在any()语句中迭代一个小列表会更快吗？

Question

在低长度迭代次数的限制中考虑以下操作,

d = (3, slice(None, None, None), slice(None, None, None))

In [215]: %timeit any([type(i) == slice for i in d])
1000000 loops, best of 3: 695 ns per loop

In [214]: %timeit any(type(i) == slice for i in d)
1000000 loops, best of 3: 929 ns per loop

设置为a list比使用生成器表达式快25%？

为什么这样设置为a list是一个额外的操作.

注意:在两次运行中,我都获得了警告:The slowest run took 6.42 times longer than the fastest. This could mean that an intermediate result is being cachedI

分析

在该特定测试中,list()结构更快,直到4发电机具有提高的性能的长度.

红线表示此事件发生的位置,黑线表示两者在性能上相等.

通过利用所有内核,代码大约需要1分钟才能在我的MacBook Pro上运行:

import timeit, pylab, multiprocessing
import numpy as np

manager = multiprocessing.Manager()
g = manager.list([])
l = manager.list([])

rng = range(1,16) # list lengths
max_series = [3,slice(None, None, None)]*rng[-1] # alternate array types
series = [max_series[:n] for n in rng]

number, reps = 1000000, 5
def func_l(d):
    l.append(timeit.repeat("any([type(i) == slice for i in {}])".format(d),repeat=reps, number=number))
    print "done List, len:{}".format(len(d))
def func_g(d):
    g.append(timeit.repeat("any(type(i) == slice for i in {})".format(d), repeat=reps, number=number))
    print "done Generator, len:{}".format(len(d))

p = multiprocessing.Pool(processes=min(16,rng[-1])) # optimize for 16 processors
p.map(func_l, series) # pool list
p.map(func_g, series) # pool gens

ratio = np.asarray(g).mean(axis=1) / np.asarray(l).mean(axis=1)
pylab.plot(rng, ratio, label='av. generator time / av. list time')
pylab.title("{} iterations, averaged over {} runs".format(number,reps))
pylab.xlabel("length of iterable")
pylab.ylabel("Time Ratio (Higher is worse)")
pylab.legend()
lt_zero = np.argmax(ratio<1.)
pylab.axhline(y=1, color='k')
pylab.axvline(x=lt_zero+1, color='r')
pylab.ion() ; pylab.show()

Answer 1

问题在于你要应用的物品的尺寸any。在更大的数据集上重复相同的过程：

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In [2]: d = ([3] * 1000) + [slice(None, None, None), slice(None, None, None)]*1000\n\nIn [3]: %timeit any([type(i) == slice for i in d])\n1000 loops, best of 3: 736 \xc2\xb5s per loop\n\nIn [4]: %timeit any(type(i) == slice for i in d)\n1000 loops, best of 3: 285 \xc2\xb5s per loop\n

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然后，使用 a list（将所有项目加载到内存中）变得更慢，并且生成器表达式的效果更好。

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