如何向量化此python代码? [英] How to vectorize this python code?

问题描述

我正在尝试使用NumPy和矢量化操作来使一段代码运行得更快.但是，我似乎对如何矢量化此代码有误解(可能是由于对矢量化的理解不完整).

I am trying to use NumPy and vectorization operations to make a section of code run faster. I appear to have a misunderstanding of how to vectorize this code, however (probably due to an incomplete understanding of vectorization).

这是带有循环的工作代码(A和B是已设置大小的2D数组，已经初始化):

Here's the working code with loops (A and B are 2D arrays of a set size, already initialized):

for k in range(num_v):
    B[:] = A[:]
    for i in range(num_v):
        for j in range(num_v):
            A[i][j] = min(B[i][j], B[i][k] + B[k][j])
return A

这是我对上述代码进行矢量化的尝试:

And here is my attempt at vectorizing the above code:

for k in range(num_v):
    B = numpy.copy(A)
    A = numpy.minimum(B, B[:,k] + B[k,:])
return A

为了测试这些，我使用了以下代码，并将上面的代码包装在称为算法"的函数中:

For testing these, I used the following, with the code above wrapped in a function called 'algorithm':

def setup_array(edges, num_v):
    r = range(1, num_v + 1)
    A = [[None for x in r] for y in r]  # or (numpy.ones((num_v, num_v)) * 1e10) for numpy
    for i in r:
        for j in r:
            val = 1e10
            if i == j:
                val = 0 
            elif (i,j) in edges:
                val = edges[(i,j)]
            A[i-1][j-1] = val 
    return A

A = setup_array({(1, 2): 2, (6, 4): 1, (3, 2): -3, (1, 3): 5, (3, 6): 5, (4, 5): 2, (3, 1): 4, (4, 3): 8, (3, 4): 6, (2, 4): -4, (6, 5): -5}, 6) 
B = []
algorithm(A, B, 6)

预期的结果，以及我用第一个代码得到的结果:

The expected outcome, and what I get with the first code is:

[[0, 2, 5, -2, 0, 10] 
 [8, 0, 4, -4, -2, 9]
 [4, -3, 0, -7, -5, 5]
 [12, 5, 8, 0, 2, 13]
 [10000000000.0, 9999999997.0, 10000000000.0, 9999999993.0, 0, 10000000000.0]
 [13, 6, 9, 1, -5, 0]]

第二个(向量化)函数返回:

The second (vectorized) function instead returns:

[[ 0. -4.  0.  0.  0.  0.]
 [ 0. -4.  0. -4.  0.  0.]
 [ 0. -4.  0.  0.  0.  0.]
 [ 0. -4.  0.  0.  0.  0.]
 [ 0. -4.  0.  0.  0.  0.]
 [ 0. -4.  0.  0. -5.  0.]]

我想念什么?

推荐答案

该问题是由该行中的数组广播引起的:

The problem is caused by array broadcasting in the line:

A = numpy.minimum(B, B[:,k] + B[k,:])

B是6乘6的大小，B [:，k]是具有6个元素的数组，B [k ,:]是具有6个元素的数组.

B is size 6 by 6, B[:,k] is an array with 6 elements, B[k,:] is an array with 6 elements.

(因为您使用的是numpy数组类型，所以B [:，k]和B [k ,:]均会返回形状为N的1列数组)

(Because you are using the numpy array type, both B[:,k] and B[k,:] return a rank-1 array of shape N)

Numpy会自动更改大小以匹配:

Numpy automatically changes the sizes to match:

1. 将第一个B [:，k]添加到B [k ,:]，以得到一个包含6个元素的中间数组结果. (这不是您想要的)
2. 第二个6元素数组通过重复行广播到6 x 6矩阵
3. 原始矩阵的最小值的三倍，并计算出此广播矩阵.

这意味着您的numpy代码等效于:

This means that your numpy code is equivalent to:

for k in range(num_v):
   B[:] = A[:]
   C=[B[i][k]+B[k][i] for i in range(num_v)]
   for i in range(num_v):
      for j in range(num_v):
         A[i][j] = min(B[i][j], C[j])

修复代码的最简单方法是使用矩阵类型而不是数组类型:

The easiest way to fix your code is to use the matrix type instead of the array type:

A = numpy.matrix(A)
for k in range(num_v):
    A = numpy.minimum(A, A[:,k] + A[k,:])

矩阵类型使用更严格的广播规则，因此在这种情况下:

The matrix type uses stricter broadcasting rules so in this case:

1. A [:，k]通过重复列扩展为6 x 6矩阵
2. A [k ,:]通过重复行扩展为6 x 6矩阵
3. 将广播的矩阵相加在一起，组成一个6 x 6的矩阵
4. 已应用最小值

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