Say I have a function f, I want to call it repeatedly n times. I.e. for n = 3, I want f ∘ f ∘ f. Is there a nice notation to do this for n not known at compile time, ideally something like f^n.

[image] Zentrik: Is there a nice notation to do this for n not known at compile time, ideally something like f^n. We had discussed supporting f^n, but there wasn’t a consensus in favor: function^n for iterated functions Ju…

I can’t think of syntaxic sugar that would allow this in Julia. The prettiest way I could think of would be a recursive function : f_rec(x, n) = (@assert n >= 0; n == 0 ? x : f_rec(f(x), n-1)) Although I heard recursive functions can actually be slow, so I don’t know if this is recommended. Anoth…

[image] Alexandre Prieur: f_rec(x, n) = (@assert n >= 0; n == 0 ? x : f_rec(f(x), n-1)) This doesn’t look right. You haven’t defined f. Something like apply(f, x, n::Integer) = ((n > 0) ? f(apply(f, x, n-1)) : x) and, yeah, probably throw an error for negative n.

Yes, I assumed it was defined before indeed.

You can do a one-liner like julia> ↑(f, n) = x -> foldl(|>, fill(f, n), init = x) ↑ (generic function with 1 method) julia> (sqrt ↑ 3)(2) 1.0905077326652577 but the fill is wasteful so it can be implemented more efficiently.

[image] Alexandre Prieur: Although I heard recursive functions can actually be slow, so I don’t know if this is recommended. Recursion can be slow for large n, and at worst you can get a stack overflow. For small n it should be fine. For large n, an iterative approach should be better, e.g. …

Maybe this: ? [image] Seven Lines of Julia (examples sought) General Usage [immagine] using IterTools using Statistics F(x)=[mean(x),prod(x)^(1/length(x)), median(x)] Fnx(x,n)=nth(iterated(F,x),n)

This looks like the winner to me. Applied to OP’s case it’s simply: using IterTools # Recursively apply function f to value x n times fn(n,x) = nth(iterated(f,x),n)

Another suggested previously is foldl(|>, Iterators.repeated(f, n)) Edit: Better (as suggested below) is either F(f,n)=foldl(∘, Iterators.repeated(f, n)) for just the iterated function, or if including the function argument, F(f,n,x) =foldl(|>, Iterators.repeated(f, n); init=x)

This is how I do it #= func_compose(func,n) Compose a function multiple times =# function func_compose(func,n) newfunc = func local loop = n while loop > 1 # help?> ∘ # "∘" can be typed by \circ<tab> newfunc = fun…

F(f,n)=foldl(∘, Iterators.repeated(f, n)) F(f,n)=foldl(∘, ntuple(i->f,n)) F(f, n)=n>1 ? ∘(f,F(f,n-1)) : f

Is there a way to repeatedly call a function

General Usage

JM_Beckers December 18, 2023, 10:44am 7

Maybe this: ?

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Is there a way to repeatedly call a function

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