How to create array of linearly spaced values from starting and ending points

Hello,

I want to create an array (not vector) of linearly space points from a vector of starting and ending points. For example, I want to do the equivalent of,

go = [1;2;3];  %starting point
st = [2;3;4];  %ending points
nGo = length(go); %number of starting points
nPoints = 10; %number of points in each row
for iGo = 1:nGo
A(iGo,:) = linspace(go,st,nPoints); %create array of linearly spaced rows
end

The problem is that the number of start points in on the order of 10000 and it is part of a fitting routine. So, I need the creation of this matrix to be as efficient as possible using minimal loops. Any suggestions?

Answer 1

[Kelly Kearney]

This method is faster than looping over linspace:

x = linspace(0,1,nPoints);
A2 = go + x.*(st - go);

Here's a timing test for nGo = 10000:

nGo = 10000;
go = randi(100, nGo, 1); 
st = go + randi(100, nGo, 1); 

nPoints = 10; %number of points in each row

tic
for ii = 1:100
    A = nan(nGo, nPoints);
    for iGo = 1:nGo
        A(iGo,:) = linspace(go(iGo),st(iGo),nPoints); %create array of linearly spaced rows
    end
end
t(1) = toc;

tic
for ii = 1:100
    x = linspace(0,1,nPoints);
    A2 = go + x.*(st - go);
end
t(2) = toc;

fprintf('Method 1: %.4f msec\nMethod 2: %.4f msec\nSpeedup: %.2f x\n', t*10, t(1)/t(2));

On my computer, I get:

Method 1: 9.6675 msec
Method 2: 0.0676 msec
Speedup: 143.07 x

Differences between A and A2 are on order of 1e-16.

[Ajay Kumar Sandula]

This method is giving error

[Walter Roberson]

I am not seeing any error when I execute ?

nGo = 10000;
go = randi(100, nGo, 1); 
st = go + randi(100, nGo, 1); 

nPoints = 10; %number of points in each row

tic
for ii = 1:100
    A = nan(nGo, nPoints);
    for iGo = 1:nGo
        A(iGo,:) = linspace(go(iGo),st(iGo),nPoints); %create array of linearly spaced rows
    end
end
t(1) = toc;

tic
for ii = 1:100
    x = linspace(0,1,nPoints);
    A2 = go + x.*(st - go);
end
t(2) = toc;

fprintf('Method 1: %.4f msec\nMethod 2: %.4f msec\nSpeedup: %.2f x\n', t*10, t(1)/t(2));

Method 1: 9.8034 msec
Method 2: 0.1712 msec
Speedup: 57.25 x

Answer 2

[Nathan Hardenberg]

I want to add the same function/functionality for Simulink. Since the given answer does not seem to work with a Simulink Matlab-function I wrote my own code. [The transposing (transpose() or ') in the beginning is nessecary since simulink seems to automatically assume a row vector.]

Since the output vector is of variable size it's important to set the output (y) as variable size an give a maximum size in the Size-field: (See also: https://de.mathworks.com/matlabcentral/answers/654813-how-do-i-resolve-an-error-about-variable-sized-data-in-my-matlab-function-block?s_tid=srchtitle)

Variable names corresponding to post:

go -> start | st -> goal | nPoints -> steps

See Comments for a slightly better Version!!!! Click Here

function y = vec_linspace(start, goal, steps)

start = start';
goal = goal';

x = linspace(0,1,steps);

% difference = (goal - start);
% 
% multip = difference'*x;
% 
% onesvec = ones(1, steps);
% startvec = start' * onesvec;
% 
% y = startvec + multip;

y = start' * ones(1, steps) + (goal - start)'*x;

[Walter Roberson]

Could you confirm that you really want to transpose the original start, and then that you want to transpose again when you use it in calculating y ? Wouldn't you get the same result if you did

function y = vec_linspace(start, goal, steps)

x = linspace(0,1,steps);

% difference = (goal - start);
% 
% multip = difference'*x;
% 
% onesvec = ones(1, steps);
% startvec = start' * onesvec;
% 
% y = startvec + multip;

y = start * ones(1, steps) + (goal - start)*x;

[Nathan Hardenberg]

Hmm, seems you are right! Your version works as well and is also better (less calculations). While debugging I had a lot of issues with matrix multiplications not working, so i probably did unnecessary transposes.

Answer 3

[Ashaq Shah]

vector = [startpoint:step:endpoint]