点击这里查看详情！

a = [[1, 2], [2, 3], [3, 4]]
a2 = zip(*a)
a2[0]

它和上面的是一样的，只是它更整洁一些 zip可以完成这项工作，但需要单个数组作为参数，*a语法将多维数组解压缩为单个数组参数

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]

如果你喜欢map-reduce风格的python, itemgetter操作符也会有帮助，而不是列表推导式，为了一点变化!

# tested in 2.4
from operator import itemgetter
def column(matrix,i):
    f = itemgetter(i)
    return map(f,matrix)

M = [range(x,x+5) for x in range(10)]
assert column(M,1) == range(1,11)

只要使用转置()，就可以像求行一样简单地求列

matrix=np.array(originalMatrix).transpose()
print matrix[NumberOfColumns]

>>> x = arange(20).reshape(4,5)
>>> x array([[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19]])

如果你想要第二列，你可以用

>>> x[:, 1]
array([ 1,  6, 11, 16])

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]