如何使用 Ramda.js 计算数组元素的相对频率？

Question

I have an array with repeating values, and I want to get the relative frequency (ie, proportions) for each repeating value.

It seems natural to me to treat this as a 2-step procedure:

1. count the occurrences of each value; and
2. divide that count by the length of the original array.

To accomplish the first step we can use R.countBy() from ramda.js :

const R = require("ramda")

const myLetters = ["a", "a", "a", "a", "b", "b", "c", "c", "c", "c"]
const counted = R.countBy(R.identity)(myLetters)
counted // => gives {"a": 4, "b": 2, "c": 4}

Now the second step would be to divide counted by the length of myLetters :

counted / R.length(myLetters) // obviously this doesn't work because it's not mapped

I'm a bit lost with how to map this correctly. My current clunky solution that I dislike:

// 1. manually calculate the length and store to a variable
const nvals = R.length(myLetters)

// 2. create a custom division function
const divide_by_length = (x) => R.divide(x, nvals)

// 3. map custom function to `counted`
R.map(divide_by_length, counted) // gives {"a": 0.4, "b": 0.2, "c": 0.4}

Although this works, there's gotta be a more straightforward way with ramda to get from counted to {"a": 0.4, "b": 0.2, "c": 0.4} .

Answer 1

You need to combine the results of counting the items in the array, with the length of the array.

You can use R.ap as the S combinator by supplying it with 2 functions. The S combinator signature is S = (f, g) => x => f(x)(g(x)) , where f and g are functions.

In your case:

f - Create a map curried with divide by the length

g - Create an object of counts

 const { ap, pipe, length, divide, __, map, countBy, identity } = R const fn = ap( pipe(length, divide(__), map), // curry a map by divide by length countBy(identity), // create the counts ) const myLetters = ["a", "a", "a", "a", "b", "b", "c", "c", "c", "c"] const counted = fn(myLetters) console.log(counted)

 <script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.28.0/ramda.min.js" integrity="sha512-t0vPcE8ynwIFovsylwUuLPIbdhDj6fav2prN9fEu/VYBupsmrmk9x43Hvnt+Mgn2h5YPSJOk7PMo9zIeGedD1A==" crossorigin="anonymous" referrerpolicy="no-referrer"></script>

Answer 2

I very much like the approach from Ori Drori, using the fact that ap (f, g) //~> (x) => f (x) (g (x)) . ( chain used for functions is the related chain (f, g) //~> (x) => f (g (x)) (x) .)

My initial thought was similar, using instead lift , which lifts a function that operates on values up to become a one that operates on containers of those values. When the containers are functions, it operates something like lift (f) (g, h) //~> (x) => f (g (x), h (x)) , although it's more generic, as lift (f) is variadic, as are the functions supplied to it and therefore also the function it generates, eg lift (f) (g, h, i, j) //~> (a, b, c) => f (g (a, b, c), h (a, b, c), i (a, b, c), j (a, b, c))

So, very similarly, I wrote:

 const frequencies = lift (map) ( pipe (length, flip (divide)), countBy (identity) ) const myLetters = ["a", "a", "a", "a", "b", "b", "c", "c", "c", "c"] console.log (frequencies (myLetters))

 <script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.28.0/ramda.min.js"></script> <script> const {lift, map, pipe, length, flip, divide, countBy, identity} = R </script>

N.netheless, it's not clear to me that a point-free approach offers any benefit here. I'm not sure which I prefer, but this non-point-free Ramda is about as readable to my mind:

 const frequencies = (letters, total = letters.length) => map (n => n / total) (countBy (identity) (letters) ) const myLetters = ["a", "a", "a", "a", "b", "b", "c", "c", "c", "c"] console.log (frequencies (myLetters))

 <script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.28.0/ramda.min.js"></script> <script> const {lift, map, pipe, length, flip, divide, countBy, identity} = R </script>

Answer 3

I think that @Scott's answer (2nd solution) caused some pennies to drop for me.

If we define two helper functions preemptively for counting and for dividing:

const myCount = R.countBy(R.identity);
const myDivideBy = (divisor, arr) => R.map(elem => elem / divisor); // I was missing this part

Then we could do:

const calcFreq = (arr) => {
    return R.pipe(myCount, myDivideBy(arr.length))(arr)
}

Which is exactly the 2-step procedure I imagined from the beginning: first count, then divide.

calcFreq(myLetters) // gives {"a": 0.4, "b": 0.2, "c": 0.4}

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A relevant post: javascript: efficient way to divide an array by a value