I need to iterate over a large (in the thousands) list of files in a network folder and "concatenate" any image sequences into just one entry with the range of images (first image, last image in sequence). I am calling this "sequence pruning" and I created the following code which does work, but it seems incredibly un-pythonic to me and runs slowly. I'm certain there is a MUCH better way to do this, so I am looking for review to help clean/speed this up.

To elaborate a bit further on the issue lets say I have this as a list of files for the input:

yet another seq.0000.png
yet another seq.0001.png
yet another seq.0002.png
yet another seq.0021.png
yet another seq.0030.png

In the end I want to return something like this:

img.001.png, [1-3]
yet another seq.0000.png, [0-30]

FYI I can safely assume that the image sequence number is always going to be a series of digits at the very end of the filename (before the extension of course). However I cannot assume that they will be perfectly sequential, as there are sometimes "gaps" between numbers.

Here's my current code, python 2.7:

import os

def split_padding(path):
    file, ext = os.path.splitext(path)
    pad_int = 0

    while file[pad_int * -1 - 1].isdigit():
        pad_int += 1

    if pad_int == 0:
        return file, '0', ext

    clean_file = file[0:pad_int * -1]
    padding = file[pad_int * -1:]

    return clean_file, padding, ext

def strip_padding(path):
    file, ext = os.path.splitext(path)

    while file[-1].isdigit():
        file = file[:-1]

    return file

def prune_files(paths):
    sequences get put into arrays like so:
    [x_folder, z_folder, [test_a.000.png, 0, 2], [test_b.000.tif, 0, 3], test_C.000.png]
    :return: [file1, file2, [first_file, seq_start, seq_end]]
    paths.sort(key=lambda s: s.lower()) # list has to be sorted for this to work

    # this odd bit of code turns all sequences into arrays of images.
    pruned_list = []
    switch = True
    for c, path in enumerate(paths):
        if c == 0:
        if not os.path.splitext(path)[1] in ['.png', '.tif', '.tiff', '.exr', '.jpg', '.jpeg']:

        test = paths[c-1]
        if strip_padding(path) == strip_padding(test):
            if switch:
                pruned_list[-1] = [pruned_list[-1]]
                switch = False
            switch = True

    # so now lets convert that to the format we want to return
    for c, item in enumerate(pruned_list):
        if type(item) == list:
            pruned_list[c] = [item[0], int(split_padding(item[0])[1]), int(split_padding(item[-1])[1])]

    return pruned_list

if __name__ == "__main__":
    test_dir = "some directory"
    print prune_files([path for path in os.listdir(test_dir) if os.path.isfile(os.path.join(test_dir, path))])
  • \$\begingroup\$ Can you assume the numbers will be sorted at all? Is it possible that 'img.003.png' comes after 'random_file.txt', or are the file names sorted alphabetically too? \$\endgroup\$
    – Daniel
    Commented May 5, 2018 at 18:59
  • \$\begingroup\$ The input list comes from os.listdir(), so I get whatever that returns. From my experience it isn't always sorted so I opted to do my own sort in the beginning of my prune_files def. I realize that was perhaps confusing in my example, so I just fixed it! \$\endgroup\$
    – Spencer
    Commented May 5, 2018 at 19:03
  • \$\begingroup\$ I'm just playing around a bit to see how performance could be improved, and noticed prune_files() currently does not take into account file extensions, only name and sequence numbers. Is that the expected behavior? \$\endgroup\$
    – Daniel
    Commented May 6, 2018 at 13:18
  • \$\begingroup\$ @Coal_ Ha that would explain an error I got yesterday. No, not really intended so I'll have to fix that. As a note I found that looping through the list and stripping out the sequence numbers, then converting to a set to remove duplicates is a very rapid way of getting rid of everything but one entry per sequence. However at that point you still don't have first and last file... \$\endgroup\$
    – Spencer
    Commented May 6, 2018 at 23:31
  • \$\begingroup\$ It would now be best to ask a new question with the changes incorporated, since you have accepted an answer. \$\endgroup\$
    – Daniel
    Commented May 8, 2018 at 21:56

1 Answer 1


Many, many problems:

  • Low level dealing with strings. Python is a high-level programming language - don't reinvent the wheel. Regex and string methods now are replacing your low-level code in my rework of it. Good rule of thumb is to avoid any code that has a lot of array indexing in Python.
  • Sorting doesn't need a function if you use sorted
  • Long comprehensions and generators are usually unreadable and unmaintainable so avoid those at all costs. Especially avoid adding needless logic in them too.
  • Extension list is a constant so you might as well extract that to the top.
  • Simpler or more standard output could have made the code much simpler but it wasn't clear if that was the requirement.
  • Hardcoding args shouldn't be done when sys.argv and argparse are so easy to use.

tl;dr fixed code. Code could be a lot simpler if the output format was a bit more standard but you can probably modify the code with minimal effort:

#!/usr/bin/env python3

import os
import re
import sys


SEQUENCE_PATTERN = r'(.*)\.([0-9]+).(.{3,4})$'

def sequences_strigifier(sequences):
    output_string = ''
    for key, seq_info in sequences.items():
        if not seq_info:
            output_string += '{}\n'.format(key)

        if seq_info['start_index'] == seq_info['end_index']:
            output_string += '{}.{}.{}\n'.format(key,

        output_string += '{}.{}.{}, [{}-{}]\n'.format(key,

    return output_string.strip()

def find_image_sequences(directory):
    sequences get put into arrays like so:
    [x_folder, z_folder, [test_a.000.png, 0, 2], [test_b.000.tif, 0, 3], test_C.000.png]
    :return: [file1, file2, [first_file, seq_start, seq_end]]
    sequences = {}

    sorted_candidate_list = sorted(os.listdir(directory))

    for candidate_path in sorted_candidate_list:
        full_candidate_path = os.path.join(directory, candidate_path)
        if not os.path.isfile(full_candidate_path):
            sequences[candidate_path] = None

        matches = re.match(SEQUENCE_PATTERN, candidate_path)
        if not matches:
            sequences[candidate_path] = None

        filename = matches.group(1)
        sequence_index = matches.group(2)
        extension = matches.group(3)

        if not extension in FILE_EXTS:
            sequences[candidate_path] = None

        if not filename in sequences:
            sequences[filename] = {
               'ext': extension,
               'start_index_str': sequence_index,
               'start_index': int(sequence_index),
               'end_index': int(sequence_index),

        sequences[filename]['end_index'] = int(sequence_index)

    return sequences_strigifier(sequences)

if __name__ == '__main__':
    if len(sys.argv) < 2:
        print('Usage: {} <dirname>'.format(sys.argv[0]))

  • \$\begingroup\$ Just a note that for a large set, the stringifier will be pretty slow so I would either print directly out of there or use join()/StringIO. \$\endgroup\$ Commented May 7, 2018 at 18:33
  • \$\begingroup\$ @Coal_ re usage format: Indeed. fixed! Good point on the exit as well. \$\endgroup\$ Commented May 8, 2018 at 13:33
  • \$\begingroup\$ Thanks Srdjan, lots of improvement! Sorry if I was unclear about the usage of this code, I do not want a string as the output so the stringifier part is unnecessary (dictionary works for me). And sys.arv is not needed because this being used as a module within my larger program, and will be called directly. I took your code with some tweaks and inserted it into my original Q, let me know what you think. The use of a dictionary for 'sequences', and regex, definitely cleans things up! However this runs only a tad faster than my original code, any ideas to speed it up? \$\endgroup\$
    – Spencer
    Commented May 8, 2018 at 19:30
  • 1
    \$\begingroup\$ Please see What to do when someone answers. I have rolled back Rev 5 → 3. \$\endgroup\$ Commented May 9, 2018 at 3:49
  • 1
    \$\begingroup\$ For the record I posted this regex problem on SO and got some good answers here: stackoverflow.com/questions/50311032/… \$\endgroup\$
    – Spencer
    Commented May 12, 2018 at 22:40

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