by

Where communities thrive


  • Join over 1.5M+ people
  • Join over 100K+ communities
  • Free without limits
  • Create your own community
People
Repo info
Activity
  • Jan 31 2019 22:17
    CrossEye commented #2779
  • Jan 31 2019 21:04
    ArturAralin commented #2779
  • Jan 31 2019 20:08
    CrossEye commented #2779
  • Jan 31 2019 18:56
    buzzdecafe commented #2631
  • Jan 31 2019 18:09
    ArturAralin commented #2779
  • Jan 31 2019 16:18
    CrossEye commented #2779
  • Jan 31 2019 16:10
    CrossEye commented #2631
  • Jan 31 2019 16:06
    CrossEye commented #2777
  • Jan 31 2019 14:44
    ArturAralin opened #2779
  • Jan 31 2019 07:39
    inferusvv commented #2631
  • Jan 31 2019 03:07
    sespinozj commented #2771
  • Jan 31 2019 02:33
    machad0 commented #2771
  • Jan 31 2019 02:26
    JeffreyChan commented #2777
  • Jan 30 2019 14:30
    CrossEye closed #2777
  • Jan 30 2019 12:13
    vanyadymousky updated the wiki
  • Jan 30 2019 01:42
    JeffreyChan commented #2777
  • Jan 29 2019 21:06
    vanyadymousky updated the wiki
  • Jan 29 2019 16:28
    CrossEye commented #2777
  • Jan 29 2019 15:50
    mbostock commented #2772
  • Jan 29 2019 15:48
    CrossEye commented #2772
Tobias Pflug
@gilligan
@joneshf good god - i wanted to "quickly skim" the discussion. I should take some time for that ;)
Scott Sauyet
@CrossEye
And I misread @niloy's suggestion. To extend contains to objects consistent with has seems contradictory. To do it instead on the values is more likely, but then becomes subject to the discussions @joneshf mentioned, #1429.
Tobias Pflug
@gilligan
@CrossEye ah, actually i got it wrong in the same way w/o realizing
Scott Sauyet
@CrossEye
@gilligan: That was the original request. But when the extension to contains was proposed, it was changed.
Jethro Larson
@jethrolarson
Anyone else think this is useful?
const findLens = pred => R.lens(
  R.find(pred)
  , (v, ar) => R.adjust(R.always(v), R.findIndex(pred, ar), ar)
)
BTW I was looking for something to replace a value at an index in array and I didn't see anything in ramda. R.adjust(R.always(x) was the closest I found
David Chambers
@davidchambers
@jethrolarson, it sounds as though R.update fits the bill.
Niloy Mondal
@niloy
is there a version of splitEvery that works with predicate?
Tobias Pflug
@gilligan
splitWhen has been or will be merged to master soon I think
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.