如果你有一个数组

a = [[1, 2], [2, 3], [3, 4]]

然后像这样提取第一列:

[row[0] for row in a]

结果是这样的:

[1, 2, 3]

另一种使用矩阵的方法

>>> from numpy import matrix
>>> a = [ [1,2,3],[4,5,6],[7,8,9] ]
>>> matrix(a).transpose()[1].getA()[0]
array([2, 5, 8])
>>> matrix(a).transpose()[0].getA()[0]
array([1, 4, 7])

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]

如果你在Python中有一个二维数组(不是numpy)，你可以像这样提取所有的列，

data = [
['a', 1, 2], 
['b', 3, 4], 
['c', 5, 6]
]

columns = list(zip(*data))

print("column[0] = {}".format(columns[0]))
print("column[1] = {}".format(columns[1]))
print("column[2] = {}".format(columns[2]))

执行这段代码会得到，

>>> print("column[0] = {}".format(columns[0]))
column[0] = ('a', 'b', 'c')

>>> print("column[1] = {}".format(columns[1]))
column[1] = (1, 3, 5)

>>> print("column[2] = {}".format(columns[2]))
column[2] = (2, 4, 6)

假设我们有nxm矩阵(n行m列)5行4列

matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16],[17,18,19,20]]

要在python中提取列，我们可以像这样使用列表推导式

[ [row[i] for row in matrix] for in range(4) ]

你可以用矩阵的列数来替换4。 结果是

,10,14,18,5,9,13,17 [[1], [2], [3,7,11,15,19], [4,8,12,16,20]]

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

>>> A
array([[1, 2, 3, 4],
    [5, 6, 7, 8]])

>>> A[:,2] # returns the third columm
array([3, 7])

参见:"numpy。“Arange”和“重塑”来分配内存

示例:(用矩阵(3x4)的形状分配数组)

nrows = 3
ncols = 4
my_array = numpy.arange(nrows*ncols, dtype='double')
my_array = my_array.reshape(nrows, ncols)