Wil*_*dan 15 python arrays numpy
假设我有这个数组:
array = np.array([[1,2,3],[4,5,6],[7,8,9]])
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返回:
123
456
789
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我应该如何让它返回这样的东西?
111222333
111222333
111222333
444555666
444555666
444555666
777888999
777888999
777888999
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Ch3*_*teR 17
你必须np.repeat
在这里使用两次。
np.repeat(np.repeat(array, 3, axis=1), 3, axis=0)
# [[1 1 1 2 2 2 3 3 3]
# [1 1 1 2 2 2 3 3 3]
# [1 1 1 2 2 2 3 3 3]
# [4 4 4 5 5 5 6 6 6]
# [4 4 4 5 5 5 6 6 6]
# [4 4 4 5 5 5 6 6 6]
# [7 7 7 8 8 8 9 9 9]
# [7 7 7 8 8 8 9 9 9]
# [7 7 7 8 8 8 9 9 9]]
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为了好玩(因为嵌套repeat
会更有效),您可以einsum
在输入数组和ones
具有额外维度的数组上使用来创建一个多维数组,其维度以理想的顺序排列为reshape
预期的二维形状:
np.einsum('ij,ikjl->ikjl', array, np.ones((3,3,3,3))).reshape(9,9)
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通用方法是:
i,j = array.shape
k = 3 # extra rows
l = 3 # extra cols
np.einsum('ij,ikjl->ikjl', a, np.ones((i,k,j,l))).reshape(i*k,j*l)
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输出:
array([[1, 1, 1, 2, 2, 2, 3, 3, 3],
[1, 1, 1, 2, 2, 2, 3, 3, 3],
[1, 1, 1, 2, 2, 2, 3, 3, 3],
[4, 4, 4, 5, 5, 5, 6, 6, 6],
[4, 4, 4, 5, 5, 5, 6, 6, 6],
[4, 4, 4, 5, 5, 5, 6, 6, 6],
[7, 7, 7, 8, 8, 8, 9, 9, 9],
[7, 7, 7, 8, 8, 8, 9, 9, 9],
[7, 7, 7, 8, 8, 8, 9, 9, 9]])
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然而,这种方法的优点在于,可以很容易地更改顺序以获得其他模式或处理更高的维度。
其他模式的示例:
>>> np.einsum('ij,iklj->iklj', a, np.ones((3,3,3,3))).reshape(9,9)
array([[1, 2, 3, 1, 2, 3, 1, 2, 3],
[1, 2, 3, 1, 2, 3, 1, 2, 3],
[1, 2, 3, 1, 2, 3, 1, 2, 3],
[4, 5, 6, 4, 5, 6, 4, 5, 6],
[4, 5, 6, 4, 5, 6, 4, 5, 6],
[4, 5, 6, 4, 5, 6, 4, 5, 6],
[7, 8, 9, 7, 8, 9, 7, 8, 9],
[7, 8, 9, 7, 8, 9, 7, 8, 9],
[7, 8, 9, 7, 8, 9, 7, 8, 9]])
>>> np.einsum('ij,kjil->kjil', a, np.ones((3,3,3,3))).reshape(9,9)
array([[1, 1, 1, 4, 4, 4, 7, 7, 7],
[2, 2, 2, 5, 5, 5, 8, 8, 8],
[3, 3, 3, 6, 6, 6, 9, 9, 9],
[1, 1, 1, 4, 4, 4, 7, 7, 7],
[2, 2, 2, 5, 5, 5, 8, 8, 8],
[3, 3, 3, 6, 6, 6, 9, 9, 9],
[1, 1, 1, 4, 4, 4, 7, 7, 7],
[2, 2, 2, 5, 5, 5, 8, 8, 8],
[3, 3, 3, 6, 6, 6, 9, 9, 9]])
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