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')當我運行上面的 python 程序時，我希望 and 操作比 or 操作慢（因為我了解到短路會減少執行時間）。然而，結果不僅完全相反，而且還在波動。我能理解波動?。ㄒ驗楹笈_進程）?？蔀槭裁唇Y果卻相反！怎么了？這是一個示例結果。"and" operation took: 5.200342893600464 seconds"or" operation took: 5.3243467807769775 seconds
2 回答
萬千封印
TA貢獻1891條經驗 獲得超3個贊
這是一個有趣的問題，所以我決定深入調查您的主要顧慮。
# 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))
pvalue=1.8293386245013954e-103
實際上，與“和”操作相比，“或”具有統計意義并且不同
RISEBY
TA貢獻1856條經驗 獲得超5個贊
當我在我的機器上運行您的代碼時，它有時打印出來的速度也True and True比 快True or True。
出現這種現象的原因是您dt.time()的代碼以“微秒”（即 1000 納秒）為尺度測量時間，但是，這個微秒尺度太稀疏，無法測量每次執行 or所花費的時間。在大多數情況下， or所花費的時間小于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量每個灰塵顆粒并將測量值相加。由于測量誤差巨大，結果失真。
為了進一步調查，如果我們執行下面的代碼（d記錄所花費的時間及其頻率）：
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% 的嘗試測量時間（8565832 / 10000000等于0.8565832和8569860 / 10000000等于0.8569860）都失敗了，因為它只測量了0納秒。大約 14.3% 的嘗試測量時間（1432066 / 10000000等于0.1432066和1428228/10000000等于0.1428228）測量的是1000納秒。1000而且，不用說，嘗試測量時間的其余部分（不到 0.1%）也導致了納秒的銷售。我們可以看到，微秒級太稀疏，無法衡量每次執行所花費的時間。
但是我們仍然可以使用普通的刻度尺。通過收集灰塵顆粒并使用尺子測量灰塵球。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尺 dt.perf_counter()，它可以精確測量出每一個灰塵顆粒的大小，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
當然，True or True比True and True！
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