计算大序列的过零点的结果不同

Question

这个问题从寻找提供的答案源于此有关的问题计数的数量过零点.提供了几个解决问题的答案,但NumPy appproach在时间上摧毁了其他人.

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,更好的生成器表达式和NumPy解决方案.

问题:为什么NumPy解决方案提供的结果与其他三种解决方案不同？(哪个是正确的？)

以下是计算过零次数的结果:

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

用于获得结果的代码:

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)

Answer 1

您的解决方案的零处理方式不同.numpy.diff解决方案仍将返回从-1到0或1到0的差异,将其计为零交叉,而您的迭代解决方案则不会,因为它们使用小于零的乘积作为其标准.相反,测试<= 0,数字将是相同的.