[英]Why can cosine similarity between two vectors be negative?
I have 2 vectors with 11 dimentions.我有 2 个 11 维的向量。
a <- c(-0.012813841, -0.024518383, -0.002765056, 0.079496744, 0.063928973,
0.476156960, 0.122111977, 0.322930189, 0.400701256, 0.454048860,
0.525526219)
b <- c(0.64175768, 0.54625694, 0.40728261, 0.24819750, 0.09406221,
0.16681692, -0.04211932, -0.07130129, -0.08182200, -0.08266852,
-0.07215885)
cosine_sim <- cosine(a,b)
which returns:返回:
-0.05397935
I used cosine()
from lsa
package.我使用了
lsa
package 的cosine()
。
for some values i am getting negative cosine_sim like the given one.对于某些值,我会像给定的那样得到负 cosine_sim。 I am not sure how the similarity can be negative.
我不确定相似性如何可能是负面的。 It should be between 0 and 1.
它应该在 0 和 1 之间。
Can anyone explain what is going on here.谁能解释这里发生了什么。
The nice thing about R is that you can often dig into the functions and see for yourself what is going on. R 的好处是您可以经常深入研究函数并亲自查看发生了什么。 If you type
cosine
(without any parentheses, arguments, etc.) then R prints out the body of the function.如果您键入
cosine
(不带任何括号,arguments 等),则 R 会打印出 function 的主体。 Poking through it (which takes some practice), you can see that there is a bunch of machinery for computing the pairwise similarities of the columns of the matrix (ie, the bit wrapped in the if (is.matrix(x) && is.null(y))
condition, but the key line of the function is通过它(需要一些练习),您可以看到有一堆机器用于计算矩阵列的成对相似性(即,包裹在
if (is.matrix(x) && is.null(y))
条件,但 function 的关键行是
crossprod(x, y)/sqrt(crossprod(x) * crossprod(y))
Let's pull this out and apply it to your example:让我们将其提取出来并将其应用于您的示例:
> crossprod(a,b)/sqrt(crossprod(a)*crossprod(b))
[,1]
[1,] -0.05397935
> crossprod(a)
[,1]
[1,] 1
> crossprod(b)
[,1]
[1,] 1
So, you're using vectors that are already normalized, so you just have crossprod
to look at.因此,您使用的是已经标准化的向量,因此您只需查看
crossprod
即可。 In your case this is equivalent to在你的情况下,这相当于
> sum(a*b)
[1] -0.05397935
(for real matrix operations, crossprod
is much more efficient than constructing the equivalent operation by hand). (对于真正的矩阵运算,
crossprod
比手动构造等效运算要高效得多)。
As @Jack Maney's answer says, the dot product of two vectors (which is length(a)*length(b)*cos(a,b)) can be negative...正如@Jack Maney 的回答所说,两个向量的点积(长度(a)*length(b)*cos(a,b))可以是负数......
For what it's worth, I suspect that the cosine
function in lsa
might be more easily/efficiently implemented for matrix arguments as as.dist(crossprod(x))
...对于它的价值,我怀疑
lsa
中的cosine
function 对于矩阵 arguments as.dist(crossprod(x))
可能更容易/更有效地实现...
edit : in comments on a now-deleted answer below, I suggested that the square of the cosine-distance measure might be appropriate if one wants a similarity measure on [0,1] -- this would be analogous to using the coefficient of determination (r^2) rather than the correlation coefficient (r) -- but that it might also be worth going back and thinking more carefully about the purpose/meaning of the similarity measures to be used...编辑:在下面对现在已删除的答案的评论中,我建议如果想要在 [0,1] 上进行相似性测量,余弦距离测量的平方可能是合适的——这类似于使用确定系数(r^2) 而不是相关系数 (r) - 但也可能值得回顾并更仔细地考虑要使用的相似性度量的目的/含义......
The cosine
function returns cosine
function 返回
crossprod(a, b)/sqrt(crossprod(a) * crossprod(b))
In this case, both the terms in the denominator are 1, but crossprod(a, b)
is -0.05.在这种情况下,分母中的两项均为 1,但
crossprod(a, b)
为 -0.05。
The cosine function can take on negative values.余弦 function可以取负值。
While cosine of two vectors can take any value between -1 and +1, cosine similarity (in dicument retreival) used to take values from the [0,1] interval.虽然两个向量的余弦可以取 -1 和 +1 之间的任何值,但余弦相似度(在 dicument retreival 中)用于从 [0,1] 区间取值。 The reason is simple: in the WordxDocument matrix there are no negative values, so the maximum angle of two vectors is 90 degrees, for wich the cosine is 0.
原因很简单:WordxDocument 矩阵中没有负值,所以两个向量的最大夹角为 90 度,余弦为 0。
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