根据两个向量的差填充numpy矩阵 [英] Populate numpy matrix from the difference of two vectors

问题描述

是否可以从函数构造numpy矩阵?在这种情况下，该函数特别是两个向量的绝对差:S[i,j] = abs(A[i] - B[j]).一个使用常规python的最小工作示例:

Is it possible to construct a numpy matrix from a function? In this case specifically the function is the absolute difference of two vectors: S[i,j] = abs(A[i] - B[j]). A minimal working example that uses regular python:

import numpy as np

A = np.array([1,3,6])
B = np.array([2,4,6])
S = np.zeros((3,3))

for i,x in enumerate(A):
    for j,y in enumerate(B):
        S[i,j] = abs(x-y)

给予:

[[ 1.  3.  5.]
 [ 1.  1.  3.]
 [ 4.  2.  0.]]

最好有一个看起来像这样的结构

It would be nice to have a construction that looks something like:

def build_matrix(shape, input_function, *args)

我可以在其中传递带有参数的输入函数，并保留numpy的速度优势.

where I can pass an input function with it's arguments and retain the speed advantage of numpy.

推荐答案

我建议您看一下numpy的广播功能:

I recommend taking a look into numpy's broadcasting capabilities:

In [6]: np.abs(A[:,np.newaxis] - B)
Out[6]: 
array([[1, 3, 5],
       [1, 1, 3],
       [4, 2, 0]])

http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html

然后，您可以简单地将函数编写为:

Then you could simply write your function as:

In [7]: def build_matrix(func,args):
   ...:     return func(*args)
   ...: 

In [8]: def f1(A,B):
   ...:     return np.abs(A[:,np.newaxis] - B)
   ...: 

In [9]: build_matrix(f1,(A,B))
Out[9]: 
array([[1, 3, 5],
       [1, 1, 3],
       [4, 2, 0]])

对于大型阵列，这应该比您的解决方案要快得多.

This should also be considerably faster than your solution for larger arrays.

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