Get some set of array based on condition from two arrays of array in Javascript

Question

I have two arrays of array in Javascript like

var array1 = [[10, 2], [11, 4], [12, 30], [13, 17], [14, 28]];
var array2 = [[8, 13], [9, 19], [10, 6], [11, 7], [12, 1]];

I want to get the set of arrays from array1 which match the first element of each array of the array2

in my example case both array1 and array2 have array with first element as 10 11 and 12, so it should return

[[10, 2], [11, 4], [12, 30]];

is there any easy and efficient way using pure javscript or lodash, underscor framework or something like that. Without iterate over and match one by one of this two array ?

Nina Scholz · Accepted Answer · 2016-10-05 17:07:43Z

4

In ES6, you could use Set.

var array1 = [[10, 2], [11, 4], [12, 30], [13, 17], [14, 28]],
    array2 = [[8, 13], [9, 19], [10, 6], [11, 7], [12, 1]],
    set = new Set(array2.map(a => a[0])),
    result = array1.filter(a => set.has(a[0]));

console.log(result);

Version with an object as hash table

var array1 = [[10, 2], [11, 4], [12, 30], [13, 17], [14, 28]],
    array2 = [[8, 13], [9, 19], [10, 6], [11, 7], [12, 1]],
    result = array1.filter(function (a) {
        return this[a[0]];
    }, array2.reduce(function (r, a) { 
        r[a[0]] = true;
        return r;
    }, Object.create(null)));

console.log(result);

edited Oct 5, 2016 at 17:07

answered Oct 5, 2016 at 16:58

Nina Scholz

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2 Comments

Mostafiz Over a year ago

this works too, upvoted but I actually do not want to bound in ES6, it should work every version

Nina Scholz Over a year ago

@Mostafiz, please have a look to version 2.

acontell · Accepted Answer · 2016-10-05 17:16:48Z

3

You can use lodash _.intersectWith function in order to solve this problem in an inline.

_.intersectionWith(array1, array2, function(a, b) {
    return a[0] === b[0];
});

I don't know about performance cos I haven't had the chance to have a look at the source code of this function. Anyway, I like it for its simplicity. Here's the fiddle in case you want to check it out.

answered Oct 5, 2016 at 17:16

acontell

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Comments

Felix Kling · Accepted Answer · 2016-10-05 17:11:29Z

If you can make use of Set, then you can compute the a set of numbers to look for first and use .filter to only get the arrays whose first element is in that set:

var haystack = new Set(array2.map(x => x[0]));
var newArray = array1.filter(x => haystack.has(x[0]));

Of course you can also use the lodash or underscore versions of .map and .filter.

Alternatives to using Set would be:

Create an array of numbers instead and use indexOf to test existence. That will scale linearly with the number of elements:
```
var haystack = array2.map(x => x[0]);
var newArray = array1.filter(x => haystack.indexOf(x[0]) > -1);
```

Create an object with number -> true entries to test existence with in, hasOwnProperty or just object access:

var haystack = array2.reduce((obj, x) => (obj[x[0]] = true, obj), {});
var newArray = array1.filter(x => haystack[x[0]]);

Which one performs better depends on the number of elements you have and the environment the code is running in.

Nenad Vracar · Accepted Answer · 2016-10-05 16:55:40Z

1

You can do this with filter() and find()

var array1 = [
  [10, 2],
  [11, 4],
  [12, 30],
  [13, 17],
  [14, 28]
];
var array2 = [
  [8, 13],
  [9, 19],
  [10, 6],
  [11, 7],
  [12, 1]
];

var result = array1.filter(function(ar) {
  return array2.find(function(e) {
    return e[0] == ar[0]
  })
})

console.log(result)

answered Oct 5, 2016 at 16:55

Nenad Vracar

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Comments

Redu · Accepted Answer · 2016-10-05 17:45:24Z

0

I would do this with a Map anf filter combo in ES6 as follows;

var array1 = [[10, 2], [11, 4], [12, 30], [13, 17], [14, 28]],
    array2 = [[8, 13], [9, 19], [10, 6], [11, 7], [12, 1]],
         m = new Map(array2),
    array3 = array1.filter(a => m.has(a[0]));
console.log(array3);

If you need backwards compatibility other answers are pretty good.

edited Oct 5, 2016 at 17:45

answered Oct 5, 2016 at 17:16

Redu

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