In list of items, remove items with a lower value compared to items with the same sub string in python

You can use a dictionary to keep track of the highest version for each string and then iterate over the string and only extract the entries with the highest version or version 0. The code below assumes that v0 always exists and that the file format is always jpg, but can be modified if not.

files = [
    '[10998321023D123T][v0].jpg',
    '[10998321023D123T][v12321].jpg',
    '[10998321023D123T][v62221].jpg',
    '[10DFSA783212131T][v0].jpg',
    '[10DFSA783212131T][v32112].jpg',
    '[10DFSA783212131T][v54541].jpg',
]

highest_version = {}
for file in files:
    # extract string and version from file name
    string, version, extension = file.split(']') 
    string = string[1:]
    version = int(version[2:])

    if string not in highest_version or version>highest_version[string]:
        highest_version[string] = version

output = []
for string, version in highest_version.items():
    output.append('['+string+'][v0].jpg')
    output.append('['+string+'][v'+str(version)+'].jpg')

print(output)

So I imagine you want to look at the version numbers (v100101 ex.) and see if it’s smaller than some threshold.

If I understand correctly your data structure is a list of strings, where the name and version is seperated by blocks, followed by the extension.

So if the version numbers are fixed length you can do the following.

Set a threshold, then construct a new list for items lower than the threshold.

threshold: int = 54541

new_list: list[str] = []
for item in original:
  name, version, extension = item.split(“]”)

  name: str = name[1:]
  version: int = int(version[2:])
  extension: str = extension[1:]

  if version == 0 or version >= threshold:
    continue

  new_list.append(f”[{name}][v{version}].{extension}”)

You could also use list comprehension if you want to be fancy.

Also I would use some kind of a data structure (like a tuple) to store the name and version seperately.

Note that you can use other methods besides split, but I’m on my phone right now and cant remember it.

Hope this helps.

Consider each string in the list to be comprised of a key and a value where the key is the part between the first pair of square brackets and the value is in the second pair of square brackets.

For the value you can ignore the leading ‘v’.

Build a dictionary from the keys and their associated values.

Then work through the dictionary to reconstruct your list as follows:

_list = [
    '[10998321023D123T][v0].jpg',
    '[10998321023D123T][v12321].jpg',
    '[10998321023D123T][v62221].jpg',
    '[10DFSA783212131T][v0].jpg',
    '[10DFSA783212131T][v32112].jpg',
    '[10DFSA783212131T][v54541].jpg'
]

def get_key_and_value(s):
    k, v, *_ = s.split(']')
    return k, int(v[2:])

td = dict()

for e in _list:
    k, v = get_key_and_value(e)
    td.setdefault(k, []).append(v)

output = []

for k, v in td.items():
    for m in min(v), max(v):
        output.append(f'{k}][v{m}].jpg')

print(output)

Output:

['[10998321023D123T][v0].jpg', '[10998321023D123T][v62221].jpg', '[10DFSA783212131T][v0].jpg', '[10DFSA783212131T][v54541].jpg']