Concatenate matrix numbers linspace
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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.
采纳的回答
Azzi Abdelmalek
2012-8-13
编辑:Azzi Abdelmalek
2012-8-29
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
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
2012-8-29
sorry, did'nt read your comment, yes what it does is puting 2^n-1 points instead of n
更多回答(5 个)
Sean de Wolski
2012-8-13
编辑:Sean de Wolski
2012-8-13
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
0 个评论
Amazing Trans
2012-8-14
编辑:Amazing Trans
2012-8-14
3 个评论
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.
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