“针对”迭代与直接索引性能

Question

刚得到一个我想理解的奇怪结果。 我有一个约325k行（列表）的数据集，每个约有90个项目（字符串，浮点数-没关系）。 说，如果我想对所有项目进行一些处理，则可以使用2个“ for”遍历它们：

for eachRow in rows:
    for eachItem in eachRow:
        # do something

在我的系统中，此代码执行了41秒。 但是，如果我用一系列索引访问项（eachRow [0]，eachRowm [1]以及到目前为止的eachRow [89]）代替嵌套循环，则执行时间将降至25秒。

for eachRow in rows:
    eachRow[0]  # do something with this item
    eachRow[1]  # do something with this item
    ..
    eachRow[89] # do something with this item

当然，编写这样的代码不是一个好主意-我只是在寻找一种提高数据处理性能的方法，偶然发现了这种奇怪的方法。 任何意见？

Answer 1

展开似乎确实在性能上有一点优势，但是可以忽略不计，因此，除非您的do_something函数实际上几乎什么都不做，否则您应该看不出区别。 我很难过，我相信使用其他方法的等效行为可能占60％，尽管我总是愿意为从未想到的一些实现细节感到惊讶。

tl;博士总结，因为我不耐烦，所以使用32500而不是325000：

do_nothing easy 3.44702410698
do_nothing indexed 3.99766016006
do_nothing mapped 4.36127090454
do_nothing unrolled 3.33416581154
do_something easy 5.4152610302
do_something indexed 5.95649385452
do_something mapped 6.20316290855
do_something unrolled 5.2877831459
do_more easy 16.6573209763
do_more indexed 16.8381450176
do_more mapped 17.6184959412
do_more unrolled 16.0713188648

CPython 2.7.3，代码：

from timeit import Timer

nrows = 32500
ncols = 90
a = [[1.0*i for i in range(ncols)] for j in range(nrows)]

def do_nothing(x):
    pass

def do_something(x):
    z = x+3
    return z

def do_more(x):
    z = x**3+x**0.5+4
    return z

def easy(rows, action):
    for eachRow in rows:
        for eachItem in eachRow:
            action(eachItem)

def mapped(rows, action):
    for eachRow in rows:
        map(action, eachRow)

def indexed(rows, action):
    for eachRow in rows:
        for i in xrange(len(eachRow)):
            action(eachRow[i])

def unrolled(rows, action):
    for eachRow in rows:
        action(eachRow[0])
        action(eachRow[1])
        action(eachRow[2])
        action(eachRow[3])
        action(eachRow[4])
        action(eachRow[5])
        action(eachRow[6])
        action(eachRow[7])
        action(eachRow[8])
        action(eachRow[9])
        action(eachRow[10])
        action(eachRow[11])
        action(eachRow[12])
        action(eachRow[13])
        action(eachRow[14])
        action(eachRow[15])
        action(eachRow[16])
        action(eachRow[17])
        action(eachRow[18])
        action(eachRow[19])
        action(eachRow[20])
        action(eachRow[21])
        action(eachRow[22])
        action(eachRow[23])
        action(eachRow[24])
        action(eachRow[25])
        action(eachRow[26])
        action(eachRow[27])
        action(eachRow[28])
        action(eachRow[29])
        action(eachRow[30])
        action(eachRow[31])
        action(eachRow[32])
        action(eachRow[33])
        action(eachRow[34])
        action(eachRow[35])
        action(eachRow[36])
        action(eachRow[37])
        action(eachRow[38])
        action(eachRow[39])
        action(eachRow[40])
        action(eachRow[41])
        action(eachRow[42])
        action(eachRow[43])
        action(eachRow[44])
        action(eachRow[45])
        action(eachRow[46])
        action(eachRow[47])
        action(eachRow[48])
        action(eachRow[49])
        action(eachRow[50])
        action(eachRow[51])
        action(eachRow[52])
        action(eachRow[53])
        action(eachRow[54])
        action(eachRow[55])
        action(eachRow[56])
        action(eachRow[57])
        action(eachRow[58])
        action(eachRow[59])
        action(eachRow[60])
        action(eachRow[61])
        action(eachRow[62])
        action(eachRow[63])
        action(eachRow[64])
        action(eachRow[65])
        action(eachRow[66])
        action(eachRow[67])
        action(eachRow[68])
        action(eachRow[69])
        action(eachRow[70])
        action(eachRow[71])
        action(eachRow[72])
        action(eachRow[73])
        action(eachRow[74])
        action(eachRow[75])
        action(eachRow[76])
        action(eachRow[77])
        action(eachRow[78])
        action(eachRow[79])
        action(eachRow[80])
        action(eachRow[81])
        action(eachRow[82])
        action(eachRow[83])
        action(eachRow[84])
        action(eachRow[85])
        action(eachRow[86])
        action(eachRow[87])
        action(eachRow[88])
        action(eachRow[89])


def timestuff():
    for action in 'do_nothing do_something do_more'.split():
        for name in 'easy indexed mapped unrolled'.split():
            t = Timer(setup="""
from __main__ import {} as fn
from __main__ import {} as action
from __main__ import a
""".format(name, action),
                      stmt="fn(a, action)").timeit(10)
            print action, name, t

if __name__ == '__main__':
    timestuff()

（请注意，我并没有打扰使比较完全公平，因为我只是在尝试估算变化的可能范围，即订单是否统一。）

Answer 2

抱歉，这是我的错。 我的系统出了点问题（这不是独立的Python解释器，而是大型系统中内置的）。 重新启动整个系统后，我得到了正确的结果-两种变体大约需要2.8秒。 我觉得我好笨。 由于不相关，正在寻找删除我的问题的方法。

Answer 3

与其他为此计时的响应者不同，我发现计时的差异很大。 首先，我的代码：

import random
import string
import timeit

r = 1000
outer1 = [[[''.join([random.choice(string.ascii_letters) for j in range(10)])] for k in range(90)] for l in range(r)]
outer2 = [[[''.join([random.choice(string.ascii_letters) for j in range(10)])] for k in range(90)] for l in range(r)]
outer3 = [[[''.join([random.choice(string.ascii_letters) for j in range(10)])] for k in range(90)] for l in range(r)]

def x1(L):
    for outer in L:
        for inner in L:
            inner = inner[:-1]

def x2(L):
    for outer in L:
        for y in range(len(outer)):
            outer[y] = outer[y][:-1]

def x3(L):
    for x in range(len(L)):
        for y in range(len(L[x])):
            L[x][y] = L[x][y][:-1]

print "x1 =",timeit.Timer('x1(outer1)', "from __main__ import x1,outer1").timeit(10)
print "x2 =",timeit.Timer('x2(outer2)', "from __main__ import x2,outer2").timeit(10)
print "x3 =",timeit.Timer('x3(outer3)', "from __main__ import x3,outer3").timeit(10)

请注意，我正在运行这10次。 每个列表中填充了3000个项目，每个项目包含90个项目，每个项目都是由10个字母组成的随机字符串。

代表性的结果：

x1 = 8.0179214353
x2 = 0.118051644801
x3 = 0.150409681521

不使用索引（x1）的函数执行时间比仅对内部循环（x2）使用索引的函数执行时间长66倍 。 奇怪的是，仅对内部循环（x2）使用索引的函数比对外部循环和内部循环（x3）使用索引的函数的性能要好。