Get the highest String Version number in Python

Question

I am trying to get the highest version of a string in Python. I was trying to sort the list but that of course doesnt work as easily as Python will sort the string representation.

For that I am trying to work with regex but it somehow doesnt match.

The Strings look like this:

topic_v10_ext2
topic_v20_ext2
topic_v2_ext2
topic_v5_ext2
topic_v7_ext2

My Regex looks like this.

version_no = re.search("(?:_v([0-9]+))?", v.name)

I was thinking about saving the names in a list and look for the highest v_xx in the list to return. Also for now I am doing this in two FOR loops. Which runs in 2*O(log(n)) which is not optimal I believe. How can I get the highest version in a fast and simple way?

Answer 1

You can use sorted or list.sort with key:

sorted(l, key=lambda x:int(x.split('_')[1][1:]), reverse=True)
['topic_v20_ext2',
 'topic_v10_ext2',
 'topic_v7_ext2',
 'topic_v5_ext2',
 'topic_v2_ext2']

• x.split('_'): returns splitted str, e.g.: ['topic', 'v20', 'ext2']
• Since the version is the key to the sorting, select it by x.split('_')[1]
• Selected V20 has unwanted character 'V', thus reselect it by slicing [1:] to get all the digits.
• Finally, convert digits to int for numerical ordering.

Also, sorted by default returns ascending order of sort. Since you require descending order, use reverse=True.

Answer 2

It could also work with regular expressions, as first tried:

import re
v = 'topic_v7_ext2'
version_no = re.search("^[^_]*_v([0-9]+)", v)
print(version_no.group(1))

That expression searches for pattern from the beginning of the string (^), takes all characters different from _ (I hope your topics can't have one, else both answers are wrong), then finds the '_v' and takes the version number.
There is no need to match _ext, so it doesn't matter if it's there or not!