Comparing Vector Elements (of 2 vectors)
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Hello,
I have two vectors (one with random numbers 0-1, and the other which is essentially a set of strictly increasing numbers--up to just above 1)... (Maybe this isn't the best way to explain my problem but what I"ve said is true)
It'll be easier to give an example (for me) than to try to form my question more rigorously, so excuse my rough edges for a sec here..
Imagine two vectors which we'd like to compare the elements of:
P = [1 1.5 3 4 5.5 6 7 8 9 10 ];
R = [0.1014 1.6421 9.0088 5.7822 5.2368 0.3141 8.8437 3.1157 1.8910 2.9378];
I would like a new matrix which gives the index value for each time something like: P(i) <R(Any element)<P(i+1)...
The result for above would be:
comp = [ 1 2 9 5 5 1 8 3 2 2] I picked the index of P where R was ~(just above) P...
If this is unclear (which it totally is, I'm sorry) please ask me to clarify.
Many thanks, Michael
3 个评论
Matt Kindig
2013-4-3
编辑：Matt Kindig
2013-4-3
Why is comp(1)=1, when R(1)=0.1014 is less than P(1)?
Similarly, why is comp(5)=5, not 4, since R(5)=5.2368 > P(4) but is < P(5)?
Michael
2013-6-10
Matt Kindig
2013-6-10
Did you see my answer below?
采纳的回答
Matt Kindig
2013-4-3
编辑：Matt Kindig
2013-6-10
I think this should do what you want:
Rc = mat2cell(R, 1, ones(size(R)));
comp = cellfun(@(x) find(P<x,1,'last'), Rc, 'UniformOutput', false);
tf = cellfun(@isempty, comp, 'UniformOutput', true); %get empties
comp(tf) = {0}; %default value if none found
comp = cell2mat(comp);
Or an even more direct way:
Rc = mat2cell(R, 1, ones(size(R)));
comp = cell2mat(cellfun(@(x) sum(P<x), Rc, 'UniformOutput', false));
Note that both of these assume P is sorted (which is, of course, the only way your question makes sense.).
2 个评论
Matt Kindig
2013-6-10
I can't believe I overlooked the most obvious way!
One liner:
[~, comp]= histc(R, P);
Michael
2013-6-13
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