取矢量的功率有效的方式 [英] efficient way to take powers of a vector

问题描述

我写了一个code，它采用数字勒让德多项式高达一些高n阶。例如：

I wrote a code that numerically uses Legendre polynomials up to some high n-th order. For example:

....
case 8 
p = (6435*x.^8-12012*x.^6+6930*x.^4-1260*x.^2+35)/128; return
case 9 
...

如果向量 X 长此会变慢。我看到有 X ^ 4 和 X之间说一个的性能差异。* X，* X * X ，我想我可以用它来改善我的code。我使用 timeit ，发现了：

If the vectorx is long this can become slow. I saw that there is a performance difference between say x.^4 and x.*x.*x.*x and thought I could use this to improve my code. I've used timeit and found that for:

x=linspace(0,10,1e6);
f1= @() power(x,4)
f2= @() x.4;
f3= @() x.^2.^2
f4= @() x.*x.*x.*x

F4 为更快由因子2 比其他人。然而，当我去 X ^ 6 有之间的差异很小为（x，* X * X）^ 2 和 X。* X，* X，* X，* X * X （而所有其他选项都是慢）。

f4 is faster by a factor 2 than the rest. However when I go to x.^6 there is very little difference between (x.*x.*x).^2 and x.*x.*x.*x.*x.*x (while all other options are slower).

有没有去告诉这将是采取矢量的功率最有效的方法是什么？
你能解释为什么在性能如此大的差异？

Is there away to tell what will be the most efficient way to take a power of a vector? Can you explain why there is such a big difference in performance?

推荐答案

这是不完全回答你的问题，但它可能会解决你的问题：

This is not exactly an answer to your question, but it may solve your problem:

x2 = x.*x; % or x.^2 or power(x,2), whichever is most efficient
p = ((((6435*x2-12012)*x2+6930)*x2-1260)*x2+35)/128

这样，你做电源只有一次，并只与指数2.这招可以应用到所有的勒让德多项式（在奇数阶多项式有一个 X2 是通过替换X ）。

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