Determining Ties in Mode of Matrix

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Bradley Flynn
Bradley Flynn 2016-2-13
评论： Bradley Flynn 2016-2-15
Let's say I have a matrix
A = [1 2 3 4
2 3 4 1
1 2 3 4
2 5 5 1]
and I take the mode of A
[M, F, C] = mode(A)
How would I go about selecting only the values in M where there was a tie and then replacing only those values? I've read that
cellfun('length',C) > 1
would tell you what values in M would have a tie, but I'm not sure how to incorporate that into replacing only the certain values where ties occurred.

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回答（1 个）

Image Analyst
Image Analyst 2016-2-13
This will do it:
A = [1 2 3 4
2 3 4 1
1 2 3 4
2 5 5 1]
% Find the mode of A. It will be 1.
modeOfA = mode(A(:))
% Find elements where the mode is
modeLocations = A == modeOfA
% Replace mode with 99
A(modeLocations) = 99
If you want it done all in one cryptic, harder to follow, but compact single line of code, you can do this:
A(A==mode(A(:))) = 99
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Image Analyst
Image Analyst 2016-2-13
Tell us what you would like to see as an output. What do you want to replace the mode value with?
If it's 99, then try this code:
A = [1 2 3 4
2 3 4 1
1 2 3 4
2 5 5 1]
for col = 1 : size(A, 2)
thisColumn = A(:, col);
% Find the mode of A. It will be 1.
modeOfA = mode(thisColumn)
% Find elements where the mode is
modeLocations = thisColumn == modeOfA
% Replace mode with 99
A(modeLocations, col) = 99
end
Bradley Flynn
Bradley Flynn 2016-2-15
I'm trying to code my own knn and I want a good way to resolve ties without a for-loop. I did some thinking and came up with this if you're interested:
[preds, F, C]= mode(class);
if ~(all(cellfun('length',C)==1))
keep = cellfun('length',C)==1;
remove = cellfun('length',C) > 1;
preds = preds.*keep;
updated = remove.*knnclassifier(xTr,yTr,xTe,k-1);
preds = preds + updated;
end

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