# Python code for counting number of zero crossings in an array

arraysdistributionperformancepython

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.

#### Best Solution

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
``````