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

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
– Anton Protopopov, Commented Apr 1, 2016 at 4:29

K. Tom · Accepted Answer · 2016-04-01 03:55:19Z

6

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

answered Apr 1, 2016 at 3:55

K. Tom

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John · Accepted Answer · 2016-04-01 03:57:14Z

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 1, 2016 at 3:57

John

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2 Comments

Jobs Over a year ago

I definitely did not know that! Thanks.

Reti43 Over a year ago

The only thing to take care of is the type. a = np.array([[100, 200], [50, 150]], dtype=np.uint8); a*a will overflow.

Anton Protopopov · Accepted Answer · 2016-04-01 04:27:46Z

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 1, 2016 at 4:27

Anton Protopopov

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