Get the indices of an array after sorting

Question

Your challenge today is to write a program or function which takes a list l and gives the positions in l at which each successive element of l sorted appears.

In other words, output the index of the smallest value, followed by the index of the second smallest value, etc.

You can assume that the input array will contain only positive integers, and will contain at least one element.

Test cases:

Input                  | Output (1-indexed)
[7, 4, 5]              | [2, 3, 1]
[1, 2, 3]              | [1, 2, 3]
[2, 6, 1, 9, 1, 2, 3]  | [3, 5, 1, 6, 7, 2, 4]
[4]                    | [1]

When two or more elements with the same value appear, their indices should appear next to each other from smallest to largest.

This is code-golf, fewest bytes wins!

Answer 1

Jelly, 1 byte

Ụ

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Answer 2

Dyalog APL, 1 byte

⍋

Dyalog APL has a built in operator function (thank you Zacharý for clearing this up) to do this.

Example

⍋11 2 4 15
    2 3 1 4  
{⍵[⍋⍵]}11 4 2 15
    2 4 11 15

Here I'm indexing into the list by the sorted indices to return the list in ascending order.

Answer 3

Haskell, 43 42 bytes

1-indexed:

import Data.List
map snd.sort.(`zip`[1..])

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-1 byte thanks to @ØrjanJohansen!

Answer 4

Python 2, 56 bytes

This solution is 0-indexed. This abuses the fact that sorted() creates a copy of the original list.

l=input()
for k in sorted(l):a=l.index(k);print a;l[a]=0

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Answer 5

Javascript (ES6), 39 bytes

-2 bytes thanks to @powelles

This only works in browsers where Array.prototype.sort is stable.

a=>[...a.keys()].sort((b,c)=>a[b]-a[c])

1-indexed version (47 bytes):

a=>a.map((_,i)=>i+1).sort((b,c)=>a[b-1]-a[c-1])

Example code snippet:

f=
a=>[...a.keys()].sort((b,c)=>a[b]-a[c])
console.log("7,4,5 => "+f([7,4,5]))
console.log("1,2,3 => "+f([1,2,3]))
console.log("2,6,1,9,1,2,3 => "+f([2,6,1,9,1,2,3]))
console.log("4 -> "+f([4]))

Answer 6

Python 2, 48 bytes

lambda x:sorted(range(len(x)),key=x.__getitem__)

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Answer 7

Perl 6,  27  21 bytes

*.pairs.sort(*.value)».key

Test it

->\x{sort {x[$_]},^x}

Test it

Inspired by a Python answer

Expanded:

->    # pointy block lambda
  \x  # declare sigilless parameter
{
  sort
    { x[$_] },  # sort by the value in ｢x｣ at the given position
    ^x          # Range up-to the number of elements in ｢x｣
}

Answer 8

Bash + coreutils, 20

nl|sort -nk2|cut -f1

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Answer 9

R, 5 bytes

There is a builtin function for this.

order

Answer 10

Swift 4, 82 bytes

func f(l:[Int]){var l=l;for k in l.sorted(){let a=l.index(of:k)!;print(a);l[a]=0}}

Test Suite.

Explanation

In Swift, l.sorted() creates a sorted copy of the original Array. We loop through the sorted elements in the list and after printing each item's index in the original Array with let a=l.index(of:k)!;print(a), and then, in order to keep the correct indexes in the Array, we assign l[a] to 0, because it does not affect our normal output.

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Take note that this is 0-indexed, since it is a port of my Python solution. If you want it to be 1-indexed, replace print(a) with print(a+1) or Try it online!.

Answer 11

Ruby, 40 bytes

->a{a.zip([*1..a.size]).sort.map &:last}

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Answer 12

MATL, 2 bytes

&S

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Input and output are implicit.

Answer 13

Octave, 17 bytes

@(i)[~,y]=sort(i)

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Octave is like MATLAB but with inline assignment, making things possible that gives the folks at Mathworks HQ headaches. It doesn't matter what you call y, but you can't do without that dummy variable, as far as I know.

Answer 14

MY, 3 bytes

MY also has a builtin for this!

⎕⍋↵

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How?

Evaluated input, grade up, then output with a newline.

Indexed however you set the index, with ⌶/0x48. (Can even be some weird integer like -1 or 2, the default is 1).

Answer 15

J, 2 bytes

/:

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Zero-based indexing.

Answer 16

Java 8, 128 + 19 = 147 bytes

Based on Mr. Xcoder's solution. 0-based. Lambda takes input as an Integer[] and returns Integer[]. Byte count includes lambda expression and required import.

import java.util.*;

l->{Integer o[]=l.clone(),s[]=l.clone(),i=0;for(Arrays.sort(s);i<l.length;)l[o[i]=Arrays.asList(l).indexOf(s[i++])]=0;return o;}

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Ungolfed lambda

l -> {
    Integer
        o[] = l.clone(),
        s[] = l.clone(),
        i = 0
    ;
    for (Arrays.sort(s); i < l.length; )
        l[o[i] = Arrays.asList(l).indexOf(s[i++])] = 0;
    return o;
}

Notes

I use Integer[] instead of int[] to allow use of Arrays.asList, which has no primitive versions. Integer is preferred to Long because values are used as array indices and would require casting.

This ended up being shorter than my best procedural-style List solution because of the cost of class and method names.

This also beat a solution I tried that streamed the inputs, mapped to (value, index) pairs, sorted on values, and mapped to indices, mostly because of the baggage needed to collect the stream.

Acknowledgments

• -5 bytes thanks to Nevay

Answer 17

Common Lisp, 82 bytes

(lambda(l)(loop as i in(sort(copy-seq l)'<)do(setf(elt l(print(position i l)))0)))

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Answer 18

MATLAB / Octave, 29 bytes

[~,y]=sort(input(''));disp(y)

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Answer 19

Clojure, 39 bytes

#(map key(sort-by val(zipmap(range)%)))

Answer 20

CJam, 12 bytes

{ee{1=}$0f=}

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Answer 21

JavaScript (ES6), 69 bytes

0-indexed. Works for lists containing up to 65,536 elements.

a=>[...a=a.map((n,i)=>n<<16|i)].sort((a,b)=>a-b).map(n=>a.indexOf(n))

Test cases

let f =

a=>[...a=a.map((n,i)=>n<<16|i)].sort((a,b)=>a-b).map(n=>a.indexOf(n))

console.log(JSON.stringify(f([7, 4, 5])))
console.log(JSON.stringify(f([1, 2, 3])))
console.log(JSON.stringify(f([2, 6, 1, 9, 1, 2, 3])))
console.log(JSON.stringify(f([4])))

Answer 22

Python 3 with Numpy, 38 26 bytes

12 bytes saved thanks to Jo King (no need to give the function a name)

import numpy
numpy.argsort

Output is 0-based.

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Answer 23

Python 3, 52 bytes

0-indexed. Based on Bruce Forte's Haskell answer here and G B's Ruby answer here.

lambda l:list(zip(*sorted(zip(l,range(len(l))))))[1]

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Answer 24

Husk, 10 7 bytes

This is a direct port of my Haskell answer, also 1-indexed:

m→O`z,N

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Ungolfed/Explained

Code        Description               Example
         -- implicit input            [2,6,1]
      N  -- natural numbers           [1,2,3,..]
   `z,   -- zip, but keep input left  [(2,1),(6,2),(1,3)]
  O      -- sort                      [(1,3),(2,1),(6,2)]
m→       -- keep only indices         [3,1,2]

Answer 25

Java (OpenJDK 8), 72 bytes

l->l.stream().sorted().map(i->{int j=l.indexOf(i);l.set(j,0);return j;})

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Takes a List<Integer>, returns a Stream<Integer> containing the results.

We get a Stream based off the initial list, sort it, then map each number to it's index in the list. In order to accommodate duplicate elements, we set the original element in the list to 0.

Answer 26

SmileBASIC, 67 bytes

DEF I(A)DIM B[0]FOR I=1TO LEN(A)PUSH B,I
NEXT
SORT A,B
RETURN B
END

Very simple, all it does is generate a list of numbers from 1 to (length of array) and sort this by the same order as the input.

Answer 27

05AB1E, 4 bytes

ā<Σè

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Answer 28

Pari/GP, 16 bytes

a->vecsort(a,,1)

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Answer 29

PHP, 54 bytes

<?php function i($a){asort($a);return array_keys($a);}

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This is zero-indexed. Simply sorts the array and returns the keys.

Answer 30

Tcl, 21 bytes

(0-indexed)

puts [lsort -indi $L]

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Answer 31

Ruby, 34 bytes

->a{(0...a.size).sort_by{|i|a[i]}}

0-indexed, 1-indexing requires one more net byte:

->a{(1..a.size).sort_by{|i|a[i-1]}}

Also, a slightly golfed version of G B's answer is 37 bytes:

->a{a.each_with_index.sort.map &:pop}

Answer 32

Kotlin,  44  34 bytes

^{not today, crossed out 44}

{l->(1..l.size).sortedBy{l[it-1]}}

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1-indexed because that's what the examples give. A 0-indexed version looks like this for 30 bytes:

{l->l.indices.sortedBy{l[it]}}

Answer 33

Japt, 6 5 bytes

0-indexed

ð ñ@v

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Alternative, 6 bytes

í ñÎmÌ

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