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在 Python 中計(jì)時(shí)短路會(huì)產(chǎn)生意想不到的結(jié)果

在 Python 中計(jì)時(shí)短路會(huì)產(chǎn)生意想不到的結(jié)果

呼喚遠(yuǎn)方 2023-03-01 16:18:17
import time as dt success = Truecan_test = True time = 0for i in range(10000000):  start = dt.time()  if success and can_test:    stop = dt.time()    time+= stop-startprint(f'"and" operation took: {time} seconds')time = 0for i in range(10000000):  start = dt.time()  if success or can_test:    stop = dt.time()    time += stop-startprint(f'"or" operation took: {time} seconds')當(dāng)我運(yùn)行上面的 python 程序時(shí)，我希望 and 操作比 or 操作慢（因?yàn)槲伊私獾蕉搪窌?huì)減少執(zhí)行時(shí)間）。然而，結(jié)果不僅完全相反，而且還在波動(dòng)。我能理解波動(dòng)?。ㄒ?yàn)楹笈_(tái)進(jìn)程）?？蔀槭裁唇Y(jié)果卻相反！怎么了？這是一個(gè)示例結(jié)果。"and" operation took: 5.200342893600464 seconds"or" operation took: 5.3243467807769775 seconds
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萬千封印

TA貢獻(xiàn)1891條經(jīng)驗(yàn) 獲得超3個(gè)贊

這是一個(gè)有趣的問題，所以我決定深入調(diào)查您的主要顧慮。


# required modules line_profiler, matplotlib, seaborn abd scipy

import time as dt 

from line_profiler import LineProfiler

import matplotlib.pyplot as plt

import seaborn as sns

from scipy import stats


success = True

can_test = True 

def and_op():

    for x in range(2000):

        s = success and can_test

def or_op():

    for x in range(2000):

        s = success or can_test

or_op_list = []

for x in range(0,1000):

    lp = LineProfiler()

    lp_wrapper = lp(or_op)

    lp_wrapper()

    lstats = lp.get_stats()

    total_time = 0

    for v in lstats.timings.values():

        for op in v:

            total_time += op[-1]

            final = op[-1]

        operator = final/total_time

    or_op_list.append(operator)


and_op_list = []

for x in range(0,1000):

    lp = LineProfiler()

    lp_wrapper = lp(and_op)

    lp_wrapper()

    lstats = lp.get_stats()

    total_time = 0

    for v in lstats.timings.values():

        for op in v:

            total_time += op[-1]

            final = op[-1]

        operator = final/total_time

    and_op_list.append(operator)

sns.kdeplot(and_op_list, label = 'AND')

sns.kdeplot(or_op_list, label = 'OR')

plt.show()

print(stats.ttest_ind(and_op_list,or_op_list, equal_var = False))

http://img1.sycdn.imooc.com//63ff0a7700018e2003700248.jpg

pvalue=1.8293386245013954e-103

實(shí)際上，與“和”操作相比，“或”具有統(tǒng)計(jì)意義并且不同


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反對(duì) 回復(fù) 2023-03-01
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RISEBY

TA貢獻(xiàn)1856條經(jīng)驗(yàn) 獲得超5個(gè)贊

當(dāng)我在我的機(jī)器上運(yùn)行您的代碼時(shí)，它有時(shí)打印出來的速度也True and True比 快True or True。


出現(xiàn)這種現(xiàn)象的原因是您dt.time()的代碼以“微秒”（即 1000 納秒）為尺度測(cè)量時(shí)間，但是，這個(gè)微秒尺度太稀疏，無法測(cè)量每次執(zhí)行 or所花費(fèi)的時(shí)間。在大多數(shù)情況下， or所花費(fèi)的時(shí)間小于1 微秒。if success and can_test:if success or can_test:if success and can_test:if success or can_test:


因此，在下面的代碼部分中：


for i in range(10000000):

   start = dt.time()

       if success and can_test:  # a dust particle

       stop = dt.time()

       time += stop - start  # measured by a normal scale ruler

for i in range(10000000):

   start = dt.time()

       if success or can_test:  # a dust particle

       stop = dt.time()

       time += stop - start  # measured by a normal scale ruler

您的代碼所做的就像用普通刻度尺測(cè)量每個(gè)灰塵顆粒并將測(cè)量值相加。由于測(cè)量誤差巨大，結(jié)果失真。


為了進(jìn)一步調(diào)查，如果我們執(zhí)行下面的代碼（d記錄所花費(fèi)的時(shí)間及其頻率）：


import time as dt

from pprint import pprint


success = True

can_test = True


time = 0

d = {}

for i in range(10000000):

    start = dt.time_ns()

    if success and can_test:  # a dust particle

        stop = dt.time_ns()

        diff_time = stop - start  # measurement by a normal scale ruler

        d[diff_time] = d.get(diff_time, 0) + 1

        time += diff_time

print(f'"and" operation took: {time} ns')

print('"and" operation time distribution:')

pprint(d)

print()


time = 0

d = {}

for i in range(10000000):

    start = dt.time_ns()

    if success or can_test:  # a dust particle

        stop = dt.time_ns()

        diff_time = stop - start  # measurement by a normal scale ruler

        d[diff_time] = d.get(diff_time, 0) + 1

        time += diff_time

print(f'"or" operation took: {time} ns')

print('"or" operation time distribution:')

pprint(d)

它將打印如下：


"and" operation took: 1467442000 ns

"and" operation time distribution:

{0: 8565832,

 1000: 1432066,

 2000: 136,

 3000: 24,

 4000: 12,

 5000: 15,

 6000: 10,

 7000: 12,

 8000: 6,

 9000: 7,

 10000: 6,

 11000: 3,

 12000: 191,

 13000: 722,

 14000: 170,

 15000: 462,

 16000: 23,

 17000: 30,

 18000: 27,

 19000: 10,

 20000: 12,

 21000: 11,

 22000: 61,

 23000: 65,

 24000: 9,

 25000: 2,

 26000: 2,

 27000: 3,

 28000: 1,

 29000: 4,

 30000: 4,

 31000: 2,

 32000: 2,

 33000: 2,

 34000: 3,

 35000: 3,

 36000: 5,

 37000: 4,

 40000: 2,

 41000: 1,

 42000: 2,

 43000: 2,

 44000: 2,

 48000: 2,

 50000: 3,

 51000: 3,

 52000: 1,

 53000: 3,

 54000: 1,

 55000: 4,

 58000: 1,

 59000: 2,

 61000: 1,

 62000: 4,

 63000: 1,

 84000: 1,

 98000: 1,

 1035000: 1,

 1043000: 1,

 1608000: 1,

 1642000: 1}


"or" operation took: 1455555000 ns

"or" operation time distribution:

{0: 8569860,

 1000: 1428228,

 2000: 131,

 3000: 31,

 4000: 22,

 5000: 8,

 6000: 8,

 7000: 6,

 8000: 3,

 9000: 6,

 10000: 3,

 11000: 4,

 12000: 173,

 13000: 623,

 14000: 174,

 15000: 446,

 16000: 28,

 17000: 22,

 18000: 31,

 19000: 9,

 20000: 11,

 21000: 8,

 22000: 42,

 23000: 72,

 24000: 7,

 25000: 3,

 26000: 1,

 27000: 5,

 28000: 2,

 29000: 2,

 31000: 1,

 33000: 1,

 34000: 2,

 35000: 4,

 36000: 1,

 37000: 1,

 38000: 2,

 41000: 1,

 44000: 1,

 45000: 2,

 46000: 2,

 47000: 2,

 48000: 2,

 49000: 1,

 50000: 1,

 51000: 2,

 53000: 1,

 61000: 1,

 64000: 1,

 65000: 1,

 942000: 1}

我們可以看到大約 85.7% 的嘗試測(cè)量時(shí)間（8565832 / 10000000等于0.8565832和8569860 / 10000000等于0.8569860）都失敗了，因?yàn)樗粶y(cè)量了0納秒。大約 14.3% 的嘗試測(cè)量時(shí)間（1432066 / 10000000等于0.1432066和1428228/10000000等于0.1428228）測(cè)量的是1000納秒。1000而且，不用說，嘗試測(cè)量時(shí)間的其余部分（不到 0.1%）也導(dǎo)致了納秒的銷售。我們可以看到，微秒級(jí)太稀疏，無法衡量每次執(zhí)行所花費(fèi)的時(shí)間。


但是我們?nèi)匀豢梢允褂闷胀ǖ目潭瘸?。通過收集灰塵顆粒并使用尺子測(cè)量灰塵球。所以我們可以試試下面的代碼：


import time as dt


success = True

can_test = True


start = dt.time()

for i in range(10000000):  # getting together the dust particles

    if success and can_test:  # a dust particle

        pass

stop = dt.time()

time = stop - start  # measure the size of the dustball

print(f'"and" operation took: {time} seconds')


start = dt.time()

for i in range(10000000):  # getting together the dust particles

    if success or can_test:  # a dust particle

        pass

stop = dt.time()

time = stop - start  # measure the size of the dustball

print(f'"or" operation took: {time} seconds')

它將打印如下：


"and" operation took: 0.6261420249938965 seconds

"or" operation took: 0.48876094818115234 seconds

或者，我們可以使用一把細(xì)尺 dt.perf_counter()，它可以精確測(cè)量出每一個(gè)灰塵顆粒的大小，如下所示：


import time as dt


success = True

can_test = True


time = 0

for i in range(10000000):

    start = dt.perf_counter()

    if success and can_test:  # a dust particle

        stop = dt.perf_counter()

        time += stop - start  # measured by a fine-scale ruler

print(f'"and" operation took: {time} seconds')


time = 0

for i in range(10000000):

    start = dt.perf_counter()

    if success or can_test:  # a dust particle

        stop = dt.perf_counter()

        time += stop - start  # measured by a fine-scale ruler

print(f'"or" operation took: {time} seconds')

它將打印如下：


"and" operation took: 1.6929048989996773 seconds

"or" operation took: 1.3965214280016083 seconds

當(dāng)然，True or True比True and True！


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