for loop question, how to put the for loop answer in new matrix

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Dear all, I got some problems with my code. I suppose it is basic. However, I can't find where is the problem. I want to find a position in my matrix. First, I would like to find out the number>=9 positions for each column. When I get the position, I would like to get the nearest one and the far one. And I use k=x1-x2, so I can get the distances for each column. if none of the numbers can fit into the statement( x1-x2 empty vector), then skip. In the end, I want to put the information into a new matrix called 'final'. Thank you for your help!
A = [88,2,77,4,5;6,2,9,5,0;6,7,3,4,5;6,7,8,5,6;7,8,9,10,6;7,8,9,99,15;45,55,2,2,2;67,66,56,87,1];
for j = 1:size(A)

Accepted Answer

dpb on 2 Dec 2021
Edited: dpb on 2 Dec 2021
A number of issues with the above code --
  1. Don't use line as a variable, it's a builtin important plotting function,
  2. size(A) returns a 2-vector, iterating over it is not what you want here, you want only number of columns in A
  3. While not fatal, you didn't preallocate the output array final which is inefficient altho for such small array won't be noticeable.
Try something more like
nC=size(A,2); % number columns in A
magicNo=9; % make variable; don't bury in code so can change easily
res=nan(nC,1); % output vector preallocation
for c=1:nC % iterate over columns
C=find(A(:,c)>=magicNo); % get vector of locations
if isempty(C), continue, end % skip if none found, leaves output NaN
res(c)=range(C); % save the range of found locations
  1 Comment
tengteng QQ
tengteng QQ on 4 Dec 2021
Sorry for replying late. I really appreciate your help.
Though, may I ask you how can I begin to learn coding in Matlab. I am a real beginner and seems my coding is not appropriate? (like it is not a good idea in your view). And I am feeling so struggle when I have an idea but it is not connected with my coding skill.

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