Need help to find min(abs(dL==0)) for every column individually

Hi i have two matrix and they are gradually

L1 =11953x1 double

L2 = 11953x201 double

then i have substracted from L2 to L1 columnwise

dL =L2(:,1:201)- L1;

i have to find the indices of min(abs(dL==0)) for every column and find the value of that indices [minimum absolute values which are near to zero] .and have to put them in a matrix of

X_f = 1x 201

{ i am a beginner , any help will be highly appriciateable}

[Amit Bhowmick]

dL =L2- L1;
 [dL_min,X_f]=min(abs(dL),[],1)

[md mayeen uddin]

this also works !! thanks a lot bro

Answer 1

[Image Analyst]

Try this:

rows = 5;
columns = 8;
L1 = randi(20, rows, 1) % Create sample data.
L2 = randi(20, rows, columns)
dL = L2 - L1
for col = 1 : columns
	% Find min value in this column, and it's row number (index).
	[minValue(col), rowOfMinValue(col)] = min(abs(dL(:, col)));
end
% Report to command window:
rowOfMinValue

Be aware that this code finds the first min in each column only. If you expect the min to occur in more than one row, it will need some modification to give you a cell array or a larger 2-D array.

[md mayeen uddin]

it works . thanks a lot from the core of my heart

Answer 2

[Amit Bhowmick]

Check it out

dL =L2- L1;
 [dL_min,X_f]=min(abs(dL),[],1)