Putting a list in the same order as another list

Question

There\'s a bunch of questions that are phrased similarly, but I was unable to find one that actually mapped to my intended semantics.

There are two lists, A

Answer 1

You can easily do this with numpy by sorting both lists (to get a mapping between the two lists) and by inverting one of the sorting permutations:

import numpy as np

a = [7, 14, 0, 9, 19, 9]
b = [45, 42, 0, 1, -1, 0]

a = np.array(a)
b = np.array(b)

ai = np.argsort(a)
bi = np.argsort(b)
aiinv = np.empty(ai.shape,dtype=int)
aiinv[ai] = np.arange(a.size)  # inverse of ai permutation

b_new = b[bi[aiinv]]
# array([ 0, 42, -1,  0, 45,  1])

numpy.argsort gives the indices (permutation) that will sort your array. This needs to be inverted to be used inside b, which can be done by the inverted assignment

aiinv[ai] = np.arange(a.size)

Answer 2

a = [7, 14, 0, 9, 19, 9]
b = [45, 42, 0, 1, -1, 0]
print zip(*sorted(zip(sorted(b), sorted(enumerate(a), key=lambda x:x[1])), key=lambda x: x[1][0]))[0]
#or, for 3.x:
print(list(zip(*sorted(zip(sorted(b), sorted(enumerate(a), key=lambda x:x[1])), key=lambda x: x[1][0])))[0])

result:

(0, 42, -1, 0, 45, 1)

You sort a, using enumerate to keep track of each item's original index. You zip the result with sorted(b), then re-sort the whole thing based on a's original indices. Then you call zip once more to extract just b's values.