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?

Please show the exact data format of the input strings. Are they in a file, a list, or some other data structure? Is the value after ext part of the version number? — Rory Daulton
– Rory Daulton, Commented Feb 22, 2019 at 14:24
What is meant by "string representation"? Why can't the topic_v… be strings? — PrinceOfCreation
– PrinceOfCreation, Commented Feb 22, 2019 at 14:25
Your regex will match literally anywhere because you've made the entire thing optional with the ? at the end. — glibdud
– glibdud, Commented Feb 22, 2019 at 14:26
@RoryDaulton no, only the _vXX is part of the Version. But the ext is optional and not always present. — Derek Haynes
– Derek Haynes, Commented Feb 22, 2019 at 14:38
@PrinceOfCreation They are a different Object which has a .name Key which can be read as a String in a for loop. — Derek Haynes
– Derek Haynes, Commented Feb 22, 2019 at 14:38

Chris · Accepted Answer · 2019-02-22 15:02:44Z

8

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.

edited Feb 22, 2019 at 15:02

answered Feb 22, 2019 at 14:27

Chris

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5 Comments

Derek Haynes Over a year ago

Though mind explaining the x:int(x.split('_')[1][1:] part a bit once you have the time?

Chris Over a year ago

@DerekHaynes I've edited the answer. Please let me know if any part is unclear.

Derek Haynes Over a year ago

Yeah man! Best man of the day! This explained it really clear for me and for many to come. Thanks a lot.

Derek Haynes Over a year ago

Is there also a way to look for a regex inside the split like: int(x.name.split('_')[1] "Regex to start and end at the number"). The version should be in the format of "v_00" but maybe something arrives like "version_00"

Chris Over a year ago

@DerekHaynes That can be a completely different scenario since it contains _ and thus split may not work. If you were to extract digits, you can use lambda x: re.findall('\d+', x)[0].

B. Go · Accepted Answer · 2019-02-23 18:08:50Z

1

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!

edited Feb 23, 2019 at 18:08

answered Feb 22, 2019 at 15:18

B. Go

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