Is there a faster way to separate the minimum and maximum from two arrays?

Question

In [3]: f1 = rand(100000) In [5]: f2 = rand(100000) # Obvious method: In [12]: timeit fmin = np.amin((f1, f2), axis=0); fmax = np.amax((f1, f2), axis=0) 10 loops, best of 3: 59.2 ms per loop In [13]: timeit fmin, fmax = np.sort((f1, f2), axis=0) 10 loops, best of 3: 30.8 ms per loop In [14]: timeit fmin = np.where(f2 < f1, f2, f1); fmax = np.where(f2 < f1, f1, f2) 100 loops, best of 3: 5.73 ms per loop In [36]: f1 = rand(1000,100,100) In [37]: f2 = rand(1000,100,100) In [39]: timeit fmin = np.amin((f1, f2), axis=0); fmax = np.amax((f1, f2), axis=0) 1 loops, best of 3: 6.13 s per loop In [40]: timeit fmin, fmax = np.sort((f1, f2), axis=0) 1 loops, best of 3: 3.3 s per loop In [41]: timeit fmin = np.where(f2 < f1, f2, f1); fmax = np.where(f2 < f1, f1, f2) 1 loops, best of 3: 617 ms per loop 

How, maybe, is there a way to do both where commands in one step with a return?

Why isn't amin implemented the same way as where if it is much faster?

Answer 1

Using numpy built into the maximum and minimum element is faster than where . The notes in the numpy docs for maximum confirm this:

Equivalent to np.where (x1> x2, x1, x2), but faster and does the right broadcast.

The line you want for the first test will look something like this:

 fmin = np.minimum(f1, f2); fmax = np.maximum(f1, f2) 

My own results show that it is pretty fast. Note that minimum and maximum will work on any n-dimensional array if the two arguments have the same shape.

 Using amax 3.506 Using sort 1.830 Using where 0.635 Using numpy maximum, minimum 0.178