Row-by-row analysis after establishing a threshold value

Hi community,

I have a matrix of 3 columns X a lot of rows. My goal is to do an analysis of each row depending on a general threshold value. Values contained in cells ranged from 0 to 1, and I want to find in which of these 3 columns is the higher value, with the condition that the higher value must be over the threshold value. The analysis I plan to do need to test several possible threshold values.

However, there are a couple of situations that may occur:

1) All values are under the threshold value: In this case I need a zero as output. 2) There is more than one value over the threshold value: I need a zero as output as well.

For example, given a threshold value of 0.5, applied to this matrix:

[0.55 0.45 0.55;0.65 0.75 0.85;0.35 0.95 0.45;0.85 0.15 0.25;0.35 0.25 0.15;0.45 0.45 0.35]

The output expected is:

[0;0;2;1;0;0]

In the first two rows there is a "multiple assignment" situation (there is more than one value over the threshold value), then a 0 is the final assignment. Rows 3 and 4 have only one value over threshold value, and they are located in columns 2 and 1, respectively. Finally, rows 5 and 6 contain all values under threshold value, in this manner a 0 must be assigned (this would be an "unassigned" situation).

Thank you very much for your collaboration!!!!

[madhan ravi]

What are the threshold values?

[Diego Gómez]

It is each value that is intended to be tested. The values contained in cells ranged from 0 to 1, so the different values that are intended to be tested are between 0 and 1 too. Thanks!!

Answer 1

[Guillaume]

Another way to obtain the same:

A = [0.55 0.45 0.55;0.65 0.75 0.85;0.35 0.95 0.45;0.85 0.15 0.25;0.35 0.25 0.15;0.45 0.45 0.35];
t = 0.5;

abovethresh = A >= t;
[~, idx] = max(abovethresh, [], 2);
idx = idx .* (sum(abovethresh, 2) == 1)

[Diego Gómez]

It works wonderful Guillaume. Thank you very much!!!!

Answer 2

[dpb]

A=[0.55 0.45 0.55;0.65 0.75 0.85;0.35 0.95 0.45;0.85 0.15 0.25;0.35 0.25 0.15;0.45 0.45 0.35];
T=0.5;
iUnOv=all(A<T,2) | sum(A>T,2)>1;     % those that aren't to consider in max()
A(iUnOv,:)=0;                        % clear them
[~,B(~iUnOv)]=max(A(~iUnOv,:),[],2);     % find max location for the others

>> B
B =
     0
     0
     2
     1
     0
     0

Look up "logical addressing"...

NB: The above excludes ==T(hreshold) from both tests. Include the '=' in the appropriate test condition if is to be inclusive on either side.

[Diego Gómez]

Hi dpb. Thank you very much for your answer. One observation: I´m having a problem regarding the ix input in the last row (Undefined function or variable ix). I guess probably I´m doing something wrong but now is not working this last command. Any help would be very valuable. Thank you very much.

[dpb]

Oh, I recast two variables into one when I posted the Answer from initial code at command line and didn't follow thru completely, my bad.

I fixed the LH side, but not the RH side. What was ix is iUnOv now. Corrected the Answer, too...