dustynachos - 1 month ago 10x

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 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

self.t=datetime.datetime.now()

def zcd(data,t):

sign_array=numpy.sign(data)

t('sign data')

out=[]

current = sign_array[0]

count=0

for i in sign_array[1:]:

if i!=current:

out.append(count)

current=i

count=0

else: count+=1

t('zcd data')

return out

def main():

t = timer()

data = numpy.fromfile('deci.dat',dtype=numpy.float32)

t('open data')

zcd(data,t)

if __name__=='__main__':

main()

Answer

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]
```

Output:

```
> 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.

Source (Stackoverflow)

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