所有列从一个矩阵到一个新的列表:

N = len(matrix) 
column_list = [ [matrix[row][column] for row in range(N)] for column in range(N) ]

如果你想抓取多个列，可以使用slice:

 a = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
    print(a[:, [1, 2]])
[[2 3]
[5 6]
[8 9]]

我更喜欢下一个提示: 将矩阵命名为matrix_a并使用column_number，例如:

import numpy as np
matrix_a = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])
column_number=2

# you can get the row from transposed matrix - it will be a column:
col=matrix_a.transpose()[column_number]

我认为你想从一个数组中提取一个列，比如下面的数组

import numpy as np
A = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])

现在如果你想要得到格式中的第三列

D=array[[3],
[7],
[11]]

然后你需要首先把数组变成一个矩阵

B=np.asmatrix(A)
C=B[:,2]
D=asarray(C)

现在你可以做基于元素的计算就像你在excel中做的一样。

array = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]

col1 = [val[1] for val in array]
col2 = [val[2] for val in array]
col3 = [val[3] for val in array]
col4 = [val[4] for val in array]
print(col1)
print(col2)
print(col3)
print(col4)

Output:
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]
[4, 8, 12, 16]

[matrix[i][column] for i in range(len(matrix))]