有效地检测python中的符号变化 [英] Efficiently detect sign-changes in python

问题描述

我想确切地做这个人所做的事情:

I want to do exactly what this guy did:

Python-计算符号更改

但是我需要对其进行优化以使其超快运行.简而言之，我想采用一个时间序列，并告诉它每次穿越时都越过零(更改符号).我想记录两次过零之间的时间.由于这是真实数据(32位浮点数)，我怀疑我每个人都有一个正好为零的数字，所以这并不重要.我目前有一个计时程序，所以我将对您的成绩进行计时，看看谁赢了.

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.

我的解决方案给出了(微秒):

My solution gives (micro seconds):

open data       8384
sign data       8123
zcd data        415466

如您所见，过零检测器是最慢的部分.这是我的代码.

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()

推荐答案

有关:

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]

输出:

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

即，zero_crossings将包含元素之前的索引，该元素发生零交叉.如果要在元素之后，只需将1添加到该数组.

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

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