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Perelyn-sama

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# Getting the frequency of an element in an array in JavaScript

First of all, what is frequency?
According to Wikipedia, it's" the rate at which something occurs over a particular period of time or in a given sample."
There are multiple ways to get the frequency of an element in an array. In this article, we'll focus on one of those ways by using some of the higher-order functions in Javascript.

## Let's begin

For instance, if we have an array of letters:

``````const letters = ["a", "b", "c", "a", "b", "c", "a", "b"];
``````

To get the frequency of each element, we will first need to create an empty `object` named `count`:

``````const count = {};
``````

Next, we'll utilize one those higher order functions we were talking about:

``````letters.forEach(e => count[e] ? count[e]++ : count[e] = 1 );
``````

What the above code does is simply check, if each element of `letters` is present in `count`. Initially `count` is empty and does not contain any of `letters` elements, so when the ternary operator runs on an element of `letters` to check if its present in `count` for the first time, it's going to work with the falsy condition which is to assign `letters` elements to 1 in the `count` object.
So the count object will then become:

``````console.log(count) // {"a" : 1, "b" : 1, "c" : 1};
``````

Afterward, when the ternary operator checks if `count` has an element of `letters` that has already been checked, the ternary operator will then run the truthy condition which is to increment the value of the elements in `letters` if they are already present in `count`, so `count` will now become:

``````console.log(count) // {"a" : 3, "b" : , "c" : 2};
``````

And just like that ladies and gentlemen, we've been able to get the frequency of each element of `letters`.

## Conclusion

Here's the complete code:

``````const letters = ["a", "b", "c", "a", "b", "c", "a", "b"];
const count = {};
letters.forEach(e => count[e] ? count[e]++ : count[e] = 1 );

console.log(count) // {"a" : 1, "b" : 1, "c" : 1}
``````

P.S: This code can be used for both strings and numbers.

Kai

There's no need for `nums.map(e => Number(e))`. Here, you are transforming numbers to... well, numbers. You can skip this step. If you really want to convert them to strings you should use either `String(e)` or `e.toString()`.

Perelyn-sama

I'm sorry It's a typo, I originally wanted to write String(e). Thanks for pointing out to me

Zen

You can change `Number(e)` with `+e`

keogami

Thats too hacky and non intuitive i.e. not readable. As a simple rule you must avoid being clever because readable code is always better than clever code 😊

Perelyn-sama

True, simple code is good code.

Perelyn-sama

Thanks for the pointer.

Rishit Khandelwal

We could just use `Array.reduce`

Perelyn-sama

Yeah, there's more that one way to skin a cat

Rishit Khandelwal
``````const count=(i,w)=>i.reduce((a,v)=>v==w?a+1:a,0)
``````

Perelyn-sama

Thanks!

bugb
``````const letters = ["a", "b", "c", "a", "b", "c", "a", "b"];
const countLetters = letters.reduce((m,n)=>({...m, [n]:-~m[n]}),{})

// For nums, it is the same
``````

Perelyn-sama

Wow thanks, that's awesome and shorter

bugb

no problems, keep learning !

Why would you need to do this any differently for numbers vs strings?? The first example (the one that you show as being "for string elements") works just fine if you pass in the array of numbers.

Perelyn-sama

Wow, I never realized that. Thanks for the pointer, I'll the remove the 'For numbers' part.

keogami

Ngl the "for numbers" part is stupid

Perelyn-sama

That I agree with, thanks for your feedback

Angela Inniss

Thanks for this. This is good for beginners like me as it says in the tags. Some of the other suggestions are cool too but a bit more confusing :)