How to check if two Maps have the same key set in JavaScript

Question

Suppose to have two Map objects, how to check if their keys sets are the same?

For example:

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

in this case both A and B maps have the same key set (which is ['x', 'y']), while the key set of C is different since it has the extra key z.

@SuperDJ being specific to Map I don't think this question is same, mainly the way of answers will be different — Koushik Chatterjee
– Koushik Chatterjee, Commented Aug 27, 2018 at 9:48
@FrancescoBorzì may I inquire as to the reason why you accepted my long and clunky answer instead of these others that look shorter / cleaner? (Only thing I can come up with is the complexity of the answer, mine looks more like self-explaining pseudo code) — Fabian N.
– Fabian N., Commented Aug 27, 2018 at 10:01

CertainPerformance · Accepted Answer · 2018-08-26 20:26:15Z

5

Check that each map's size is the same, and then iterate over the keys of one Map and check that the key exists in the other as well. Utilizing Array.prototype.every.call means that there's no need to create an intermediate array:

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

const sameKeySet = (m1, m2) => (
  m1.size === m2.size
  && Array.prototype.every.call(m1.keys(), key => m2.has(key))
);
console.log(sameKeySet(A, B));
console.log(sameKeySet(A, C));

edited Aug 26, 2018 at 20:26

answered Aug 26, 2018 at 20:06

CertainPerformance

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3 Comments

Fabian N. Over a year ago

Well, that looks certainly cleaner than mine. Is there a reason to use Array.prototype instead of []. Also can't you drop the (... && ...) brackets around the condition?

CertainPerformance Over a year ago

Utilizing Array.prototype.every.call means that there's no need to create an intermediate array, whereas something like [...m1.keys()] would unnecessarily create a new Array from the iterator. Yes, it would be syntactically valid to drop the parentheses, but because it's a multi-line expression, I think the parentheses make the code clearer to read.

Fabian N. Over a year ago

Ah ok, I was thinking about [].every.call hoping that an empty array wouldn't be that much of a problem and it wouldn't look "so scary" as the Array.prototype =) [actually I only wanted to golf it down to two lines]

Fabian N. · Accepted Answer · 2018-08-26 20:10:36Z

1

You could check the size and then iterate over the keys of one map and check that the other one has them too.

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

function sameKeys(a, b) {
  if (a.size != b.size) {
    return false;
  }

  for (let key in a.keys()) {
    if (!b.has(key)) {
      return false;
    }
  }

  return true;
}

console.log(sameKeys(A, B));
console.log(sameKeys(A, C));

answered Aug 26, 2018 at 20:10

Fabian N.

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Comments

Dez · Accepted Answer · 2018-08-26 20:12:26Z

1

You can convert the keys of a Map into an array by spreading the iterator returned by the keys() method:

const aKeys = [...A.keys()];

Then you just will have to compare all the keys arrays. For the case you show up, you can simple do:

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

const aKeys = [...A.keys()];
const bKeys = [...B.keys()];
const cKeys = [...C.keys()];

console.log(aKeys.sort().toString() == bKeys.sort().toString());
console.log(aKeys.sort().toString() == cKeys.sort().toString());
console.log(bKeys.sort().toString() == cKeys.sort().toString());

answered Aug 26, 2018 at 20:12

Dez

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Comments

amrender singh · Accepted Answer · 2018-08-26 20:13:20Z

Basically you need to check for two things:

Size of both the maps, if they are unequal than simply return false.
If size is same than check if all the keys of map1 are present in map2,if they are than return true else return false.

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

const D = new Map();
C.set('x', 121);
C.set('z', 232);


function isSame(a,b){
  if(a.size != b.size)
    return false;
 for(const [key, value] of a.entries()){
    if(!b.has(key))
      return false;
  }
  return true;
}
console.log(isSame(A,B));
console.log(isSame(A,C));
console.log(isSame(A,D));

Nina Scholz · Accepted Answer · 2018-08-26 20:33:14Z

1

You could check the size and take the prototype of has and the second map as thisArg for checking all keys with Array#some.

This works for any types, because it does not mutate the type of the keys.

const
    compare = (a, b) => a.size === b.size && [...a.keys()].some(Map.prototype.has, b),
    a = new Map([['x', 123], ['y', 345]]);
    b = new Map([['y', 567], ['x', 789]]);
    c = new Map([['x', 121], ['y', 232], ['z', 434]]);

console.log(compare(a, b));
console.log(compare(a, c));

edited Aug 26, 2018 at 20:33

answered Aug 26, 2018 at 20:26

Nina Scholz

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Comments

Fabian N. · Accepted Answer · 2018-08-27 10:07:14Z

0

You can construct a new Map with both of their entries, and then compare size. Anyway you need to check the size of both of them and if its same then only you should proceed for this.

map1.size.size === map2.size &&

new Map([...map1, ...map2])).size === map1.size //or map2.size

Let's create a working example:

const A = new Map();
A.set('x', 123);
A.set('y', 345);

const B = new Map();
B.set('y', 567);
B.set('x', 789);

const C = new Map();
C.set('x', 121);
C.set('y', 232);
C.set('z', 434);

let compareMap = (m1, m2) => (
  m1.size === m2.size &&
  (new Map([...m1, ...m2])).size === m1.size
)

console.log('Compare A & B: ', compareMap(A, B));
console.log('Compare A & C: ', compareMap(A, C));
console.log('Compare B & C: ', compareMap(B, C));

edited Aug 27, 2018 at 10:07

Fabian N.

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answered Aug 27, 2018 at 7:06

Koushik Chatterjee

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2 Comments

Fabian N. Over a year ago

In case you didn't know: Code-Snippets have an inbuild autoformat button: i.sstatic.net/RwGHH.png

Koushik Chatterjee Over a year ago

Thanks for the alignment edit. However, people should focus on the content (at least when it's 2-liner) but yes, alignment dose matter to a better have visibility of the actual content!

Keith Whittingham · Accepted Answer · 2021-07-29 19:45:17Z

0

Here's a one liner wrapped in a TypeScript'ized function

function sameKeys(a: Map<string, string>, b: Map<string, string>): boolean {
    return a.size === b.size && [...a.keys()].every(key => b.has(key))
}

answered Jul 29, 2021 at 19:45

Keith Whittingham

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