Efficiently detect sign-changes in python

Question

I want to do exactly what this guy did:

Python - count sign changes

However I need to optimize it to run super fast. In brief I want to take a time series an

Answer 1

Another way that might suit certain applications is to extend the evaluation of the expression np.diff(np.sign(a)).

If we compare how this expression reacts to certain cases:

1. Rising crossing without zero: np.diff(np.sign([-10, 10])) returns array([2])
2. Rising crossing with zero: np.diff(np.sign([-10, 0, 10])) returns array([1, 1])
3. Falling crossing without zero: np.diff(np.sign([10, -10])) returns array([-2])
4. Falling crossing with zero: np.diff(np.sign([10, 0, -10])) returns array([-1, -1])

So we have to evaluate np.diff(...) for the returned patterns in 1. and 2:

sdiff = np.diff(np.sign(a))
rising_1 = (sdiff == 2)
rising_2 = (sdiff[:-1] == 1) & (sdiff[1:] == 1)
rising_all = rising_1
rising_all[1:] = rising_all[1:] | rising_2

and for the cases 3. and 4.:

falling_1 = (sdiff == -2) #the signs need to be the opposite
falling_2 = (sdiff[:-1] == -1) & (sdiff[1:] == -1)
falling_all = falling_1
falling_all[1:] = falling_all[1:] | falling_2

After this we can easily find the indices with

indices_rising = np.where(rising_all)[0]
indices_falling = np.where(falling_all)[0]
indices_both = np.where(rising_all | falling_all)[0]

This approach should be reasonable fast because it can manage without using a "slow" loop.

This combines the approach of several other answers.