I am new to Python coding so here a question. I want to find files that are called "untitled" with any kind of extension, e.g. jpg, indd, psd. Then rename them to the date of the current day.
I have tried the following:
import os
for file in os.listdir("/Users/shirin/Desktop/Artez"):
if file.endswith("untitled.*"):
print(file)
When I run the script, nothing happens.
You might find the glob function more useful in this situation:
import glob
for file in glob.glob("/Users/shirin/Desktop/Artez/untitled.*"):
print(file)
Your function does not print anything as there are probably no files ending with .* in the name. The glob.glob() function will carry out the file expansion for you.
You can then use this to do your file renaming as follows:
import glob
import os
from datetime import datetime
current_day = datetime.now().strftime("%Y-%m-%d")
for source_name in glob.glob("/Users/shirin/Desktop/Artez/untitled.*"):
path, fullname = os.path.split(source_name)
basename, ext = os.path.splitext(fullname)
target_name = os.path.join(path, '{}{}'.format(current_day, ext))
os.rename(source_name, target_name)
A Python datetime object can be used to get you a suitable timestamp.
Python string comparator does not support wildcards. You can search for "untitled." anywhere in the text:
import os
for file in os.listdir("/Users/shirin/Desktop/Artez"):
if "untitled." in file:
print(file)
keep in mind that this will include any file that has "untitled." at any location of the file.
try with this approach
import os
directoryPath = '/Users/shirin/Desktop/Artez'
lstDir = os.walk(directoryPath)
for root, dirs, files in lstDir:
for fichero in files:
(filename, extension) = os.path.splitext(fichero)
if filename.find('untitle') != -1: # == 0 if starting with untitle
os.system('mv '+directoryPath+filename+extension+' '+directoryPath+'$(date +"%Y_%m_%d")'+filename+extension)
import os
for file in os.listdir("/Users/shirin/Desktop/Artez"):
if(file.startswith("untitled")):
os.rename(file, datetime.date.today().strftime("%B %d, %Y") + "." + file.split(".")[-1])
Related
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))
How can i apply a script for file modification in all subfolder (in python)?
I created a little script for rename some pictures but my program only change pictures in the script folder and not in subfolder.
from PIL import Image
from os import *
import sys
from os.path import basename
import os
#from PIL.ExifTags import TAGS
from datetime import datetime
extension = ''
#path='/home/pi/Desktop/testrename'
folder_path = "/home/pi/Desktop/testrename/"
l=[]
import PIL.Image
from os import walk
#from path import path
import glob
EXIF_DATETIME = 36867
def renamefinaljpeg() :
glob.glob ('*/.jpeg')
if len(fname) < 20 :
try :
old = PIL.Image.open(fname)._getexif()[EXIF_DATETIME]
old2 = old.split(' ')
os.rename (fname, "yes" + old2[0]+' '+fname)
print('fait')
except :
pass
print('pas jpeg')
def renamefinaljpg() :
glob.glob ('*/.jpg')
if len(fname) < 20 :
try :
old = PIL.Image.open(fname)._getexif()[EXIF_DATETIME]
old2 = old.split(' ')
os.rename (fname, "yes" + old2[0]+' '+fname)
print('fait')
except :
pass
print('pas jpg')
rootDir = "/home/pi/Desktop/testrename/"
for dirName, subdirlist, fileList in os.walk(rootDir) :
for fname in fileList :
print(fname)
try :
renamefinaljpg() or renamefinaljpeg()
except :
pass
print('passe')
The image are renamed in the main directory but not in directory tree (but they are read)
Thank you for your help.
It looks like you've already got a lot of ideas for how to do it in your script. Let's use os.walk since that's probably the most straightforward. os.walk will iterate over all of our directories recursively and give us the list of filenames contained in each one. To filter the files to only .jpg we can use fnmatch.fnmatch.
import fnmatch
import os
import sys
from PIL import Image
folder_path = '/home/pi/Desktop/testrename/old'
# Give special numbers a specific name so it's easier to remember what it
# actually means.
EXIF_DATETIME = 36867
def renamefinal(dir_path, filename):
try:
old = Image.open(file_path)._getexif()[EXIF_DATETIME]
date, time = old.split(' ', maxsplit=1)
new_filename = date + ' ' + filename
os.rename(
# Source file name, including directory and filename
os.path.join(dir_path, filename),
# Destination file name, including date
os.path.join(dir_path, new_filename))
print(
'jpg renommé ({}): {} to {}'.format(
dir_path, filename, new_filename))
except:
# Including filename in our output so we know what to check if
# something goes wrong.
print('pas jpg ({}): {}'.format(dir_path, filename))
for path, dirs, files in os.walk(folder_path):
for filename in files:
if not fnmatch.fnmatch(filename, '*.jpg'):
# Go to the next file, skipping the rest of the loop for this file.
continue
if 30 <= len(filename):
continue
renamefinal(path, filename)
Some more that might help:
If you want to take arguments to your script check out the argparse module.
Try to catch exceptions more specifically, for example using except KeyError: instead of except: when you know that EXIF_DATETIME may not be in the exif data - also try putting your try: ... except: ... block only around the lines that might actually fail.
Check out the logging module instead of using print to show information about what your script is doing.
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))
I have 700 files in a single folder. I need to find files that have "h10v03" as part of the name and copy them to a different folder using python.
Heres an example of one of the files: MOD10A1.A2000121.h10v03.005.2007172062725.hdf
I appreciate any help.
Something like this would do the trick.
import os
import shutil
source_dir = "/some/directory/path"
target_dir = "/some/other/directory/path"
part = "h10v03"
files = [file for file in os.listdir(source_dir)
if os.path.isfile(file) and part in file]
for file in files:
shutil.copy2(os.path.join(source_dir, file), target_dir)
Does it need to be python?
A unix shell does that for you quite fine:
cp ./*h10v03* /other/directory/
In python I would suggest you take a look at os.listdir() and shutil.copy()
EDIT:
some untested code:
import os
import shutil
src_dir = "/some/path/"
target_dir = "/some/other/path/"
searchstring = "h10v03"
for f in os.listdir(src_dir):
if searchstring in f and os.path.isfile(os.path.join(src_dir, f)):
shutil.copy2(os.path.join(src_dir, f), target_dir)
print "COPY", f
with the glob module (untested):
import glob
import os
import shutil
for f in glob.glob("/some/path/*2000*h10v03*"):
print f
shutil.copy2(f, os.path.join("/some/target/dir/", os.path.basename(f)))
Firstly, find all the items in that folder with os.listdir. Then you can use the count() method of string to determine if it has your string. Then you can use shutil to copy the file.
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.***