Linspace应用于数组

问题描述

给出一个像a = [ -1; 0; 1];这样的数组.对于每个a(i)，我需要使用linspace(min(a(i),0),max(a(i),0),3);计算线性间隔的矢量，其中每个linspace-vector都应存储在矩阵中:

Given an array like a = [ -1; 0; 1];. For each a(i), I need to compute a linearly spaced vector with linspace(min(a(i),0),max(a(i),0),3);, where each linspace-vector should be stored into a matrix:

A = [-1 -0.5 0;
      0   0  0;
      0  0.5 1];

通过for循环，我可以这样做:

With a for loop, I can do this like so:

for i=1:3
    A(i) = linspace(min(a(i),0),max(a(i),0),3);
 end

如何在不使用循环的情况下实现这一目标?

How can I achieve this without using loops?

推荐答案

我能想到的最快方法是计算步长，并使用隐式二进制扩展从中构造向量.

The fastest way I can think of is calculating the step-size, construct the vector from that using implicit binary expansion.

 a = [ -1; 0; 1];
 n = 3;
 stepsizes = (max(a,0)-min(a,0))/(n-1);
 A = min(a,0) + (0:(n-1)).*stepsizes;

时间:

使用timeit(@SO)的几个结果(使用timeit(@SO)并从要计时的块中删除注释):

A couple of timeit results using (use timeit(@SO) and remove comments from the blocks to be timed):

function SO()
n = 1e3;
m = 1e5;
a = randi(9,m,1)-4;

% %Wolfie
% aminmax = [min(a, 0), max(a,0)]';
% A = interp1( [0,1], aminmax, linspace(0,1,n) )';

% %Nicky
% stepsizes = (max(a,0)-min(a,0))/(n-1);
% A = min(a,0) + (0:(n-1)).*stepsizes;

% %Loop
% A = zeros(m,n);
% for i=1:m
%     A(i,:) = linspace(min(a(i),0),max(a(i),0),n);
% end

%Arrayfun:
A = cell2mat(arrayfun(@(x) linspace(min(x,0),max(x,0),n),a,'UniformOutput',false));

那么时间是:

• 狼:2.2243秒
• 矿山:0.3643秒
• 标准循环:1.0953 s
• arrayfun:2.6298 s

• Wolfie: 2.2243 s
• Mine: 0.3643 s
• Standard loop: 1.0953 s
• arrayfun: 2.6298 s

这篇关于Linspace应用于数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！