functional programming - How to convert simple average function in javascript to pointfree form using Ramda? - Stack Overflow

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How do I convert the following simple average function to pointfree form (using Ramda)?

var _average = function(xs) {
  return R.reduce(R.add, 0, xs) / xs.length;
};

I've been this for a while now, but the R.divide function is throwing me off since the numerator and the denominator requires evaluation first

How do I convert the following simple average function to pointfree form (using Ramda)?

var _average = function(xs) {
  return R.reduce(R.add, 0, xs) / xs.length;
};

I've been this for a while now, but the R.divide function is throwing me off since the numerator and the denominator requires evaluation first

• javascript
• functional-programming
• currying
• pointfree
• ramda.js

Share Improve this question Follow asked Sep 16, 2016 at 15:46 ChadChad 2,08166 gold badges2828 silver badges4444 bronze badges 3

• 2 May be worth a read mail.haskell/pipermail/beginners/2011-June/007266.html – Xotic750 Commented Sep 16, 2016 at 16:13
• 1 Thanks for raising a great point regarding readability. And it is definitely good to remember that " If the point-free style isn't easy to write, it's probably also not easy to read." But as an exercise, how would you answer the question in case you had to. – Chad Commented Sep 16, 2016 at 16:22
• I don't honestly don't know and it looks like a headache. :) – Xotic750 Commented Sep 16, 2016 at 16:34

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4 Answers 4

Sorted by: Reset to default 8

Using R.converge:

//    average :: Array Number -> Number
const average = R.converge(R.divide, [R.sum, R.length]);

Using R.lift (which a more generally applicable function than R.converge):

//    average :: Array Number -> Number
const average = R.lift(R.divide)(R.sum, R.length);

Here's one way to do it:

let xs = [5, 5];
let average = R.pose(R.apply(R.divide), R.juxt([R.sum, R.length]));
console.log(average(xs));

<script src="//cdn.jsdelivr/ramda/latest/ramda.min.js"></script>

Basically, R.juxt maps the array values into R.sum and R.length which gives you an array with the sum of the array and the length of the array. The result is applied to R.divide.

You can try the below one

var _sum = function(xs) {
  return R.reduce(R.add, 0, xs);
};

var _average = function(xs) {
  return R.divide(_sum(xs), xs.length);
};

console.log(_average([3,4,5,6]));

or simply

var _average = function(xs) {
  return R.divide(R.reduce(R.add, 0, xs), xs.length);
};

console.log(_average([3,4,5,6]));

Simpler one:

R.mean([1,3]) // returns 2

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