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

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md mayeen uddin el 3 de Jul. de 2021
Comentada: md mayeen uddin el 3 de Jul. de 2021
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}
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Amit Bhowmick el 3 de Jul. de 2021
dL =L2- L1;
[dL_min,X_f]=min(abs(dL),[],1)
md mayeen uddin el 3 de Jul. de 2021
this also works !! thanks a lot bro

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Image Analyst el 3 de Jul. de 2021
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.
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md mayeen uddin el 3 de Jul. de 2021
it works . thanks a lot from the core of my heart

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### Más respuestas (1)

Amit Bhowmick el 3 de Jul. de 2021
Check it out
dL =L2- L1;
[dL_min,X_f]=min(abs(dL),[],1)
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