考虑这样一个问题，给定一个矩阵（多维数组，<span style="color:#E53333;">numpy.ndarray()</span>），如何<span style="color:#E53333;">shuffle</span>这个矩阵（也就是对其行进行全排列），如何随机地选择其中的k行，这叫组合，实现一种某一维度空间的切片。例如五列中选三列（全部三列的排列数），便从原有的五维空间中降维到三维空间，因为是全部的排列数，故不会漏掉任何一种可能性。
涉及的函数主要有：
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- np.random.permutation()
- itertools.combinations()
- itertools.permutations()
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# 1. 对0-5之间的数进行一次全排列
>>>np.random.permutation(6)
array([3, 1, 5, 4, 0, 2])
# 2. 创建待排矩阵
>>>A = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
# 3. shuffle矩阵A
>>>p = np.random.permutation(A.shape[0])
>>>p
array([1, 2, 0])
>>>A[p, :]
array([[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[ 1, 2, 3, 4]])
的实现
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>>>from itertools import combinations
>>>combins = [c for c in combinations(range(5), 2)]
>>>len(combins)
10
>>>combins # 而且是按序排列
[(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
任取其中的k（k=2）行
>>>c = [c for c in combinations(range(A.shape[0]), 2)]
>>>A[c[0], :] # 一种排列
array([[1, 2, 3, 4],
[5, 6, 7, 8]])
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>>>from itertools import permutations >>>pertumations(range(5), 2)>>>perms = permutations(range(5), 2) >>>perms [(0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 3), (2, 4), (3, 0), (3, 1), (3, 2), (3, 4), (4, 0), (4, 1), (4, 2), (4, 3)] >>>len(perms) 20
注意combinations和<a href="https://www.361way.com/python-itertools/5705.html" target="_blank" rel="noopener">product</a>（笛卡尔积）的区别，后者一般传入的是两个数组，而前者是通过顺序排列，且传入的是排除字符的长度位参数。
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