Question

这个问题源于查看有关计算 this 数量的zero crossings问题的答案。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loop and simple generator，better generator expression和NumPy solution。

问题：为什么NumPy解决方案提供的结果与其他三种不同？（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，数字将是等效的。

Answer 2

我得到与循环相同的结果：

((array[:-1] * array[1:]) < 0).sum()

此：

small = np.array([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)

for name, array in [('small', small), ('med', med), ('big', big)]:
    print('loop ', name, zero_crossings_loop(array))
    print('Numpy', name, ((array[:-1] * array[1:]) < 0).sum())

打印：

loop  small 8
Numpy small 8
loop  med 44901
Numpy med 44901
loop  big 4496911
Numpy big 4496911

<强> UDPATE

此版本避免了零问题：

def numpy_zero_crossings2(array):
    nonzero_array = array[np.nonzero(array)]
    return ((nonzero_array[:-1] * nonzero_array[1:]) < 0).sum()

它给出了与@djsutton的答案相同的结果：

>>> numpy_zero_crossings2(big) == numpy_zero_crossings(big)     
True

但看起来有点快：

%timeit numpy_zero_crossings2(big)
1 loops, best of 3: 194 ms per loop

VS

%timeit numpy_zero_crossings(big)
1 loops, best of 3: 227 ms per loop

Answer 3

当数据元素等于零时，迭代和numpy解决方案在计算交叉时表现不佳。对于数据[1,0，-1]，迭代解决方案给出了0个交叉点，而numpy解决方案给出了2个交叉点，这两个交叉点似乎都不正确。

一种解决方案是将数据元素丢弃为零。在NumPy中你可能会尝试像

这样的东西

def numpy_zero_crossings(data):
    return (np.diff(np.sign(data)[np.nonzero(data)]) != 0).sum()

然而，这引入了数组的另一次迭代，因此它会增加另一个O（n）的运行时间

计算大序列的过零点的结果不同

3 个答案: