I have a specific path with folders and files. I want to filter out files with pdf, docx,jpg extensions.
Already, I have a script to list all files. So I got stuck. Anyone who could help me out to filter out those files.
The code is below.
import os
path = r'C:\Users\PacY\Documents'
FileList = []
extensions = ['.pdf', '.docx', '.jpg']
for FileList in os.listdir(path):
print("\nFiles: ", FileList)
listdir() gives file names as strings and string has function endswith() (ends with) which can get single string (ie. filename.endswith('.pdf')) or tuple of strings (ie. filename.endswith( ('.pdf', '.docx', '.jpg') ))
import os
path = r'C:\Users\PacY\Documents'
extensions = ('.pdf', '.docx', '.jpg') # has to be tuple instead of list
for filename in os.listdir(path):
if filename.endswith( extensions ):
print("filename:", filename)
To make sure you can also convert name to lower() to recognize also .PDF, .Pdf, etc.
if filename.lower().endswith( extensions ):
import os
path = r'C:\Users\PacY\Documents'
extensions = ('.pdf', '.docx', '.jpg') # has to be tuple instead of list
filtered_filenames = []
for filename in os.listdir(path):
if filename.lower().endswith( extensions ):
#print("filename:", filename)
filtered_filenames.append(filename)
#filtered_filenames.append( os.path.join(path, filename) ) # full path
print(filtered_filenames)
by the way:
it works also with extensions which have more dots - like .pdf.zip or popular on linux .tar.gz
I would use the split method. Here an example printing all the files in a folder that have one of those extensions.
import os
path = r'C:\Users\PacY\Documents'
FileList = []
extensions = ['pdf', 'docx', 'jpg']
FileList = os.listdir(path)
for file in FileList:
if file.split('.')[-1] in extensions:
print(file)
import os
from os import path
print(os.path.realpath)
extension = '.epub'
dir_path = os.path.dirname(os.path.realpath(__file__))
# Making list of files with .epub extension
files = []
for directory, subdir_list, file_list in os.walk(dir_path):
for name in file_list:
if list(name.split("."))[1] == extension.strip("."):
files.append(name)
Above code returns list of all the files with requested extension for the directory and all subdirectories.
i would write code in bit different way where you can use pathlib.
import pathlib
path = r'C:\Users\PacY\Documents'
extensions = ['pdf', 'docx', 'jpg']
list(pathlib.Path(path).glob(extensions))
I am trying to copy all jpeg files from a directory (with multiple subdirectories) to a single directory. There are multiple files with the same name, so I am trying to rename the files using the name of the parent directory. For example: c:\images\tiger\image_00001.jpg will be moved to a new folder and renamed to c:\images\allimages\tiger_image_00001.jpg. I tried the code below, but nothing happens. The folder gets created, but the files do not move. This is what I have so far:
import os
path = 'source/'
os.mkdir('source/allimages/')
extensions = ['.jpeg']
for folder, _, filenames in os.walk(path):
for filename in filenames:
if folder == path or folder == os.path.join(path, 'allimages'):
continue
folder = folder.strip(path)
extension = os.path.splitext(os.path.splitext(filename)[0])[-1].lower()
if extension in extensions:
infilename = os.path.join(path, folder, filename)
newname = os.path.join(path, 'all_files', "{}-{}".format(folder.strip('./')))
os.rename(infilename, newname)
I would recommend having a function dedicated to resolving a unique filename. A while loop should do the trick. This should work.
import os
import shutil
def resolve_path(filename, destination_dir):
dest = os.path.join(destination_dir, filename)
*base_parts, extension = filename.split('.')
base_name = '.'.join(base_parts)
duplicate_num = 1
while os.path.exists(dest):
new_base = base_name + str(duplicate_num).zfill(5)
new_filename = "{}.{}".format(new_base, extension)
dest = os.path.join(destination_dir, new_filename)
duplicate_num += 1
return dest
That is such that the following is the result....
>>> with open('/path/to/file.extension', 'w') as f:
>>> pass # just create the file
>>> resolve_path('file.extension', '/path/to/')
'/path/to/file00001.extension'
Then put it together with traversing the source...
def consolidate(source, destination, extension='.jpg'):
if not os.path.exists(destination):
os.makedirs(destination)
for root, dirs, files in os.walk(source):
for f in files:
if f.lower().endswith(extension):
source_path = os.path.join(root, f)
destination_path = resolve_path(f, destination)
shutil.copyfile(source_path, destination_path)
You're calling splitext on its own output, which doesn't get what you want:
In [4]: os.path.splitext(os.path.splitext('foo.bar')[0])[-1]
Out[4]: ''
You just want extension = os.path.splitext(filename)[-1].lower(), or if you don't want the dot, then extension = os.path.splitext(filename)[-1].lower()[1:].
(Edited) more seriously, there's a problem with folder.strip(path): this will remove all characters in path from folder. For instance 'source/rescue'.strip('source/') == ''. What you want is folder.replace(path, '').
I am trying the script below to rename all files in a folder.It is working fine,But when i am trying to run it outside the folder.It shows error.
import os
path=os.getcwd()
path=os.path.join(path,'it')
filenames = os.listdir(path)
i=0
for filename in filenames:
os.rename(filename, "%d.jpg"%i)
i=i+1
'it' is the name of the folder in which files lie.
Error:FileNotFoundError: [Errno 2] No such file or directory: '0.jpg' -> '0.jpg'
Print is showing names of files
When you do os.listdir(path) you get the filenames of files in the folder, but not the complete paths to those files. When you call os.rename you need the path to the file rather than just the filename.
You can join the filename to its parent folder's path using os.path.join.
E.g. os.path.join(path, file).
Something like this might work:
for filename in filenames:
old = os.path.join(path, filename)
new = os.path.join(path, "%d.jpg"%i)
os.rename(old, new)
i=i+1
You need to mention complete or relative path to file.
In this case, it should be
path + '/' + filename
or more generally,
newpath = os.path.join(path, filename)
I am trying to write out the filepath for files with specific file extensions to a text file. There are some files that have different extensions but the same file name, and I am assuming these are duplicates and only want to retain one entry. Here is what I have for code - it is not writing anything out to the file. What am I missing?
import os
path = r'S:\Photogr\ASC'
file_ext_lst = ['.2dm','.2de','.3dm','.3de','.dgn']
txtfile = r'D:\test\microstation_filenames_paths.txt'
for dirpath, dirnames, filenames in os.walk(path):
for filename in filenames:
fullPath = os.path.join(dirpath, filename)
name = os.path.splitext(filename)[0]
if filename[-4:] in file_ext_lst:
with open(txtfile,'r+') as f:
for line in f:
if name not in line:
f.write(fullPath +'\n')
f.close()
The following code writes duplicate file names and paths to a text file.
import os
# path = r'S:\Photogr\ASC'
path = 'temp'
file_ext_lst = ['.2dm','.2de','.3dm','.3de','.dgn']
txtfile = r'D:\test\microstation_filenames_paths.txt'
found = dict()
for dirpath, _, filenames in os.walk(path):
for filename in filenames:
fullPath = os.path.join(dirpath, filename)
name,ext = os.path.splitext(filename)
if ext not in file_ext_lst:
continue
if name not in found:
found[name] = fullPath
with open('unique.txt', 'w') as outf:
print >>outf, 'Unique files:'
for name,path in found.iteritems():
print >>outf, '{:<10} {}'.format(name,path)
Disclaimer: I haven't tried creating some sample files and testing the code - if my first two suggestions do not help, I can try and look further!
You can remove f.close() after with open(), Python does that automatically for you.
You can also simplify the with open() block to:
with open(txtfile,'r+') as f:
if name not in f.read():
f.write(fullPath +'\n')
On another note: Opening and writing to your text file could happen a lot, which would be very slow - I would suggest storing your candidates in an array first and writing that to the text file only after the os.walk() part.
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.