Python to zip files with dynamic name

Question

Good day, I have a directory with 12 files. I would like to create 4 separate zip files with 3 files in each. The hitch is I would like to dynamically name the zip file based on the file names. (All 3 for each zip start off the same)

ie.

IT-12123.txt
IT-12123.pdf
IT-12123.xls

IT-23232.txt
IT-23232.pdf
IT-23232.xls

I would like to create IT-23232.zip (with all of those files in that zip) also a zip called IT-12123.zip (with all of those files in that zip)

Thank you

Can you please share your code? Are you able to zip files with non dynamic name or do you want someone to write that code as well for you? — DevB2F, Dec 19 '17 at 20:07
Welcome to [SO]! Stack Overflow is a question-and-answer site. At [SO], readers, such as yourself, ask questions while other readers try to answer those questions. Your post has a lot of good information, but it is missing the one essential ingredient: a question! What, precisely, are you asking? (By the way, please visit take the [tour] and read [ask] when you have a moment.) — Robᵩ, Dec 19 '17 at 20:09
See this: https://stackoverflow.com/questions/678236/how-to-get-the-filename-without-the-extension-from-a-path-in-python?noredirect=1&lq=1 — Ashutosh Bharti, Dec 19 '17 at 20:53

score 0 · Answer 1 · answered Dec 19 '17 at 21:25

Here is a simple solution if your files are all located in the same folder.

import itertools
import glob
import os
import zipfile 
# move to the directory where the files are located
os.chdir("/mydir")
# find all files starting with IT
files = glob.glob('IT*')
files_no_ext = []
for file in files:
    # remove extension from files names
    files_no_ext.append(os.path.splitext(file)[0])
# group no extension files by name
grouped_files= itertools.groupby(files_no_ext)
for name, group in grouped_files:
    zipped = zipfile.ZipFile(name+'.zip', 'w')
    to_zip = glob.glob(name+'*')
    for file in to_zip:
        zipped.write(file)
    zipped.close()

Python to zip files with dynamic name

1 Answers1