Madhava–Leibniz系列: 我只需要创建一个从元素k到元素n的子序列: 从这个系列中创建并返回越快越好。你知道吗
我从“莱布尼兹”一线性函数开始,用试错法一步一步地发展到“马达瓦”函数,速度快了一倍。你知道吗
更新:感谢Andrej Kesely的建议和两个破纪录的功能:“madhavaèu leibniz”(比上一个记录提高20%)和“madhavaèu leibnizèu starmap”(另外15%)。你知道吗
最快的时间现在属于一个新的函数“madhava\u leibniz”。这是一个2.8比第一个'莱布尼兹'版本快!你知道吗
现在我的脚本输出:
('3', '7', '3')
macOS version 10.14.5
Darwin-18.6.0-x86_64-i386-64bit
Python ('v3.7.3:ef4ec6ed12', 'Mar 25 2019 16:52:21') Clang 6.0 (clang-600.0.57)
Executing in 64bit
chunks * elements = m * n = 4 * 25,000,000 = 100,000,000
Time Pi 3.141592653589793 Error of calculation function
21.997 3.141592643589326 1.0000467121074053e-08 leibniz
11.489 3.141592643589326 1.0000467121074053e-08 madhava
9.269 3.141592643589326 1.0000467121074053e-08 madhava_leibniz
7.834 3.141592643589326 1.0000467121074053e-08 madhava_leibniz_starmap
时间以秒为单位。你知道吗
我选择了4作为块的数量,所以稍后我可以针对多处理/多线程优化这些函数。n=250\u 000\u 000,但我建议在调试期间设置n=25\u 000\u 000,以缩短测试时间
Pi计算和精度只是为了测试所创建的子序列。时间才是最重要的。你知道吗
我仍然在学习Python,可能错过了一些让它更快的方法。你能推荐一个比“madhavaèu leibniz”更快的版本吗?
剧本:
import math
import time
import platform
import sys
from itertools import cycle, starmap
from operator import truediv
# Functions with '*leibniz*' name must be Pythonic stanza (one-liner)
# but the statement can span multiple lines if over "Maximum Line Length"
# [79 characters in PEP 8 -- Style Guide for Python Code]
def madhava_leibniz_starmap(k, n):
return starmap(truediv, zip(cycle([1, -1] if k & 1 else [1, -1]),
range(2*k+1, 2*n+2, 2)))
def madhava_leibniz(k, n):
return [s / d for s, d in zip(cycle([1, -1] if k & 1 else [1, -1]),
range(2*k+1, 2*n+2, 2))]
def leibniz(k, n):
return [[1.0, -1.0][i % 2] / (2 * i + 1) for i in range(k, n+1)]
# Functions without '*leibniz*' but with 'madhava' pattern in the name are
# optimized for speed any way you like. Can be multiple statements.
def madhava(k, n):
series = [0.0] * (n - k + 1)
first_divisor = 2 * k + 1
last_divisor_plus_1 = 2 * n + 2
i = 0
if k & 1:
for divisor in range(first_divisor, last_divisor_plus_1, 4):
series[i] = -1 / divisor
i += 2
i = 1
for divisor in range(first_divisor + 2, last_divisor_plus_1, 4):
series[i] = 1 / divisor
i += 2
else:
for divisor in range(first_divisor, last_divisor_plus_1, 4):
series[i] = 1 / divisor
i += 2
i = 1
for divisor in range(first_divisor + 2, last_divisor_plus_1, 4):
series[i] = -1 / divisor
i += 2
return series
def report(function, m, n): # Test a function: time and values in data series
t = time.time()
series = []
for i in range(m):
series += function(i * n, (i + 1) * n - 1)
p = sum(series) * 4.0
print('{:6.3f}{:19.15f} {} {}'.format(time.time() - t, p, math.pi-p,
function.__name__))
if len(series) != n * m:
print('Error! actual length {}, requested {}'.format(len(series), n*m))
exit(1)
sign = [1.0, -1.0]
for i, a in enumerate(series):
e = sign[i % 2] / (2*i + 1)
if a != e:
print('Error! @ {} actual value {}, expected {}'.format(i, a, e))
exit(1)
if __name__ == '__main__': # Testing ... #####################################
def main():
print(platform.node())
(mac_ver, _, _) = platform.mac_ver()
print(platform.python_version_tuple())
if mac_ver is not None and mac_ver != "":
print("macOS version", mac_ver)
print(platform.platform())
print("Python", platform.python_build(), platform.python_compiler())
print("Executing in", "64bit" if sys.maxsize > 2 ** 32 else "32bit")
m = 4
n = 25_000_000
print('\nchunks * elements = m * n = {:,} * {:,} = {:,}\n'.format(m, n,
m*n))
print('Time Pi%18.15f Error of calculation function' % math.pi)
for f in [leibniz, madhava, madhava_leibniz, madhava_leibniz_starmap]:
report(f, m, n)
main()
我使用
itertools.cycle
稍微更改了madhava()
函数:我的机器上的原始版本(AMD 2400G,Ubuntu 18.04):
使用
itertools.cycle
的版本:另一个版本,使用
itertools.starmap
和operator.truediv
:在我的电脑上,这会产生:
比列表理解快一秒。你知道吗
相关问题 更多 >
编程相关推荐