朱莉娅：以量化的方式

Question

学习序列的产品，从Python的传递到朱莉娅，我想转换的旧代码，我有，就是计算这个表达的序列的产品：

我在Python中有两个版本的代码，一个使用for循环实现，另一个使用广播。该for循环的版本是：

import numpy as np 
A = np.arange(1.,5.,1) 
G = np.array([[1.,2.],[3.,4.]]) 

def calcF(G,A): 
    N = A.size 
    print A 
    print N 
    F = [] 
    for l in range(N): 
     F.append(G/A[l]) 
     print F[l] 
     for j in range(N): 
      if j != l: 
       F[l]*=((G - A[l])/(G + A[j]))*((A[l] - A[j])/(A[l] + A[j])) 
    return F 

F= calcF(G,A) 
print F 

而且矢量版本我已经从我的问题here响应了解到，这个函数：

def calcF_vectorized(G,A): 
    # Get size of A 
    N = A.size 

    # Perform "(G - A[l])/(G + A[j]))" in a vectorized manner 
    p1 = (G - A[:,None,None,None])/(G + A[:,None,None]) 

    # Perform "((A[l] - A[j])/(A[l] + A[j]))" in a vectorized manner 
    p2 = ((A[:,None] - A)/(A[:,None] + A)) 

    # Elementwise multiplications between the previously calculated parts 
    p3 = p1*p2[...,None,None] 

    # Set the escaped portion "j != l" output as "G/A[l]" 
    p3[np.eye(N,dtype=bool)] = G/A[:,None,None] 

    Fout = p3.prod(1) 

    # If you need separate arrays just like in the question, split it 
    return np.array_split(Fout,N) 

我试图天真地翻译了Python for循环代码朱莉娅：

function JuliacalcF(G,A) 
    F = Array{Float64}[] 
    for l in eachindex(A) 
     push!(F,G/A[l]) 
     println(A[i]) 
     for j in eachindex(A) 
      if j!=l 
       F[l]*=((G - A[l])/(G + A[j]))*((A[l] - A[j])/(A[l] + A[j])) 
      end 
     end 
    end 
    #println(alpha) 
    return F 
end 
A = collect(1.0:1.0:5.0) 
G = Vector{Float64}[[1.,2.],[3.,4.]] 
println(JuliacalcF(G,A)) 

但是，有没有办法做到这一点很巧妙地作为转播numpy g矢量化版本？

Answer 1

另外，请看More-Dots和Loop Fusion，其中矢量化是用示例来描述的。