How can I square every element in an n by n numpy array?

Question

I don't want any for loop and was wondering if there is a function I can use.

@Jobs I think you mean that list becomes `[[1,4,9],[4,9,16]]` — Anton Protopopov, Apr 01 '16 at 04:29
https://docs.scipy.org/doc/numpy-dev/user/quickstart.html#basic-operations — Warren Weckesser, Apr 01 '16 at 04:36

score 6 · Answer 1 · answered Apr 01 '16 at 03:55

6

If A was the numpy array, I would just type in A*A.

answered Apr 01 '16 at 03:55

K. Tom

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score 5 · Answer 2 · answered Apr 01 '16 at 03:57

5

As K. Tom suggested you can do A * A you can also do A ** 2

import numpy as np

array = np.array([1,2,3])

print array * array #[1 4 9]
print array ** 2    #[1 4 9]

answered Apr 01 '16 at 03:57

John

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I definitely did not know that! Thanks. – Jobs Apr 01 '16 at 03:58
1

The only thing to take care of is the type. `a = np.array([[100, 200], [50, 150]], dtype=np.uint8); a*a` will overflow. – Reti43 Apr 01 '16 at 03:59

score 2 · Answer 3 · answered Apr 01 '16 at 04:27

2

You could use np.square or np.power:

l =  [[1,2,3], [2,3,4]]

In [5]: np.power(l, 2)
Out[5]:
array([[ 1,  4,  9],
       [ 4,  9, 16]], dtype=int32)

In [6]: np.square(l)
Out[6]:
array([[ 1,  4,  9],
       [ 4,  9, 16]], dtype=int32)

answered Apr 01 '16 at 04:27

Anton Protopopov

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How can I square every element in an n by n numpy array?

3 Answers3