Cosine distance range interpretation
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I am trying to use the cosine distance in pdist2. I am confused about it's output. As far as I know it should be between 0 and 1. Since Matlab uses 1-(cosine), then 1 would be the highest variability while 0 would be the lowest. However the output seems to range from 0.5 to 1.5 or something along that!
Can somebody please advise me on how to interpret its output and why ?
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dpb
2013-12-13
Looking at the m-file, it doesn't appear to do what it says, precisely...
...
case 'cos' % Cosine
[X,Y,flag] = normalizeXY(X,Y);
...
case 'cor' % Correlation
X = bsxfun(@minus,X,mean(X,2));
Y = bsxfun(@minus,Y,mean(Y,2));
[X,Y,flag] = normalizeXY(X,Y);
...
case 'spe'
X = tiedrank(X')'; % treat rows as a series
Y = tiedrank(Y')';
X = X - (p+1)/2; % subtract off the (constant) mean
Y = Y - (p+1)/2;
[X,Y,flag] = normalizeXY(X,Y);
...
case {'cos' 'cor' 'spe'} % Cosine, Correlation, Rank Correlation
% This assumes that data have been appropriately preprocessed
for i = 1:ny
d = zeros(nx,1,outClass);
for q = 1:p
d = d + (X(:,q).*Y(i,q));
end
...
There's some other normalization and ordering but no cos() in sight. The difference between the various alternatives seems only in the precondition of the input values before the distance computation for the three cases here.
I don't have time at the moment to try to actually read this more thoroughly; perhaps the above will give you some klews...
采纳的回答
Roger Stafford
2013-12-14
编辑：Roger Stafford
2013-12-14
Your statement, "Since Matlab uses 1-(cosine), then 1 would be the highest variability while 0 would be the lowest", is not true. The cosine difference as defined by matlab can range anywhere between 0 and 2. The cosine of the included angle between two vectors can range from -1 up to +1, so one minus cosine would range from 2 down to 0. The cosine distance would be zero for two vectors pointing in the same direction while it would be 2 if they pointed in opposite directions. The formula for computing the cosine distance between two vectors v1 and v2 is:
d = 1 - dot(v1,v2)/norm(v1)/norm(v2);
Suppose you have v1 = [6,-8], and v2 = [-9,12] which are pointed in exactly opposite directions. The computation would go:
d = 1 - ((6)*(-9)+(-8)*(12))/...
sqrt((6)^2+(-8)^2)/sqrt((-9)^2+(12)^2)
= 1 - (-150)/10/15 = 1 - (-1) = 1 + 1 = 2
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