comparing ECMA6 sets for equality

问题:

How do you compare two javascript sets? I tried using == and === but both return false.

a = new Set([1,2,3]);
b = new Set([1,3,2]);
a == b; //=> false
a === b; //=> false

These two sets are equivalent, because by definition, sets do not have order (at least not usually). I've looked at the documentation for Set on MDN and found nothing useful. Anyone know how to do this?

回答1:

Try this:

var a = new Set([1,2,3]);
var b = new Set([1,3,2]);

alert(eqSet(a, b)); // true

function eqSet(as, bs) {
    if (as.size !== bs.size) return false;
    for (var a of as) if (!bs.has(a)) return false;
    return true;
}

A more functional approach would be:

var a = new Set([1,2,3]);
var b = new Set([1,3,2]);

alert(eqSet(a, b)); // true

function eqSet(as, bs) {
    return as.size === bs.size && all(isIn(bs), as);
}

function all(pred, as) {
    for (var a of as) if (!pred(a)) return false;
    return true;
}

function isIn(as) {
    return function (a) {
        return as.has(a);
    };
}

The all function works for all iterable objects (e.g. Set and Map).

If Array.from was more widely supported then we could have implemented the all function as:

function all(pred, as) {
    return Array.from(as).every(pred);
}

Hope that helps.

回答2:

You can also try:

var a = new Set([1,2,3]);
var b = new Set([1,3,2]);

isSetsEqual = (a, b) => a.size === b.size && [...a].every(value => b.has(value));

console.log(isSetsEqual(a,b)) 

回答3:

lodash provides _.isEqual(), which does deep comparisons. This is very handy if you don't want to write your own. As of lodash 4, _.isEqual() properly compares Sets.

const _ = require("lodash");

let s1 = new Set([1,2,3]);
let s2 = new Set([1,2,3]);
let s3 = new Set([2,3,4]);

console.log(_.isEqual(s1, s2)); // true
console.log(_.isEqual(s1, s3)); // false

回答4:

The other answer will work fine; here is another alternative.

// Create function to check if an element is in a specified set.
function isIn(s)          { return elt => s.has(elt); }

// Check if one set contains another (all members of s2 are in s1).
function contains(s1, s2) { return [...s2] . every(isIn(s1)); }

// Set equality: a contains b, and b contains a
function eqSet(a, b)      { return contains(a, b) && contains(b, a); }

// Alternative, check size first
function eqSet(a, b)      { return a.size === b.size && contains(a, b); }

However, be aware that this does not do deep equality comparison. So

eqSet(Set([{ a: 1 }], Set([{ a: 1 }])

will return false. If the above two sets are to be considered equal, we need to iterate through both sets doing deep quality comparisons on each element. We stipulate the existence of a deepEqual routine. Then the logic would be

// Find a member in "s" deeply equal to some value
function findDeepEqual(s, v) { return [...s] . find(m => deepEqual(v, m)); }

// See if sets s1 and s1 are deeply equal. DESTROYS s2.
function eqSetDeep(s1, s2) {
  return [...s1] . every(a1 => {
    var m1 = findDeepEqual(s2, a1);
    if (m1) { s2.delete(m1); return true; }
  }) && !s2.size;
}

What this does: for each member of s1, look for a deeply equal member of s2. If found, delete it so it can't be used again. The two sets are deeply equal if all the elements in s1 are found in s2, and s2 is exhausted. Untested.

You may find this useful: http://www.2ality.com/2015/01/es6-set-operations.html.

回答5:

None of these solutions bring “back” the expected functionality to a data structure such as set of sets. In its current state, the Javascript Set is useless for this purpose because the superset will contain duplicate subsets, which Javascript wrongly sees as distinct. The only solution I can think of is converting each subset to Array, sorting it and then encoding as String (for example JSON).

Solution

var toJsonSet = aset /* array or set */ => JSON.stringify([...new Set(aset)].sort()); 
var fromJsonSet = jset => new Set(JSON.parse(jset));

Basic usage

var toJsonSet = aset /* array or set */ => JSON.stringify([...new Set(aset)].sort()); 
var fromJsonSet = jset => new Set(JSON.parse(jset));

var [s1,s2] = [new Set([1,2,3]), new Set([3,2,1])];
var [js1,js2] = [toJsonSet([1,2,3]), toJsonSet([3,2,1])]; // even better

var r = document.querySelectorAll("td:nth-child(2)");
r[0].innerHTML = (toJsonSet(s1) === toJsonSet(s2)); // true
r[1].innerHTML = (toJsonSet(s1) == toJsonSet(s2)); // true, too
r[2].innerHTML = (js1 === js2); // true
r[3].innerHTML = (js1 == js2); // true, too

// Make it normal Set:
console.log(fromJsonSet(js1), fromJsonSet(js2)); // type is Set

<style>td:nth-child(2) {color: red;}</style>

<table>
<tr><td>toJsonSet(s1) === toJsonSet(s2)</td><td>...</td></tr>
<tr><td>toJsonSet(s1) == toJsonSet(s2)</td><td>...</td></tr>
<tr><td>js1 === js2</td><td>...</td></tr>
<tr><td>js1 == js2</td><td>...</td></tr>
</table>

Ultimate test: set of sets

var toSet = arr => new Set(arr);
var toJsonSet = aset /* array or set */ => JSON.stringify([...new Set(aset)].sort()); 
var toJsonSet_WRONG = set => JSON.stringify([...set]); // no sorting!

var output = document.getElementsByTagName("code"); 
var superarray = [[1,2,3],[1,2,3],[3,2,1],[3,6,2],[4,5,6]];
var superset;

Experiment1:
    superset = toSet(superarray.map(toSet));
    output[0].innerHTML = superset.size; // incorrect: 5 unique subsets
Experiment2:
    superset = toSet([...superset].map(toJsonSet_WRONG));
    output[1].innerHTML = superset.size; // incorrect: 4 unique subsets
Experiment3:
    superset = toSet([...superset].map(toJsonSet));
    output[2].innerHTML = superset.size; // 3 unique subsets
Experiment4:
    superset = toSet(superarray.map(toJsonSet));
    output[3].innerHTML = superset.size; // 3 unique subsets

code {border: 1px solid #88f; background-color: #ddf; padding: 0 0.5em;}

<h3>Experiment 1</h3><p>Superset contains 3 unique subsets but Javascript sees <code>...</code>.<br>Let’s fix this... I’ll encode each subset as a string.</p>
<h3>Experiment 2</h3><p>Now Javascript sees <code>...</code> unique subsets.<br>Better! But still not perfect.<br>That’s because we didn’t sort each subset.<br>Let’s sort it out...</p>
<h3>Experiment 3</h3><p>Now Javascript sees <code>...</code> unique subsets. At long last!<br>Let’s try everything again from the beginning.</p>
<h3>Experiment 4</h3><p>Superset contains 3 unique subsets and Javascript sees <code>...</code>.<br><b>Bravo!</b></p>

回答6:

Comparing two objects with ==, ===

When using == or === operator to compare two objects, you will always get false unless those object reference the same object. For example:

var a = b = new Set([1,2,3]); // NOTE: b will become a global variable
a == b; // <-- true: a and b share the same object reference

Otherwise, == equates to false even though the object contains the same values:

var a = new Set([1,2,3]);
var b = new Set([1,2,3]);
a == b; // <-- false: a and b are not referencing the same object

You may need to consider manual comparison

In ECMAScript 6, you may convert sets to arrays beforehand so you can spot the difference between them:

function setsEqual(a,b){
    if (a.size !== b.size)
        return false;
    let aa = Array.from(a); 
    let bb = Array.from(b);
    return aa.filter(function(i){return bb.indexOf(i)<0}).length==0;
}

NOTE: Array.from is one of the standard ECMAScript 6 features but it is not widely supported in modern browsers. Check the compatibility table here : https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from#Browser_compatibility

回答7:

I follow this approach in tests :

let setA = new Set(arrayA);
let setB = new Set(arrayB);
let diff = new Set([...setA].filter(x => !setB.has(x)));
expect([...diff].length).toBe(0);

回答8:

I created a quick polyfill for Set.prototype.isEqual()

Set.prototype.isEqual = function(otherSet) {
    if(this.size !== otherSet.size) return false;
    for(let item of this) if(!otherSet.has(item)) return false;
    return true;
}

Github Gist - Set.prototype.isEqual

回答9:

1) Check if sizes are equal . If not, then they are not equal.

2) iterate over each elem of A and check in that exists in B. If one fails return unequal

3) If the above 2 conditions fails that means they are equal.

let isEql = (setA, setB) => {
  if (setA.size !== setB.size)
    return false;
  
  setA.forEach((val) => {
    if (!setB.has(val))
      return false;
  });
  return true;
}

let setA = new Set([1, 2, {
  3: 4
}]);
let setB = new Set([2, {
    3: 4
  },
  1
]);

console.log(isEql(setA, setB));

2) Method 2

let isEql = (A, B) => {
  return JSON.stringify([...A].sort()) == JSON.stringify([...B].sort());
}

let res = isEql(new Set([1, 2, {3:4}]), new Set([{3:4},1, 2]));
console.log(res);

回答10:

Based on the accepted answer, assuming support of Array.from, here is a one-liner:

function eqSet(a, b) {
    return a.size === b.size && Array.from(a).every(b.has.bind(b));
}

回答11:

If sets contains only primitive data types or object inside sets have reference equality, then there is simpler way

const isEqualSets = (set1, set2) => (set1.size === set2.size) && (set1.size === new Set([...set1, ...set2]).size);