finding the index of a particular point in a matrix.

Hi everyone, I hope you can help me. let's say I have 5by5 matrix. e.g. A=

 0     5     2    18     8
15     0     1     4     1
1     4     0     2     3
6     7     7     0     8
10    11    12     1     0

I want to first find the minimum value vertically, while ignoring the zero, then see if it is the maximum horizontally. for this example, A(3,2)=4 satisfy the conditions. then the final output at the end that I'd like the code to give is the row and column indexes. So for the above example, Idx=3 and Idy =2 . not to mention there could be more than one point that satisfies these conditions in the matrix.

so, to break it down: 1. find the min in columns, while ignoring the zeros. 2. find whether (min part 1) is the max in its row. 3. if yes, output is the x and y indexes e.g. [Idx, Idy].

Thanks a lot in advance

** Really guys big thank you and I am very appreciative.. I see this sentenced caused confusion " not to mention there could be more than one point that satisfies these conditions in the matrix." so, please let me clarify

what I meant is I want the could to output both points not just the 1st time. so let's say that I have 2 points in a matrix satisfy the rules. for example A(3,2) and A(5,1) just example, then [Idx,Idy] should output both. So there cannot be more than one max per row or min per column, however, another point in completely different column and row that apply the rule is a possibility and should be outputted as well. I really apologies for my confusion.

Additionally, since you are awesome people, I'll be a little greedy and add, what about if A now is a 3D matrix with 5by5by10 where this computation is required across the 3D dimension. meaning [Idy,Idy] size each is gby10. g here is # of points that satisfy the conditions.

Again Thank you so much!

[Madhav]

If you assign NaN to your zeroes, A(A==0) = NaN; then min() and max() functions can be used directly on individual rows and columns for your problem.

[Andrei Bobrov]

Hi Omar!

Please write result for matrix:

A = [ 0     4     2    18     8
     15     0     1     4     1
      4     4     0     2     3
      3     4     2     0     4
     10    11    12     1     0];

[Omar Alamoudi]

Hi Andrei,

I get your point completely :) but my matrix A will not have the same value per column/ row because each point in the column represents a connection strength value from j -> I . j here is the column and I is the row. so the outflow goes from j to I and the inflow goes from I to j. I don't bother you more with my project detail. but I assure you that each point in one column/row will have different value.

Answer 1

[Stephen23]

One simple solution:

A = [
   0     5     2    18     8
  15     0     1     4     1
   1     4     0     2     3
   6     7     7     0     8
  10    11    12     1     0];
S = size(A);
B = A + 0./+(A~=0);
% Linear indices of first min value in each column:
[~,idr] = min(B,[],1);
mni = sub2ind(S,idr,1:S(2));
% Linear indices of all max values in each row:
mxi = find(bsxfun(@eq,A,max(B,[],2)));
% Get matching linear indices, convert to row/col:
I = intersect(mni,mxi);
[row,col] = ind2sub(S,I)

giving:

row =  3
col =  2

[Omar Alamoudi]

Thank you! All answers are correct. but because you were the 1st!

can I ask what if I want the analysis to be conducted for a range, let's say the 3 lowest values rather than the abs min. same thing for the max, a range of the highest 3 values. How big of a difference is that? basically I am less strict in my rules allowing a point to be outputted if it is within the range of the 3 lowest/highest values per column/row respectively.

Thank you!

[Stephen23]

Of course you can get the minimum three (or N) values, the easiest way is to use sort and obtain its output indices. Important is to use sort's optional input to specify the dimension to act along. For example:

S = size(A);
B = A + 0./+(A~=0);
% Linear indices of the lowest N values in each column:
N = 3;
[~,idx] = sort(B,1);
min_idr = idx(1:N,:)
min_idc = repmat(1:S(2),N,1)
min_idx = sub2ind(S,min_idr,min_idc)
% Linear indices of the highest N values in each row:
etc.

[Omar Alamoudi]

Very important clarification, I don't need at each row to find the linear indices of the highest N values. I only need to scan the rows that my min values occurred in from the previous step. Here is the summary my problem:

I do need to scan each column for the lowest N values and see at what rows they are and then only scan these row and see if each one of the N values is the maximum (within the highest N values) in their respective row. if previous statement is true for any of the N value, output is the x and y indices of that value.

Can you please show me how to do this? :) Should I make this in a new question? ^_^

[Stephen23]

That is what my code above does. Did you try it? I did the first part for you. The second part is basically the same, just for each row. Then you can use intersect, just as I used in my answer.

"I am aware of the sort function and I was afraid to use it because it misses up my indices which is what I want "

I have no idea why you think that sort will "mess up" any indices. In the code I showed it returns the row indices of the three lowest values (which is apparently what you want), and does not change any of the other matrices. Lets have a look:

>> idx
idx =
   3   3   2   5   2
   4   1   1   3   3
   5   4   4   2   1
   2   5   5   1   4
   1   2   3   4   5

That looks like the correct row indices of the values sorted along each column from lowest to highest. When we take the three first rows of indices then you can get the three lowest values in each column. How exactly is this "messed up" ?

[Omar Alamoudi]

I am sorry I was talking in general that I was avoiding using sort because of that.. but your code is great. I apologies I should have been more clear! can you please show me how to do the 2nd part? Thanks!

[Stephen23]

Try something like this:

S = size(A);
B = A + 0./+(A~=0);
% Linear indices of lowest N values in each column:
N = 2;
[~,idx] = sort(B,1);
min_idr = idx(1:N,:)
min_idc = repmat(1:S(2),N,1)
min_idx = sub2ind(S,min_idr,min_idc)
% Linear indices of the highest N values in each row:
[~,idx] = sort(B,2);
max_idc = idx(:,end-N:end-1)
max_idr = repmat((1:S(1))',1,N)
max_idx = sub2ind(S,max_idr,max_idc)
% Get matching linear indices, convert to row/col:
I = intersect(min_idx(:),max_idx(:));
[row,col] = ind2sub(S,I)

[Omar Alamoudi]

THANK YOU SO MUCH SIR!!

[az]

Hello Stephen Cobeldick,

Suppose I Have a column vector wcich has multiple maxima and minima, I just want to find the index of N number of them ( prominent maxia and minima).

How thw code will work please, I think I able to calculate minima by this code but not maxima.

And if I use something like ;findpeak' then it gives a lots of local max and min as for a noisy signal. I ant to get the prominant max & min indices of the 2nd colm of the attachment.

Thank you

[Walter Roberson]

Threshold

Minimum height difference between a peak and its neighbors, specified as the comma-separated pair consisting of 'Threshold' and a nonnegative real scalar. Use this argument to have findpeaks return only those peaks that exceed their immediate neighboring values by at least the value of 'Threshold'.

MinPeakDistance

Minimum peak separation, specified as the comma-separated pair consisting of 'MinPeakDistance' and a positive real scalar. When you specify a value for 'MinPeakDistance', the algorithm chooses the tallest peak in the signal and ignores all peaks within 'MinPeakDistance' of it. The function then repeats the procedure for the tallest remaining peak and iterates until it runs out of peaks to consider.

Answer 2

[Walter Roberson]

A = [
   0     5     2    18     8
  15     0     1     4     1
   1     4     0     2     3
   6     7     7     0     8
  10    11    12     1     0];

B = A; B(A==0) = NaN;
[mB, midx] = min(B);
col_matches = all(mB.' >= B(midx,:) | isnan(B(midx,:)),2);  %row vector
row_idx = midx(col_matches);  %automatically column vector
col_idx = find(col_matches(:));  %force column vector

[Stephen23]

midx is not defined.

[Walter Roberson]

Good catch, I missed copying back one line.