Concatenate matrix numbers linspace

Hi there,

I have a matrix variable x = [0 1 2 3]

I want to generates linearly spaced vectors in between the numbers into a variable. My problem here is concatenate the numbers into p the next time n increases.

I know i should be using linspace to generate number for eg:

for i = 1:(length(x)-1)
p = linspace(x(i),x(i+1),0.5)
end

the results i want is:

p = 0 0.5 1 1.5 2 2.5 3

Hope someone can shed some light here.

[Azzi Abdelmalek]

what do you mean by

 p = linspace(x(i),x(i+1),0.5)

Answer 1

[Azzi Abdelmalek]

try this

 x = [0.25 1 1.5 2 2.4 2.6]

 x=unique(sort([x x(1:length(x)-1)+diff(x)/2]) )

if you want put 2^n-1 samples between each value use this function

 function y=linspace_n(x,n)
 for k=1:n
 x=unique(sort([x x(1:length(x)-1)+diff(x)/2]) )
 end
 y=x

[Sean de Wolski]

This won't work if you have duplicate values in x.

[Azzi Abdelmalek]

yes Wolski, but i have used "unique" to remove duplication

[Sean de Wolski]

I get that. But your answer will fail with x = [0 1 2 3 2 1 0]; where x has unique values!

[Matt Fig]

Regardless, this function does not do what you say it does, Azzi.

>> x = [1 2 3];y = linspace_n(x,3).' % 3 points between each?

y =

      1.0000
      1.1250
      1.2500
      1.3750
      1.5000
      1.6250
      1.7500
      1.8750
      2.0000
      2.1250
      2.2500
      2.3750
      2.5000
      2.6250
      2.7500
      2.8750
      3.0000

[Matt Fig]

Better to use one of these. First a vectorized version:

function y = linspace_n2(x,N)
% Puts N values linear interpolated between each element of x.
% Author - Matt Fig
L = length(x);
y = bsxfun(@plus,bsxfun(@times,(0:N)',diff(x)/(N+1)),x(1:L-1));
y = [reshape(y,1,(L-1)*(N+1)),x(L)];

Or one could even go with a simplistic:

function y = linspace_n3(x,N)
% Puts N values linear interpolated between each element of x.
% Author - Matt Fig
y = [];
R = (0:N);
D = diff(x)/(N+1);

for ii = 1:length(x)-1
    y = [y x(ii) + R.*D(ii)];
end
y = [y x(end)];

[Azzi Abdelmalek]

sorry, did'nt read your comment, yes what it does is puting 2^n-1 points instead of n

Answer 2

[Matt Fig]

p = 0:.5:3;

or

p = linspace(0,3,7)

EDIT.

I think I misunderstood your problem. Do you mean like this:

x = [0.25 1 1.5 2 2.4 2.6]
x(2,1:end-1) = diff(x)/2+x(1:end-1);
x = reshape(x(1:end-1),1,[])

Answer 3

[Amazing Trans]

Sorry, i had to be more clear. x is not linearly equal, which means x can be this as well

x = [0.25 1 1.5 2 2.4 2.6]

Thanks!

Answer 4

[Sean de Wolski]

Here is a terrible solution:

x = 0:3; %sample x
x = [x(1:end-1); x(2:end)]; %each start/end pair
nAdd = 2; %add two elements between
xnew = interp1((1:2)',x,linspace(1,2,2+nAdd)); %interpolate (2d linspace)
xnew = vertcat(reshape(xnew(1:end-1,:),[],1),x(end)) %keep non-duplicate parts

Answer 5

[Amazing Trans]

Alright here is something more challenging:

If i have x = [0 1 2 4 1]

I want x to be spaced equally n time. n = 2 therefore results = [0 0.25 0.5 1 1.25 1.5 2 2.6667 3.3333 4 3 2 1]

the other function i would like is increase by 0.1 or n where: x = [ 0 1 2] results = [0 0.1 0.2 0.3...0.9 1 1.1 1.2 1.3...1.9 2]

My first initial thought was to use linspace with for loop. and then results is a matrix where when the i increase it will concatenate into the results matrix... I'm not sure on the ??? part. The easier the code the better as I am transferring the matlab code to VB later as well. Of course this code below does not work for results(start).the error i get is: "Subscript indices must either be real positive integers or logicals." Any idea what to modify here?

for eg:

x = [0 3 1]
start = 0;
for i = 1:(length(x)-1)
y = linspace(x(i), x(i+1),points);
results(start) = y
start = start + points + 2;
end

[Amazing Trans]

I guess i will do this for n points in between x1 & x2:

start = 0;
for i = 1: (length(x)-1)
    y=linspace(x(i),x(i+1),n+2)
    for j = 1:length(y)
        results(start+j) = y(j)
    end   
    start = start +length(y) -1
end

[Matt Fig]

Please close out this question by selecting a best answer then post a new question and link back to this one.

[Sean de Wolski]

My answer handles your first scenario!

Answer 6

[Amazing Trans]

Thanks very much everybody. Problem solved.