Sorting list based on values from another list

Question

I have a list of strings like this:

X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"] Y = [ 0,   1,   1,   0,   1,   2,   2,   0,   1 ] 

What is the shortest way of sorting X using values from Y to get the following output?

["a", "d", "h", "b", "c", "e", "i", "f", "g"] 

The order of the elements having the same "key" does not matter. I can resort to the use of for constructs but I am curious if there is a shorter way. Any suggestions?

Answer 1

Shortest Code

[x for _, x in sorted(zip(Y, X))] 

Example:

X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"] Y = [ 0,   1,   1,    0,   1,   2,   2,   0,   1]  Z = [x for _,x in sorted(zip(Y,X))] print(Z)  # ["a", "d", "h", "b", "c", "e", "i", "f", "g"] 

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Generally Speaking

[x for _, x in sorted(zip(Y, X), key=lambda pair: pair[0])] 

Explained:

1. zip the two lists.
2. create a new, sorted list based on the zip using sorted().
3. using a list comprehension extract the first elements of each pair from the sorted, zipped list.

For more information on how to set\use the key parameter as well as the sorted function in general, take a look at this.

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Answer 2

Zip the two lists together, sort it, then take the parts you want:

>>> yx = zip(Y, X) >>> yx [(0, 'a'), (1, 'b'), (1, 'c'), (0, 'd'), (1, 'e'), (2, 'f'), (2, 'g'), (0, 'h'), (1, 'i')] >>> yx.sort() >>> yx [(0, 'a'), (0, 'd'), (0, 'h'), (1, 'b'), (1, 'c'), (1, 'e'), (1, 'i'), (2, 'f'), (2, 'g')] >>> x_sorted = [x for y, x in yx] >>> x_sorted ['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g'] 

Combine these together to get:

[x for y, x in sorted(zip(Y, X))]