计算大序列的零交叉的结果不同 [英] Different results to counting zero-crossings of a large sequence

问题描述

此问题源于查看提供给 >零交叉点 .提供了解决该问题的几种答案，但是NumPy appproach 在时间方面破坏了其他问题.

This question stems from looking at the answers provided to this question regarding counting the number of zero crossings. Several answer were provided that solve the problem, but the NumPy appproach destroyed the others with respect to time.

但是，当我比较四个答案时，我注意到NumPy解决方案为大序列提供了不同的结果.有问题的四个答案是循环和简单生成器， NumPy解决方案.

When I compared four of the answers however I notice that the NumPy solution provides a different result for large sequences. The four answers in question are loop and simple generator, better generator expression , and NumPy solution.

问题:为什么NumPy解决方案提供的结果与其他三个解决方案不同?(哪个是正确的?)

以下是计算零交叉点数量的结果:

Here are the results for counting the number of zero crossings:

Blazing fast NumPy solution
total time: 0.303605794907 sec
Zero Crossings Small: 8
Zero Crossings Med: 54464
Zero Crossings Big: 5449071

Loop solution
total time: 15.6818780899 sec
Zero Crossings Small: 8
Zero Crossings Med: 44960
Zero Crossings Big: 4496847

Simple generator expression solution
total time: 16.3374049664 sec
Zero Crossings Small: 8
Zero Crossings Med: 44960
Zero Crossings Big: 4496847

Modified generator expression solution
total time: 13.6596589088 sec
Zero Crossings Small: 8
Zero Crossings Med: 44960
Zero Crossings Big: 4496847

以及用于获取结果的代码:

And the code used to get the results:

import time
import numpy as np

def zero_crossings_loop(sequence):
    s = 0
    for ind, _ in enumerate(sequence):
        if ind+1 < len(sequence):
            if sequence[ind]*sequence[ind+1] < 0:
                s += 1
    return s

def print_three_results(r1, r2, r3):
    print 'Zero Crossings Small:', r1
    print 'Zero Crossings Med:', r2
    print 'Zero Crossings Big:', r3
    print '\n'

small = [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]
med = np.random.randint(-10, 10, size=100000)
big = np.random.randint(-10, 10, size=10000000)

print 'Blazing fast NumPy solution'
tic = time.time()
z1 = (np.diff(np.sign(small)) != 0).sum()
z2 = (np.diff(np.sign(med)) != 0).sum()
z3 = (np.diff(np.sign(big)) != 0).sum()
print 'total time: {0} sec'.format(time.time()-tic)
print_three_results(z1, z2, z3)

print 'Loop solution'
tic = time.time()
z1 = zero_crossings_loop(small)
z2 = zero_crossings_loop(med)
z3 = zero_crossings_loop(big)
print 'total time: {0} sec'.format(time.time()-tic)
print_three_results(z1, z2, z3)

print 'Simple generator expression solution'
tic = time.time()
z1 = sum(1 for i, _ in enumerate(small) if (i+1 < len(small)) if small[i]*small[i+1] < 0)
z2 = sum(1 for i, _ in enumerate(med) if (i+1 < len(med)) if med[i]*med[i+1] < 0)
z3 = sum(1 for i, _ in enumerate(big) if (i+1 < len(big)) if big[i]*big[i+1] < 0)
print 'total time: {0} sec'.format(time.time()-tic)
print_three_results(z1, z2, z3)

print 'Modified generator expression solution'
tic = time.time()
z1 = sum(1 for i in xrange(1, len(small)) if small[i-1]*small[i] < 0)
z2 = sum(1 for i in xrange(1, len(med)) if med[i-1]*med[i] < 0)
z3 = sum(1 for i in xrange(1, len(big)) if big[i-1]*big[i] < 0)
print 'total time: {0} sec'.format(time.time()-tic)
print_three_results(z1, z2, z3)

推荐答案

您的解决方案对零的处理方式有所不同. numpy.diff解决方案仍将返回从-1到0或1到0的diff，并将其视为零交叉，而您的迭代解决方案则不会，因为它们使用乘积小于零作为标准.取而代之的是测试<= 0，这些数字将是等效的.

Your solutions differ in their treatment of zero. The numpy.diff solution will still return a diff going from -1 to 0 or 1 to 0, counting those as a zero crossing, while your iterative solutions don't because they use the product being less than zero as their criterion. Instead, test for <= 0, and the numbers will be equivalent.

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