Numpy：从两个向量（或一个自身）的笛卡尔积中创建矩阵，同时将函数应用于所有对

Question

为了给出一些解释，我想创建一个协方差矩阵，其中每个元素由内核函数k(x, y)定义，我想为单个向量执行此操作。它应该是这样的：

# This is given
x = [x1, x2, x3, x4, ...]

# This is what I want to compute
result = [[k(x1, x1), k(x1, x2), k(x1, x3), ...],
          [k(x2, x1), k(x2, x2), ...],
          [k(x3, x1), k(x3, x2), ...],
          ...]

但是当然这应该在numpy数组中完成，理想情况下，由于性能的原因，不需要进行Python交互。如果我不关心性能，我可能只会写：

result = np.zeros((len(x), len(x)))

for i in range(len(x)):
  for j in range(len(x)):
     result[i, j] = k(x[i], x[j])

但我觉得必须有一种更惯用的方式来编写这种模式。

Answer 1

如果k在2D阵列上运行，您可以使用np.meshgrid。但是，这会产生额外的内存开销。另一种方法是创建与2D相同的np.meshgrid网格视图，就像这样 -

def meshgrid1D_view(x):
    shp = (len(x),len(x))
    mesh1 = np.broadcast_to(x,shp)
    mesh2 = np.broadcast_to(x[:,None],shp)    
    return mesh1, mesh2

样品运行 -

In [140]: x
Out[140]: array([3, 5, 6, 8])

In [141]: np.meshgrid(x,x)
Out[141]: 
[array([[3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8]]), array([[3, 3, 3, 3],
        [5, 5, 5, 5],
        [6, 6, 6, 6],
        [8, 8, 8, 8]])]
In [142]: meshgrid1D(x)
Out[142]: 
(array([[3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8],
        [3, 5, 6, 8]]), array([[3, 3, 3, 3],
        [5, 5, 5, 5],
        [6, 6, 6, 6],
        [8, 8, 8, 8]]))

这有什么用？

它有助于提高内存效率，从而提高性能。让我们测试大型阵列，看看差异 -

In [143]: x = np.random.randint(0,10,(10000))

In [144]: %timeit np.meshgrid(x,x)
10 loops, best of 3: 171 ms per loop

In [145]: %timeit meshgrid1D(x)
100000 loops, best of 3: 6.91 µs per loop

Answer 2

另一种解决方案是让numpy进行广播：

import numpy as np

def k(x,y):
    return x**2+y

def meshgrid1D_view(x):
    shp = (len(x),len(x))
    mesh1 = np.broadcast_to(x,shp)
    mesh2 = np.broadcast_to(x[:,None],shp)    
    return mesh1, mesh2

x = np.random.randint(0,10,(10000))
b=k(a[:,None],a[None,:])

def sol0(x):
    k(x[:,None],x[None,:])

def sol1(x):
    x,y=np.meshgrid(x,x)
    k(x,y)

def sol2(x):
    x,y=meshgrid1D_view(x)
    k(x,y)

%timeit sol0(x)
165 ms ± 1.67 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit sol1(x)
655 ms ± 6.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit sol2(x)
341 ms ± 2.91 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

您会发现这样做效率更高，而且代码更少。