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]])

Answer 1

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

Answer 2

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.

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)