Convert real-valued numpy array to binary array by sign

Question

I am looking for a fast way to compute the following:

import numpy as np
a = np.array([-1,1,2,-4,5.5,-0.1,0])

Now I want to cast a to an array of binary values such that it has a 1 for every positive entry of a and a 0 otherwise. So the result I want is this:

array([ 0.,  1.,  1.,  0.,  1.,  0.,  0.])

One way to achieve this would be

np.array([x if x >=0 else 0 for x in np.sign(a)])
array([ 0.,  1.,  1.,  0.,  1.,  0.,  0.])

But I am hoping someone can point out a faster solution.

%timeit np.array([x if x >=0 else 0 for x in np.sign(a)])
100000 loops, best of 3: 11.4 us per loop

EDIT: timing the great solutions from the answers

%timeit (a > 0).astype(int)
100000 loops, best of 3: 3.47 us per loop

What are you going to do with the result? I ask because the boolean array created by a > 0 might work without converting it to type int or int8. — Warren Weckesser
– Warren Weckesser, Commented Sep 22, 2015 at 21:23
@WarrenWeckesser Good point. I'll be computing distances, like hamming, so I'll check if i can make that work faster with boolean arrays. — benbo
– benbo, Commented Sep 22, 2015 at 21:25

Salvador Dali · Accepted Answer · 2015-09-22 21:42:44Z

3

You can do this using mask:

(a > 0).astype(int)

I do not know how to properly use timeit, but even

import numpy as np
from datetime import datetime
n = 50000000
a = np.random.rand(1, n).ravel()
startTime = datetime.now()
np.array([ x if x >=0 else 0 for x in np.sign(a)])
print datetime.now() - startTime

startTime = datetime.now()
(a > 0).astype(int)
print datetime.now() - startTime
pass

shows dramatic difference of 26 seconds vs 0.5 second.

P.S. based on your comment

I'll be computing distances, like hamming

you do not really need to have an integer array and a > 0 will be enough. It will save you memory and make things slightly faster.

edited Sep 22, 2015 at 21:42

answered Sep 22, 2015 at 20:56

Salvador Dali

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Comments

Alex Riley · Accepted Answer · 2015-09-22 21:15:01Z

You can check where a is greater than 0 and cast the boolean array to an integer array:

>>> (a > 0).astype(int)
array([0, 1, 1, 0, 1, 0, 0])

This should be significantly faster than the method proposed in the question (especially over larger arrays) because it avoids looping over the array at the Python level.

Faster still is to simply view the boolean array as the int8 dtype - this prevents the need to create a new array from the boolean array:

>>> (a > 0).view(np.int8)
array([0, 1, 1, 0, 1, 0, 0], dtype=int8)

Timings:

>>> b = np.random.rand(1000000)
>>> %timeit np.array([ x if x >=0 else 0 for x in np.sign(b)])
1 loops, best of 3: 420 ms per loop

>>> %timeit (b > 0).astype(int)
100 loops, best of 3: 4.63 ms per loop

>>> %timeit (b > 0).view(np.int8)
1000 loops, best of 3: 1.12 ms per loop

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