How to deep compare/match objects in an array, based on an array of strings property in each object?

Question

The array with objects:

var myArr = [ { user: 'jane', properties: ['prop1', 'prop2'] }
              { user: 'bob', properties: ['prop1', 'prop4'] }
              { user: 'dave', properties: ['prop1', 'prop2', 'prop3'] }
              { user: 'tim', properties: ['prop4', 'prop2', 'prop3'] } 
];

Is there a way in pure javascript or perhaps using lodash to obtain the users that have at least one property in common? The output should indicate which user matches with which user, without duplicates.

I've tried a number of things, but unfortunately was not able to get the desired output.

can you give a sample output? would it be an array or an object? — Jayce444
– Jayce444, Commented Oct 28, 2019 at 5:27
@Jayce444 Any output is good enough, as long as the output shows which users match with which users (matching pairs). The precise structure does not matter. — Erwin Lengkeek
– Erwin Lengkeek, Commented Oct 28, 2019 at 6:12

Robby Cornelissen · Accepted Answer · 2019-10-28 05:38:18Z

3

The solution below uses a Set to determine whether matching properties exist, combined with a reduce() operation to output matching user pairs:

const array = [{ user: 'jane', properties: ['prop1', 'prop2'] },
  { user: 'bob', properties: ['prop1', 'prop4'] },
  { user: 'dave', properties: ['prop1', 'prop2', 'prop3'] },
  { user: 'tim', properties: ['prop4', 'prop2', 'prop3'] }];
  
const result = array.reduce((acc, val1, i, arr) => [
  ...acc,
  ...arr.slice(i + 1).filter(val2 => {
    const properties = [...val1.properties, ...val2.properties];
    return new Set(properties).size < properties.length;
  }).map(val2 => [val1.user, val2.user])
], []);

console.log(result);

answered Oct 28, 2019 at 5:38

Robby Cornelissen

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1 Comment

Erwin Lengkeek Over a year ago

This is precisely what I was looking for. I hadn't considered this approach. Thank you @Robby Cornelissen

Code Maniac · Accepted Answer · 2019-10-28 05:43:22Z

0

You can use Map and Object, count the user occurrence for each properties, if it is same as length of array then there is atleast one property which occurs in all the users

let matcher = (arr) => {
  let mapper = {}
  arr.forEach(v => {
  let { user, properties } = v
    properties.forEach(v => {
      if (mapper[v]) {
        mapper[v].set(user, (mapper[v].get(user) || 0) + 1)
      } else {
        mapper[v] = new Map()
        mapper[v].set(user, 1)
      }
    })
  })
  return Object.entries(mapper).some(([_, v]) => {
    let total = [...v].reduce((a, [_, b]) => a + b, 0)
    return total === arr.length
  })
}

let arr = [{user: 'jane',properties: ['prop1', 'prop2']}, {user: 'bob',properties: ['prop1', 'prop4']}, {user: 'dave',properties: ['prop1', 'prop2', 'prop3']}, {user: 'tim',properties: ['prop4', 'prop2', 'prop3']}];

console.log(matcher(arr))

let arr2 = [{user: 'jane',properties: ['prop1', 'prop2']},{user: 'bob',properties: ['prop1', 'prop4']}]

console.log(matcher(arr2))

answered Oct 28, 2019 at 5:43

Code Maniac

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2 Comments

Erwin Lengkeek Over a year ago

This does not answer this question, as it is of particular importance to know which users match with which users.

Code Maniac Over a year ago

Some can be replaced with filter and you will get the prop which is common along all the users

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