Question

刚刚得到一个我想要了解的奇怪结果。我有一个大约325k行（列表）的数据集，每个行约90个项目（字符串，浮点数等 - 它并不重要）。比如，如果我想对所有项目进行一些处理，那么我可以使用2“for”来迭代它们：

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（eachRow [0]，eachRowm [1]和到目前为止每个Row [89]）替换嵌套循环，则执行时间下降到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，编写这样的代码并不是一个好主意 - 我只是在寻找一种方法来提高数据处理性能并意外地发现了这种奇怪的方法。有什么意见吗？

Answer 1

进行展开似乎确实有轻微的性能优势，但它可以忽略不计，所以除非你的do_something函数几乎什么也没做，否则你不应该看到差异。我很难相信使用不同方法的等效行为可能达到60％，尽管我总是愿意对我从未想过的一些实现细节感到惊讶。

tl; dr摘要，使用32500而不是325000因为我不耐烦了：

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，我并不打算使比较完全公平，因为我只是试图衡量变化的可能规模，即顺序是否统一的变化。）

Answer 2

抱歉，伙计们，这是我的错。我的系统出了问题（这不是一个独立的Python解释器，而是一个内置于大系统中）。重新启动整个系统后，我得到了正确的结果 - 两种变体大约2.8秒。我觉得我好笨。寻找一种方法来删除我的问题，因为无关紧要。

Answer 3

与计时的其他响应者不同，我看到时间差异很大。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)

注意我正在运行这10次。每个列表都填充了3000个项目，每个项目包含90个项目，每个项目都是10个字母的随机字符串。

代表性成果：

x1 = 8.0179214353
x2 = 0.118051644801
x3 = 0.150409681521

不使用索引（x1）的函数执行的时间比执行内部循环（x2）索引的时间长<66次。奇怪的是，仅对内循环（x2）使用索引的函数比对外循环和内循环（x3）使用索引的函数执行得更好。

“for”迭代与直接索引性能

3 个答案: