How to get the intersection of two sets while recognizing equal set values/items not only by reference but by their equal structures and entries too?

Question

I have two deal with two Set instances.

const set1 = new Set([
  { name: 'a' },
  { name: 'b', lastname: 'bb' },
  { name: 'c' },
  { name: 'd' },
]);

const set2 = new Set([
  { name: 'b' },
  { name: 'd' },
]);

Any object within a set will feature several and also distinct keys and values. The goal is to find structurally equal objects (same keys and values) in both sets, which is ... The intersection of equal data items in/of set1 and set2.

In the following example the expected result is [ { name: 'd' } ] ...

console.log([...set1].filter(item => set2.has(item)));

... but it logs an empty array / [] instead.

An object features more than 20 keys so one has to compare them one by one, which can not be done in a hard coded way.

How could one achieve a generic approach for an intersection of two lists of structurally equal data items?

Answer 1

You can do something like this:

const set1 = new Set([
    {name: 'a'},
    {name: 'b', lastname: 'bb'},
    {name: 'c'},
    {name: 'd'}
]);

const set2 = new Set([
    {name: 'b'},
    {name: 'd'}
]);

set1.forEach((value) => {
    if (![...set2].some((o) => Object.entries(o).every(([k, v], _, arr) => (Object.keys(value).length === arr.length && value[k] === v)))) {
        set1.delete(value);
    }
})

console.log([...set1]);

What this does, is to iterate through set1 and if the item at the current iteration is not the same as any item in set2 (![...set2].some(..)), it is deleted. The items are considered the same if they have the same number of keys and if the values at the same key are strictly equal.

This only works if the values of the objects in the sets are primitives, if they are not, you'll have to change value[k] === v to an appropriate comparison.

Answer 2

One could write a generic solution which compares pure, thus JSON conform, data structures regardless of any object's nesting depth/level and (creation time) key order.

Such a function would be self recursive for Array item (order matters) and Object property (key order does not matter) comparison. Otherwise values are compared strictly.

function isDeepDataStructureEquality(a, b) {
  let isEqual = Object.is(a, b);

  if (!isEqual) {
    if (Array.isArray(a) && Array.isArray(b)) {

      isEqual = (a.length === b.length) && a.every(
        (item, idx) => isDeepDataStructureEquality(item, b[idx])
      );
    } else if (
      a && b
      && (typeof a === 'object')
      && (typeof b === 'object')
    ) {
      const aKeys = Object.keys(a);
      const bKeys = Object.keys(b);

      isEqual = (aKeys.length === bKeys.length) && aKeys.every(
        (key, idx) => isDeepDataStructureEquality(a[key], b[key])
      );
    }
  }
  return isEqual;
}

const objA = { // `objA` equals `objB`.
  name: 'foo',
  value: 1,
  obj: {
    z: 'z',
    y: 'y',
    a: {
      name: 'bar',
      value: 2,
      obj: {
        x: 'x',
        w: 'w',
        b: 'b',
      },
      arr: ['3', 4, 'W', 'X', {
        name: 'baz',
        value: 3,
        obj: {
          k: 'k',
          i: 'i',
          c: 'c',
        },
        arr: ['5', 6, 'B', 'A'],
      }],
    },
  },
  arr: ['Z', 'Y', 1, '2'],
};

const objB = { // `objB` equals `objA`.
  arr: ['Z', 'Y', 1, '2'],
  obj: {
    z: 'z',
    y: 'y',
    a: {
      obj: {
        x: 'x',
        w: 'w',
        b: 'b',
      },
      arr: ['3', 4, 'W', 'X', {
        obj: {
          k: 'k',
          i: 'i',
          c: 'c',
        },
        name: 'baz',
        value: 3,
        arr: ['5', 6, 'B', 'A'],
      }],
      name: 'bar',
      value: 2,
    },
  },
  name: 'foo',
  value: 1,
};

const objC = { // `objC` equals neither `objA` nor `objB`.
  arr: ['Z', 'Y', 1, '2'],
  obj: {
    z: 'z',
    y: 'y',
    a: {
      obj: {
        x: 'x',
        w: 'w',
        b: 'b',
      },
      arr: ['3', 4, 'W', 'X', {
        obj: {
          k: 'k',
          i: 'i',
          c: 'C', // the single difference to `objA` and `objB`.
        },
        name: 'baz',
        value: 3,
        arr: ['5', 6, 'B', 'A'],
      }],
      name: 'bar',
      value: 2,
    },
  },
  name: 'foo',
  value: 1,
};

console.log(
  'isDeepDataStructureEquality(objA, objB) ?..',
  isDeepDataStructureEquality(objA, objB)
);
console.log(
  'isDeepDataStructureEquality(objA, objC) ?..',
  isDeepDataStructureEquality(objA, objC)
);
console.log(
  'isDeepDataStructureEquality(objB, objC) ?..',
  isDeepDataStructureEquality(objB, objC)
);
console.log(
  'isDeepDataStructureEquality(objB, objA) ?..',
  isDeepDataStructureEquality(objB, objA)
);

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Based on the above implementation of isDeepDataStructureEquality one can solve the OP's task, that actually looks for the intersection of two list structures, by additionally providing a getIntersectionOfDeeplyEqualDataStructures functionality ...

function getIntersectionOfDeeplyEqualDataStructures(a, b) {
  return [...(a ?? [])]
    .reduce((collector, sourceItem) => {

      const { target, intersection } = collector;

      const targetIndex = target.findIndex(targetItem =>
        isDeepDataStructureEquality(targetItem, sourceItem)
      );
      if (targetIndex >= 0) {
        // collect the intersection of
        // both, source (a) and target (b).

        intersection.push(target[targetIndex]);
      }
      return collector;

    }, {

      target: [...(b ?? [])],
      intersection: [],

    }).intersection;
}

const set1 = new Set([
  { name: 'a' },
  { name: 'b', lastname: 'bb' },
  { name: 'c' },
  { name: 'd' }
]);
const set2 = new Set([
  { name: 'b' },
  { name: 'd' },
]);
console.log(
  "getIntersectionOfDeeplyEqualDataStructures(set1, set2) ...",
  getIntersectionOfDeeplyEqualDataStructures(set1, set2)
);

const set3 = new Set([
  { name: 'a' },
  { name: 'b', lastname: 'bb' },
  { name: 'c' },
  {
    name: 'd',
    list: ['foo', 1, null, false, 0, {
      foo: { bar: { baz: 'bizz', buzz: '' } }
    }],
  },
]);
const set4 = new Set([
  {
    list: ['foo', 1, null, false, 0, {
      foo: { bar: { buzz: '', baz: 'bizz' } }
    }],
    name: 'd',
  },
  { name: 'C' },
  { lastname: 'bb', name: 'b' },
  { name: 'aa' }
]);
console.log(
  "getIntersectionOfDeeplyEqualDataStructures(set3, set4) ...",
  getIntersectionOfDeeplyEqualDataStructures(set3, set4)
);

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<script>
  function isDeepDataStructureEquality(a, b) {
    let isEqual = Object.is(a, b);

    if (!isEqual) {
      if (Array.isArray(a) && Array.isArray(b)) {

        isEqual = (a.length === b.length) && a.every(
          (item, idx) => isDeepDataStructureEquality(item, b[idx])
        );
      } else if (
        a && b
        && (typeof a === 'object')
        && (typeof b === 'object')
      ) {
        const aKeys = Object.keys(a);
        const bKeys = Object.keys(b);

        isEqual = (aKeys.length === bKeys.length) && aKeys.every(
          (key, idx) => isDeepDataStructureEquality(a[key], b[key])
        );
      }
    }
    return isEqual;
  }
</script>

Edit

As for Titus' approach ...

set1.forEach(value => {
  if (
    ![...set2].some(o =>
      Object.entries(o).every(([k, v], _, arr) =>
        (Object.keys(value).length === arr.length && value[k] === v)
      )
    )
  ) {
    set1.delete(value);
  }
});

... which works for flat objects only, though already agnostic to key insertion order, one could optimize the code by ...

• ... not creating the keys array of the most outer currently processed object again and again with every nested some and every iteration.

  • thus, something like ... const valueKeys = Object.keys(value); ... before the if clause, already helps improving the code.

• ... inverting the nested some and every logic which does result in a more efficient way of ... deleting every flat data-item from the processed set which does not equal any flat data-item from the secondary set.

On top of that, one could implement a function statement which not only helps code-reuse but also makes the implementation independent from outer scope references.

For instance, the primary set which is operated and going to be mutated can be accessed as such a function's third parameter. But most important for outer scope independency is the also available thisArg binding for any set's forEach method. Thus any function statement or function expression can access e.g. the other/secondary set via this in case the latter was passed as the forEach's 2nd parameter.

Also an improved wording supports a better readability of the code ...

//the function naming of cause is exaggerated.
function deleteItemFromSourceWhichDoesNotEqualAnyItemFromBoundTarget(sourceItem, _, sourceSet) {
  const targetSet = this;

  const sourceKeys = Object.keys(sourceItem);
  if (
    // ... for any data-item from the (bound) target-set ...
    [...targetSet].every(targetItem =>

      // ... which does not equal the currently processed data-item from the source-set ...
      Object.entries(targetItem).some(([targetKey, targetValue], _, targetEntries) =>
        sourceKeys.length !== targetEntries.length || sourceItem[targetKey] !== targetValue
      )
    )
  ) {
    // ... delete the currently processed data-item from the source-set.
    sourceSet.delete(sourceItem);
  }
}

const set1 = new Set([
  { name: 'a' },                        // - to be kept.
  { name: 'b', lastname: 'bb' },        // - to be kept.
  { name: 'c' },                        // - to be deleted.
  { name: 'd', nested: { name: 'a' } }, // - to be kept, but fails ...
]);                                     //   ... due to not being flat.
const set2 = new Set([
  { name: 'd', nested: { name: 'a' } }, // - should equal, but doesn't.
  { name: 'a' },                        // - does equal.
  { lastname: 'bb', name: 'b' },        // - does equal.
  { name: 'e' },                        // - doesn't equal.
]);

// `set1` is going to be mutated.
set1.forEach(deleteItemFromSourceWhichDoesNotEqualAnyItemFromBoundTarget, set2);

console.log(
  'mutated `set1` now (almost) being equal to the intersection of initial `set1` and `set2` ...',
  [...set1]
);

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Answer 3

const set1 = new Set([
    {name: 'a'},
    {name: 'b', lastname: 'bb'},
    {name: 'c'},
    {name: 'd'}
]);

const set2 = new Set([
    {name: 'b'},
    {name: 'd'}
]);

const names = [...set2].map(s2 => s2.name);

console.log([...set1].filter(item => names.includes(item.name)));

const set1 = new Set([
    {name: 'a'},
    {name: 'b', lastname: 'bb'},
    {name: 'c'},
    {name: 'd'},
    {name: 'e'}
]);

const set2 = new Set([
    {name: 'c', lastname: 'ccc'},
    {name: 'd'},
    {name: 'b', lastname: 'cc'},
    {name: 'e'}
]);

console.log([...set1].filter(item => {
    const s2Arr = [...set2];
    const itemKeys = Object.keys(item);
    for(let i = 0; i < s2Arr.length; i++){
        const s2Obj = s2Arr[i];
        const s2ObjKeys = Object.keys(s2Obj);
        if(s2ObjKeys.length == itemKeys.length){
            let oneSame = true;
            for(let j = 0; j < s2ObjKeys.length; j++){
                const s2ObjKey = s2ObjKeys[j];
                if(item[s2ObjKey] != s2Obj[s2ObjKey]){
                    oneSame = false;
                }
            }
            if(oneSame)
              return true;
        }
    }
    return false;
}));