Sorting a list in Python using the result from sorting another list [duplicate]

Answer 1

Zip the lists together, sort, unzip the lists:

together = zip(list_1, list_2)
sorted_together =  sorted(together)

list_1_sorted = [x[0] for x in sorted_together]
list_2_sorted = [x[1] for x in sorted_together]

What's happening here is that zip creates a list of tuples, with the elements you want the list to be sorted by being the first elements:

>>> a = [1,3,7,3,2]
>>> b = ["one","two",'three','four','five']
>>> zip(a,b)
[(1, 'one'), (3, 'two'), (7, 'three'), (3, 'four'), (2, 'five')]

Then when you sort them, they elements stay paired:

>>> sorted(zip(a,b))
[(1, 'one'), (2, 'five'), (3, 'four'), (3, 'two'), (7, 'three')]

Then all that's left is to unpack these lists.

Answer 2

You can use zip:

>>> list_1 = ['string1', 'string2', 'string3']
>>> list_2 = [5, 2, 8]
>>> s = sorted(zip(list_2, list_1), reverse=True)
>>> list_1_sorted = [e[1] for e in s]
>>> list_2_sorted = [e[0] for e in s]
>>> list_1_sorted
['string3', 'string1', 'string2']
>>> list_2_sorted
[8, 5, 2]
>>> 

Answer 3

@Ignacio's answer is the best, but just in case you need to sort the lists in-place without making new lists, you can try this:

import itertools
list_enumerate = itertools.count()

list_2.sort(reverse=True, key=lambda k: list_1[next(list_enumerate)])
list_1.sort(reverse=True)
print list_1
print list_2

Note that I do not think there is any guarantee that the key function is called for each list item in order (which is necessary for this to work), so this is a risky method to use.

Answer 4

Schwartzian transform

list_1_sorted, list_2_sorted = zip(*sorted(zip(list_1, list_2),
  key=operator.itemgetter(0), reverse=True))