I'd like to browse through the current folder and all its subfolders and get all the files with .htm|.html extensions. I have found out that it is possible to find out whether an object is a dir or file like this:
import os
dirList = os.listdir("./") # current directory
for dir in dirList:
if os.path.isdir(dir) == True:
# I don't know how to get into this dir and do the same thing here
else:
# I got file and i can regexp if it is .htm|html
and in the end, I would like to have all the files and their paths in an array. Is something like that possible?
You can use os.walk() to recursively iterate through a directory and all its subdirectories:
for root, dirs, files in os.walk(path):
for name in files:
if name.endswith((".html", ".htm")):
# whatever
To build a list of these names, you can use a list comprehension:
htmlfiles = [os.path.join(root, name)
for root, dirs, files in os.walk(path)
for name in files
if name.endswith((".html", ".htm"))]
I had a similar thing to work on, and this is how I did it.
import os
rootdir = os.getcwd()
for subdir, dirs, files in os.walk(rootdir):
for file in files:
#print os.path.join(subdir, file)
filepath = subdir + os.sep + file
if filepath.endswith(".html"):
print (filepath)
Hope this helps.
In python 3 you can use os.scandir():
def dir_scan(path):
for i in os.scandir(path):
if i.is_file():
print('File: ' + i.path)
elif i.is_dir():
print('Folder: ' + i.path)
dir_scan(i.path)
Use newDirName = os.path.abspath(dir) to create a full directory path name for the subdirectory and then list its contents as you have done with the parent (i.e. newDirList = os.listDir(newDirName))
You can create a separate method of your code snippet and call it recursively through the subdirectory structure. The first parameter is the directory pathname. This will change for each subdirectory.
This answer is based on the 3.1.1 version documentation of the Python Library. There is a good model example of this in action on page 228 of the Python 3.1.1 Library Reference (Chapter 10 - File and Directory Access).
Good Luck!
Slightly altered version of Sven Marnach's solution..
import os
folder_location = 'C:\SomeFolderName'
file_list = create_file_list(folder_location)
def create_file_list(path):
return_list = []
for filenames in os.walk(path):
for file_list in filenames:
for file_name in file_list:
if file_name.endswith((".txt")):
return_list.append(file_name)
return return_list
There are two ways works for me.
1. Work with the `os` package and use `'__file__'` to replace the main
directory when the project locates
import os
script_dir = os.path.dirname(__file__)
path = 'subdirectory/test.txt'
file = os.path.join(script_dir, path)
fileread = open(file,'r')
2. By using '\\' to read or write the file in subfolder
fileread = open('subdirectory\\test.txt','r')
from tkinter import *
import os
root = Tk()
file = filedialog.askdirectory()
changed_dir = os.listdir(file)
print(changed_dir)
root.mainloop()
I need to iterate through all .asm files inside a given directory and do some actions on them.
How can this be done in a efficient way?
Python 3.6 version of the above answer, using os - assuming that you have the directory path as a str object in a variable called directory_in_str:
import os
directory = os.fsencode(directory_in_str)
for file in os.listdir(directory):
filename = os.fsdecode(file)
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
Or recursively, using pathlib:
from pathlib import Path
pathlist = Path(directory_in_str).glob('**/*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Use rglob to replace glob('**/*.asm') with rglob('*.asm')
This is like calling Path.glob() with '**/' added in front of the given relative pattern:
from pathlib import Path
pathlist = Path(directory_in_str).rglob('*.asm')
for path in pathlist:
# because path is object not string
path_in_str = str(path)
# print(path_in_str)
Original answer:
import os
for filename in os.listdir("/path/to/dir/"):
if filename.endswith(".asm") or filename.endswith(".py"):
# print(os.path.join(directory, filename))
continue
else:
continue
This will iterate over all descendant files, not just the immediate children of the directory:
import os
for subdir, dirs, files in os.walk(rootdir):
for file in files:
#print os.path.join(subdir, file)
filepath = subdir + os.sep + file
if filepath.endswith(".asm"):
print (filepath)
You can try using glob module:
import glob
for filepath in glob.iglob('my_dir/*.asm'):
print(filepath)
and since Python 3.5 you can search subdirectories as well:
glob.glob('**/*.txt', recursive=True) # => ['2.txt', 'sub/3.txt']
From the docs:
The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell, although results are returned in arbitrary order. No tilde expansion is done, but *, ?, and character ranges expressed with [] will be correctly matched.
Since Python 3.5, things are much easier with os.scandir() and 2-20x faster (source):
with os.scandir(path) as it:
for entry in it:
if entry.name.endswith(".asm") and entry.is_file():
print(entry.name, entry.path)
Using scandir() instead of listdir() can significantly increase the
performance of code that also needs file type or file attribute
information, because os.DirEntry objects expose this information if
the operating system provides it when scanning a directory. All
os.DirEntry methods may perform a system call, but is_dir() and
is_file() usually only require a system call for symbolic links;
os.DirEntry.stat() always requires a system call on Unix but only
requires one for symbolic links on Windows.
Python 3.4 and later offer pathlib in the standard library. You could do:
from pathlib import Path
asm_pths = [pth for pth in Path.cwd().iterdir()
if pth.suffix == '.asm']
Or if you don't like list comprehensions:
asm_paths = []
for pth in Path.cwd().iterdir():
if pth.suffix == '.asm':
asm_pths.append(pth)
Path objects can easily be converted to strings.
Here's how I iterate through files in Python:
import os
path = 'the/name/of/your/path'
folder = os.fsencode(path)
filenames = []
for file in os.listdir(folder):
filename = os.fsdecode(file)
if filename.endswith( ('.jpeg', '.png', '.gif') ): # whatever file types you're using...
filenames.append(filename)
filenames.sort() # now you have the filenames and can do something with them
NONE OF THESE TECHNIQUES GUARANTEE ANY ITERATION ORDERING
Yup, super unpredictable. Notice that I sort the filenames, which is important if the order of the files matters, i.e. for video frames or time dependent data collection. Be sure to put indices in your filenames though!
You can use glob for referring the directory and the list :
import glob
import os
#to get the current working directory name
cwd = os.getcwd()
#Load the images from images folder.
for f in glob.glob('images\*.jpg'):
dir_name = get_dir_name(f)
image_file_name = dir_name + '.jpg'
#To print the file name with path (path will be in string)
print (image_file_name)
To get the list of all directory in array you can use os :
os.listdir(directory)
I'm not quite happy with this implementation yet, I wanted to have a custom constructor that does DirectoryIndex._make(next(os.walk(input_path))) such that you can just pass the path you want a file listing for. Edits welcome!
import collections
import os
DirectoryIndex = collections.namedtuple('DirectoryIndex', ['root', 'dirs', 'files'])
for file_name in DirectoryIndex(*next(os.walk('.'))).files:
file_path = os.path.join(path, file_name)
I really like using the scandir directive that is built into the os library. Here is a working example:
import os
i = 0
with os.scandir('/usr/local/bin') as root_dir:
for path in root_dir:
if path.is_file():
i += 1
print(f"Full path is: {path} and just the name is: {path.name}")
print(f"{i} files scanned successfully.")
Get all the .asm files in a directory by doing this.
import os
path = "path_to_file"
file_type = '.asm'
for filename in os.listdir(path=path):
if filename.endswith(file_type):
print(filename)
print(f"{path}/{filename}")
# do something below
I don't understand why some answers are complicated. This is how I would do it with Python 2.7. Replace DIRECTORY_TO_LOOP with the directory you want to use.
import os
DIRECTORY_TO_LOOP = '/var/www/files/'
for root, dirs, files in os.walk(DIRECTORY_TO_LOOP, topdown=False):
for name in files:
print(os.path.join(root, name))
This is what I have:
glob(os.path.join('src','*.c'))
but I want to search the subfolders of src. Something like this would work:
glob(os.path.join('src','*.c'))
glob(os.path.join('src','*','*.c'))
glob(os.path.join('src','*','*','*.c'))
glob(os.path.join('src','*','*','*','*.c'))
But this is obviously limited and clunky.
pathlib.Path.rglob
Use pathlib.Path.rglob from the pathlib module, which was introduced in Python 3.5.
from pathlib import Path
for path in Path('src').rglob('*.c'):
print(path.name)
If you don't want to use pathlib, use can use glob.glob('**/*.c'), but don't forget to pass in the recursive keyword parameter and it will use inordinate amount of time on large directories.
For cases where matching files beginning with a dot (.); like files in the current directory or hidden files on Unix based system, use the os.walk solution below.
os.walk
For older Python versions, use os.walk to recursively walk a directory and fnmatch.filter to match against a simple expression:
import fnmatch
import os
matches = []
for root, dirnames, filenames in os.walk('src'):
for filename in fnmatch.filter(filenames, '*.c'):
matches.append(os.path.join(root, filename))
For python >= 3.5 you can use **, recursive=True :
import glob
for f in glob.glob('/path/**/*.c', recursive=True):
print(f)
If recursive is True (default is False), the pattern ** will match any files and zero
or more directories and subdirectories. If the pattern is followed by
an os.sep, only directories and subdirectories match.
Python 3 Demo
Similar to other solutions, but using fnmatch.fnmatch instead of glob, since os.walk already listed the filenames:
import os, fnmatch
def find_files(directory, pattern):
for root, dirs, files in os.walk(directory):
for basename in files:
if fnmatch.fnmatch(basename, pattern):
filename = os.path.join(root, basename)
yield filename
for filename in find_files('src', '*.c'):
print 'Found C source:', filename
Also, using a generator alows you to process each file as it is found, instead of finding all the files and then processing them.
I've modified the glob module to support ** for recursive globbing, e.g:
>>> import glob2
>>> all_header_files = glob2.glob('src/**/*.c')
https://github.com/miracle2k/python-glob2/
Useful when you want to provide your users with the ability to use the ** syntax, and thus os.walk() alone is not good enough.
Starting with Python 3.4, one can use the glob() method of one of the Path classes in the new pathlib module, which supports ** wildcards. For example:
from pathlib import Path
for file_path in Path('src').glob('**/*.c'):
print(file_path) # do whatever you need with these files
Update:
Starting with Python 3.5, the same syntax is also supported by glob.glob().
import os
import fnmatch
def recursive_glob(treeroot, pattern):
results = []
for base, dirs, files in os.walk(treeroot):
goodfiles = fnmatch.filter(files, pattern)
results.extend(os.path.join(base, f) for f in goodfiles)
return results
fnmatch gives you exactly the same patterns as glob, so this is really an excellent replacement for glob.glob with very close semantics. An iterative version (e.g. a generator), IOW a replacement for glob.iglob, is a trivial adaptation (just yield the intermediate results as you go, instead of extending a single results list to return at the end).
You'll want to use os.walk to collect filenames that match your criteria. For example:
import os
cfiles = []
for root, dirs, files in os.walk('src'):
for file in files:
if file.endswith('.c'):
cfiles.append(os.path.join(root, file))
Here's a solution with nested list comprehensions, os.walk and simple suffix matching instead of glob:
import os
cfiles = [os.path.join(root, filename)
for root, dirnames, filenames in os.walk('src')
for filename in filenames if filename.endswith('.c')]
It can be compressed to a one-liner:
import os;cfiles=[os.path.join(r,f) for r,d,fs in os.walk('src') for f in fs if f.endswith('.c')]
or generalized as a function:
import os
def recursive_glob(rootdir='.', suffix=''):
return [os.path.join(looproot, filename)
for looproot, _, filenames in os.walk(rootdir)
for filename in filenames if filename.endswith(suffix)]
cfiles = recursive_glob('src', '.c')
If you do need full glob style patterns, you can follow Alex's and
Bruno's example and use fnmatch:
import fnmatch
import os
def recursive_glob(rootdir='.', pattern='*'):
return [os.path.join(looproot, filename)
for looproot, _, filenames in os.walk(rootdir)
for filename in filenames
if fnmatch.fnmatch(filename, pattern)]
cfiles = recursive_glob('src', '*.c')
Consider pathlib.rglob().
This is like calling Path.glob() with "**/" added in front of the given relative pattern:
import pathlib
for p in pathlib.Path("src").rglob("*.c"):
print(p)
See also #taleinat's related post here and a similar post elsewhere.
import os, glob
for each in glob.glob('path/**/*.c', recursive=True):
print(f'Name with path: {each} \nName without path: {os.path.basename(each)}')
glob.glob('*.c') :matches all files ending in .c in current directory
glob.glob('*/*.c') :same as 1
glob.glob('**/*.c') :matches all files ending in .c in the immediate subdirectories only, but not in the current directory
glob.glob('*.c',recursive=True) :same as 1
glob.glob('*/*.c',recursive=True) :same as 3
glob.glob('**/*.c',recursive=True) :matches all files ending in .c in the current directory and in all subdirectories
In case this may interest anyone, I've profiled the top three proposed methods.
I have about ~500K files in the globbed folder (in total), and 2K files that match the desired pattern.
here's the (very basic) code
import glob
import json
import fnmatch
import os
from pathlib import Path
from time import time
def find_files_iglob():
return glob.iglob("./data/**/data.json", recursive=True)
def find_files_oswalk():
for root, dirnames, filenames in os.walk('data'):
for filename in fnmatch.filter(filenames, 'data.json'):
yield os.path.join(root, filename)
def find_files_rglob():
return Path('data').rglob('data.json')
t0 = time()
for f in find_files_oswalk(): pass
t1 = time()
for f in find_files_rglob(): pass
t2 = time()
for f in find_files_iglob(): pass
t3 = time()
print(t1-t0, t2-t1, t3-t2)
And the results I got were:
os_walk: ~3.6sec
rglob ~14.5sec
iglob: ~16.9sec
The platform: Ubuntu 16.04, x86_64 (core i7),
Recently I had to recover my pictures with the extension .jpg. I ran photorec and recovered 4579 directories 2.2 million files within, having tremendous variety of extensions.With the script below I was able to select 50133 files havin .jpg extension within minutes:
#!/usr/binenv python2.7
import glob
import shutil
import os
src_dir = "/home/mustafa/Masaüstü/yedek"
dst_dir = "/home/mustafa/Genel/media"
for mediafile in glob.iglob(os.path.join(src_dir, "*", "*.jpg")): #"*" is for subdirectory
shutil.copy(mediafile, dst_dir)
based on other answers this is my current working implementation, which retrieves nested xml files in a root directory:
files = []
for root, dirnames, filenames in os.walk(myDir):
files.extend(glob.glob(root + "/*.xml"))
I'm really having fun with python :)
For python 3.5 and later
import glob
#file_names_array = glob.glob('path/*.c', recursive=True)
#above works for files directly at path/ as guided by NeStack
#updated version
file_names_array = glob.glob('path/**/*.c', recursive=True)
further you might need
for full_path_in_src in file_names_array:
print (full_path_in_src ) # be like 'abc/xyz.c'
#Full system path of this would be like => 'path till src/abc/xyz.c'
Johan and Bruno provide excellent solutions on the minimal requirement as stated. I have just released Formic which implements Ant FileSet and Globs which can handle this and more complicated scenarios. An implementation of your requirement is:
import formic
fileset = formic.FileSet(include="/src/**/*.c")
for file_name in fileset.qualified_files():
print file_name
Another way to do it using just the glob module. Just seed the rglob method with a starting base directory and a pattern to match and it will return a list of matching file names.
import glob
import os
def _getDirs(base):
return [x for x in glob.iglob(os.path.join( base, '*')) if os.path.isdir(x) ]
def rglob(base, pattern):
list = []
list.extend(glob.glob(os.path.join(base,pattern)))
dirs = _getDirs(base)
if len(dirs):
for d in dirs:
list.extend(rglob(os.path.join(base,d), pattern))
return list
Or with a list comprehension:
>>> base = r"c:\User\xtofl"
>>> binfiles = [ os.path.join(base,f)
for base, _, files in os.walk(root)
for f in files if f.endswith(".jpg") ]
If the files are on a remote file system or inside an archive, you can use an implementation of the fsspec AbstractFileSystem class. For example, to list all the files in a zipfile:
from fsspec.implementations.zip import ZipFileSystem
fs = ZipFileSystem("/tmp/test.zip")
fs.glob("/**") # equivalent: fs.find("/")
or to list all the files in a publicly available S3 bucket:
from s3fs import S3FileSystem
fs_s3 = S3FileSystem(anon=True)
fs_s3.glob("noaa-goes16/ABI-L1b-RadF/2020/045/**") # or use fs_s3.find
you can also use it for a local filesystem, which may be interesting if your implementation should be filesystem-agnostic:
from fsspec.implementations.local import LocalFileSystem
fs = LocalFileSystem()
fs.glob("/tmp/test/**")
Other implementations include Google Cloud, Github, SFTP/SSH, Dropbox, and Azure. For details, see the fsspec API documentation.
Just made this.. it will print files and directory in hierarchical way
But I didn't used fnmatch or walk
#!/usr/bin/python
import os,glob,sys
def dirlist(path, c = 1):
for i in glob.glob(os.path.join(path, "*")):
if os.path.isfile(i):
filepath, filename = os.path.split(i)
print '----' *c + filename
elif os.path.isdir(i):
dirname = os.path.basename(i)
print '----' *c + dirname
c+=1
dirlist(i,c)
c-=1
path = os.path.normpath(sys.argv[1])
print(os.path.basename(path))
dirlist(path)
That one uses fnmatch or regular expression:
import fnmatch, os
def filepaths(directory, pattern):
for root, dirs, files in os.walk(directory):
for basename in files:
try:
matched = pattern.match(basename)
except AttributeError:
matched = fnmatch.fnmatch(basename, pattern)
if matched:
yield os.path.join(root, basename)
# usage
if __name__ == '__main__':
from pprint import pprint as pp
import re
path = r'/Users/hipertracker/app/myapp'
pp([x for x in filepaths(path, re.compile(r'.*\.py$'))])
pp([x for x in filepaths(path, '*.py')])
In addition to the suggested answers, you can do this with some lazy generation and list comprehension magic:
import os, glob, itertools
results = itertools.chain.from_iterable(glob.iglob(os.path.join(root,'*.c'))
for root, dirs, files in os.walk('src'))
for f in results: print(f)
Besides fitting in one line and avoiding unnecessary lists in memory, this also has the nice side effect, that you can use it in a way similar to the ** operator, e.g., you could use os.path.join(root, 'some/path/*.c') in order to get all .c files in all sub directories of src that have this structure.
This is a working code on Python 2.7. As part of my devops work, I was required to write a script which would move the config files marked with live-appName.properties to appName.properties. There could be other extension files as well like live-appName.xml.
Below is a working code for this, which finds the files in the given directories (nested level) and then renames (moves) it to the required filename
def flipProperties(searchDir):
print "Flipping properties to point to live DB"
for root, dirnames, filenames in os.walk(searchDir):
for filename in fnmatch.filter(filenames, 'live-*.*'):
targetFileName = os.path.join(root, filename.split("live-")[1])
print "File "+ os.path.join(root, filename) + "will be moved to " + targetFileName
shutil.move(os.path.join(root, filename), targetFileName)
This function is called from a main script
flipProperties(searchDir)
Hope this helps someone struggling with similar issues.
Simplified version of Johan Dahlin's answer, without fnmatch.
import os
matches = []
for root, dirnames, filenames in os.walk('src'):
matches += [os.path.join(root, f) for f in filenames if f[-2:] == '.c']
Here is my solution using list comprehension to search for multiple file extensions recursively in a directory and all subdirectories:
import os, glob
def _globrec(path, *exts):
""" Glob recursively a directory and all subdirectories for multiple file extensions
Note: Glob is case-insensitive, i. e. for '\*.jpg' you will get files ending
with .jpg and .JPG
Parameters
----------
path : str
A directory name
exts : tuple
File extensions to glob for
Returns
-------
files : list
list of files matching extensions in exts in path and subfolders
"""
dirs = [a[0] for a in os.walk(path)]
f_filter = [d+e for d in dirs for e in exts]
return [f for files in [glob.iglob(files) for files in f_filter] for f in files]
my_pictures = _globrec(r'C:\Temp', '\*.jpg','\*.bmp','\*.png','\*.gif')
for f in my_pictures:
print f
import sys, os, glob
dir_list = ["c:\\books\\heap"]
while len(dir_list) > 0:
cur_dir = dir_list[0]
del dir_list[0]
list_of_files = glob.glob(cur_dir+'\\*')
for book in list_of_files:
if os.path.isfile(book):
print(book)
else:
dir_list.append(book)
I modified the top answer in this posting.. and recently created this script which will loop through all files in a given directory (searchdir) and the sub-directories under it... and prints filename, rootdir, modified/creation date, and size.
Hope this helps someone... and they can walk the directory and get fileinfo.
import time
import fnmatch
import os
def fileinfo(file):
filename = os.path.basename(file)
rootdir = os.path.dirname(file)
lastmod = time.ctime(os.path.getmtime(file))
creation = time.ctime(os.path.getctime(file))
filesize = os.path.getsize(file)
print "%s**\t%s\t%s\t%s\t%s" % (rootdir, filename, lastmod, creation, filesize)
searchdir = r'D:\Your\Directory\Root'
matches = []
for root, dirnames, filenames in os.walk(searchdir):
## for filename in fnmatch.filter(filenames, '*.c'):
for filename in filenames:
## matches.append(os.path.join(root, filename))
##print matches
fileinfo(os.path.join(root, filename))
Here is a solution that will match the pattern against the full path and not just the base filename.
It uses fnmatch.translate to convert a glob-style pattern into a regular expression, which is then matched against the full path of each file found while walking the directory.
re.IGNORECASE is optional, but desirable on Windows since the file system itself is not case-sensitive. (I didn't bother compiling the regex because docs indicate it should be cached internally.)
import fnmatch
import os
import re
def findfiles(dir, pattern):
patternregex = fnmatch.translate(pattern)
for root, dirs, files in os.walk(dir):
for basename in files:
filename = os.path.join(root, basename)
if re.search(patternregex, filename, re.IGNORECASE):
yield filename
I needed a solution for python 2.x that works fast on large directories.
I endet up with this:
import subprocess
foundfiles= subprocess.check_output("ls src/*.c src/**/*.c", shell=True)
for foundfile in foundfiles.splitlines():
print foundfile
Note that you might need some exception handling in case ls doesn't find any matching file.
This question already has answers here:
Find all files in a directory with extension .txt in Python
(25 answers)
Closed 2 months ago.
I am trying to find all the .c files in a directory using Python.
I wrote this, but it is just returning me all files - not just .c files:
import os
import re
results = []
for folder in gamefolders:
for f in os.listdir(folder):
if re.search('.c', f):
results += [f]
print results
How can I just get the .c files?
try changing the inner loop to something like this
results += [each for each in os.listdir(folder) if each.endswith('.c')]
Try "glob":
>>> import glob
>>> glob.glob('./[0-9].*')
['./1.gif', './2.txt']
>>> glob.glob('*.gif')
['1.gif', 'card.gif']
>>> glob.glob('?.gif')
['1.gif']
KISS
# KISS
import os
results = []
for folder in gamefolders:
for f in os.listdir(folder):
if f.endswith('.c'):
results.append(f)
print results
There is a better solution that directly using regular expressions, it is the standard library's module fnmatch for dealing with file name patterns. (See also glob module.)
Write a helper function:
import fnmatch
import os
def listdir(dirname, pattern="*"):
return fnmatch.filter(os.listdir(dirname), pattern)
and use it as follows:
result = listdir("./sources", "*.c")
for _,_,filenames in os.walk(folder):
for file in filenames:
fileExt=os.path.splitext(file)[-1]
if fileExt == '.c':
results.append(file)
For another alternative you could use fnmatch
import fnmatch
import os
results = []
for root, dirs, files in os.walk(path)
for _file in files:
if fnmatch.fnmatch(_file, '*.c'):
results.append(os.path.join(root, _file))
print results
or with a list comprehension:
for root, dirs, files in os.walk(path)
[results.append(os.path.join(root, _file))\
for _file in files if \
fnmatch.fnmatch(_file, '*.c')]
or using filter:
for root, dirs, files in os.walk(path):
[results.append(os.path.join(root, _file))\
for _file in fnmatch.filter(files, '*.c')]
Change the directory to the given path, so that you can search files within directory. If you don't change the directory then this code will search files in your present directory location:
import os #importing os library
import glob #importing glob library
path=raw_input() #input from the user
os.chdir(path)
filedata=glob.glob('*.c') #all files with .c extenstions stores in filedata.
print filedata
import os, re
cfile = re.compile("^.*?\.c$")
results = []
for name in os.listdir(directory):
if cfile.match(name):
results.append(name)
The implementation of shutil.copytree is in the docs. I mofdified it to take a list of extentions to INCLUDE.
def my_copytree(src, dst, symlinks=False, *extentions):
""" I modified the 2.7 implementation of shutils.copytree
to take a list of extentions to INCLUDE, instead of an ignore list.
"""
names = os.listdir(src)
os.makedirs(dst)
errors = []
for name in names:
srcname = os.path.join(src, name)
dstname = os.path.join(dst, name)
try:
if symlinks and os.path.islink(srcname):
linkto = os.readlink(srcname)
os.symlink(linkto, dstname)
elif os.path.isdir(srcname):
my_copytree(srcname, dstname, symlinks, *extentions)
else:
ext = os.path.splitext(srcname)[1]
if not ext in extentions:
# skip the file
continue
copy2(srcname, dstname)
# XXX What about devices, sockets etc.?
except (IOError, os.error), why:
errors.append((srcname, dstname, str(why)))
# catch the Error from the recursive copytree so that we can
# continue with other files
except Error, err:
errors.extend(err.args[0])
try:
copystat(src, dst)
# except WindowsError: # cant copy file access times on Windows
# pass
except OSError, why:
errors.extend((src, dst, str(why)))
if errors:
raise Error(errors)
Usage: For example, to copy only .config and .bat files....
my_copytree(source, targ, '.config', '.bat')
this is pretty clean.
the commands come from the os library.
this code will search through the current working directory and list only the specified file type. You can change this by replacing 'os.getcwd()' with your target directory and choose the file type by replacing '(ext)'. os.fsdecode is so you don't get a bytewise error from .endswith(). this also sorts alphabetically, you can remove sorted() for the raw list.
import os
filenames = sorted([os.fsdecode(file) for file in os.listdir(os.getcwd()) if os.fsdecode(file).endswith(".(ext)")])
Here's yet another solution, using pathlib (and Python 3):
from pathlib import Path
gamefolder = "path/to/dir"
result = sorted(Path(gamefolder).glob("**.c"))
Notice the double asterisk (**) in the glob() argument. This will search the gamefolder as well as its subdirectories. If you only want to search the gamefolder, use a single * in the pattern: "*.c". For more details, see the documentation.
If you replace '.c' with '[.]c$', you're searching for files that contain .c as the last two characters of the name, rather than all files that contain a c, with at least one character before it.
Edit: Alternatively, match f[-2:] with '.c', this MAY be computationally cheaper than pulling out a regexp match.
Just to be clear, if you wanted the dot character in your search term, you could've escaped it too:
'.*[backslash].c' would give you what you needed, plus you would need to use something like:
results.append(f), instead of what you had listed as results += [f]
This function returns a list of all file names with the specified extension that live in the specified directory:
import os
def listFiles(path, extension):
return [f for f in os.listdir(path) if f.endswith(extension)]
print listFiles('/Path/to/directory/with/files', '.txt')
If you want to list all files with the specified extension in a certain directory and its subdirectories you could do:
import os
def filterFiles(path, extension):
return [file for root, dirs, files in os.walk(path) for file in files if file.endswith(extension)]
print filterFiles('/Path/to/directory/with/files', '.txt')
You can actually do this with just os.listdir
import os
results = [f for f in os.listdir(gamefolders/folder) if f.endswith('.c')]