# Sorting list based on values from another list

**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:**

`zip`

the two`list`

s.- create a new, sorted
`list`

based on the`zip`

using`sorted()`

. - 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.*

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))]`

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 numpypeople = 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/