Looping and counting on arrays

Question

I have an array of objects with known keys. And I want to count how many objects have a particular key with a certain value. Currently I made it up with this:

counter = () => {
        if (Array) {
            let a = 0;
            for (let i = 0; i < Array.length; i++) {
                if (Array[i].key === '1' || Array[i].key === '2' || Array[i].key === '3' || Array[i].key === '4' || Array[i].key === '5' || Array[i].key === '6' || Array[i].key === '7') {
                    a++;
                }
            }
            return a;
        }
    }

I tried reduce() and find() while converting arrays to objects and vice versa but with no luck yet.
I suppose there should be more elegant way to perform such a simple task in plain vanilla JS ES6 or ES7. Ideally a oneliner. No lodashing please.

All other questions here target searching in elements, when mine touch one level deeper.

Possible duplicate of How to count the number of certain element in an array? — A l w a y s S u n n y
– A l w a y s S u n n y, Commented Sep 6, 2018 at 16:20

Pedro Simão · Accepted Answer · 2018-09-06 16:20:15Z

2

You can try to use the javascript filter function: Something like:

var count = array.filter(function(e){
                  return (e.key === '1' || e.key === '2' || e.key === '3' || e.key === '4' || e.key === '5' || e.key === '6' || e.key === '7')

}).length;

Basically, this filter your array of objects by the keys you want, at the end it calculates the array length.

answered Sep 6, 2018 at 16:20

Pedro Simão

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Comments

Jonas Wilms · Accepted Answer · 2018-09-06 16:56:04Z

1

Sure, here is that oneliner:

 array.reduce((count, el) => count + +["1", "2", "3", "4", "5", "6", "7"].includes(el.key), 0)

This uses includes to check wether el.key is one of the numbers, then the resulting boolean is turned into a number with the unary plus (true -> 1, false -> 0), and that is added to count for each element, resulting in the total count.

edited Sep 6, 2018 at 16:56

answered Sep 6, 2018 at 16:19

Jonas Wilms

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2 Comments

Igniter Over a year ago

reduce() looks powerful, could you explain what's going on here?

Jonas Wilms Over a year ago

@igniter added a short explanation.

Ranjith Eswaran · Accepted Answer · 2018-09-06 16:34:56Z

1

Try using

var count = b.filter(element => {return element.key == requiredKey;}).length;

count will have the number of objects with same key

answered Sep 6, 2018 at 16:34

Ranjith Eswaran

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Comments

mhodges · Accepted Answer · 2018-09-06 16:42:10Z

1

You can do this with .filter() fairly easily, like so:

The idea is that you want to filter down to only the objects that match, and then use a .length to get the count at the end. We can know if an element matches by checking if the object's key property value is included in the list of keys you are searching for.

var test = [
{ key: '1' },
{ key: '3' },
{ key: '2' },
{ key: '7' },
{ key: '5' },
{ key: '8' },
]
var keysToSearch = ['1', '2', '3', '4', '5', '6', '7'];

function counter (objs, searchKeys) {
  return objs.filter(obj => searchKeys.includes(obj.key)).length;
}

console.log(counter(test, keysToSearch));

EDIT

Here's a solution that does the same thing as the above, but doesn't create a new array with .filter().

var test = [
{ key: '1' },
{ key: '3' },
{ key: '2' },
{ key: '7' },
{ key: '5' },
{ key: '8' },
]
var keysToSearch = ['1', '2', '3', '4', '5', '6', '7'];

function counter (objs, searchKeys) {
  var count = 0;
  for (var obj of objs) {
    if (searchKeys.includes(obj.key)) { count++; }
  }
  return count;
}

console.log(counter(test, keysToSearch));

edited Sep 6, 2018 at 16:42

answered Sep 6, 2018 at 16:20

mhodges

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5 Comments

charlietfl Over a year ago

Why create a new array to do this?

mhodges Over a year ago

@charlietfl which array are you referring to?

charlietfl Over a year ago

filter() returns new array. Also OP is starting with array of objects , not a single object and also wants to match values

mhodges Over a year ago

@charlietfl Oh I missed the array of objects part. I will fix. Also updated with an answer that creates no new arrays

mhodges Over a year ago

@charlietfl Updated. I think I got it all now. Thank you for your comment, I didn't fully understand the question at first

Orelsanpls · Accepted Answer · 2018-09-07 08:10:29Z

0

Do not create new array, one line, loop on keys to reduce the loop iterations.

Reduce work like :

array.reduce((accumulator, oneArrayValue) => {
  // Function where treatment are done
  return theAccumulatorValueForNextLoopIteration;
}, initialValueOfTheAccumulator);

function counter(obj, keys) {
  // void 0 equals undefined
  // So it's equivalent to typeof x !== 'undefined'

  // tmp is the accumulator, it's value is setted up at 0
  // then it becomes what you return after every loop
  return keys.reduce((tmp, x) => obj[x] !== void 0 ? tmp + 1 : tmp, 0);
}

console.log(counter({
  1: 'hey',
  2: 'how',
  8: 'you',
  12: 'doing',
  15: 'baguette',
}, [
  '1',
  '2',
  '3',
  '4',
  '5',
  '6',
  '7',
]));

edited Sep 7, 2018 at 8:10

answered Sep 6, 2018 at 16:27

Orelsanpls

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1 Comment

Igniter Over a year ago

Could you please explain what 1) 'void 0' and 2) 'tmp, 0' mean here in plain English?

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