1、nparray显然更适合数据分析与科学计算
import numpy as np
a=np.array([[[1,2,3],[4,5,6]],[[7,8,9],[10,11,12]]])
b=[[[1,2,3],[4,5,6]],[[7,8,9],[10,11,12]]]
a*5
Out[53]:
array([[[ 5, 10, 15],
[20, 25, 30]],
[[35, 40, 45],
[50, 55, 60]]])
b*5
Out[55]:
[[[1, 2, 3], [4, 5, 6]],
[[7, 8, 9], [10, 11, 12]],
[[1, 2, 3], [4, 5, 6]],
[[7, 8, 9], [10, 11, 12]],
[[1, 2, 3], [4, 5, 6]],
[[7, 8, 9], [10, 11, 12]],
[[1, 2, 3], [4, 5, 6]],
[[7, 8, 9], [10, 11, 12]],
[[1, 2, 3], [4, 5, 6]],
[[7, 8, 9], [10, 11, 12]]]
2、nparray可以使用元组作为index，内置的list方法不可以。
a[1,1]
Out[60]: array([10, 11, 12])
b[1,1]
Traceback (most recent call last):
File "<ipython-input-61-346be69afb77>", line 1, in <module>
b[1,1]
TypeError: list indices must be integers or slices, not tuple
3、如果使用第一种方法实际三个列表是同一个指针指向的位置
board=[['_']*3]*3
board
Out[67]: [['_', '_', '_'], ['_', '_', '_'], ['_', '_', '_']]
board[1][2]='o'
board
Out[69]: [['_', '_', 'o'], ['_', '_', 'o'], ['_', '_', 'o']]
board=[['_']*3 for i in range(3)]
board
Out[71]: [['_', '_', '_'], ['_', '_', '_'], ['_', '_', '_']]
board[1][2]='x'
board
Out[73]: [['_', '_', '_'], ['_', '_', 'x'], ['_', '_', '_']]
