将numpy数组复制到另一个数组的一部分

rle*_*827 17 python arrays numpy

如果我运行以下内容:

import numpy as np
a = np.arange(9)
a = a.reshape((3,3))
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我会得到这个:

a = [[0 1 2]
     [3 4 5]
     [6 7 8]]
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如果我像这样创建一个更大的数组:

b = np.zeros((5,5))
b = [[ 0.  0.  0.  0.  0.]
     [ 0.  0.  0.  0.  0.]
     [ 0.  0.  0.  0.  0.]
     [ 0.  0.  0.  0.  0.]
     [ 0.  0.  0.  0.  0.]]
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如何有效地复制ab这样的数组?

# border of 0 surrounding a to be filled in with other data later
b = [[ 0.  0.  0.  0.  0.]
     [ 0.  1.  2.  3.  0.]
     [ 0.  4.  5.  6.  0.]
     [ 0.  7.  8.  9.  0.]
     [ 0.  0.  0.  0.  0.]]
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我正在寻找一个内置的功能,numpy如果它存在.

fal*_*tru 21

您可以指定b[1:4, 1:4]表示该部分:

>>> import numpy as np
>>> a = np.arange(9)
>>> a = a.reshape((3, 3))
>>> b = np.zeros((5, 5))
>>> b[1:4, 1:4] = a
>>> b
array([[ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  1.,  2.,  0.],
       [ 0.,  3.,  4.,  5.,  0.],
       [ 0.,  6.,  7.,  8.,  0.],
       [ 0.,  0.,  0.,  0.,  0.]])

>>> b[1:4,1:4] = a + 1  # If you really meant `[1, 2, ..., 9]`
>>> b
array([[ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  2.,  3.,  0.],
       [ 0.,  4.,  5.,  6.,  0.],
       [ 0.,  7.,  8.,  9.,  0.],
       [ 0.,  0.,  0.,  0.,  0.]])
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NaN*_*NaN 5

作为替代方案,如果您想要一个非零的填充值,则可以使用此选项

>>> a = np.arange(9.).reshape(3,3)
>>> np.pad(a, 1, 'constant', constant_values=0)
array([[ 0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  1.,  2.,  0.],
       [ 0.,  3.,  4.,  5.,  0.],
       [ 0.,  6.,  7.,  8.,  0.],
       [ 0.,  0.,  0.,  0.,  0.]])
>>> np.pad(a, 1, 'constant', constant_values=5)
array([[ 5.,  5.,  5.,  5.,  5.],
       [ 5.,  0.,  1.,  2.,  5.],
       [ 5.,  3.,  4.,  5.,  5.],
       [ 5.,  6.,  7.,  8.,  5.],
       [ 5.,  5.,  5.,  5.,  5.]])
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