Array expansion in numpy

Question

What I'd like to do is take an input integer array, and expand its data into indices (e.g., [2, 1] -> [2, 2, 1]). I apologize if the terminology is off -- I wasn't sure of the best way to describe this, as such, it is possible that this is a duplicate.

Here is an example of the current method I have in place:

>>> def expand(a):
...     b = np.empty(a.sum(), dtype=np.int32)
...     idx = 0
...     for i in a:
...         for j in range(i):
...             b[idx] = i
...             idx += 1
...     return b
... 
>>> a = np.array([3, 2, 1, 4])
>>> expand(a)
array([3, 3, 3, 2, 2, 1, 4, 4, 4, 4], dtype=int32)

This method is called within a nested for loop that I'd like to squeeze additional performance out of. Below is a current timing call:

>>> a = np.random.randint(0, 1000, 1000)
>>> %timeit expand(a)
10 loops, best of 3: 86.9 ms per loop

Is there a different approach that could be used to lower the expense of the method?

Your code gives [2, 1] -> [2, 2, 1] not [2, 1] -> [0, 0, 1] as per your example. Which is correct? — John Zwinck
– John Zwinck, Commented Aug 19, 2014 at 0:47
Embarassingly, the code is actually whats wrong for what I had intended. The solution presented by @hpaulj is correct given the code in the question however, so I will close this and open a new, fixed, question in a little bit. Sorry for the confusion — daniel
– daniel, Commented Aug 19, 2014 at 1:33

hpaulj · Accepted Answer · 2014-08-19 01:22:33Z

9

The np.repeat should do most of what you want:

a.repeat(a)

I timeit at 5ms v your 88.

Your first example would be

arange(2).repeat([2,1])

edited Aug 19, 2014 at 1:22

answered Aug 19, 2014 at 0:43

hpaulj

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