simple Example for your Help.

import os
from os import path

path_to_directory = "C:\\X\\Data"

for f in os.listdir(path_to_directory):
    name , extension = path.splitext(f)
    print(name)

Output

foo
bar

First I would extract filename itself. I'd split it from the extension. You can go easy way by doing:

path = "Planning_Group_20180108.ind"
filename, ext = path.split(".")

It is assuming that path is actually only a filename and extension. If I'd want to stay safe and platform independent, I'd use os module for that:

fullpath = "this/could/be/a/full/path/Planning_Group_20180108.ind"
path, filename = os.path.split(fullpath)

And then extract "root" and extension:

root, ext = os.path.splitext(filename)

That should leave me with Planning_Group_20180108 as root. To discard "_20180108" we need to split string by "_" delimiter, going from the right end, and do it only once. I would use .rsplit() method of string, which lets me specify delimiter, and number of times I want to make splits.

what_i_want, the_rest = root.rsplit("_", 1)

what_i_want should contain left side of Planning_Group_20180108 cut in place of first "_" counting from right side, so it should be Planning_Group

The more compact way of writing the same, but not that easy to read, would be:

what_i_want = os.path.splitext(os.path.split("/my/path/to/Planning_Group_20180108.ind")[1])[0].rsplit("_", 1)

PS. You may skip the part with extracting root and extension if you're sure, that extension will not contain underscore. If you're unsure of that, this step will be necessary. Also you need to think of case with multiple extensions, like /path/to/file/which_has_a.lot.of.periods.and_extentions. In that case would you like to get which_has_a.lot.of.periods.and, or which_has? Think of it while planning your app. If you need latter, you may want to extract root by doing filename.split(".", 1) instead of using os.path.splitext()

reference:

os.path.split(path),

os.path.splitext(path)

str.rsplit(sep=None, maxsplit=-1)

Besides split and rsplit, there is partition/rpartition. It separates string once, but the way question was asked, it may apply as well.

Example:

Use with AI>>> "MATCHES__STRING".partition("__")
('MATCHES', '__', 'STRING')

>>> "MATCHES__STRING".partition("__")[::2]
('MATCHES', 'STRING')

And a bit faster then split("_",1):

Use with AI$ python -m timeit "'validate_field_name'.split('_', 1)[-1]"
2000000 loops, best of 5: 136 nsec per loop

$ python -m timeit "'validate_field_name'.partition('_')[-1]"
2000000 loops, best of 5: 108 nsec per loop

Timeit lines are based on this answer

filename="LN0001_07272010_3.dat"
newfilename=filename.split("_")[0]+"_JY_"+filename.split("_")[1]

linearr=[]
for line in open(filename).readlines():
     linearr.append(line+"|")

f=open(newfilename, "w")
for line in linearr:
     f.write(line)
f.close()

If every entry block starts with a colon, you can just split by that:

with  open('entries.txt') as fp:
    contents = fp.read()
    for entry in contents.split(':'):
        # do something with entry  

awk '{f="file" NR; print $0 " -|"> f}' RS='-\\|'  input-file

Explanation (edited):

RS is the record separator, and this solution uses a gnu awk extension which allows it to be more than one character. NR is the record number.

The print statement prints a record followed by " -|" into a file that contains the record number in its name.

I think that list is not suitable for the link. Insted, use dict.

First import required modules.

from pathlib import Path

Initiate variable.

image_dict = {}  # handle name and link
all_file = list(Path().iterdir())  # get all file in current directory

Populate image_dict.

for f in all_file:
    extension = str(f).split(".")[-1]
    filename = str(f).split("." + extension)[0]

    if extension in ["png", "jpeg"]:  # In case of there is other file that's no an image
        part_name = filename.split("_")  # Get keywords from filename

        for pn in part_name:  # Append a link from each keyword
            if image_dict.get(pn, None) is not None:
                image_dict[pn].append(f.resolve())
            else:
                image_dict[pn] = [f.resolve()]

print(image_dict)  # This is the result