return a single vector after a for loop and not individual results

I am working on a large set of data (600x300) but will illustrate with the reduced set below; For each row, I would like the code to determine the first number less than 5 and return the location of that number (which would be the column number), if there is no number less than 5 like in row 2, it returns nothing. This seems to work fine but unfortunately only returns individual results yet i would want the code to return a vector (in this case 3x1) with all results in one variable. I have tried several options like creating a zero matrix and assigning new results to it among others but have not been successful, shall be grateful for your help.

Divergence=[9,8,5,6,7;10,11,13,15,14;1,2,18,19,10]
for row=1:3
    rowdata=Divergence(row,:)
    place=find(rowdata<=5,1)
 end

[Matt J]

I would like the code to determine the first number less than 5

Here you say "less than", but your code says "less than or equal to".

[Anita Nabatanzi]

or sorry i did mean less than or equal to

Answer 1

[Matt J]

For a fully vectorized solution:

[maxval,place]=max(Divergence<=5,[],2);
place(maxval==0)=nan;

Or, to do the same with a loop,

[m,n]=size(Divergence);

place=nan(m,1); %PREALLOCATE

for row=1:m
    rowdata=Divergence(row,:);
    tmp=find(rowdata<=5,1);
    if ~isempty(tmp)
        place(row)=tmp;
    end
end

[Anita Nabatanzi]

hi Matt, i've added this to the for loop but the solution is incorrect. Ideally for this scenario the vector result would be [3;Nan;1]

[Matt J]

With the first of the two solutions I gave you (which is best for large data sets) there shouldn't be any loop. Those lines of code are the whole thing. Example:

Divergence =

     9     7    12     1    10
    13    13    15     9    12
    14    12     4     8    14
    11     2    10     6    15
     9     7    13    13     6
     9     1     7     4     2
    10     2     7     2     9
     6     6     9    14    12
    15    15    15     6     5
     4    13    15     5    10
     

 >> [maxval,place]=max(Divergence<=5,[],2);  place(maxval==0)=nan

place =

     4
   NaN
     3
     2
   NaN
     2
     2
   NaN
     5
     1

[Anita Nabatanzi]

oh i get you! Thanks, without the for loop it works fine. shall read about the max function then.

Great, from your feedback, I noticed the line i missed out within the for loop was in case the matrix was empty, what would the code do. Do you suggest it wise to carry on with the for loop for my large data or your first lines are a better option. Thanks for the help.

[Matt J]

The vectorized solution will be faster for large data sets. Compare:

Divergence=randi(18,5000,5000);

tic;
[m,n]=size(Divergence);

place=nan(m,1); %PREALLOCATE

for row=1:m
    rowdata=Divergence(row,:);
    tmp=find(rowdata<=5,1);
    if ~isempty(tmp)
        place(row)=tmp;
    end
end
toc %Elapsed time is 0.275612 seconds.

tic
[maxval,place]=max(Divergence<=5,[],2);
place(maxval==0)=nan;
toc %Elapsed time is 0.025761 seconds.

[Anita Nabatanzi]

Thats a significant time lag! Thanks Matt