dustynachos dustynachos - 1 year ago 241
Python Question

Efficiently detect sign-changes in python

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 and tell every time it crosses crosses zero (changes sign). I want to record the time in between zero crossings. Since this is real data (32 bit float) I doubt I'll every have a number which is exactly zero, so that is not important. I currently have a timing program in place so I'll time your results to see who wins.

My solution gives (micro seconds):

open data 8384
sign data 8123
zcd data 415466

As you can see the zero-crossing detector is the slow part. Here's my code.

import numpy, datetime

class timer():
def __init__(self):
self.t0 = datetime.datetime.now()
self.t = datetime.datetime.now()
def __call__(self,text='unknown'):
print text,'\t',(datetime.datetime.now()-self.t).microseconds

def zcd(data,t):
t('sign data')
current = sign_array[0]
for i in sign_array[1:]:
if i!=current:
else: count+=1
t('zcd data')
return out

def main():
t = timer()
data = numpy.fromfile('deci.dat',dtype=numpy.float32)
t('open data')

if __name__=='__main__':

Answer Source

What about:

import numpy
a = [1, 2, 1, 1, -3, -4, 7, 8, 9, 10, -2, 1, -3, 5, 6, 7, -10]
zero_crossings = numpy.where(numpy.diff(numpy.sign(a)))[0]


> zero_crossings
array([ 3,  5,  9, 10, 11, 12, 15])

i.e. zero_crossings will contain the indices of elements after which a zero crossing occurs. If you want the elements before, just add 1 to that array.

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