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"]

---

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 .

---

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))]

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/

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!)

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')

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']

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() .

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]

Answer 9

zip, sort by the second column, return the first column.

zip(*sorted(zip(X,Y), key=operator.itemgetter(1)))[0]

Answer 10

This is an old question but some of the answers I see posted don't actually work because zip is not scriptable. Other answers didn't bother to import operator and provide more info about this module and its benefits here.

There are at least two good idioms for this problem. Starting with the example input you provided:

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

Using the " Decorate-Sort-Undecorate " idiom

This is also known as the Schwartzian_transform after R. Schwartz who popularized this pattern in Perl in the 90s:

# Zip (decorate), sort and unzip (undecorate).
# Converting to list to script the output and extract X
list(zip(*(sorted(zip(Y,X)))))[1]                                                                                                                       
# Results in: ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')

Note that in this case Y and X are sorted and compared lexicographically. That is, the first items (from Y ) are compared; and if they are the same then the second items (from X ) are compared, and so on. This can create unstable outputs unless you include the original list indices for the lexicographic ordering to keep duplicates in their original order.

Using the operator module

This gives you more direct control over how to sort the input, so you can get sorting stability by simply stating the specific key to sort by. See more examples here .

import operator    

# Sort by Y (1) and extract X [0]
list(zip(*sorted(zip(X,Y), key=operator.itemgetter(1))))[0]                                                                                                 
# Results in: ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')

Answer 11

A quick one-liner.

list_a = [5,4,3,2,1]
list_b = [1,1.5,1.75,2,3,3.5,3.75,4,5]

Say you want list a to match list b.

orderedList =  sorted(list_a, key=lambda x: list_b.index(x))

This is helpful when needing to order a smaller list to values in larger. Assuming that the larger list contains all values in the smaller list, it can be done.

Answer 12

I have created a more general function, that sorts more than two lists based on another one, inspired by @Whatang's answer.

def parallel_sort(*lists):
    """
    Sorts the given lists, based on the first one.
    :param lists: lists to be sorted

    :return: a tuple containing the sorted lists
    """

    # Create the initially empty lists to later store the sorted items
    sorted_lists = tuple([] for _ in range(len(lists)))

    # Unpack the lists, sort them, zip them and iterate over them
    for t in sorted(zip(*lists)):
        # list items are now sorted based on the first list
        for i, item in enumerate(t):    # for each item...
            sorted_lists[i].append(item)  # ...store it in the appropriate list

    return sorted_lists

Answer 13

You can create a pandas Series , using the primary list as data and the other list as index , and then just sort by the index:

import pandas as pd
pd.Series(data=X,index=Y).sort_index().tolist()

output:

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

Answer 14

Here is Whatangs answer if you want to get both sorted lists (python3).

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

Zx, Zy = zip(*[(x, y) for x, y in sorted(zip(Y, X))])

print(list(Zx))  # [0, 0, 0, 1, 1, 1, 1, 2, 2]
print(list(Zy))  # ['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']

Just remember Zx and Zy are tuples. I am also wandering if there is a better way to do that.

Warning: If you run it with empty lists it crashes.

Answer 15

This function should work for arrays.

def sortBoth(x,y,reverse=False):
    '''
    Sort both x and y, according to x. 
    '''
    xy_sorted=array(sorted(zip(x,y),reverse=reverse)).T
    return xy_sorted[0],xy_sorted[1]

Answer 16

I think most of the solutions above will not work if the 2 lists are of different sizes or contain different items. The solution below is simple and should fix those issues:

import pandas as pd

list1 = ['B', 'A', 'C']  # Required sort order
list2 = ['C', 'A']       # Items to be sorted according to list1

result = pd.merge(pd.DataFrame(list1), pd.DataFrame(list2))
print(list(result[0]))

output:

['A', 'C']

• Note: Any item not in list1 will be ignored since the algorithm will not know what's the sort order to use.

Answer 17

list1 = ['a','b','c','d','e','f','g','h','i']
list2 = [0,1,1,0,1,2,2,0,1]

output=[]
cur_loclist = []

To get unique values present in list2

list_set = set(list2)

To find the loc of the index in list2

list_str = ''.join(str(s) for s in list2)

Location of index in list2 is tracked using cur_loclist

[0, 3, 7, 1, 2, 4, 8, 5, 6]

for i in list_set:
cur_loc = list_str.find(str(i))

while cur_loc >= 0:
    cur_loclist.append(cur_loc)
    cur_loc = list_str.find(str(i),cur_loc+1)

print(cur_loclist)

for i in range(0,len(cur_loclist)):
output.append(list1[cur_loclist[i]])
print(output)

Answer 18

You can get the index of one sorted list and apply it to another, eg

idx = sorted(range(len(Y)), key=lambda i: Y[i])
ans = [X[i] for i in idx]

Answer 19

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

You can do so in one line:

X, Y = zip(*sorted(zip(Y, X)))

Answer 20

Most of the solutions above are complicated and I think they will not work if the lists are of different lengths or do not contain the exact same items. The solution below is simple and does not require any imports.

list1 = ['B', 'A', 'C']  # Required sort order
list2 = ['C', 'B']       # Items to be sorted according to list1

result = list1
for item in list1:
    if item not in list2: result.remove(item)

print(result)

Output:

['B', 'C']

• Note: Any item not in list1 will be ignored since the algorithm will not know what's the sort order to use.

Answer 21

For java what I got using selection sort is this:-

static void Sorting_list_based_on_values_from_another_list(){

    String []  X = {"a", "b", "c", "d", "e", "f", "g", "h", "i" };
    int []  Y = { 0,   1,   1,   0,   1,   2,   2,   0,   1 };


    for(int i=0;i<=Y.length;i++){
        for (int j=i;j<Y.length;j++){
           if(Y[i]>Y[j]){
               int n=Y[i];
               Y[i] = Y[j];
               Y[j] = n;

            // swpping alphabets
               String x=X[i];
               X[i]=X[j];
               X[j]=x;
           }
           }
    }

    for (int a:Y) {
        System.out.print(a +" ");
    }
    System.out.println("");
    for (String a:X) {
        System.out.print(a+" ");
    }
}

Answer 22

I think that the title of the original question is not accurate. If you have 2 lists of identical number of items and where every item in list 1 is related to list 2 in the same order (eg a = 0, b = 1, etc.) then the question should be 'How to sort a dictionary?', not 'How to sorting list based on values from another list?'. The solution below is the most efficient in this case:

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

dict1 = dict(zip(X,Y))
result = sorted(dict1, key=dict1.get)
print(result)

Result:

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