function - 重复应用函数特定次数

Question

如果您有一个函数，是否有简单或内置的方法来应用它n次，或者直到结果是特定的。因此，例如，如果您想应用 sqrt作用4次，效果:

julia> sqrt(sqrt(sqrt(sqrt(11231))))
1.791229164345863

你可以输入如下内容:

repeatf(sqrt, 11231, 4)

Answer 1

我很喜欢定义 ^运算符(operator)处理 Function s 和 Int秒

julia> (^)(f::Function, i::Int) = i==1 ? f : x->(f^(i-1))(f(x))
^ (generic function with 1 method)

julia> (sqrt^1)(2)
1.4142135623730951

julia> (sqrt^2)(2)
1.189207115002721

julia> (sqrt^3)(2)
1.0905077326652577

正如@DNF 指出的，因为 julia 没有尾调用优化，
最好反复执行此操作；

    julia> function (∧)(f::Function, i::Int)
               function inner(x)
                  for ii in i:-1:1
                     x=f(x)
                  end
               x
               end
           end


After warmup:

    julia> @time((sqrt ∧ 1_000)(20e300)) #Iterative
      0.000018 seconds (6 allocations: 192 bytes)
    1.0
    
    julia> @time((sqrt ^ 1_000)(20e300)) #Recursive
      0.000522 seconds (2.00 k allocations: 31.391 KB)
    1.0
    
    #########
    
    julia> @time((sqrt ∧ 10_000)(20e300)) #Iterative
      0.000091 seconds (6 allocations: 192 bytes)
    1.0
    
    
    julia> @time((sqrt ^ 10_000)(20e300)) #Recursive
      0.003784 seconds (20.00 k allocations: 312.641 KB)
    1.0
    
    #########
    
    julia> @time((sqrt ∧ 30_000)(20e300)) # Iterative
      0.000224 seconds (6 allocations: 192 bytes)
    1.0

    julia> @time((sqrt ^ 30_000)(20e300)) #Recursive
      0.008128 seconds (60.00 k allocations: 937.641 KB)
    1.0
    
    
    #############
   
    julia> @time((sqrt ∧ 100_000)(20e300)) #Iterative
      0.000393 seconds (6 allocations: 192 bytes)
    1.0

    julia> @time((sqrt ^ 100_000)(20e300)) #Recursive
    ERROR: StackOverflowError:
     in (::##5#6{Base.#sqrt,Int64})(::Float64) at ./REPL[1]:1 (repeats 26667 times)



The overhead isn't too bad in this case, but that `StackOverflowError` at the end is a kicker.