Javascript compare multiple arrays of objects

Question

How to compare several arrays of objects and add new properties with the number of occurrences found by the object and the indices of the array where the object was found? Comparison of objects should be performed using the property name.

Example:

var arrays = [
  [
    {
      name: 'aa',
      value: 1
    },
    {
      name: 'ab',
      value: 2
    },
    {
      name: 'ac',
      value: 3
    },
    {
      name: 'aa',
      value: 1
    }
  ],
  [
    {
      name: 'aa',
      value: 1
    },
    {
      name: 'ab',
      value: 2
    },
  ],
  [
    {
      name: 'ac',
      value: 3
    },
    {
      name: 'aa',
      value: 1
    }
  ]
]

After execution, an object from the specified array must have the following properties:

[
  [
    {
      name: 'aa',
      value: 1,
      occurrences: 3,
      where: [0, 1, 2]
    },
    {
      name: 'ab',
      value: 2,
      occurrences: 2,
      where: [0, 1]
    },
    {
      name: 'ac',
      value: 3,
      occurrences: 2,
      where: [0, 2]
    },
    {
      name: 'aa',
      value: 1,
      occurrences: 3,
      where: [0, 1, 2]
    }
  ],
  [
    {
      name: 'aa',
      value: 1,
      occurrences: 3,
      where: [0, 1, 2]
    },
    {
      name: 'ab',
      value: 2,
      occurrences: 2,
      where: [0, 1]
    }
  ],
  [
    {
      name: 'ac',
      value: 3,
      occurrences: 2,
      where: [0, 2]
    },
    {
      name: 'aa',
      value: 1,
      occurrences: 3,
      where: [0, 1, 2]
    }
  ]
]

Basically, I want to check if an object with a specific property exists namein other arrays.

This is the solution that comes to my mind: 1. Scroll through the array that has the most objects. 2. Scrolling through each object
 3. Scroll through the remaining arrays and apply. Array.prototype.find()
But it will take a lot of time, since each of my arrays will have at least 500 objects ...

Answer 1

array#reduce, .

arrays, Object.assign() where occurrences.

var arrays = [ [ { name: 'aa', value: 1 }, { name: 'ab', value: 2 }, { name: 'ac', value: 3 }, { name: 'aa', value: 1 } ], [ { name: 'aa', value: 1 }, { name: 'ab', value: 2 }, ], [ { name: 'ac', value: 3 }, { name: 'aa', value: 1 } ] ];

var result = arrays.reduce((res, arr, index) => {
  arr.forEach(({name,value}) => {
    res[name] =  res[name] || {occurrences: 0};
    res[name]['where'] = res[name]['where'] || [];
    if(!res[name]['where'].includes(index)){
      res[name]['where'].push(index);
      res[name].occurrences += 1;
    }
  });
  return res;
},{});

arrays.forEach(arr => arr.forEach(obj => Object.assign(obj, result[obj.name])));
console.log(arrays);

Hide result

Answer 2

, ), + .

, , prop ( ):

let f = (ai, a,v,i,m) => {
    if (!a[v.id]) {
    a[v.id] = {id: v.id, v: v.name, count: 1, at: [ai]};
    } else {
    a[v.id].count += 1;
    a[v.id].at.push(ai);
    }
    return a;
};
let r = [[{id: 'aa', value: 42}], [{id: 'ba', value: 11}, {id: 'aa', value: 42}]]
.reduce ((a, v, i) => v.reduce (f.bind (null, i),a), {}); 
console.log (r);

- , O (n), (, 1000 1000 200 x 5000).