Python 以numpy为单位按索引拾取
假设两个数组Python 以numpy为单位按索引拾取,python,arrays,numpy,Python,Arrays,Numpy,假设两个数组 ind = array([[1, 3, 2, 4, 0], [0, 1, 3, 2, 4], [3, 4, 2, 0, 1]]) x = array([[[24, 97, 28, 57, 59], [97, 67, 94, 77, 50], [56, 89, 25, 55, 76], [88, 21, 1, 50, 24]], [[54, 83, 64, 81, 12],
ind =
array([[1, 3, 2, 4, 0],
[0, 1, 3, 2, 4],
[3, 4, 2, 0, 1]])
x =
array([[[24, 97, 28, 57, 59],
[97, 67, 94, 77, 50],
[56, 89, 25, 55, 76],
[88, 21, 1, 50, 24]],
[[54, 83, 64, 81, 12],
[89, 49, 15, 26, 97],
[94, 97, 32, 55, 79],
[24, 63, 63, 15, 40]],
[[41, 99, 84, 64, 21],
[12, 9, 85, 43, 28],
[75, 98, 48, 10, 0],
[93, 94, 37, 22, 63]]])
我想根据第一个数组对第二个数组重新排序。(第一个数组是索引)
所以,可能结果如下
数组([[97,57,28,59,24],
[67, 77, 94, 50, 97],
[89, 55, 25, 76, 56],
[21, 50, 1, 24, 88]],
[[54, 83, 81, 64, 12],
[89, 49, 26, 15, 97],
[94, 97, 55, 32, 79],
[24, 63, 15, 63, 40]],
[[64, 21, 84, 41, 99],
[43, 28, 85, 12, 9],
[10, 0, 48, 75, 98],
[22, 63, 37, 93, 94]]])
#x[0]被ind[0]重新排序,依此类推。
这有可能吗?使用
沿轴取U很容易:
>>> np.take_along_axis(x, ind[:, None, :], 2)
array([[[97, 57, 28, 59, 24],
[67, 77, 94, 50, 97],
[89, 55, 25, 76, 56],
[21, 50, 1, 24, 88]],
[[54, 83, 81, 64, 12],
[89, 49, 26, 15, 97],
[94, 97, 55, 32, 79],
[24, 63, 15, 63, 40]],
[[64, 21, 84, 41, 99],
[43, 28, 85, 12, 9],
[10, 0, 48, 75, 98],
[22, 63, 37, 93, 94]]])
如果您使用的是1.15 numpy之前的版本,您可以执行以下操作:
>>> m,n,k = x.shape
>>> m,n,k = np.ogrid[:m, :n, :k]
>>> x[m,n,ind[:, None, :]]
array([[[97, 57, 28, 59, 24],
[67, 77, 94, 50, 97],
[89, 55, 25, 76, 56],
[21, 50, 1, 24, 88]],
[[54, 83, 81, 64, 12],
[89, 49, 26, 15, 97],
[94, 97, 55, 32, 79],
[24, 63, 15, 63, 40]],
[[64, 21, 84, 41, 99],
[43, 28, 85, 12, 9],
[10, 0, 48, 75, 98],
[22, 63, 37, 93, 94]]])
这个解决方案很好!但我使用的是NUMPY1.14,它不支持沿轴取。。。(take
函数中的axis
参数与take\u沿轴
函数看起来非常不同)