Sorting list based on values from another list

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Sorting list based on values from another list

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?

Asked By: Legend

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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"]

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 setuse the key parameter as well as the sorted function in general, take a look at this.


Answered By: Whatang

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))]
Answered By: Ned Batchelder

Answer #3:

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/

Answered By: Tom

Answer #4:

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)

As Wenmin Mu and Jack Peng have pointed out, this assumes that the values in X are all distinct. That’s easily managed with an index list:

>>> Z = ["A", "A", "C", "C", "C", "F", "G", "H", "I"]
>>> Z_index = list(range(len(Z)))
>>> Z_index.sort(key=keydict.get)
>>> Z = [Z[i] for i in Z_index]
>>> Z
['A', 'C', 'H', 'A', 'C', 'C', 'I', 'F', 'G']

Since the decorate-sort-undecorate approach described by Whatang is a little simpler and works in all cases, it’s probably better most of the time. (This is a very old answer!)

Answered By: senderle

Answer #5:

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')
Answered By: pylang

Answer #6:

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']
Answered By: nackjicholson

Answer #7:

I like having a list of sorted indices. That way, I can sort any list in the same order as the source list. Once you have a list of sorted indices, a simple list comprehension will do the trick:

X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0,   1,   1,    0,   1,   2,   2,   0,   1]
sorted_y_idx_list = sorted(range(len(Y)),key=lambda x:Y[x])
Xs = [X[i] for i in sorted_y_idx_list ]
print( "Xs:", Xs )
# prints: Xs: ["a", "d", "h", "b", "c", "e", "i", "f", "g"]

Note that the sorted index list can also be gotten using numpy.argsort().

Answered By: 1-ijk

Answer #8:

Another alternative, combining several of the answers.

zip(*sorted(zip(Y,X)))[1]

In order to work for python3:

list(zip(*sorted(zip(B,A))))[1]
Answered By: TMC

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