Sort list with multiple criteria in python

Question

I'm currently trying to sort the list of files that were made from version numbers. For example:

0.0.0.0.py
1.0.0.0.py
1.1.0.0.py

All of them are stored in the list. My idea was to use a sortlist method in conjunction with a lambda expression. The lambda expression must first remove the extensions .py, and then break the string into dots. What distinguishes each number to an integer and sorts by them.

I know how to do this in C #, but I have no idea how to do this with python. One of the problems is how can I sort by several criteria? And how to include this lambda expression?

Can anybody help me?

Many thanks!

Answer 1

You can use the key argument of the function sorted:

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 breaks the file name into parts, removes the last part, and converts the rest to integers. It then sorteduses this value as a sorting criterion.

Answer 2

Ask the function keyto return a list of items. In this case, the sorting is lexicographic.

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

Answer 3

# read list in from memory and store as variable file_list
sorted(file_list, key = lambda x: map(int, x.split('.')[:-1]))

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