random.shuffle 使用注意
有关random.shuffle函数的局限性
刚刚踩的坑，记录一下，万一对别人有帮助呢。
本文重点： random.shuffle千万不要用于二维numpy.array（也就是矩阵）!!!
首先上证据
如下代码：
import random
import numpy as np
a = np.array([[1,2,3,4],
[5,6,7,8]])
random.shuffle(a)
print(a)
可能对应输出：
[[1 2 3 4]
[1 2 3 4]]
得到一个错误输出！
查看random.shuffle源码：
def shuffle(self, x, random=None):
"""Shuffle list x in place, and return None.
Optional argument random is a 0-argument function returning a
random float in [0.0, 1.0); if it is the default None, the
standard random.random will be used.
"""
if random is None:
randbelow = self._randbelow
for i in reversed(range(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = randbelow(i+1)
x[i], x[j] = x[j], x[i]
else:
_int = int
for i in reversed(range(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = _int(random() * (i+1))
x[i], x[j] = x[j], x[i]
只需要关注其中交换元素操作为 x[i], x[j] = x[j], x[i] （好骚的操作，我之前不知道有这种写法）。
自己写个代码测试一下，这种交换方式对于二维numpy.array会发生什么事情：
import numpy as np
a = np.array([[1,2,3,4],
[5,6,7,8]])
a[0], a[1] = a[1], a[0]
print(a)
输出：
[[5 6 7 8]
[5 6 7 8]]
显然这种方式不适合numpy.array的行交换（但是二维list就可以使用这种交换方式。可以自行证明，至于原因暂时不知道，求解答）。
打乱numpy.array正确的姿势当然是使用numpy自带的numpy.random.shuffle()
def shuffle(x): # real signature unknown; restored from __doc__
"""
shuffle(x)
Modify a sequence in-place by shuffling its contents.
This function only shuffles the array along the first axis of a
multi-dimensional array. The order of sub-arrays is changed but
their contents remains the same.
Parameters
----------
x : array_like
The array or list to be shuffled.
Returns
-------
None
Examples
--------
>>> arr = np.arange(10)
>>> np.random.shuffle(arr)
>>> arr
[1 7 5 2 9 4 3 6 0 8]
Multi-dimensional arrays are only shuffled along the first axis:
>>> arr = np.arange(9).reshape((3, 3))
>>> np.random.shuffle(arr)
>>> arr
array([[3, 4, 5],
[6, 7, 8],
[0, 1, 2]])
"""
pass
注释中Multi-dimensional arrays are only shuffled along the first axis: 多维向量只是沿着第一个坐标轴进行重新排序。