How to find the distance between vectors whose coordinates are stored in different arrays.
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Hi,
I get the error:??? Cell contents assignment to a non-cell array object. at this line in my code:
for k=1:N
A{k}=zeros(length(m),2);
end
That's my first problem. Assuming that I figure out how to solve my first problem, I have another problem: once I have stored values in A{N}, I would like to compute
A{1}(1,:)-A{2}(1,:), A{1}(2,:)-A{2}(2,:), A{1}(3,:)-A{2}(3,:), ..., A{1}(n,:)-A{2}(n,:)
A{2}(1,:)-A{3}(1,:), A{2}(2,:)-A{3}(2,:), A{2}(3,:)-A{3}(3,:), ... , A{2}(n,:)-A{3}(n,:)
A{3}(1,:)-A{4}(1,:), A{3}(2,:)-A{4}(2,:), A{3}(3,:)-A{4}(3,:), ... , A{3}(n,:)-A{4}(n,:)
.
.
.
A{N-1}(1,:)-A{N}(1,:), A{N-1}(2,:)-A{N}(2,:), A{N-1}(3,:)-A{N}(3,:),..., A{N-1}(n,:)-A{N}(n,:)
and store these values into an n X N matrix.
I know I'm gonna find some help here, I always do.
Please help me solve my two problems.
Thanks.
1 件のコメント
Matt J
2013 年 3 月 11 日
Your example doesn't lead to an n X N matrix. Each A{i}(j,:) is a 1x2 vector, so this leads to a N-1 x 2*n matrix.
回答 (1 件)
Matt J
2013 年 3 月 11 日
編集済み: Matt J
2013 年 3 月 11 日
Your first problem.
A=cell(N,1);
for k=1:N
A{k}=zeros(length(m),2);
end
5 件のコメント
Matt J
2013 年 3 月 11 日
Errors like these are a job for the debugger. Set a breakpoint at line 55 in ImagExperiment and rerun your code. When the code halts at line 55, run
z=cellfun(@(c) size(c), A,'uni',0); z{:}
at the K>> prompt and show us the output.
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