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

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Christopher Saltonstall on 25 Oct 2017
Edited: Ashaq Shah on 8 Jan 2022
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?

Kelly Kearney on 25 Oct 2017
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.
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 5 Jul 2021
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

Nathan Hardenberg on 5 Jul 2021
Edited: Nathan Hardenberg on 5 Jul 2021
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
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;
##### 2 CommentsShowHide 1 older comment
Nathan Hardenberg on 5 Jul 2021
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.

Ashaq Shah on 8 Jan 2022
Edited: Ashaq Shah on 8 Jan 2022
vector = [startpoint:step:endpoint]

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