Keeping track of original indicies when sorting a list of lists by length

Question

You can sort a list of lists by length as follows:

l1 = [1,2,3]
l2 = [1,2,3]
l3 = [1,2]
lists = [l1, l2, l3]
sorted_lists = sorted(lists, key=len)
print sorted_lists  #[[1,2], [1,2,3], [1,2,3]]

I can't figure out how to keep track of the indicies to then match up the contents of sorted_lists with the original list names l1, l2 and l3.

This gets close, but I'm not sure how the solution can be implemented when sorting by length.

Answer 1

It is very possible. Just modify the key a bit to specify the right predicate on which len is to be applied.

>>> lists = [l1, l2, l3]
>>> lists = sorted(enumerate(lists), key=lambda x: len(x[1])) # enumerate is a tuple of (index, elem), sort by len(elem)
[(2, [1, 2]), (0, [1, 2, 3]), (1, [1, 2, 3])]

Answer 2

Using arg.sort() from numpy with list comprehension can be other way:

import numpy

new_list = [(index, lists[index])for index in numpy.argsort(lists)]
print(new_list)

Output:

[(2, [1, 2]), (0, [1, 2, 3]), (1, [1, 2, 3])]