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  • Jan 31 2019 22:17
    CrossEye commented #2779
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    CrossEye commented #2772
Niloy Mondal
@niloy
ok
Asaf Romano
@asaf-romano
Any ideas on how to write the lamda here in point-free style: http://bit.ly/1PYf0ul ?
I was messing around with converge, zipObj and unapply an flip, but things got nasty when I had to unapply everything: zipObj, difference and flip(difference)
Jayson Harshbarger
@Hypercubed
Does anyone know of a better way to write this (_) => R.comparator(R.useWith(_, [_getter, _getter])). I want to take a binary function and return a new binary function that first applies _getter to each argument.
Tobias Pflug
@gilligan
Hm, i am somehow stuck right now - I want to write a function that wraps functions which might throw and returns either Right(result) or Left(errorObject) : https://gist.github.com/gilligan/66a1ac0431a0e77fdaae - this is not quite right. Can anyone help out ?
Tobias Pflug
@gilligan
@raine ah yes right
driving me bonkers that i am not getting my stuff right though
Tobias Pflug
@gilligan
wrapEither(R.inc)(0); // --> Either { value: '1[object Object]' }
x__X
Tobias Pflug
@gilligan
D'OH
arguments is not available in arrow functions
Stefano Vozza
@svozza
probably best off using rest parameters anyway
Jayson Harshbarger
@Hypercubed
FYI... I'm working on replacing an utility library I created a few years ago with a Ramda based version: https://github.com/Hypercubed/florida
It's kind of like a hyperactive version of a lens... but only the getter. Any feedback would be appreciated.
Hardy Jones
@joneshf
@Hypercubed it sounds like you want something like: http://hackage.haskell.org/package/base-4.8.1.0/docs/Data-Function.html#v:on
Given the type (b -> b -> c) -> (a -> b) -> a -> a -> c it almost writes itself.
const on = curry((f, g, x, y) => f(g(x), g(y)));
Since you need a value of type c, you only have one place to get it, the f function. Well, the f function needs two values of type b, and you can only get those from the g function. The g function needs a value of type a, and you have two of those: x, and y. So you be fair and send both values x, and y to g, then send that to f, and your function wrote itself.
Hardy Jones
@joneshf
There's only two other implementations of that function so you've got a 1 in 3 chance of getting it right :)
And once you add in the law on(f, identity) == f you have no other choice for an implementation.
This message was deleted
Jakub Korzeniowski
@kujon
Hi All,
my brain isn't in a good shape today and I got stuck on trying to do the following:
I want something like R.ap, but a variant which is not going to do every permutation, but only apply first function to the first argument in the list, the second function to the second argument in the list, and so on:
R.ap([R.multiply(2), R.add(3)], [1,2]); // => [2, 5]
any ideas?
Hardy Jones
@joneshf
zipWith(curry((f, x) => f(x)), [R.multiply(2), R.add(3)], [1,2])
maybe avoid the curry in this case?
zipWith((f, x) => f(x), [R.multiply(2), R.add(3)], [1,2])
Jakub Korzeniowski
@kujon
that is one glorious solution, much appreciated!
Hardy Jones
@joneshf
Use a data type that encodes this behavior is another possibility.
const ZipArray = xs => ({
  toArray: xs,
  map(f) {
    return ZipArray(R.map(f, xs));
  },
  ap({toArray: ys}) {
    if (xs.length == 0 || ys.length == 0) {
      return ZipArray([]);
    } else {
      const y = R.head(ys);
      const x = R.head(xs);
      const yss = R.tail(ys);
      const xss = R.tail(xs);

      return ZipArray(R.prepend(x(y), ZipArray(xss).ap(ZipArray(yss)).toArray));
    }
  },
});

R.ap(ZipArray([R.multiply(2), R.add(3)]), ZipArray([1,2])).toArray //=> [2, 5]
Or I guess you could clean it up...
const ZipArray = xs => ({
  toArray: xs,
  map(f) {
    return ZipArray(R.map(f, xs));
  },
  ap({toArray: ys}) {
    return ZipArray(R.zipWith((f, x) => f(x), xs, ys));
  },
});

R.ap(ZipArray([R.multiply(2), R.add(3)]), ZipArray([1,2])).toArray //=> [2, 5]
probably better that way, since tco.
Sean Sartell
@ssartell
Hey everyone, is it possible to do recursion with strictly pointfree style syntax? I'm guessing not, but thought I'd ask here anyways.
David Chambers
@davidchambers
@ssartell, does the Y combinator make this possible?
Scott Christopher
@scott-christopher
var Y = f => (x => f(y => x(x)(y)))(x => f(y => x(x)(y)));
var factorial = Y(go => n => n === 0 ? 1 : n * go(n - 1));
factorial(6); // 720
I'm sure it's probably possible to define the function passed to Y in a points-free fashion, but it wouldn't be pretty.
David Chambers
@davidchambers
@scott-christopher, that's cool!
Scott Christopher
@scott-christopher
Scott Sauyet
@CrossEye
The Y-combinator allows recursion even where the language doesn't explicitly account for it. But it had little to do with being points-free.
s/had/has
There should be no issues doing points-free recursion, but I'm not going to try an example on my phone.
David Chambers
@davidchambers

@CrossEye, have you ever found yourself writing code like this?

const f = R.pipe(
  a,
  b,
  c,
  d,
  e,
  x => f(x),
  g,
  h
);

I was wondering whether the Y combinator could obviate the need for the lambda.

Scott Sauyet
@CrossEye
I'm afraid it won't.
David Chambers
@davidchambers
No worries. :)
Scott Christopher
@scott-christopher
It's a silly example, but to show that the Y combinator allows for points-free recursion:
var min10 = Y(R.ifElse(R.compose(R.gte(R.__, 10), R.nthArg(1)),
                       R.nthArg(1),
                       R.useWith(R.call, [R.identity, R.inc])));

min10(12); // 12
min10(-3); // 10
the use of R.nthArg might be cheating a little, depending on your rules of points-free golf :D
Scott Christopher
@scott-christopher
it does also emphasise that you're probably not going to have something readable by the time you manage to get the recursive function to be points-free.
Scott Sauyet
@CrossEye
Yes, I was once again missing the point. It sounded easy to do recursion points-free until I tried it. Of course is not possible to do naively, but something like the Y-comb would allow it. But as always with the Y-combinator, it feels mostly like a trick. You prove that it can be done, but you don't use it for anything real.
Hardy Jones
@joneshf
function loop() {
  return loop;
}
seems pretty free of points
Tobias Pflug
@gilligan
and arguments