如何在数组项的第二个对象中查找第一个数组项并从第二个对象返回匹配项

Question

How to find first array items in second object of array items and return matched items from second object using JavaScript.

First Array:

const firstArr = ["foo", "bar"];

Second Object:

const secondObj = {
  "items": [{
    "name": "First item",
    "labels":["foo"]
  },
  {
    "name": "Second item",
    "labels":["foo", "bar"]
  },
  {
    "name": "Third item",
    "labels":["baz"]
  }]
}

Expected:

{
  "items": [{
    "name": "First item",
    "labels":["foo"]
  },
  {
    "name": "Second item",
    "labels":["foo", "bar"]
  }]
}

I tried something like the following;

function updatedVersion() {
  var thirdArr = [];
  for (var array of firstArr) {
    if (secondObj) {
      for (var obj of secondObj.items) {
        for (var label of obj.labels) {
          if (array === label) {
            thirdArr.push(obj);
          }
        }
      }
    }
  }
  return thirdArr;
};

Answer 1

You can filter .data with ( Array#filter ) to produce an array of objects based on membership ( Array#some ) by the .labels array ( Array#includes ), then re-build the result object you want afterwards or re-attach the filtered array to .items :

 const firstArr = ["foo", "bar"]; const secondObj = { "items": [{ "name": "First item", "labels":["foo"] }, { "name": "Second item", "labels":["foo", "bar"] }, { "name": "Third item", "labels":["baz"] }] }; secondObj.items = secondObj.items.filter(({labels}) => labels.some(e => firstArr.includes(e)) ); console.log(secondObj);

Note that we're traversing firstArr many times for a complexity of

O(data.length * firstArr.length * max(data[_].labels.length))

If you make needles a Set and use .has you have

O(firstArr.length + data.length * max(data[_].labels.length))

assuming constant time lookups on the set. This is premature optimization if your structures are small, and the cost of the Set allocation adds overhead.

Similarly, you could make each labels array a Set and cut costs there if it makes sense to do so for your application.

Answer 2

 const firstArr = ["foo", "bar"]; const secondObj = { "items": [{ "name": "First item", "labels":["foo"] }, { "name": "Second item", "labels":["foo", "bar"] }, { "name": "Third item", "labels":["baz"] }] }; const filtered = {"items": secondObj["items"].filter(item=>item.labels.filter(val =>firstArr.includes(val)).length)} console.log(filtered)

Answer 3

Edit: As @ggorlen stated, using Array#some is more efficient than using Array#includes for each search item.

 const firstArr = ['foo', 'bar']; const secondObj = { items: [ { name: 'First item', labels: ['foo'], }, { name: 'Second item', labels: ['foo', 'bar'], }, { name: 'Third item', labels: ['baz'], }, ], }; const filter = (o, arr) => ({ items: o.items.filter((item) => item.labels.some((label) => firstArr.indexOf(label) + 1) ), }); console.log(filter(secondObj, firstArr));