Question

I have a list containing version strings, such as things:

versions_list = ["1.1.2", "1.0.0", "1.3.3", "1.0.12", "1.0.2"]

I would like to sort it, so the result would be something like this:

versions_list = ["1.0.0", "1.0.2", "1.0.12", "1.1.2", "1.3.3"]

The order of precedence for the digits should obviously be from left to right, and it should be descending. So 1.2.3 comes before 2.2.3 and 2.2.2 comes before 2.2.3.

How do I do this in Python?

Answer 1

Split each version string to compare it as a list of integers:

versions_list.sort(key=lambda s: map(int, s.split('.')))

Gives, for your list:

['1.0.0', '1.0.2', '1.0.12', '1.1.2', '1.3.3']

In Python3 map no longer returns a list, So we need to wrap it in a list call.

versions_list.sort(key=lambda s: list(map(int, s.split('.'))))

The alternative to map here is a list comprehension. See this post for more on list comprehensions.

versions_list.sort(key=lambda s: [int(u) for u in s.split('.')])