These are chat archives for ramda/ramda

8th
Oct 2015
Hardy Jones
@joneshf
Oct 08 2015 01:24 UTC
@davidchambers it might help to make it point free.
compose(filter(p), filter(q)) === compose(filter(q), filter(p))
Jethro Larson
@jethrolarson
Oct 08 2015 01:26 UTC
Doesn't it makes sense? If you take all the people then get rid of the blondes then the brunettes or get right of the brunettes then blondes, either way you got a bunch of red heads.
David Chambers
@davidchambers
Oct 08 2015 01:26 UTC
That looks very much like:
compose(add(4), add(2)) === compose(add(2), add(4))
Hardy Jones
@joneshf
Oct 08 2015 01:27 UTC
Oh, I see @CrossEye wrote that above :)
I was thrown off by the escaped characters
Scott Sauyet
@CrossEye
Oct 08 2015 01:27 UTC
@crosseye tried to write that above...
David Chambers
@davidchambers
Oct 08 2015 01:28 UTC
@jethrolarson, it undoubtedly makes sense, I'm just looking for an intellectual-sounding name for the law. ;)
Scott Sauyet
@CrossEye
Oct 08 2015 01:29 UTC
@jethrolarson, and why this prejudice against the baldies? :smile:
Hardy Jones
@joneshf
Oct 08 2015 01:29 UTC
well it's the function composition that's commutative, not the filter.
David Chambers
@davidchambers
Oct 08 2015 01:30 UTC

Does

compose(add(4), add(2)) === compose(add(2), add(4))

demonstrate associativity or commutativity?

Hardy Jones
@joneshf
Oct 08 2015 01:30 UTC
or I guess more formally, filter is commutative under function composition
I'm pretty sure that's a free theorem though.
Scott Sauyet
@CrossEye
Oct 08 2015 01:30 UTC
demonstrates commutativity, but as @joneshf said, it's more about composition than about the filter itself
associativity has to do with how you group operations.
Scott Christopher
@scott-christopher
Oct 08 2015 01:31 UTC
associativity would be:
compose(filter(f), compose(filter(g), filter(h))) === compose(compose(filter(f), filter(g)), filter(h))
Hardy Jones
@joneshf
Oct 08 2015 01:31 UTC
meaning the proof is probably not bad...
Jethro Larson
@jethrolarson
Oct 08 2015 01:31 UTC
Filter Commutativity Law: filter is commutative under composition
David Chambers
@davidchambers
Oct 08 2015 01:32 UTC

But

compose(add(4), add(2)) === compose(add(2), add(4))

is my attempt to encode

(2 + 3) + 4 === 2 + (3 + 4)

Scott Christopher
@scott-christopher
Oct 08 2015 01:32 UTC
also:
compose(filter(f), filter(g)) === filter(both(f, g))
Hardy Jones
@joneshf
Oct 08 2015 01:33 UTC
you'd have to prove four cases: x satisfies p and x satisfies q, x satisfies p and x not satisfies q, x not satisfies p and x satisfies q, and x not satisfies p and x not satisfies q.
but the middle two cases are the same WLOG.
David Chambers
@davidchambers
Oct 08 2015 01:33 UTC
@scott-christopher, that's another one!
Jethro Larson
@jethrolarson
Oct 08 2015 01:33 UTC
That one is probably more useful, since it's less iterations
mirrors
compose(map(f), map(g)) == map(compose(f,g))
Hardy Jones
@joneshf
Oct 08 2015 01:34 UTC
@scott-christopher that's even better, since you can use the commutativity of both as a lemma in the filter commutativity proof!
compose(filter(p), filter(q)) === filter(both(p, q)) === filter(both(q, p)) === compose(filter(q), filter(p))
David Chambers
@davidchambers
Oct 08 2015 01:36 UTC

Filter Commutativity Law: filter is commutative under composition

I think we can all agree on this one. It doesn't make sense to talk about associativity, because we can't write both

(a filter b) filter c

and

a filter (b filter c)

It doesn't make sense to talk about commutativity (except under function composition) because we can't write both

a filter b

and

b filter a
Thank you for a stimulating conversation, folks. As always, I'm learning. :)
Hardy Jones
@joneshf
Oct 08 2015 01:38 UTC

why not:

compose(filter(p), compose(filter(q), filter(r))) === compose(compose(filter(p), filter(q)), filter(r))

?

which then reduces to unnesting the composes.
Jethro Larson
@jethrolarson
Oct 08 2015 01:39 UTC
TIL that creating a pure stream implementation is harder than it looks
Hardy Jones
@joneshf
Oct 08 2015 01:39 UTC
@jethrolarson oh?
Sounds interesting :)
David Chambers
@davidchambers
Oct 08 2015 01:39 UTC
@joneshf, I don't understand your question.
Jethro Larson
@jethrolarson
Oct 08 2015 01:39 UTC
I can gist my WIP if you want to see my failings
Hardy Jones
@joneshf
Oct 08 2015 01:40 UTC
@davidchambers to say that filter is associative under function composition.
though, I'm not sure if you even need to say that.
Jethro Larson
@jethrolarson
Oct 08 2015 01:43 UTC
Scott Sauyet
@CrossEye
Oct 08 2015 01:49 UTC
@joneshf: I'm trying to decide the same thing. It's easy enough to find functions which don't commute under composition. But I can't think of ones that don't associate. I'm pretty sure that it's a simple property of composition.
Does this do it?
(f o (g o h))(x) = (f(\y -> g(h(y)))(x) = f(g(h(x)) = (f(g(\y -> h(y)))(x) = ((f o g) o h)(x) 
Scott Sauyet
@CrossEye
Oct 08 2015 01:55 UTC
So function composition is always associative, and filter has nothing to do with it.
Jethro Larson
@jethrolarson
Oct 08 2015 02:20 UTC
Oh, duh lol
Its not certainly commutative tho
Scott Sauyet
@CrossEye
Oct 08 2015 02:22 UTC
no, of course not
Jethro Larson
@jethrolarson
Oct 08 2015 02:23 UTC
Sometimes i get stuck in the abstract and forget the basics
Scott Sauyet
@CrossEye
Oct 08 2015 02:23 UTC
n => n + 1 and n => n * n don't commute (unless n = 0)
Jethro Larson
@jethrolarson
Oct 08 2015 02:25 UTC
I commute. Doing so now
Scott Sauyet
@CrossEye
Oct 08 2015 02:34 UTC
I go to bed. Doing so now. Good night.
Niloy Mondal
@niloy
Oct 08 2015 11:05 UTC
Any particular reason why R.__ uses 2 underscore instead of 1?
Scott Christopher
@scott-christopher
Oct 08 2015 11:09 UTC
Not 100% sure, but if I were to guess I'd say it was to prevent confusion with underscore/lodash.
Niloy Mondal
@niloy
Oct 08 2015 11:15 UTC
Kinda sucks that it requires typing one extra character since the underscore would already be in the R namespace
Raine Virta
@raine
Oct 08 2015 11:17 UTC
I kind of like how __ stands out more
Niloy Mondal
@niloy
Oct 08 2015 11:20 UTC
R.lt has the parameter list wrong, const lessThan10 = R.lt(10); lessThan10(9) // false :(
Raine Virta
@raine
Oct 08 2015 11:21 UTC
that has been a topic of discussion
currently you need to use the placeholder
Niloy Mondal
@niloy
Oct 08 2015 11:21 UTC
awww okay, consensus is keep it the current way?
Martin Algesten
@algesten
Oct 08 2015 11:22 UTC
ramda always curries left to right
R.bla(a,b,c,d)
is always curried a -> b -> c -> d
Niloy Mondal
@niloy
Oct 08 2015 11:22 UTC
yes sir, thats why I said the parameter list is wrong
I mean parameter order
Martin Algesten
@algesten
Oct 08 2015 11:24 UTC
it's that way to be entirely consistent when not using partial application.
R.div(10,2) is another one.
Niloy Mondal
@niloy
Oct 08 2015 11:24 UTC
ok, I get it
Raine Virta
@raine
Oct 08 2015 11:25 UTC
ramda/ramda#1363
ramda/ramda#388
Martin Algesten
@algesten
Oct 08 2015 11:29 UTC

i struggle making my model code functional. are there any good examples around of something simple, like the TODO-list that is used in lots of front end code. it needs to show me:

1. how to model a state without mutation
2. how to persist this state using an asynchronous back end
3. how to draw this state in a view

especially 2 is where i get stuck.

Raine Virta
@raine
Oct 08 2015 11:30 UTC
Martin Algesten
@algesten
Oct 08 2015 11:31 UTC
thanks @raine
Raine Virta
@raine
Oct 08 2015 11:31 UTC
offers solutions to at least 1. and 3.
Martin Algesten
@algesten
Oct 08 2015 11:31 UTC
xD .... ok it's a start
Hardy Jones
@joneshf
Oct 08 2015 15:08 UTC
@jethrolarson hmm, since you're kind of stuck, maybe this is a good time to learn about the cofree comonad. Then your stream is Stream a = Cofree Identity a
Plus you get tons of free stuff.
Like, since Stream a is built on Cofree, you know it's a Functor, an Extend and a Comonad.
And since Identity a is a Monoid you also have that Stream a is an Apply, an Applicative, a Chain, and a Monad.
Eric Gjertsen
@ericgj
Oct 08 2015 15:48 UTC
@algesten , re your question "2. how to persist this state using an asynchronous back end", the repo that raine linked to has examples of this using a variety of techniques. Take a look at the zip-codes example, zip-codes-future, file-uploader, and autocomplete. https://github.com/paldepind/functional-frontend-architecture/tree/master/examples
Martin Algesten
@algesten
Oct 08 2015 15:50 UTC
yes. i realised that it would be there. it's a great resource. i need to plan some time in to go through the examples in detail. much appreciated!
Raine Virta
@raine
Oct 08 2015 16:04 UTC
yep, it's good stuff. after sitting down to understand the whole architecture I created https://www.habbo.com/mosaic/ as a small experiment
David Chambers
@davidchambers
Oct 08 2015 16:07 UTC
I like it, @raine!
Martin Algesten
@algesten
Oct 08 2015 16:07 UTC
trippy :)
Eric Gjertsen
@ericgj
Oct 08 2015 16:18 UTC
also the Elm Architecture document is a good reference, most of those functional-frontend-arch examples translate its concept of Effects: https://github.com/evancz/elm-architecture-tutorial/ (making heavy use of ramda-fantasy futures I should add...)
Jethro Larson
@jethrolarson
Oct 08 2015 17:26 UTC
@joneshf I'm open to that. You have a resource that'll teach it without getting too academic?
Jethro Larson
@jethrolarson
Oct 08 2015 18:42 UTC
I read a couple articles, I feel dumber than before I started