Python numpy.square vs **

Question

Is there a difference between numpy.square and using the ** operator on a Numpy array?

From what I can see it yields the same result.

Any differences in efficiency of execution?

An example for clarification:

In [1]: import numpy as np

In [2]: A = np.array([[2, 2],[2, 2]])

In [3]: np.square(A)
Out[3]: 
array([[4, 4],
       [4, 4]])

In [4]: A ** 2
Out[4]: 
array([[4, 4],
       [4, 4]])

Asked By: Skeppet

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Source

Answer 1

For most appliances, both will give you the same results.
Generally the standard pythonic a*a or a**2 is faster than the numpy.square() or numpy.pow(), but the numpy functions are often more flexible and precise.
If you do calculations that need to be very accurate, stick to numpy and probably even use other datatypes float96.

For normal usage a**2 will do a good job and way faster job than numpy.
The guys in this thread gave some good examples to a similar questions.

Answered By: foehnx

Answer 2

You can check the execution time to get clear picture of it

In [2]: import numpy as np
In [3]: A = np.array([[2, 2],[2, 2]])
In [7]: %timeit np.square(A)
1000000 loops, best of 3: 923 ns per loop
In [8]: %timeit A ** 2
1000000 loops, best of 3: 668 ns per loop

Answered By: saimadhu.polamuri

Answer 3

@saimadhu.polamuri and @foehnx/@Lumos

On my machine, currently, NumPy performs faster than **.

In [1]: import numpy as np

In [2]: A = np.array([[1,2],[3,4]])

In [3]: %timeit A ** 2

256 ns ± 0.922 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [4]: %timeit np.square(A)

240 ns ± 0.759 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Answered By: Tom

Python numpy.square vs **

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