How to push same element into an empty array

Question

Hi I was wondering how I could get bobs and tina same followers into an empty array mutualfollowers. I am getting output Both followers have undefined. Seem like the name is not passing through. Please advise.

    let bobsFollowers = ['grey', 'mary', 'james', 'ash'];
    let tinasFollowers = ['grey', 'mary', 'rex'];
    let mutualFollowers = [];
    
    for(let i = 0; i<bobsFollowers.length; i++){
      for(let k = 0; k<tinasFollowers.length; k++){
        if (bobsFollowers[i] === tinasFollowers[k]) {
          tinasFollowers.push(mutualFollowers);
          console.log('Both followers have ' + mutualFollowers[k]);
        }
      }
    }

let mutualFollowers = bobsFollowers.filter(f => tinasFollowers.includes(f)); should do it. — user5734311
– user5734311, Commented Jun 15, 2021 at 8:05
You are pushing into the wrong array. Push in mutualFollowers mutualFollowers.push(tinasFollowers[k]) . — Tushar Shahi
– Tushar Shahi, Commented Jun 15, 2021 at 8:06
Does this answer your question? Simplest code for array intersection in javascript — Hans-Martin Mosner
– Hans-Martin Mosner, Commented Jun 15, 2021 at 8:10

Nina Scholz · Accepted Answer · 2021-06-15 08:07:14Z

3

You need to push to mutualFollowers. And for making the loops a bit more performat, you could leave the inner loop on found.

let bobsFollowers = ['grey', 'mary', 'james', 'ash'];
let tinasFollowers = ['grey', 'mary', 'rex'];
let mutualFollowers = [];

for (let i = 0; i < bobsFollowers.length; i++) {
  for (let k = 0; k < tinasFollowers.length; k++) {
    if (bobsFollowers[i] === tinasFollowers[k]) {
      mutualFollowers.push(bobsFollowers[i]);
      break;
    }
  }
}

console.log(mutualFollowers);

To get just the common items, you could take a Set and filter with Set#has.

const
    bobsFollowers = ['grey', 'mary', 'james', 'ash'],
    tinasFollowers = ['grey', 'mary', 'rex'],
    common = bobsFollowers.filter(
        Set.prototype.has,
        new Set(tinasFollowers)
    );

console.log(common);

answered Jun 15, 2021 at 8:07

Nina Scholz

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6 Comments

Kim Over a year ago

I am getting output as 'Both followers have undefined' twice

Nina Scholz Over a year ago

it looks like you add undefined to the array. maybe the index is wrong.

ikhvjs Over a year ago

Nina, could you please elaborate more about how it works with the filter and set together? I don't understand why you pass Set.prototype.has and new Set in the filter() function.

Nina Scholz Over a year ago

the prototype delivers the methos and the set is thisArg of the filter method. please ahave a look here: Array#filter

Nina Scholz Over a year ago

yes. if filter does not have thisArg, you could bind it and take the returned function, like Set.prototype.has.bind(new Set(tinasFollowers)).

|

Jcl · Accepted Answer · 2021-06-15 08:12:56Z

The problem in your code is that you are pushing the mutualFollowers empty array into tinasFollowers one.

You need to push the actual element, so replace this:

tinasFollowers.push(mutualFollowers);
// You are pushing the "mutualFollowers" array as an item into the "tinasFollowers" array

With this:

mutualFollowers.push(tinasFollowers[k]);
// You are pushing the "tinasFollowers[k]" item (the one you are checking for equality to the bobsFollowers[i] item) into the "mutualFollowers" array

There's a much more simple way to do what you want to do though:

let bobsFollowers = ['grey', 'mary', 'james', 'ash'];
let tinasFollowers = ['grey', 'mary', 'rex'];
let mutualFollowers = bobsFollowers.filter(value => tinasFollowers.includes(value));

Which is a way of saying: "add all elements in bobFollowers, which are in tinasFollowers"

PS: what you are doing is called a "two-array intersection" (your result is the elements that intersect on both arrays), and there are better and more optimized algorithms around, if you want to search more

Nico Nisaw · Accepted Answer · 2021-06-15 08:41:15Z

0

hope this piece of code helps, it is just same as yours, just modified a bit. Using the JS filter method is much better for this task.

  let bobsFollowers = ['grey', 'mary', 'james', 'ash'];
  let tinasFollowers = ['grey', 'mary', 'rex'];
  let mutualFollowers = [];
    

  bobsFollowers.forEach((index, name) => {
    tinasFollowers.forEach((test_index, test_name) => {
      if (tinasFollowers[test_name] == bobsFollowers[name]) {
        mutualFollowers.push(tinasFollowers[test_name])
      }
    })
  })

  console.log(mutualFollowers)

answered Jun 15, 2021 at 8:41

Nico Nisaw

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Comments

Rixio Barrios · Accepted Answer · 2023-12-20 15:12:50Z

0

This is giving me this output: [ 'grey' ] [ 'grey', 'mary' ]

I assume that is the desired output.

let bobsFollowers = ['grey', 'mary', 'james', 'ash'];
let tinasFollowers = ['grey', 'mary', 'rex'];
let mutualFollowers = [];

for (let i = 0; i < bobsFollowers.length; i++) {
    for (let k = 0; k < tinasFollowers.length; k++) {
        if (bobsFollowers[i] == tinasFollowers[k]) {
            mutualFollowers.push(tinasFollowers[k]);
            console.log(mutualFollowers);
        }
    }
}

answered Dec 20, 2023 at 15:12

Rixio Barrios

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