获取不重复的数组组合答案

【问题标题】：Get combinations of array without self repetition获取不重复的数组组合
【发布时间】：2021-02-16 20:24:23
【问题描述】：

我有以下形状为(3,3) 的数组points，它们是3D 点

[[-0.20592405  0.33034652  1.0430001 ]
 [-0.23757319  0.29612657  1.03800011]
 [-0.304515    0.34571081  1.20400012]]

我需要找出这三点的所有组合。我正在执行以下操作：

comb = np.array(list(itertools.product(points.tolist(), repeat=3)))

我得到了

[[[-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]
  [-0.304515    0.34571081  1.20400012]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]
  [-0.20592405  0.33034652  1.0430001 ]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]
  [-0.23757319  0.29612657  1.03800011]]

 [[-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]
  [-0.304515    0.34571081  1.20400012]]]

这给了我所有的组合，但我想丢弃重复的组合。所以，假设 3D 点 a、b、c 我得到 [[a, a, a], [a, a, b],..,[a, b, c],.., [c, c, c]] 的组合。我只想保留具有所有不同点的组合，即[a, b, c]、[b, c, a] 等。

【问题讨论】：

用combinations代替product?
排列应该做：np.array(list(itertools.permutations(points, r=3)))
@QuangHoang 你是对的，那行得通！如果您将此作为答案发布，我会接受。
@MustafaAydın 我试图避免重复点，这意味着顺序无关紧要。

标签： python numpy combinations itertools

【解决方案1】：

itertools.combinations，只有一个：

l1=[[-0.20592405,  0.33034652,  1.0430001 ],
    [-0.23757319,  0.29612657,  1.03800011],
    [-0.304515,    0.34571081,  1.20400012]]
comb = np.array(list(itertools.combinations(l1, 3)))
#array([[[-0.20592405,  0.33034652,  1.0430001 ],
#        [-0.23757319,  0.29612657,  1.03800011],
#        [-0.304515  ,  0.34571081,  1.20400012]]])

与itertools.permutations:

comb = np.array(list(itertools.permutations(l1, 3)))

comb 的值：

array([[[-0.20592405,  0.33034652,  1.0430001 ],
        [-0.23757319,  0.29612657,  1.03800011],
        [-0.304515  ,  0.34571081,  1.20400012]],

       [[-0.20592405,  0.33034652,  1.0430001 ],
        [-0.304515  ,  0.34571081,  1.20400012],
        [-0.23757319,  0.29612657,  1.03800011]],

       [[-0.23757319,  0.29612657,  1.03800011],
        [-0.20592405,  0.33034652,  1.0430001 ],
        [-0.304515  ,  0.34571081,  1.20400012]],

       [[-0.23757319,  0.29612657,  1.03800011],
        [-0.304515  ,  0.34571081,  1.20400012],
        [-0.20592405,  0.33034652,  1.0430001 ]],

       [[-0.304515  ,  0.34571081,  1.20400012],
        [-0.20592405,  0.33034652,  1.0430001 ],
        [-0.23757319,  0.29612657,  1.03800011]],

       [[-0.304515  ,  0.34571081,  1.20400012],
        [-0.23757319,  0.29612657,  1.03800011],
        [-0.20592405,  0.33034652,  1.0430001 ]]])

【讨论】：