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importing filename and using regex to change the filename and save back

I have files with filenames as "lsud-ifgufib-1234568789.png" I want to rename this file as digits only which are followed by last "-" and then save them back in the folder.
Basically I want the final filename to be the digits that are followed by "-".
~ path = 'C:/Users/abc/downloads'
for filename in os.listdir(path):
r = re.compile("(\d+)")
newlist = filter(r.match, filename)
print(newlist)
~
How do I proceed further?

Assumptions:
You want to rename files if the file has a hyphen before the number.
The file may or may not have an extention.
If the file has an extention, preserve it.
Then would you please try the following:
import re, os
path = 'C:/Users/abc/downloads'
for filename in os.listdir(path):
m = re.search(r'.*-(\d+.*)', filename)
if m:
os.rename(os.path.join(path, filename), os.path.join(path, m.group(1)))

You could try a regex search followed by a path join:
import re
import os
path = 'C:/Users/abc/downloads'
for filename in os.listdir(path):
os.rename(filename, os.path.join(path, re.search("\d+(?=\D+?$)", filename).group()))

import re
import pathlib
fileName = "lsud-ifgufib-1234568789.png"
_extn = pathlib.Path(fileName).suffix
_digObj = re.compile(r'\d+')
digFileName = ''.join(_digObj.findall(fileName))
replFileName = digFileName + _extn

Importing for loop to skip data file not exist

I have about 500 Excel files in the format: data_1, data_2 ... data_500
However, not all file are there. File like data_3 is not in the folder.
I want to import all available data into dataframe.
However, the my code below will stop when it hit a name of file not in the list, say data_3
Can you please help me to skip these record?
Thank you,
HN
for i in range(500):
filename='data_'+ str(i) + 'xlsx'
output = pd.read_excel('PATH' + filename)
THE KEY IS CHECK IN FULL PATH IN glob.glob
import glob
for i in xlx_file_list:
filename = 'Excel_Sample' + str(i) + '.xlsx' #; print(filename)
full_path = 'D:\Python...\\' + filename #; print(full_path)
if full_path not in glob.glob('D:\Python...\*'):
print(filename, ' not in folder')
continue
outfile = pd.read_excel(full_path, sheet_name='data_sheet')
print(outfile)

Hi in your sample probably PATH is a variable, not a string, 'PATH'+filename cannot work.
i suggest to use os.path.join() to compose file path, don't use string composition for this.
There are two way to solve this problem:
Generate all names and see if the file exists:
import os
for i in range(500):
filename='data_'+ str(i) + 'xlsx'
if os.path.exists(filename)
output = pd.read_excel(filename)
or generate only the correct filename list:
import glob
for filename in glob.glob('data_*.xlsx'):
output = pd.read_excel(filename)

How to sort a list by creation date? (Python - OS library)

I'd like to rename a list of video cuts in a folder (in my case /Desktop/Cuts/)
from:
xxx.mp4, yyy.mp4, ..., zzz.mp4
to:
1.mp4, 2.mp4, ..., 33.mp4.
1.mp4 will be the video that was first created in the folder. 2.mp4 will be the second, etc.
What I am doing now (please see code below) renames the video cuts in /Cuts/, but according to an order that I don't understand, and that is definitely NOT creation date. (it does not seem alphabetical, either)
import os
path = '/Users/ntelmaut/Desktop/Cuts/'
i = 1
for file in os.listdir(path):
new_file_name = "{}.mp4".format(i)
os.rename(path + file, path + new_file_name)
#print path + file
i = i+1
I have tried make a variable out of os.listdir(path) and to apply sorted() function, but to no avail yet (I am a newbie to Python - sorry).
Would love a pointer, here.
Thank you!

Solution
import glob
import os
dirpath = '/Users/ntelmaut/Desktop/Cuts/'
files = glob.glob(os.path.join(path, '*.mp4'))
files.sort(key=os.path.getmtime)
for i, filename in enumerate(files, 1):
try:
os.rename(filename, "{0}.mp4".format(i))
except OSError as e:
print("OSError: ", e)

If I understand correctly this is what you want.
import os
path = '/Users/ntelmaut/Desktop/Cuts/'
def sort_key(file):
return os.stat(file).st_ctime
files = os.listdir(path)
sorted_files = sorted((os.path.join(path, f) for f in files), key=sort_key)
i = 1
for file in sorted_files:
new_file_name = f"{i}.mp4"
os.rename(file, new_file_name)
i += 1
A cleaner way would be to use the pathlib module.
from pathlib import Path
p = Path('/Users/ntelmaut/Desktop/Cuts/')
def sort_key(path):
return path.stat().st_ctime
sorted_files = sorted(p.iterdir(), key=sort_key)
for idx, file in enumerate(sorted_files, 1):
new_name = f"{idx}.mp4"
file.rename(p / new_name)

Exporting multiple files with different filenames

Lets say I have n files in a directory with filenames: file_1.txt, file_2.txt, file_3.txt .....file_n.txt. I would like to import them into Python individually and then do some computation on them, and then store the results into n corresponding output files: file_1_o.txt, file_2_o.txt, ....file_n_o.txt.
I've figured out how to import multiple files:
import glob
import numpy as np
path = r'home\...\CurrentDirectory'
allFiles = glob.glob(path + '/*.txt')
for file in allFiles:
# do something to file
...
...
np.savetxt(file, ) ???
Not quite sure how to append the _o.txt (or any string for that matter) after the filename so that the output file is file_1_o.txt

Can you use the following snippet to build the output filename?
parts = in_filename.split(".")
out_filename = parts[0] + "_o." + parts[1]
where I assumed in_filename is of the form "file_1.txt".
Of course would probably be better to put "_o." (the suffix before the extension) in a variable so that you can change at will just in one place and have the possibility to change that suffix more easily.
In your case it means
import glob
import numpy as np
path = r'home\...\CurrentDirectory'
allFiles = glob.glob(path + '/*.txt')
for file in allFiles:
# do something to file
...
parts = file.split(".")
out_filename = parts[0] + "_o." + parts[1]
np.savetxt(out_filename, ) ???
but you need to be careful, since maybe before you pass out_filename to np.savetxt you need to build the full path so you might need to have something like
np.savetxt(os.path.join(path, out_filename), )
or something along those lines.
If you would like to combine the change in basically one line and define your "suffix in a variable" as I mentioned before you could have something like
hh = "_o." # variable suffix
..........
# inside your loop now
for file in allFiles:
out_filename = hh.join(file.split("."))
which uses another way of doing the same thing by using join on the splitted list, as mentioned by #NathanAck in his answer.

import os
#put the path to the files here
filePath = "C:/stack/codes/"
theFiles = os.listdir(filePath)
for file in theFiles:
#add path name before the file
file = filePath + str(file)
fileToRead = open(file, 'r')
fileData = fileToRead.read()
#DO WORK ON SPECIFIC FILE HERE
#access the file through the fileData variable
fileData = fileData + "\nAdd text or do some other operations"
#change the file name to add _o
fileVar = file.split(".")
newFileName = "_o.".join(fileVar)
#write the file with _o added from the modified data in fileVar
fileToWrite = open(newFileName, 'w')
fileToWrite.write(fileData)
#close open files
fileToWrite.close()
fileToRead.close()

how to get list of files from a directory into text file using python

This question is how to get list of files from a directory into text file using python.
Result in the text file should exactly be like this:
E:\AA\a.jpg
E:\AA\b.jpg
...
How to correct the code below:
WD = "E:\\AA"
import glob
files = glob.glob ('*.jpg')
with open ('infiles.txt', 'w') as in_files:
in_files.write(files +'\n')

glob.glob() returns a list. You have to iterate through it.
WD = "E:\\AA"
import glob
files = glob.glob ('*.jpg')
with open ('infiles.txt', 'w') as in_files:
for eachfile in files: in_files.write(eachfile+'\n')

Input directory path : WD = "E://AA"
You can assign specific file extention that you needed eg: path = WD+'/*.jpg',
if you need all file list then give '' eg: path = WD+'/'
import glob
w_dir = WD + "/*.jpg"
with open("infiles.txt","wb")as fp:
for path in [filepath for filepath in glob.glob(w_dir)]:
fp.write(path+"\n")

Without path, glob.glob returns list of filename (No directory part). To get full path you need to call os.path.abspath(filename) / os.path.realpath(filename) / os.path.join(WD, filename)
>>> glob.glob('*.png')
['gnome-html.png', 'gnome-windows.png', 'gnome-set-time.png', ...]
>>> os.path.abspath('gnome-html.png')
'/usr/share/pixmaps/gnome-html.png'
With path, glob.glob return list of filename with directory part.
>>> glob.glob('/usr/share/pixmaps/*.png')
['/usr/share/pixmaps/gnome-html.png', '/usr/share/pixmaps/gnome-windows.png', '/usr/share/pixmaps/gnome-set-time.png', ...]
import glob
import os
WD = r'E:\AA'
files = glob.glob(os.path.join(WD, '*.jpg'))
with open('infiles.txt', 'w') as in_files:
in_files.writelines(fn + '\n' for fn in files)
or
import glob
import os
WD = r'E:\AA'
os.chdir(WD)
files = glob.glob('*.jpg')
with open('infiles.txt', 'w') as in_files:
in_files.writelines(os.path.join(WD, fn) + '\n' for fn in files)

Here is a two line simple solution:
import os
filee = open('all_names.txt','w')
given_dir = 'the_dierctory'
[filee.write(os.path.join(os.path.dirname(os.path.abspath(__file__)),given_dir,i)+'\n') for i in os.listdir(given_dir)]
where given_dir is the directory name. The output is a text file (all_names.txt) where each line in the file is the full path to all files and directories in the given_dir.