dir):
root = Path(root)
for folder in dirs:
folder = root.joinpath(folder)
# add empty folders to the zip
if not list(folder.iterdir()):
_log.debug(_sep)
folder_name = f"{str(folder.relative_to(src_dir))}/"
_log.debug("empty dir: '%s'", folder_name)
if dry:
continue
zip_out.writestr(ZipInfo(folder_name), "")
for file in files:
file = root.joinpath(file)
_log.debug(_sep)
_log.debug("adding: '%s'", str(file))
should_skip = None
for suffix in file.suffixes:
if suffix in skip_suffixes:
should_skip = suffix
break
if should_skip:
_log.debug("skipped [%s]: %s", should_skip, str(file))
continue
arcname = str(file.relative_to(src_dir))
_log.debug("arcname: '%s'", arcname)
if dry:
continue
zip_out.write(str(file), arcname=arcname)
if not dry:
dst_zip.write_bytes(tmp_zip_path.read_bytes())
tmp_zip_path.unlink()
if __name__ == '__main__':
import logging
logging.basicConfig(level=logging.DEBUG, format="%(asctime)s | %(levelname)8s | %(module)25s:%(lineno)-5s | %(message)s")
zip_dir("/tmp/opera_profile", "opera_profile.zip", skip_suffixes=[".log"], dry=True)
I have a C++/Obj-C background and I am just discovering Python (been writing it for about an hour).
I am writing a script to recursively read the contents of text files in a folder structure.
The problem I have is the code I have written will only work for one folder deep. I can see why in the code (see #hardcoded path), I just don't know how I can move forward with Python since my experience with it is only brand new.
Python Code:
import os
import sys
rootdir = sys.argv[1]
for root, subFolders, files in os.walk(rootdir):
for folder in subFolders:
outfileName = rootdir + "/" + folder + "/py-outfile.txt" # hardcoded path
folderOut = open( outfileName, 'w' )
print "outfileName is " + outfileName
for file in files:
filePath = rootdir + '/' + file
f = open( filePath, 'r' )
toWrite = f.read()
print "Writing '" + toWrite + "' to" + filePath
folderOut.write( toWrite )
f.close()
folderOut.close()
Make sure you understand the three return values of os.walk:
for root, subdirs, files in os.walk(rootdir):
has the following meaning:
root: Current path which is "walked through"
subdirs: Files in root of type directory
files: Files in root (not in subdirs) of type other than directory
And please use os.path.join instead of concatenating with a slash! Your problem is filePath = rootdir + '/' + file - you must concatenate the currently "walked" folder instead of the topmost folder. So that must be filePath = os.path.join(root, file). BTW "file" is a builtin, so you don't normally use it as variable name.
Another problem are your loops, which should be like this, for example:
import os
import sys
walk_dir = sys.argv[1]
print('walk_dir = ' + walk_dir)
# If your current working directory may change during script execution, it's recommended to
# immediately convert program arguments to an absolute path. Then the variable root below will
# be an absolute path as well. Example:
# walk_dir = os.path.abspath(walk_dir)
print('walk_dir (absolute) = ' + os.path.abspath(walk_dir))
for root, subdirs, files in os.walk(walk_dir):
print('--\nroot = ' + root)
list_file_path = os.path.join(root, 'my-directory-list.txt')
print('list_file_path = ' + list_file_path)
with open(list_file_path, 'wb') as list_file:
for subdir in subdirs:
print('\t- subdirectory ' + subdir)
for filename in files:
file_path = os.path.join(root, filename)
print('\t- file %s (full path: %s)' % (filename, file_path))
with open(file_path, 'rb') as f:
f_content = f.read()
list_file.write(('The file %s contains:\n' % filename).encode('utf-8'))
list_file.write(f_content)
list_file.write(b'\n')
If you didn't know, the with statement for files is a shorthand:
with open('filename', 'rb') as f:
dosomething()
# is effectively the same as
f = open('filename', 'rb')
try:
dosomething()
finally:
f.close()
If you are using Python 3.5 or above, you can get this done in 1 line.
import glob
# root_dir needs a trailing slash (i.e. /root/dir/)
for filename in glob.iglob(root_dir + '**/*.txt', recursive=True):
print(filename)
As mentioned in the documentation
If recursive is true, the pattern '**' will match any files and zero or more directories and subdirectories.
If you want every file, you can use
import glob
for filename in glob.iglob(root_dir + '**/**', recursive=True):
print(filename)
Agree with Dave Webb, os.walk will yield an item for each directory in the tree. Fact is, you just don't have to care about subFolders.
Code like this should work:
import os
import sys
rootdir = sys.argv[1]
for folder, subs, files in os.walk(rootdir):
with open(os.path.join(folder, 'python-outfile.txt'), 'w') as dest:
for filename in files:
with open(os.path.join(folder, filename), 'r') as src:
dest.write(src.read())
TL;DR: This is the equivalent to find -type f to go over all files in all folders below and including the current one:
for currentpath, folders, files in os.walk('.'):
for file in files:
print(os.path.join(currentpath, file))
As already mentioned in other answers, os.walk() is the answer, but it could be explained better. It's quite simple! Let's walk through this tree:
docs/
└── doc1.odt
pics/
todo.txt
With this code:
for currentpath, folders, files in os.walk('.'):
print(currentpath)
The currentpath is the current folder it is looking at. This will output:
.
./docs
./pics
So it loops three times, because there are three folders: the current one, docs, and pics. In every loop, it fills the variables folders and files with all folders and files. Let's show them:
for currentpath, folders, files in os.walk('.'):
print(currentpath, folders, files)
This shows us:
# currentpath folders files
. ['pics', 'docs'] ['todo.txt']
./pics [] []
./docs [] ['doc1.odt']
So in the first line, we see that we are in folder ., that it contains two folders namely pics and docs, and that there is one file, namely todo.txt. You don't have to do anything to recurse into those folders, because as you see, it recurses automatically and just gives you the files in any subfolders. And any subfolders of that (though we don't have those in the example).
If you just want to loop through all files, the equivalent of find -type f, you can do this:
for currentpath, folders, files in os.walk('.'):
for file in files:
print(os.path.join(currentpath, file))
This outputs:
./todo.txt
./docs/doc1.odt
The pathlib library is really great for working with files. You can do a recursive glob on a Path object like so.
from pathlib import Path
for elem in Path('/path/to/my/files').rglob('*.*'):
print(elem)
import glob
import os
root_dir = <root_dir_here>
for filename in glob.iglob(root_dir + '**/**', recursive=True):
if os.path.isfile(filename):
with open(filename,'r') as file:
print(file.read())
**/** is used to get all files recursively including directory.
if os.path.isfile(filename) is used to check if filename variable is file or directory, if it is file then we can read that file.
Here I am printing file.
If you want a flat list of all paths under a given dir (like find . in the shell):
files = [
os.path.join(parent, name)
for (parent, subdirs, files) in os.walk(YOUR_DIRECTORY)
for name in files + subdirs
]
To only include full paths to files under the base dir, leave out + subdirs.
I've found the following to be the easiest
from glob import glob
import os
files = [f for f in glob('rootdir/**', recursive=True) if os.path.isfile(f)]
Using glob('some/path/**', recursive=True) gets all files, but also includes directory names. Adding the if os.path.isfile(f) condition filters this list to existing files only
For my taste os.walk() is a little too complicated and verbose. You can do the accepted answer cleaner by:
all_files = [str(f) for f in pathlib.Path(dir_path).glob("**/*") if f.is_file()]
with open(outfile, 'wb') as fout:
for f in all_files:
with open(f, 'rb') as fin:
fout.write(fin.read())
fout.write(b'\n')
use os.path.join() to construct your paths - It's neater:
import os
import sys
rootdir = sys.argv[1]
for root, subFolders, files in os.walk(rootdir):
for folder in subFolders:
outfileName = os.path.join(root,folder,"py-outfile.txt")
folderOut = open( outfileName, 'w' )
print "outfileName is " + outfileName
for file in files:
filePath = os.path.join(root,file)
toWrite = open( filePath).read()
print "Writing '" + toWrite + "' to" + filePath
folderOut.write( toWrite )
folderOut.close()
os.walk does recursive walk by default. For each dir, starting from root it yields a 3-tuple (dirpath, dirnames, filenames)
from os import walk
from os.path import splitext, join
def select_files(root, files):
"""
simple logic here to filter out interesting files
.py files in this example
"""
selected_files = []
for file in files:
#do concatenation here to get full path
full_path = join(root, file)
ext = splitext(file)[1]
if ext == ".py":
selected_files.append(full_path)
return selected_files
def build_recursive_dir_tree(path):
"""
path - where to begin folder scan
"""
selected_files = []
for root, dirs, files in walk(path):
selected_files += select_files(root, files)
return selected_files
I think the problem is that you're not processing the output of os.walk correctly.
Firstly, change:
filePath = rootdir + '/' + file
to:
filePath = root + '/' + file
rootdir is your fixed starting directory; root is a directory returned by os.walk.
Secondly, you don't need to indent your file processing loop, as it makes no sense to run this for each subdirectory. You'll get root set to each subdirectory. You don't need to process the subdirectories by hand unless you want to do something with the directories themselves.
Try this:
import os
import sys
for root, subdirs, files in os.walk(path):
for file in os.listdir(root):
filePath = os.path.join(root, file)
if os.path.isdir(filePath):
pass
else:
f = open (filePath, 'r')
# Do Stuff
If you prefer an (almost) Oneliner:
from pathlib import Path
lookuppath = '.' #use your path
filelist = [str(item) for item in Path(lookuppath).glob("**/*") if Path(item).is_file()]
In this case you will get a list with just the paths of all files located recursively under lookuppath.
Without str() you will get PosixPath() added to each path.
This worked for me:
import glob
root_dir = "C:\\Users\\Scott\\" # Don't forget trailing (last) slashes
for filename in glob.iglob(root_dir + '**/*.jpg', recursive=True):
print(filename)
# do stuff
If just the file names are not enough, it's easy to implement a Depth-first search on top of os.scandir():
stack = ['.']
files = []
total_size = 0
while stack:
dirname = stack.pop()
with os.scandir(dirname) as it:
for e in it:
if e.is_dir():
stack.append(e.path)
else:
size = e.stat().st_size
files.append((e.path, size))
total_size += size
The docs have this to say:
The scandir() function returns directory entries along with file attribute information, giving better performance for many common use cases.