Comparing Vector Elements (of 2 vectors)

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 Kommentare

Matt Kindig
Matt Kindig am 3 Apr. 2013
Bearbeitet: Matt Kindig am 3 Apr. 2013
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
Michael am 10 Jun. 2013
You're correct! I was typing fast and came up with a bad example that I didn't think through properly!
I decided to use loops over each element in the one vector, and loop over the other vector as well! Is there something easier or quicker than this??
Thanks in advance!
Matt Kindig
Matt Kindig am 10 Jun. 2013
Did you see my answer below?

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 Akzeptierte Antwort

Matt Kindig
Matt Kindig am 3 Apr. 2013
Bearbeitet: Matt Kindig am 10 Jun. 2013

0 Stimmen

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 Kommentare

I can't believe I overlooked the most obvious way!
One liner:
[~, comp]= histc(R, P);
Michael
Michael am 13 Jun. 2013
jenyus!

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