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

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md mayeen uddin
md mayeen uddin 2021-7-3
评论： md mayeen uddin 2021-7-3
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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采纳的回答

Image Analyst
Image Analyst 2021-7-3
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.

更多回答（1 个）

Amit Bhowmick
Amit Bhowmick 2021-7-3
Check it out
dL =L2- L1;
[dL_min,X_f]=min(abs(dL),[],1)

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