Python code for counting number of zero crossings in an array

Question

I am looking to count the number of times the values in an array change in polarity (EDIT: Number of times the values in an array cross zero).

Suppose I have an array:

[80.6  120.8  -115.6  -76.1  131.3  105.1  138.4  -81.3
 -95.3  89.2  -154.1  121.4  -85.1  96.8  68.2]`

I want the count to be 8.

One solution is to run a loop and check for greater than or less than 0, and keep a history of the previous polarity.

Can we do this faster?

EDIT: My purpose is really to find something faster, because I have these arrays of length around 68554308, and I have to do these calculations on 100+ such arrays.

Answer 1

This produces the same result:

import numpy as np
my_array = np.array([80.6, 120.8, -115.6, -76.1, 131.3, 105.1, 138.4, -81.3, -95.3,  
                     89.2, -154.1, 121.4, -85.1, 96.8, 68.2])
((my_array[:-1] * my_array[1:]) < 0).sum()

gives:

8

and seems to be the fastest solution:

%timeit ((my_array[:-1] * my_array[1:]) < 0).sum()
100000 loops, best of 3: 11.6 µs per loop

Compared to the fastest so far:

%timeit (np.diff(np.sign(my_array)) != 0).sum()
10000 loops, best of 3: 22.2 µs per loop

Also for larger arrays:

big = np.random.randint(-10, 10, size=10000000)

this:

%timeit ((big[:-1] * big[1:]) < 0).sum()
10 loops, best of 3: 62.1 ms per loop

vs:

%timeit (np.diff(np.sign(big)) != 0).sum()
1 loops, best of 3: 97.6 ms per loop

Answer 2

Here's a numpy solution. Numpy's methods are generally pretty fast and well-optimized, but if you're not already working with numpy there's probably some overhead from converting the list to a numpy array:

import numpy as np
my_list = [80.6, 120.8, -115.6, -76.1, 131.3, 105.1, 138.4, -81.3, -95.3,  89.2, -154.1, 121.4, -85.1, 96.8, 68.2]
(np.diff(np.sign(my_list)) != 0).sum()
Out[8]: 8