Numpy -两个矩阵行之间的协方差

Question

我需要计算两个不同矩阵的每一行之间的协方差，即第一个矩阵的第一行与第二个矩阵的第一行之间的协方差，以此类推，直到两个矩阵的最后一行。我可以在没有NumPy的情况下使用下面的代码，我的问题是:是否有可能避免使用"for循环"?并得到相同的结果与NumPy?

m1 = np.array([[1,2,3],[2,2,2]])
m2 = np.array([[2.56, 2.89, 3.76],[1,2,3.95]])
output = []
for a,b in zip(m1,m2):
cov = np.cov(a, b)
output.append(cov[0][1])
print(output)

提前感谢!

Answer 1

如果你正在处理大数组，我会考虑这样做:

from numba import jit
import numpy as np

m1 = np.random.rand(10000, 3)
m2 = np.random.rand(10000, 3)
@jit(nopython=True) 
def nb_cov(a, b): 
return [np.cov(x)[0,1] for x in np.stack((a, b), axis=1)]

获取

的运行时

>>> %timeit nb_cov(m1, m2)
The slowest run took 94.24 times longer than the fastest. This could mean that an intermediate result is being cached.
1 loop, best of 5: 10.5 ms per loop

与

>>> %timeit [np.cov(x)[0,1] for x in np.stack((m1, m2), axis=1)]
1 loop, best of 5: 410 ms per loop  

Answer 2

您可以使用列表推导而不是for循环，并且可以通过沿着第三维连接两个数组来消除zip(如果您愿意)。

import numpy as np
m1 = np.array([[1,2,3],[2,2,2]])
m2 = np.array([[2.56, 2.89, 3.76],[1,2,3.95]])
# List comprehension on zipped arrays.
out2 = [np.cov(a, b)[0][1] for a, b in zip(m1, m2)]
print(out2)
# [0.5999999999999999, 0.0]
# List comprehension on concatenated arrays.
big_array = np.concatenate((m1[:, np.newaxis, :],
m2[:, np.newaxis, :]), axis=1)
out3 = [np.cov(X)[0][1] for X in big_array]
print(out3)
# [0.5999999999999999, 0.0]