How do I return rows of a matrix based on the value of the maximum element.

Can someone help me? I am using MATLAB R2011b and I am having problems returning rows based on the maximum value in (:, 2:end). I have two .txt files with data in the format:

A =

1901 1.0 0.9 1.4 3.0 3.9 8.3 9.5 9.4 8.8 6.6 4.1 1.7

1902 0.3 2.4 1.5 3.7 5.1 8.3 10.4 10.3 7.6 5.9 2.7 0.7

1903 0.6 0.1 0.7 2.6 4.4 7.6 9.3 10.1 7.9 4.7 -0.7 1.2

B =

1901 5.9 7.0 8.2 10.5 14.3 17.1 17.6 18.6 16.5 13.2 10.2 6.6

1902 5.4 8.6 8.2 13.5 13.8 17.9 18.8 19.3 16.8 12.9 8.9 6.1

1903 5.3 6.1 7.7 11.0 14.3 18.0 17.5 18.1 16.6 11.9 5.8 6.5

If I use:

D = [A(:,1), (B(:,2:13)-A(:,2:13))] and get:

1901 4.9 6.1 6.8 7.5 10.4 8.8 8.1 9.2 7.7 6.6 6.1 4.9

1902 5.1 6.2 6.7 9.8 8.7 9.6 8.4 9.0 9.2 7.0 6.2 5.4

1903 4.7 6.0 7.0 8.4 9.9 10.4 8.2 8.0 8.7 7.2 6.5 5.3

I expect to get:

1903 4.9 6.1 6.8 7.5 10.4 8.8 8.1 9.2 7.7 6.6 6.1 4.9

1905 4.7 6.0 7.0 8.4 9.9 10.4 8.2 8.0 8.7 7.2 6.5 5.3

[DGM]

Your statements of objective are conflicting. You said you want the maximum of (:,2:end) of both arrays, but you're taking the difference instead, and then saying you want the difference of ([1 3],2:end). Do you want the difference or the maximum? Furthermore, you're expecting to get elements in the first column that aren't in either array.

[Fil Okua]

@DGM forgive my confusion. I really want the difference of the two first matrices then return arrays of rows with the maximum value found in the right columns. Notice that the first columns of the first two matrices represent the years and they should be left as is. Could a loop be more appropriate here?

[Scott MacKenzie]

Clear as mud. Like DGM, I read your objective. I also read your clarification and still don't know what you are trying to do.

[DGM]

I still don't understand what you're aiming for. Why is the difference of the 1901 and 1903 datasets now associated with 1903 and 1905 instead? When you say "return arrays of rows with the maximum value found in the right columns" are you talking about the right column of A or B, or the right column of the difference? I'm assuming you mean A or B, since the selected rows don't correspond to the maximal values in the last column of the difference.

If you're looking for the row with the maximal value in the rightmost position, why are there 2 rows instead of one? Are you looking for the N most maximal values? Is N strictly 2 as in this example?

In any case, I doubt that using a loop would be necessary.

Let's say I assume you meant to find the difference of the rows of the arrays corresponding to the two largest values in A(:,end)

A = [1901 1.0 0.9 1.4 3.0 3.9 8.3 9.5 9.4 8.8 6.6 4.1 1.7;
	1902 0.3 2.4 1.5 3.7 5.1 8.3 10.4 10.3 7.6 5.9 2.7 0.7;
	1903 0.6 0.1 0.7 2.6 4.4 7.6 9.3 10.1 7.9 4.7 -0.7 1.2];

B = [1901 5.9 7.0 8.2 10.5 14.3 17.1 17.6 18.6 16.5 13.2 10.2 6.6;
	1902 5.4 8.6 8.2 13.5 13.8 17.9 18.8 19.3 16.8 12.9 8.9 6.1;
	1903 5.3 6.1 7.7 11.0 14.3 18.0 17.5 18.1 16.6 11.9 5.8 6.5];

N = 2; % how many maximal values
[~,idx] = sort(A(:,end),'descend'); % finding maximal values in the end column of A
ddiff = B(idx(1:N),2:end)-A(idx(1:N),2:end); % only taking the difference of those rows

C = [A(idx(1:N),1) ddiff]

Excepting the first column, that would give the result you described.

[Fil Okua]

@Scott Mackenzie I am subtracting matrix A(:,2:13) from B(:,2:13) but leaving the first column, to give me matrix D, notice that matrix D is a constract involving the first column i.e D = (A(:,1), B(:,2:13)-A(:,2:13)). Then I am looking for a maximum value in matrix D(:,2:13) using MaxVal =max(max(D(:,2:13)) then looping through matrix D to look for the MaxVal and returning all rows which hold that MaxVal. There must be a better way of doing this.

[Fil Okua]

@DGM this is a really good start. Let me see if I can replicate what you just said.

[Fil Okua]

@DGM there can be any repeated values.

[maxVal, idx] = max(D(:, 2:end))

Gives me all the maximum values but this exclues the fist column since the year column is not in the calculation.

I can see 2 values in D which are the same and they are the maximums.

[maxVal, idx] = max(max(D(:, 2:end))) will only find the first value. Thats why i was thinking that perhalps a loop may help?

[Image Analyst]

Is this a duplicate of your other conversation in the other thread???

[Fil Okua]

Not quite. The other thread is a bit more specific.

[DGM]

So something like this?

A = [1901 1.0 0.9 1.4 3.0 3.9 8.3 9.5 9.4 8.8 6.6 4.1 1.7;
	1902 0.3 2.4 1.5 3.7 5.1 8.3 10.4 10.3 7.6 5.9 2.7 0.7;
	1903 0.6 0.1 0.7 2.6 4.4 7.6 9.3 10.1 7.9 4.7 -0.7 1.2];

B = [1901 5.9 7.0 8.2 10.5 14.3 17.1 17.6 18.6 16.5 13.2 10.2 6.6;
	1902 5.4 8.6 8.2 13.5 13.8 17.9 18.8 19.3 16.8 12.9 8.9 6.1;
	1903 5.3 6.1 7.7 11.0 14.3 18.0 17.5 18.1 16.6 11.9 5.8 6.5];

D = B(:,2:end)-A(:,2:end);
a = max(D,[],2); 
a = a==max(a); % logical index of all rows which share the maximum value

C = [A(a,1) D(a,:)]

[Fil Okua]

@DGM Thanks this returns the row associated with the first maximum only.

[DGM]

It returns both rows associated with 10.4, which is the global maximum of the data difference.

"I can see 2 values in D which are the same and they are the maximums."

Which is what you were saying.

[Fil Okua]

@DGM thanks. I checked again and again. For some reason MATLAB has rounded up some field which was resulting in my confusion. All now good. Thanks 👍