What is the fastest way to get a new array given a list of indices?

See e.g. http://wesmckinney.com/blog/numpy-indexing-peculiarities/ that shows that np.take(a, indices) can be much faster than a[indices]



In [1]:

    
import numpy as np



In [2]:

    
a = np.arange(10)
indices = np.array([0, 0, 0, 1, 2, 2, 3, 4, 5, 9])
print a[indices]
print a.take(indices)
%timeit a[indices]
%timeit a.take(indices)









    



[0 0 0 1 2 2 3 4 5 9]
[0 0 0 1 2 2 3 4 5 9]
1000000 loops, best of 3: 746 ns per loop
1000000 loops, best of 3: 442 ns per loop



In [3]:

    
a = np.arange(10)
a = a[:, np.newaxis]
indices = np.array([0, 0, 0, 1, 2, 2, 3, 4, 5, 9])
print a[indices]
print a.take(indices, axis=0)
%timeit a[indices]
%timeit a.take(indices, axis=0)









    



[[0]
 [0]
 [0]
 [1]
 [2]
 [2]
 [3]
 [4]
 [5]
 [9]]
[[0]
 [0]
 [0]
 [1]
 [2]
 [2]
 [3]
 [4]
 [5]
 [9]]
1000000 loops, best of 3: 1.62 µs per loop
1000000 loops, best of 3: 936 ns per loop

np.take is really faster, even though in this context by not as much as otherwise.



In [4]:

    
a = np.arange(10, dtype = float)
a = a[:, np.newaxis]
indices = np.array([0, 0, 0, 1, 2, 2, 3, 4, 5, 9])
out = np.empty((10, 1), dtype = float)
%timeit a.take(indices, axis=0)
%timeit a.take(indices, out=out, axis=0, mode='clip')









    



1000000 loops, best of 3: 945 ns per loop
1000000 loops, best of 3: 911 ns per loop

By preallocating the output array we can save a slight overhead.