I need to change a prefix for a current file.
An example would look as follows:
from pathlib import Path
file = Path('/Users/my_name/PYTHON/Playing_Around/testing_lm.py')
# Current file with destination
print(file)
# Prefix to be used
file_prexif = 'A'
# Hardcoding wanted results.
Path('/Users/my_name/PYTHON/Playing_Around/A_testing_lm.py')
As can be seen hardcoding it is easy. However is there a way to automate this step?
There is a pseudo - idea of what I want to do:
str(file).split('/')[-1] = str(file_prexif) + str('_') + str(file).split('/')[-1]
I only want to change last element of PosixPath file. However it is not possible to change only last element of string
file.stem accesses the base name of the file without extension.
file.with_stem() (added in Python 3.9) returns an updated Path with a new stem:
from pathlib import Path
file = Path('/Users/my_name/PYTHON/Playing_Around/testing_lm.py')
print(file.with_stem(f'A_{file.stem}'))
\Users\my_name\PYTHON\Playing_Around\A_testing_lm.py
Use file.parent to get the parent of the path and file.name to get the final path component, excluding the drive and root.
from pathlib import Path
file = Path('/Users/my_name/PYTHON/Playing_Around/testing_lm.py')
file_prexif_lst = ['A','B','C']
for prefix in file_prexif_lst:
p = file.parent.joinpath(f'{prefix}_{file.name}')
print(p)
/Users/my_name/PYTHON/Playing_Around/A_testing_lm.py
/Users/my_name/PYTHON/Playing_Around/B_testing_lm.py
/Users/my_name/PYTHON/Playing_Around/C_testing_lm.py
This is my code
path = r'C:/Users/user/documents/data'
def new_name(path, keyword, ext="csv"):
return ('\\'.join(path.split("\\")[:-1]) + "\\" + path.split("\\")[-1].split(".")[0] + keyword + "." + ext)
new_path = new_name(path, '_test')
print(new_path)
>>> \C:/Users/user/documents/data_test.csv # Output
C:/Users/user/documents/data_test.csv # What I want
I get a path with a file and I would like to add a keyword to this file. Does this option also exist with Pathlib (or something like that)?
Since this path doesn't work properly on Mac, I would like to build a more universal solution.
The pathlib module (Python 3.4+) is your friend when working with paths in Python.
In special, the class pathlib.Path:
This class represents concrete paths of the system’s path flavour (instantiating it creates either a PosixPath or a WindowsPath).
Which means that you don't have to worry about OS-specific path details such as forward/backward slashes.
So the above example could be rewritten as:
from pathlib import Path
path = Path('C:/Users/user/documents/data')
def new_name(path, keyword, ext="csv"):
return Path(str(path) + keyword + '.' + ext)
new_path = new_name(path, '_test')
print(new_path)
>>> C:/Users/user/documents/data_test.csv
I'd use the os.path.join command in this case
for example:
Assume you are running a script from *C:\temp*
and want to load "file_name=inputfile.json" from ./json folder.
You need to concatenate the paths to get one absolute path to the filename: C:\temp\json\inputfile.json
Code would look like:
from os import path
file_name = "inpoutfile.json"
abs_file_path = os.path.join(os.path.dirname(__file__),"json", file_name)
print (abs_file_path)
result is:
C:\temp\json\inputfile.json
How can I rename many files from lower to upper?
For example a file named example.txt to EXAMPLE.txt
I would use pathlib for this. You'll have to modify this to suit your needs - In this example I'm just using pathlib.Path.glob to iterate through all .txt files in the current working directory:
from pathlib import Path
for path in Path(".").glob("*.txt"):
path.rename(f"{path.stem.upper()}{path.suffix}")
For easy solution, you can try:
import os
fileName = 'example.txt'
def toUpper(fileName):
tempName = fileName.split('.')
return (tempName[0].upper() + '.' + tempName[1])
os.rename(fileName, toUpper(fileName))
It won't work if your file name has multiple dots.
import re
import os
def get_name(original_file_name):
name, ext = (re.match('(.*)(\.[^.]+)', original_file_name).groups())
return name.upper() + ext
def rename_file(original_file_name):
os.rename(original_file_name, get_name(original_file_name))
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])
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.***