" POW" （电）功能 [英] "pow" (power) function

问题描述

我对那里的数字计算器有几个问题：


是否pow（x，2）简单的方形x，或者它首先计算对数

（如果指数不是整数则必要）？


是否x ** 0.5"使用与sqrt（x）相同的算法，还是使用一些

其他（可能效率较低）基于对数的算法？


谢谢，

Russ

解决方案

Russ写道：

我有一个数字计算器的几个问题：

是否pow（x，2）简单的方形x，或者它首先计算对数
（如果指数不是整数则必要）？

x ** 0.5是否为x ** 0.5。使用与sqrt（x）相同的算法，还是使用一些基于对数的其他（可能效率较低）算法？

可以尝试和timeit
111 ** 111
10736201288847422580121456504669550195985072399422 48048047759111756250761957833470224912261700936346 21466103743092986967777786330067310159463303558666 91009102601778558729553962214205731543706973022937 5357546494103400699864397711L timeit.Timer（ pow（111,111））。timeit（）
40.888447046279907 timeit.Timer（" 111 ** 111"）。timeit（）
39.732122898101807 timeit.Timer（" 111 ** 0.5" ;）。timeit（）
2.0990891456604004 timeit.Timer（" pow（111,0.5）"。）timeit（）
4.1776390075683594 timeit.Timer（" 111 ** 0.3"）。timeit （）
2.3824679851531982 timeit.Timer（" pow（111,0.3）"）。timeit（）

4.29450416 56494141

有趣的结果

似乎**计算速度更快

我没有shure哪个算法，但我假设所有的Python都是，

来调用底层C pow（）函数。


Sam

SchüleDaniel< uv ** @ rz.uni-karlsruhe.de>写道：

>>> timeit.Timer（" 111 ** 0.3"）。timeit（）2.3824679851531982>>> timeit.Timer（" pow（111,0.3）"。）timeit（）

4.2945041656494141

有趣的结果
似乎**计算得更快

也许是111 ** 0.3如果timeit使用eval，则解析速度比pow（111,0.3）快。

此外，pow（）可能会产生更多的子程序调用开销 - 更好地检查

两者的字节码版本。

I have a couple of questions for the number crunchers out there:

Does "pow(x,2)" simply square x, or does it first compute logarithms
(as would be necessary if the exponent were not an integer)?

Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some
other (perhaps less efficient) algorithm based on logarithms?

Thanks,
Russ

解决方案

Russ wrote:

I have a couple of questions for the number crunchers out there:

Does "pow(x,2)" simply square x, or does it first compute logarithms
(as would be necessary if the exponent were not an integer)?

Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some
other (perhaps less efficient) algorithm based on logarithms?

you can try and timeit

111**111 10736201288847422580121456504669550195985072399422 48048047759111756250761957833470224912261700936346 21466103743092986967777786330067310159463303558666 91009102601778558729553962214205731543706973022937 5357546494103400699864397711L timeit.Timer("pow(111,111)").timeit() 40.888447046279907 timeit.Timer("111**111").timeit() 39.732122898101807 timeit.Timer("111**0.5").timeit() 2.0990891456604004 timeit.Timer("pow(111,0.5)").timeit() 4.1776390075683594 timeit.Timer("111**0.3").timeit() 2.3824679851531982 timeit.Timer("pow(111,0.3)").timeit()

4.2945041656494141

interesting result
seems that ** computates faster

I not shure which algorithm,but I am assumeing that all Python does,is
to call the underlying C pow() function.

Sam

Schüle Daniel <uv**@rz.uni-karlsruhe.de> writes:

>>> timeit.Timer("111**0.3").timeit() 2.3824679851531982 >>> timeit.Timer("pow(111,0.3)").timeit()

4.2945041656494141

interesting result
seems that ** computates faster

Maybe "111**0.3" parses faster than pow(111,0.3), if timeit uses eval.
Also, pow() may incur more subroutine call overhead--better check
the bytecode for both versions.

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