问题描述
两个对象可以具有相同的余弦和Tanimoto系数距离测度吗,
Can two objects have identical cosine and Tanimoto coefficient distance measure, where
Tanimoto distance measure, d(x,y) = x.y / (|x|*|x|) + (|y|*|y|)- x*y
和
cosine measure, d(x,y) = x.y /(|x|* |x|) * (|y| *|y|)
推荐答案
Tanimoto相似系数定义了a>(不是不是真实距离度量)通过
The Tanimoto similarity coefficient (which is not a true distance measure) is definedby
d(x,y) = x.y / ((|x|*|x|) + (|y|*|y|)- x.y)
用于位向量x和y.
现在将其与余弦相似度系数
d(x,y) = x.y / (|x| * |y|)
分母之间的区别是x.y
项.如果x.y
为零,则Tanimoto和余弦相似系数将相同.
The denominators differ by a x.y
term. The Tanimoto and cosine similarity coefficients would be the same if x.y
is zero.
在几何上,当且仅当x
和y
垂直时,x.y
为零.
Geometrically, x.y
is zero if and only if x
and y
are perpendicular.
由于x
和y
是位向量(即每个维度中的值只能为0或1),所以x.y
等于零表示
Since x
and y
are bit vectors (i.e. whose values in each dimension can only be 0 or 1), x.y
equalling zero means
x1*y1 + x2*y2 + ... + xn*yn = 0
如果xi * yi = 1 * 1 = 1,则总和为正.为了使总和为零,没有项 xi * yi可以等于1.它们都必须等于0:
If xi*yi = 1*1 = 1, then the whole sum would be positive. For the whole sum to be zero, no term xi*yi can equal 1. They must all equal 0:
所以
x1*y1 = 0
x2*y2 = 0
...
xn*yn = 0
换句话说,如果xi为1,则yi必须为0,反之亦然.
In other words, if xi is 1, then yi must be 0, and vice versa.
因此,有很多例子中Tanimoto相似度等于余弦相似度:
So there are tons of examples where the Tanimoto similarity is equal to the cosine similarity:
x = (0,1,0,1)
y = (1,0,0,0)
例如.
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