Changing the order of an array based on another with javascript sort

Question

I have two arrays,

A = [green, blue, red];

B = [2, 1, 0]

I want to change the order of A to be in [red, blue, green]. The changes are based on the value of B Below is what I have tried,

arrangeValues();
function arrangeValues() {
    A.sort((a,b)=>{
        let orderA:any=B.indexOf(a.value);
        let orderB:any=B.indexOf(b.value);
        if (orderA==-1)
            orderA=99999;
        if (orderB==-1)
            orderB=99999;
        return orderB-orderA
    })
}

The order of b doesn't make sense with the expected output, that would be [red,blue,green] — baao
– baao, Commented Jul 7, 2018 at 14:38
Is B always the same length as A and made up of consecutive numbers starting at 0? — Ry-
– Ry- ♦, Commented Jul 7, 2018 at 14:39
Then you'll need to specify -- see the answer below, right now we can't tell what you want exactly — baao
– baao, Commented Jul 7, 2018 at 14:40

T.J. Crowder · Accepted Answer · 2018-07-07 15:15:43Z

3

I want to change the order of A to be in [red, blue, green]

The simplest way is to create a new array, using the indexes from B, and taking values from A based on where you are in B:

const A = ["green", "blue", "red"];
const B = [2, 1, 0];
const result = [];
B.forEach((entry, index) => {
  result[entry] = A[index];
});
console.log(result);

Just to be sure that really does what you want (because it was actually a typo on my part, I meant to do something different but seem to have accidentally done what you wanted), here are all six combinations:

const A = ["green", "blue", "red"];

function test(B) {
  const result = [];
  B.forEach((entry, index) => {
    result[entry] = A[index];
  });
  console.log(B.join(", "), "=>", result.join(", "));
}
for (let i = 0; i < A.length; ++i) {
  for (let j = 0; j < A.length; ++j) {
    if (j != i) {
      for (let k = 0; k < A.length; ++k) {
        if (i != k && j != k) {
          test([i, j, k]);
        }
      }
    }
  }
}

.as-console-wrapper {
  max-height: 100% !important;
}

Here's are all six doing what I meant to do (which is what zvona did, zvona's is cleaner):

const A = ["green", "blue", "red"];

function test(B) {
  const result = [];
  B.forEach((entry, index) => {
    result[index] = A[entry];
  });
  console.log(B.join(", "), "=>", result.join(", "));
}
for (let i = 0; i < A.length; ++i) {
  for (let j = 0; j < A.length; ++j) {
    if (j != i) {
      for (let k = 0; k < A.length; ++k) {
        if (i != k && j != k) {
          test([i, j, k]);
        }
      }
    }
  }
}

.as-console-wrapper {
  max-height: 100% !important;
}

edited Jul 7, 2018 at 15:15

answered Jul 7, 2018 at 14:38

T.J. Crowder

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2 Comments

kartik Over a year ago

@ T.J what is ab

T.J. Crowder Over a year ago

@saila - It's a joke: "abuse" vs. "use".

Said · Accepted Answer · 2018-07-07 14:59:25Z

1

You can do that using a simple for loop something like:

function arrangeValues(A,B) {
  var sorted = [];
  for(var i=0;i<A.length;i++){
      sorted.push(A[B[i]]);
  }
  return sorted;
}

Here is a live example:

function arrangeValues(A,B) {
  var sorted = [];
  for(var i=0;i<A.length;i++){
      sorted.push(A[B[i]]);
  }
  return sorted;
}

var A = ["green", "blue", "red"];
var B = [2, 1, 0];
var result = arrangeValues(A,B);
console.log(result);

edited Jul 7, 2018 at 14:59

answered Jul 7, 2018 at 14:44

Said

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Comments

T.J. Crowder · Accepted Answer · 2018-07-07 15:17:52Z

1

Simple way:

const A: Array<string> = ["green", "blue", "red"];
const B: Array<number> = [2, 1, 0];
const result: Array<string> = B.map((entry: number) => A[entry]);

Live Example:

const A = ["green", "blue", "red"];
const B = [2, 1, 0];
const result = B.map((entry) => A[entry]);
console.log(result);

All six combinations using this approach:

const A = ["green", "blue", "red"];
function test(B) {
  const result = B.map((entry) => A[entry]);
  console.log(B.join(", "), "=>", result.join(", "));
}
for (let i = 0; i < A.length; ++i) {
  for (let j = 0; j < A.length; ++j) {
    if (j != i) {
      for (let k = 0; k < A.length; ++k) {
        if (i != k && j != k) {
          test([i, j, k]);
        }
      }
    }
  }
}

edited Jul 7, 2018 at 15:17

T.J. Crowder

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answered Jul 7, 2018 at 14:47

Samuli Hakoniemi

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2 Comments

Ry- Over a year ago

If the original approach of sorting A based on the corresponding B was correct – and the comment “2 in B refers to green, so greens index should be 2” suggests it was – this won’t do the same thing in general. [1, 2, 0] is blue-red-green if it’s a sequence of indexes to read and red-green-blue if it’s the target index for a given element, for example.

Ry- Over a year ago

@saila: You should try them out on different arrays first, because they don’t do the same thing.

Leonid Pyrlia · Accepted Answer · 2018-07-07 14:50:45Z

0

Map the corresponding colors from A to values from B by index and sort by them:

const A = ['green', 'blue', 'red'];

const B = [2, 1, 0];

const sorted = A.sort((a,b) => {

    const [aV,bV] = [a,b].map((_,i) => B[i]);

    return aV - bV;

})

console.log(sorted);

Or if we can assume that B contains indexes, then a simple map would do the job:

const A = ['green', 'blue', 'red'];

const B = [2, 1, 0];

const result = B.map(i => A[i]);

console.log(result);

edited Jul 7, 2018 at 14:50

answered Jul 7, 2018 at 14:43

Leonid Pyrlia

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