Ari*_*ide 6 python benchmarking combinations scipy
令我感到困惑的是,combSciPy的功能似乎比天真的Python实现慢.这是解决欧拉项目问题53的两个等效程序的测量时间.
%%timeit
from scipy.misc import comb
result = 0
for n in range(1, 101):
for k in range(1, n + 1):
if comb(n, k) > 1000000:
result += 1
result
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输出:
1 loops, best of 3: 483 ms per loop
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%%timeit
from math import factorial
def comb(n, k):
return factorial(n) / factorial(k) / factorial(n - k)
result = 0
for n in range(1, 101):
for k in range(1, n + 1):
if comb(n, k) > 1000000:
result += 1
result
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输出:
10 loops, best of 3: 86.9 ms per loop
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第二个版本大约快5倍(在两台Mac上测试,python-2.7.9-1,IPython 2.3.1-py27_0).这是combSciPy功能的预期行为(源代码)吗?我究竟做错了什么?
编辑(来自Anaconda 3.7.3-py27_0发行版的SciPy):
import scipy; print scipy.version.version
0.12.0
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编辑(IPython以外的相同区别):
$ time python with_scipy.py
real 0m0.700s
user 0m0.610s
sys 0m0.069s
$ time python without_scipy.py
real 0m0.134s
user 0m0.099s
sys 0m0.015s
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看起来您可能正在错误地运行计时并测量加载scipy到内存中所需的时间.当我跑:
import timeit
from scipy.misc import comb
from math import factorial
def comb2(n, k):
return factorial(n) / factorial(k) / factorial(n - k)
def test():
result = 0
for n in range(1, 101):
for k in range(1, n + 1):
if comb(n, k) > 1000000:
result += 1
def test2():
result = 0
for n in range(1, 101):
for k in range(1, n + 1):
if comb2(n, k) > 1000000:
result += 1
T = timeit.Timer(test)
print T.repeat(3,50)
T2 = timeit.Timer(test2)
print T2.repeat(3,50)
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我明白了:
[2.2370951175689697, 2.2209839820861816, 2.2142510414123535]
[2.136591911315918, 2.138144016265869, 2.1437559127807617]
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这表明scipy 比python版本略快.
回答我自己的问题。似乎SciPy 中同一事物存在两个不同的函数。我不太清楚为什么。但另外一个 ,binom比 快 8.5 倍comb,比我的快 1.5 倍,这让人放心:
%%timeit
from scipy.special import binom
result = 0
for n in range(1, 101):
for k in range(1, n + 1):
if binom(n, k) > 1000000:
result += 1
result
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输出:
10 loops, best of 3: 56.3 ms per loop
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SciPy 0.14.0 的朋友们,这对你们也有用吗?
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