Sort list with multiple criteria in python

Question

Currently I'm trying to sort a list of files which were made of version numbers. For example:

0.0.0.0.py
1.0.0.0.py
1.1.0.0.py

They are all stored in a list. My idea was to use the sort method of the list in combination with a lambda expression. The lambda-expression should first remove the .py extensions and than split the string by the dots. Than casting every number to an integer and sort by them.

I know how I would do this in c#, but I have no idea how to do this with python. One problem is, how can I sort over multiple criteria? And how to embed the lambda-expression doing this?

Can anyone help me?

Thank you very much!

Answer 1

You can use the key argument of sorted function:

filenames = [
    '1.0.0.0.py',
    '0.0.0.0.py',
    '1.1.0.0.py'
]

print sorted(filenames, key=lambda f: map(int, f.split('.')[:-1]))

Result:

['0.0.0.0.py', '1.0.0.0.py', '1.1.0.0.py']

The lambda splits the filename into parts, removes the last part and converts the remaining ones into integers. Then sorted uses this value as the sorting criterion.

Answer 2

Have your key function return a list of items. The sort is lexicographic in that case.

l = [ '1.0.0.0.py', '0.0.0.0.py', '1.1.0.0.py',]
s = sorted(l, key = lambda x: [int(y) for y in x.replace('.py','').split('.')])
print s