Javascript - Compare two arrays. If object is found in both, change it

Question

I can check if the objects of an array are also present in another array (and display the duplicates in its own array) with the following code:

var target = ['alfredo', 'john', 'sebas', 'paul', 'lionel', 'sebas2'];
var src = ['paul', 'michael', 'sebas'];

var final = target.filter(function(val) {
  return src.indexOf(val) != -1;
});

console.log(final);

Desired Output:

However, I do not know how to get following output:

final = Array ["paul2", "michael", "sebas3"]

Objects in array final should have the same order as originally in array src but, if the object is also to be found in array target a number (starting from 2) should be attached to this duplicated object. In case the name + number exists, the number has to increase by 1 until the value is not found in target

My try so far:

var final= [];

src.forEach(function(element) {
  if(target.includes(element))
    { newSrc = element+2; final.push(newSrc) } else { final.push(element) }
});

console.log(final);

However, I do not know how to deal with the number + 1 part without recurring to loops in loops etc.

Hmmm ... I think the best answer I can give is: change your datastructure. Why not just { michael: 2, john: 1 ... } or at least clearly seperate index and username, such as username#12 ? — Jonas Wilms
– Jonas Wilms, Commented Jul 25, 2018 at 15:58
@Andreas thanks for the feedback. I updated my question to reflect my try so far — agustin
– agustin, Commented Jul 25, 2018 at 16:10
@JonasW. input comes as it is... I could maybe create a new Array reflection this structure... good idea. — agustin
– agustin, Commented Jul 25, 2018 at 16:11

Nina Scholz · Accepted Answer · 2018-07-25 16:02:25Z

2

You could count first and then map the names with the adjusted count.

var target = ['alfredo', 'john', 'sebas', 'paul', 'lionel', 'sebas2'],
    src = ['paul', 'michael', 'sebas'],
    count = target.reduce((c, s) => (k => (c[k] = (c[k] || 0) + 1, c))(s.match(/^\D+/)[0]), Object.create(null)),
    final = src.map(s => s + (count[s] ? count[s] + 1 : ''));

console.log(final);

answered Jul 25, 2018 at 16:02

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

agustin Over a year ago

Thanks Nina... this is a new set of functions that I didn't know at all. I was trying to avoid loops in loops and this is exactly perfect. I just need to slowly go through it and try to understand how reduce and map works.

Ele · Accepted Answer · 2018-07-25 16:22:57Z

1

An alternative is using the function reduce to group the names and the function some to check for strings str + number. This grouping process creates an object to leverage the fast access using keys.

After that grouping process, get the keys for building the desired output.

var target = ['alfredo', 'john', 'sebas', 'paul', 'lionel', 'sebas2'],
  src = ['paul', 'michael', 'sebas'],
  obj = src.reduce((a, str) => {
    if (target.some(s => str === s))  {
      a[str] = (a[str] || 1) + 1
      while (target.some(s => `${str}${a[str]}` === s)) a[str]++;
    } else a[str] = 0;

    return a;
  }, {}),
  result = Object.keys(obj).map(k => `${k}${obj[k] || '' }`);

console.log(result);

.as-console-wrapper {  max-height: 100% !important;top: 0;}

edited Jul 25, 2018 at 16:22

answered Jul 25, 2018 at 16:14

Ele

33.8k7 gold badges43 silver badges80 bronze badges

Collectives™ on Stack Overflow

Javascript - Compare two arrays. If object is found in both, change it

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related