So if I have an array of Bool, isprime (obtained by sieving, e.g.) of length n, [1:n][isprime] will give the same result as primes(n)? That's nice to know, though I'd call the syntax... interesting. :D Concerning the original problem: When we use do notation, an additional line x,y=p won't cost much. In lambda style, one could define a convenience function splice(f::Function)=p->f(p...) and use splice((x,y)->whatever we can do to x and y) in map or filter, without much damage to readability. Well, within reason... try map(splice(*),enumerate(1:10)). Haskell aficionados may know that the borderline between point-free and pointless is a bit fuzzy.
Am Donnerstag, 19. Juni 2014 18:25:40 UTC+2 schrieb Keith Campbell: > > You could also use a list comprehension without enumerate: > > d=[6,5,4,3,2,1,0] > n=4 > sum(d[ Bool[i<=n && d[i]!=i && d[d[i]]==i for i in 1:length(d)] ] ) > > >> 18 > > Coming from numpy, I find this syntax intuitive, aside from the > requirement to explicitly set the comprehension type to Bool. > > On Thursday, June 19, 2014 11:57:04 AM UTC-4, Matt Bauman wrote: >> >> There's also an open issue about "splitting" or "splatting" of tuples >> within function arguments that's very relevant here: Allow tuple >> destructuring in formal arguments, >> https://github.com/JuliaLang/julia/issues/6614 >> >> It's a common use case that I run into often, especially with enumerate >> and dictionaries. So you're not alone. >> >> On Thursday, June 19, 2014 3:57:33 AM UTC-4, gentlebeldin wrote: >>> >>> Ah, I see, multiple dispatch is the real cause. I'll have to live with >>> the little ideosyncrasy, then. >>> Peter's idea filter(p -> +(p...)%4 != 0, pairs) works fine... in my >>> simple example. In real life, I wanted to filter an array d of integers, >>> with a predicate >>> i<=n && d[i]!=i && d[d[i]]==i, so filter((i,v)->i<=n && v!=i && >>> d[v]==i,enumerate(d)) looked like a reasonable idea, it just doesn't work. >>> We can make it work, though that looks a bit strange, too: >>> filter([e for e in enumerate(d)]) do x >>> i,v=x >>> i<=n && v!=i && d[v]==i >>> end >>> Here, it looks as if the function filter took only one argument. It >>> works, however. In my code, I had to sum up the values of i after >>> filtering, so I chose the obscure formulation >>> sum(map(first,filter(x->x[1]!=x[2] && x[2]<=n && >>> d[x[2]]==x[1],enumerate(d)))) >>> >>>> >>>>>
