Mimic the array swaps from a sort

Question

l = [0, 1, 3, 2]

l2 = ['foo', 3, 'bar', 10]

If I say sorted(l), I will get [0, 1, 2, 3]. It will swap the last two elements.

How can I apply the same row swaps to l2? I.e., I want l2 to be ['foo', 3, 10, 'bar'].

Answer 1

You can use zip, unpack tuple and a list comprehension to achieve the results:

[y for x, y in sorted(zip(l, l2))]

Answer 2

TL;DR

>>> l, l2 = zip(*sorted(zip(l, l2)))
>>> list(l)
[0, 1, 2, 3]
>>> list(l2)
['foo', 3, 10, 'bar']

---

Explanation

1. zip both the lists together

   >>> list(zip(l, l2))
   [(0, 'foo'), (1, 3), (2, 10), (3, 'bar')]
   

2. then sort them, (since we get tuples from zip, the first elements of tuples will be compared first and only if they are same, the second element will be compared. So the sorting effectively happens with the values of l)

   >>> sorted(zip(l, l2))
   [(0, 'foo'), (1, 3), (2, 10), (3, 'bar')]
   

3. and then unzip them,

   >>> list(zip(*sorted(zip(l, l2))))
   [(0, 1, 2, 3), ('foo', 3, 10, 'bar')]
   

   you can actually unzip over l and l2, like this

   >>> l, l2 = zip(*sorted(zip(l, l2)))
   >>> l, l2
   ((0, 1, 2, 3), ('foo', 3, 10, 'bar'))
   >>> list(l)
   [0, 1, 2, 3]
   >>> list(l2)
   ['foo', 3, 10, 'bar']
   

---

Alternate approach

You can actually sort the values along with the current index and then you can reconstruct the values like this

>>> l = [0, 1, 3, 2]
>>> l2 = ['foo', 3, 'bar', 10]
>>> l_s = sorted((value, idx) for idx, value in enumerate(l))
>>> l_s
[(0, 0), (1, 1), (2, 3), (3, 2)]
>>> l = [value for value, idx in l_s]
>>> l
[0, 1, 2, 3]
>>> l2 = [l2[idx] for value, idx in l_s]
>>> l2
['foo', 3, 10, 'bar']