Directory-tree listing in Python - python

How do I get a list of all files (and directories) in a given directory in Python?

This is a way to traverse every file and directory in a directory tree:
import os
for dirname, dirnames, filenames in os.walk('.'):
# print path to all subdirectories first.
for subdirname in dirnames:
print(os.path.join(dirname, subdirname))
# print path to all filenames.
for filename in filenames:
print(os.path.join(dirname, filename))
# Advanced usage:
# editing the 'dirnames' list will stop os.walk() from recursing into there.
if '.git' in dirnames:
# don't go into any .git directories.
dirnames.remove('.git')

You can use
os.listdir(path)
For reference and more os functions look here:
Python 2 docs: https://docs.python.org/2/library/os.html#os.listdir
Python 3 docs: https://docs.python.org/3/library/os.html#os.listdir

Here's a helper function I use quite often:
import os
def listdir_fullpath(d):
return [os.path.join(d, f) for f in os.listdir(d)]

import os
for filename in os.listdir("C:\\temp"):
print filename

If you need globbing abilities, there's a module for that as well. For example:
import glob
glob.glob('./[0-9].*')
will return something like:
['./1.gif', './2.txt']
See the documentation here.

For files in current working directory without specifying a path
Python 2.7:
import os
os.listdir('.')
Python 3.x:
import os
os.listdir()

Try this:
import os
for top, dirs, files in os.walk('./'):
for nm in files:
print os.path.join(top, nm)

While os.listdir() is fine for generating a list of file and dir names, frequently you want to do more once you have those names - and in Python3, pathlib makes those other chores simple. Let's take a look and see if you like it as much as I do.
To list dir contents, construct a Path object and grab the iterator:
In [16]: Path('/etc').iterdir()
Out[16]: <generator object Path.iterdir at 0x110853fc0>
If we want just a list of names of things:
In [17]: [x.name for x in Path('/etc').iterdir()]
Out[17]:
['emond.d',
'ntp-restrict.conf',
'periodic',
If you want just the dirs:
In [18]: [x.name for x in Path('/etc').iterdir() if x.is_dir()]
Out[18]:
['emond.d',
'periodic',
'mach_init.d',
If you want the names of all conf files in that tree:
In [20]: [x.name for x in Path('/etc').glob('**/*.conf')]
Out[20]:
['ntp-restrict.conf',
'dnsextd.conf',
'syslog.conf',
If you want a list of conf files in the tree >= 1K:
In [23]: [x.name for x in Path('/etc').glob('**/*.conf') if x.stat().st_size > 1024]
Out[23]:
['dnsextd.conf',
'pf.conf',
'autofs.conf',
Resolving relative paths become easy:
In [32]: Path('../Operational Metrics.md').resolve()
Out[32]: PosixPath('/Users/starver/code/xxxx/Operational Metrics.md')
Navigating with a Path is pretty clear (although unexpected):
In [10]: p = Path('.')
In [11]: core = p / 'web' / 'core'
In [13]: [x for x in core.iterdir() if x.is_file()]
Out[13]:
[PosixPath('web/core/metrics.py'),
PosixPath('web/core/services.py'),
PosixPath('web/core/querysets.py'),

A recursive implementation
import os
def scan_dir(dir):
for name in os.listdir(dir):
path = os.path.join(dir, name)
if os.path.isfile(path):
print path
else:
scan_dir(path)

I wrote a long version, with all the options I might need: http://sam.nipl.net/code/python/find.py
I guess it will fit here too:
#!/usr/bin/env python
import os
import sys
def ls(dir, hidden=False, relative=True):
nodes = []
for nm in os.listdir(dir):
if not hidden and nm.startswith('.'):
continue
if not relative:
nm = os.path.join(dir, nm)
nodes.append(nm)
nodes.sort()
return nodes
def find(root, files=True, dirs=False, hidden=False, relative=True, topdown=True):
root = os.path.join(root, '') # add slash if not there
for parent, ldirs, lfiles in os.walk(root, topdown=topdown):
if relative:
parent = parent[len(root):]
if dirs and parent:
yield os.path.join(parent, '')
if not hidden:
lfiles = [nm for nm in lfiles if not nm.startswith('.')]
ldirs[:] = [nm for nm in ldirs if not nm.startswith('.')] # in place
if files:
lfiles.sort()
for nm in lfiles:
nm = os.path.join(parent, nm)
yield nm
def test(root):
print "* directory listing, with hidden files:"
print ls(root, hidden=True)
print
print "* recursive listing, with dirs, but no hidden files:"
for f in find(root, dirs=True):
print f
print
if __name__ == "__main__":
test(*sys.argv[1:])

Here is another option.
os.scandir(path='.')
It returns an iterator of os.DirEntry objects corresponding to the entries (along with file attribute information) in the directory given by path.
Example:
with os.scandir(path) as it:
for entry in it:
if not entry.name.startswith('.'):
print(entry.name)
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 Docs

The one worked with me is kind of a modified version from Saleh's answer elsewhere on this page.
The code is as follows:
dir = 'given_directory_name'
filenames = [os.path.abspath(os.path.join(dir,i)) for i in os.listdir(dir)]

A nice one liner to list only the files recursively. I used this in my setup.py package_data directive:
import os
[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]
I know it's not the answer to the question, but may come in handy

For Python 2
#!/bin/python2
import os
def scan_dir(path):
print map(os.path.abspath, os.listdir(pwd))
For Python 3
For filter and map, you need wrap them with list()
#!/bin/python3
import os
def scan_dir(path):
print(list(map(os.path.abspath, os.listdir(pwd))))
The recommendation now is that you replace your usage of map and filter with generators expressions or list comprehensions:
#!/bin/python
import os
def scan_dir(path):
print([os.path.abspath(f) for f in os.listdir(path)])

#import modules
import os
_CURRENT_DIR = '.'
def rec_tree_traverse(curr_dir, indent):
"recurcive function to traverse the directory"
#print "[traverse_tree]"
try :
dfList = [os.path.join(curr_dir, f_or_d) for f_or_d in os.listdir(curr_dir)]
except:
print "wrong path name/directory name"
return
for file_or_dir in dfList:
if os.path.isdir(file_or_dir):
#print "dir : ",
print indent, file_or_dir,"\\"
rec_tree_traverse(file_or_dir, indent*2)
if os.path.isfile(file_or_dir):
#print "file : ",
print indent, file_or_dir
#end if for loop
#end of traverse_tree()
def main():
base_dir = _CURRENT_DIR
rec_tree_traverse(base_dir," ")
raw_input("enter any key to exit....")
#end of main()
if __name__ == '__main__':
main()

FYI Add a filter of extension or ext file
import os
path = '.'
for dirname, dirnames, filenames in os.walk(path):
# print path to all filenames with extension py.
for filename in filenames:
fname_path = os.path.join(dirname, filename)
fext = os.path.splitext(fname_path)[1]
if fext == '.py':
print fname_path
else:
continue

If figured I'd throw this in. Simple and dirty way to do wildcard searches.
import re
import os
[a for a in os.listdir(".") if re.search("^.*\.py$",a)]

Below code will list directories and the files within the dir
def print_directory_contents(sPath):
import os
for sChild in os.listdir(sPath):
sChildPath = os.path.join(sPath,sChild)
if os.path.isdir(sChildPath):
print_directory_contents(sChildPath)
else:
print(sChildPath)

Here is a one line Pythonic version:
import os
dir = 'given_directory_name'
filenames = [os.path.join(os.path.dirname(os.path.abspath(__file__)),dir,i) for i in os.listdir(dir)]
This code lists the full path of all files and directories in the given directory name.

I know this is an old question. This is a neat way I came across if you are on a liunx machine.
import subprocess
print(subprocess.check_output(["ls", "/"]).decode("utf8"))

Easiest way:
list_output_files = [os.getcwd()+"\\"+f for f in os.listdir(os.getcwd())]

Related

How to loop and optimise data extraction - Python [duplicate]

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))

Find and copy files from directories using wildcard [duplicate]

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.

How can I check if there are any more JSON files remaining in a directory in Python? [duplicate]

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.

How can I iterate over files in a given directory?

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))

Rename multiple files in a directory in Python

I'm trying to rename some files in a directory using Python.
Say I have a file called CHEESE_CHEESE_TYPE.*** and want to remove CHEESE_ so my resulting filename would be CHEESE_TYPE
I'm trying to use the os.path.split but it's not working properly. I have also considered using string manipulations, but have not been successful with that either.
Use os.rename(src, dst) to rename or move a file or a directory.
$ ls
cheese_cheese_type.bar cheese_cheese_type.foo
$ python
>>> import os
>>> for filename in os.listdir("."):
... if filename.startswith("cheese_"):
... os.rename(filename, filename[7:])
...
>>>
$ ls
cheese_type.bar cheese_type.foo
Here's a script based on your newest comment.
#!/usr/bin/env python
from os import rename, listdir
badprefix = "cheese_"
fnames = listdir('.')
for fname in fnames:
if fname.startswith(badprefix*2):
rename(fname, fname.replace(badprefix, '', 1))
The following code should work. It takes every filename in the current directory, if the filename contains the pattern CHEESE_CHEESE_ then it is renamed. If not nothing is done to the filename.
import os
for fileName in os.listdir("."):
os.rename(fileName, fileName.replace("CHEESE_CHEESE_", "CHEESE_"))
Assuming you are already in the directory, and that the "first 8 characters" from your comment hold true always. (Although "CHEESE_" is 7 characters... ? If so, change the 8 below to 7)
from glob import glob
from os import rename
for fname in glob('*.prj'):
rename(fname, fname[8:])
I have the same issue, where I want to replace the white space in any pdf file to a dash -.
But the files were in multiple sub-directories. So, I had to use os.walk().
In your case for multiple sub-directories, it could be something like this:
import os
for dpath, dnames, fnames in os.walk('/path/to/directory'):
for f in fnames:
os.chdir(dpath)
if f.startswith('cheese_'):
os.rename(f, f.replace('cheese_', ''))
Try this:
import os
import shutil
for file in os.listdir(dirpath):
newfile = os.path.join(dirpath, file.split("_",1)[1])
shutil.move(os.path.join(dirpath,file),newfile)
I'm assuming you don't want to remove the file extension, but you can just do the same split with periods.
This sort of stuff is perfectly fitted for IPython, which has shell integration.
In [1] files = !ls
In [2] for f in files:
newname = process_filename(f)
mv $f $newname
Note: to store this in a script, use the .ipy extension, and prefix all shell commands with !.
See also: http://ipython.org/ipython-doc/stable/interactive/shell.html
Here is a more general solution:
This code can be used to remove any particular character or set of characters recursively from all filenames within a directory and replace them with any other character, set of characters or no character.
import os
paths = (os.path.join(root, filename)
for root, _, filenames in os.walk('C:\FolderName')
for filename in filenames)
for path in paths:
# the '#' in the example below will be replaced by the '-' in the filenames in the directory
newname = path.replace('#', '-')
if newname != path:
os.rename(path, newname)
It seems that your problem is more in determining the new file name rather than the rename itself (for which you could use the os.rename method).
It is not clear from your question what the pattern is that you want to be renaming. There is nothing wrong with string manipulation. A regular expression may be what you need here.
import os
import string
def rename_files():
#List all files in the directory
file_list = os.listdir("/Users/tedfuller/Desktop/prank/")
print(file_list)
#Change current working directory and print out it's location
working_location = os.chdir("/Users/tedfuller/Desktop/prank/")
working_location = os.getcwd()
print(working_location)
#Rename all the files in that directory
for file_name in file_list:
os.rename(file_name, file_name.translate(str.maketrans("","",string.digits)))
rename_files()
This command will remove the initial "CHEESE_" string from all the files in the current directory, using renamer:
$ renamer --find "/^CHEESE_/" *
I was originally looking for some GUI which would allow renaming using regular expressions and which had a preview of the result before applying changes.
On Linux I have successfully used krename, on Windows Total Commander does renaming with regexes, but I found no decent free equivalent for OSX, so I ended up writing a python script which works recursively and by default only prints the new file names without making any changes. Add the '-w' switch to actually modify the file names.
#!/usr/bin/python
# -*- coding: utf-8 -*-
import os
import fnmatch
import sys
import shutil
import re
def usage():
print """
Usage:
%s <work_dir> <search_regex> <replace_regex> [-w|--write]
By default no changes are made, add '-w' or '--write' as last arg to actually rename files
after you have previewed the result.
""" % (os.path.basename(sys.argv[0]))
def rename_files(directory, search_pattern, replace_pattern, write_changes=False):
pattern_old = re.compile(search_pattern)
for path, dirs, files in os.walk(os.path.abspath(directory)):
for filename in fnmatch.filter(files, "*.*"):
if pattern_old.findall(filename):
new_name = pattern_old.sub(replace_pattern, filename)
filepath_old = os.path.join(path, filename)
filepath_new = os.path.join(path, new_name)
if not filepath_new:
print 'Replacement regex {} returns empty value! Skipping'.format(replace_pattern)
continue
print new_name
if write_changes:
shutil.move(filepath_old, filepath_new)
else:
print 'Name [{}] does not match search regex [{}]'.format(filename, search_pattern)
if __name__ == '__main__':
if len(sys.argv) < 4:
usage()
sys.exit(-1)
work_dir = sys.argv[1]
search_regex = sys.argv[2]
replace_regex = sys.argv[3]
write_changes = (len(sys.argv) > 4) and sys.argv[4].lower() in ['--write', '-w']
rename_files(work_dir, search_regex, replace_regex, write_changes)
Example use case
I want to flip parts of a file name in the following manner, i.e. move the bit m7-08 to the beginning of the file name:
# Before:
Summary-building-mobile-apps-ionic-framework-angularjs-m7-08.mp4
# After:
m7-08_Summary-building-mobile-apps-ionic-framework-angularjs.mp4
This will perform a dry run, and print the new file names without actually renaming any files:
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1"
This will do the actual renaming (you can use either -w or --write):
rename_files_regex.py . "([^\.]+?)-(m\\d+-\\d+)" "\\2_\\1" --write
You can use os.system function for simplicity and to invoke bash to accomplish the task:
import os
os.system('mv old_filename new_filename')
This works for me.
import os
for afile in os.listdir('.'):
filename, file_extension = os.path.splitext(afile)
if not file_extension == '.xyz':
os.rename(afile, filename + '.abc')
What about this :
import re
p = re.compile(r'_')
p.split(filename, 1) #where filename is CHEESE_CHEESE_TYPE.***

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