Vectorize numpy array for loop

Question

I'm trying to figure out how to vectorize the following loop:

for i in range(1,size):
    if a[i] < a[i-1]:
        b[i] = a[i]
    else: b[i] = b[i-1]

b is a (large) array of the same size as a. I could use

numpy.where(a[1:]<a[:-1])

to replace the if statement but how do you simultaneously replace the else statement?

I don't think this can be vectorized, since each element depends on the previous one. — Eric
– Eric, Commented Dec 18, 2012 at 17:31
This is a guess, but the vectorize docs say it the vectorizing function accepts a sequence. Could you rewrite your loop as a generator instead? — Francis Avila
– Francis Avila, Commented Dec 18, 2012 at 17:51

Bi Rico · Accepted Answer · 2012-12-18 20:37:53Z

2

I think you want something like this:

import numpy as np

def foo(a, b):
    # cond is a boolean array marking where the condition is met
    cond = a[1:] < a[:-1]
    cond = np.insert(cond, 0, False)
    # values is an array of the items in from a that will be used to fill b
    values = a[cond]
    values = np.insert(values, 0, b[0])
    # labels is an array of increasing indices into values
    label = cond.cumsum()
    b[:] = values[label]

answered Dec 18, 2012 at 20:37

Bi Rico

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2 Comments

Jonno Over a year ago

This did it thanks. Now to make sure I understand it and then test it for speed.

Bi Rico Over a year ago

I hope this gives you some speedup. If you have more information about b, you might be able to get even more of a speedup. For example if you know ahead of time that b should be monotonically increasing.

Eric · Accepted Answer · 2012-12-18 17:37:39Z

1

From the docs:

numpy.where(condition[, x, y])

Return elements, either from x or y, depending on condition.

So you can simplify a loop containing:

if cond:
    a[i] = b[i]
else:
    a[i] = c[i]

to

a = numpy.where(cond, a, b)

However, I don't think your example can be vectorized, since each element depends on the previous one.

edited Dec 18, 2012 at 17:37

answered Dec 18, 2012 at 17:30

Eric

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1 Comment

Jonno Over a year ago

I think you're right. So what options are there to speed up an operation such as this if vectorizing is not possible?

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