## 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?

2020/05/13
1
391
5/13/2020 9:38:27 AM

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"]
``````

Generally Speaking

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

Explained:

1. `zip` the two `list`s.
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.

2017/08/22
507
8/22/2017 9:57:35 AM

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))]
``````
2011/07/08

Also, if you don't mind using numpy arrays (or in fact already are dealing with numpy arrays...), here is another nice solution:

``````people = ['Jim', 'Pam', 'Micheal', 'Dwight']
ages = [27, 25, 4, 9]

import numpy
people = numpy.array(people)
ages = numpy.array(ages)
inds = ages.argsort()
sortedPeople = people[inds]
``````

I found it here: http://scienceoss.com/sort-one-list-by-another-list/

2014/01/12

The most obvious solution to me is to use the `key` keyword arg.

``````>>> X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
>>> Y = [ 0,   1,   1,    0,   1,   2,   2,   0,   1]
>>> keydict = dict(zip(X, Y))
>>> X.sort(key=keydict.get)
>>> X
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
``````

Note that you can shorten this to a one-liner if you care to:

``````>>> X.sort(key=dict(zip(X, Y)).get)
``````
2011/07/08

`more_itertools` has a tool for sorting iterables in parallel:

Given

``````from more_itertools import sort_together

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

Demo

``````sort_together([Y, X])[1]
# ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')
``````
2020/03/25

I actually came here looking to sort a list by a list where the values matched.

``````list_a = ['foo', 'bar', 'baz']
list_b = ['baz', 'bar', 'foo']
sorted(list_b, key=lambda x: list_a.index(x))
# ['foo', 'bar', 'baz']
``````
2018/08/17