An Array Exercise

May 8, 2018

Depending on your language, you need some kind of ordered map to solve this problem; we use the avl trees of the Standard Prelude:

(define (rank vec)
  (let ((dict (make-dict <)))
    (do ((i 0 (+ i 1))) ((= i (vector-length vec)))
      (set! dict (dict 'insert (vector-ref vec i) (vector-ref vec i))))
    (do ((i 0 (+ i 1))) ((= i (vector-length vec)) vec)
      (vector-set! vec i (+ 1 (dict 'rank (vector-ref vec i)))))))

The first pass stores each array element in an avl tree dict. The second pass resets each element of the array to its rank in the avl tree, then returns the array. Here’s the example:

> (rank '#(10 8 15 12 6 20 1))
#(4 3 6 1 2 7 1)

You can run the program at https://ideone.com/tF9m40.

19 Responses to “An Array Exercise”

James Curtis-Smith said

May 8, 2018 at 9:34 AM

Golf solution:

perl -E 'say"@{[map{$a=$_;1+grep{$_<$a}@ARGV}@ARGV]}";' 10 8 15 12 6 20 1

Pascal Bourguignon said

May 8, 2018 at 9:48 AM


;; First idea:
;; - wrap the elements with their original position,
;; - sort the wrapping by element,
;; - replace the elements by their rank = position in the sorted vector,
;; - sort the wrapping by the original position,
;; - unwrap.
;; The principle is simple, and we obtain easily a correct solution:

(defun vector-ranks (vector)
  ;; This is O(nlogn+n+nlogn+n) = O(nlogn) in time,
  ;; and O(3n)=O(n) in space.
  ;; Original vector is not modified.
  (map 'vector (function cdr)
       (sort (map 'vector (let ((rank 0))
                            (lambda (entry)
                              (setf (cdr entry) (incf rank))
                              entry))
                  (sort (map 'vector (let ((position -1))
                                       (lambda (element) (cons (incf position) element)))
                             vector)
                        (function <)
                        :key (function cdr)))
             (function <)
             :key (function car))))

(assert (equalp (vector-ranks #(10 8 15 12 6 20 1))
                #(4 3 6 5 2 7 1)))

;; If you insist on mutation:
(defun program-to-replace-vector-elements-by-their-rank (vector)
  (replace vector (vector-ranks vector)))



;; Then, we can see that since we have the position in the sorted
;; wrapping, we can update directly the original vector without
;; re-sorting and unwrapping, thus obtaining an optimized solution:

(defun replace-vector-by-rank (vector)
  ;; This is O(nlogn+n) = O(nlogn) in time,
  ;; and O(n) in space.
  ;; Original vector is modified.
  (let ((sorted (sort (map 'vector (let ((position -1))
                                     (lambda (element) (cons (incf position) element)))
                           vector)
                      (function <)
                      :key (function cdr))))
    (loop
      :for rank :from 1
      :for (position) :across sorted
      :do (setf (aref vector position) rank))
    vector))

Pascal Bourguignon said
May 8, 2018 at 9:49 AM
Forgot the test of the second solution:
```
(assert (equalp (replace-vector-by-rank (vector 10 8 15 12 6 20 1))
                #(4 3 6 5 2 7 1)))
```
(Note we use (vector …) instead of the literal #(…) since now we are mutating this argument!).
Seán Stickle said
May 8, 2018 at 12:49 PM
Solution in J:

a =: 10 8 15 12 6 20 1
1+ (a /: a) i. a

matthew said

May 8, 2018 at 1:15 PM

>>> a = [10,8,15,12,6,20,1]
>>> n = len(a)
>>> r = sorted(range(n),key=lambda i: a[i])
>>> for i in range(n): a[r[i]] = i+1
>>> a
[4, 3, 6, 5, 2, 7, 1]

Zack said
May 8, 2018 at 1:26 PM
Here is my take on this, using Julia.

function Nums2Ranks(x::Array{Int64, 1})
n = length(x)
if length(unique(x)) != n; println(“the input array must be comprised of distinct integers!”); return []; end
y = Array{Int64}(n)
z = sort(x)
```
for i = 1:n
    y[i] = find(z .== x[i])[1]
end

return y
```
end

Testing with x = [10,8,15,12,6,20,1], as well as some non-valid input data:

Nums2Ranks(x)
7-element Array{Int64,1}:
4
3
6
5
2
7
1

Nums2Ranks([1,1,5])
the input array must be comprised of distinct integers!

0-element Array{Any,1}

mcmillhj said

May 8, 2018 at 3:16 PM

Perl6 solution using some interesting standard list functions: antipairs and kv

sub rankify(@numbers) {
  @numbers.antipairs().sort(
    { $^a.key <=> $^b.key }
  ).kv().map(
    sub ($rank, $pair) {
      $pair.value => $rank + 1
    }
  ).sort(
    { $^a.key <=> $^b.key }
  ).map({ .value });
}

say rankify([10, 8, 15, 12, 6, 20, 1]);
# (4 3 6 5 2 7 1)

kbob said

May 8, 2018 at 7:14 PM

>>> from itertools import count
>>> a = [10, 8, 15, 12, 6, 20, 1]
>>> s = dict(zip(sorted(a), count(1)))
>>> [s[e] for e in a]
[4, 3, 6, 5, 2, 7, 1]

V said

May 9, 2018 at 12:36 AM

Quick and dirty solution in Ruby.

def order_by_rank(xs)  
  xs.map { |x| 1 + xs.sort.find_index(x) }
end

# Test
order_by_rank([10,8,15,12,6,20,1])

# Outputs
[4, 3, 6, 5, 2, 7, 1]

V said
May 9, 2018 at 12:50 AM
I just realised the function name order_by_rank is misleading as the result doesn’t include the elements of the original array. rankify is a better function name. Ohh well.
sbocq said
May 10, 2018 at 5:49 PM
The old Unix way using Bash.
```
$ { tr -d '[]'|tr ',' $'\n'|cat -n|sort -n -k2|cat -n|sort -n -k2,2|cut -f1|tr -d ' '|echo "[$(paste -sd "," -)]"; } <<< "[10,8,15,12,6,20,1]"
[4,3,6,5,2,7,1]
```
Delete bracket characters from the string representation of an array.
Output one array element per line by transliterating commas to newlines.
Number all output lines to record in the first column the index of each element.
Sort lines numerically on the array elements recorded now in the second column to list them in their rank order.
Number again all output lines to record in the first column the rank of each element.
Sort lines back in the initial order using the index of each element recorded now in the second column.
Cut out the first column, which now contains the rank of each element in the initial order.
Delete extraneous white spaces in each input line.
Paste rank numbers separated by commas in a string within brackets to restore the original array representation.

Daniel said

May 10, 2018 at 8:35 PM

Here’s a solution in Python using numpy.

import numpy as np

def replace_with_ranks(array):
    array[np.argsort(array)] = np.arange(1,len(array)+1)

array = np.array([10,8,15,12,6,20,1])
replace_with_ranks(array)
print array

Output:

[4 3 6 5 2 7 1]

Daniel said

May 10, 2018 at 9:19 PM

Here’s a solution in C.

#include <stdio.h>
#include <stdlib.h>

typedef struct {
  int value;
  size_t index;
} indexed_int_t;

static int compar(const void* a, const void* b) {
  int a_value = (*(indexed_int_t*)a).value;
  int b_value = (*(indexed_int_t*)b).value;
  if (a_value < b_value) return -1;
  if (a_value > b_value) return 1;
  return 0;
}

void replace_with_ranks(int* array, size_t n) {
  indexed_int_t* indexed_array = malloc(n * sizeof(indexed_int_t));
  for (size_t i = 0; i < n; ++i) {
    indexed_array[i].value = array[i];
    indexed_array[i].index = i;
  }
  qsort(indexed_array, n, sizeof(indexed_int_t), compar);
  for (size_t i = 0; i < n; ++i) {
    array[indexed_array[i].index] = i + 1;
  }
  free(indexed_array);
}

int main(int argc, char* argv[]) {
  size_t n = argc - 1;
  int* array = malloc(n * sizeof(int));
  for (size_t i = 0; i < n; ++i) {
    array[i] = atoi(argv[i+1]);
  }
  replace_with_ranks(array, n);
  printf("[");
  for (size_t i = 0; i < n; ++i) {
    if (i > 0) printf(", ");
    printf("%d", array[i]);
  }
  printf("]\n");
  free(array);
  return EXIT_SUCCESS;
}

Example:

$ ./replace_with_ranks 10 8 15 12 6 20 1
[4, 3, 6, 5, 2, 7, 1]

Sam said

May 12, 2018 at 5:02 AM

#lang racket/base

(define (rank-sort xs)
  (sort xs < #:key car #:cache-keys? #t))

(define (ranky v)
  (define ordered
    (rank-sort
     (for/list ([j (in-vector v)]
                [i (in-naturals)])
       (cons j i))))

  (let ([v2 (make-vector (vector-length v))])
    (for ([x (in-list ordered)]
          [i (in-naturals 1)])
      (vector-set! v2 (cdr x) i))
    v2))

(ranky #(10 8 15 12 6 20 1))
;; => '#(4 3 6 5 2 7 1)

sbocq said

May 14, 2018 at 8:36 PM

Clojure/Script.

(defn ranks [v]
  (mapv (comp inc first)
        (sort-by second (map-indexed cons (sort-by second (map-indexed vector v))))))

(ranks [10 8 15 12 6 20 1])
[4 3 6 5 2 7 1]

Try online at http://clojurescript.net/.

Andrew said

May 15, 2018 at 10:48 PM

sbocq’s Clojure/Script can also be written as:

(defn ranks [v]
  (->> v
       (map-indexed vector)
       (sort-by second)
       (map-indexed cons)
       (sort-by second)
       (mapv (comp inc first))))

WWW.ХХХ.UFZSN.ХУZ said
May 18, 2018 at 2:58 PM
What ?
Jef said
May 18, 2018 at 9:33 PM
Tcl version:

proc rank {vector} {
set sorted [lsort -integer $vector]
lmap v $vector { expr [lsearch $sorted $v] + 1 }
}

% rank {10 8 15 12 6 20 1}
4 3 6 5 2 7 1
B said
May 21, 2018 at 2:30 AM
The implementation provided at https://ideone.com/tF9m40 seems to have a bug. Here’s an input to test on:

‘#(5242882 8388611 2097156 3145733 9437190 8388614 5242889 7340042 2097163 2097166)

The result should be: 5 8 1 4 10 9 6 7 2 3.

But when the ideone code is run, I get 5 8 2 5 10 9 6 7 3 4 .

You can see it in action at https://ideone.com/9dh6b0.

Programming Praxis