48

If I run the following:

import numpy as np
a = np.arange(9)
a = a.reshape((3,3))

I will get this:

a = [[0 1 2]
     [3 4 5]
     [6 7 8]]

If I create a larger array like this:

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.]]

How do I efficiently copy a into b to get an array like this?

# border of 0 surrounding a to be filled in with other data later
b = [[ 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.]]

I am looking for a function built into numpy if it exists.

Hendy Irawan
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rlee827
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2 Answers2

55

You can specify b[1:4, 1:4] to denote the part:

>>> 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.]])
falsetru
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10

Just as an alternative, should you want a different pad value other than zero, you can use this option

>>> 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.]])
NaN
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