Concatenate matrix numbers linspace

2012 8 13
6 个回答
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0 个投票

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.

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 采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2012-8-13
编辑：Azzi Abdelmalek 2012-8-29

0 个投票

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

6 个评论
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Sean de Wolski
Sean de Wolski 2012-8-14
This won't work if you have duplicate values in x.
Azzi Abdelmalek
Azzi Abdelmalek 2012-8-14
yes Wolski, but i have used "unique" to remove duplication
Sean de Wolski
Sean de Wolski 2012-8-14
I get that. But your answer will fail with x = [0 1 2 3 2 1 0]; where x has unique values!
Matt Fig
Matt Fig 2012-8-14
编辑：Matt Fig 2012-8-14
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
Matt Fig 2012-8-14
编辑：Matt Fig 2012-8-14
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
Azzi Abdelmalek 2012-8-29
sorry, did'nt read your comment, yes what it does is puting 2^n-1 points instead of n

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更多回答（5 个）

Matt Fig
Matt Fig 2012-8-13
编辑：Matt Fig 2012-8-13

0 个投票

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,[])
Amazing Trans
Amazing Trans 2012-8-13

0 个投票

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!
Sean de Wolski
Sean de Wolski 2012-8-13
编辑：Sean de Wolski 2012-8-13

0 个投票

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
Amazing Trans
Amazing Trans 2012-8-14
编辑：Amazing Trans 2012-8-14

0 个投票

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

3 个评论
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Amazing Trans
Amazing Trans 2012-8-14
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
Matt Fig 2012-8-14
Please close out this question by selecting a best answer then post a new question and link back to this one.
Sean de Wolski
Sean de Wolski 2012-8-14
My answer handles your first scenario!

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Amazing Trans
Amazing Trans 2012-8-14

0 个投票

Thanks very much everybody. Problem solved.

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