>From what you are saying isn't the Jaccard distance for the multi-class
case equivalent to the (1-hammingloss). Where the hamming loss is the
average of places where the two vectors are different.
I want to understand what do your examples represent? Could you give an
example where the dimension is y is 2 x 3 because I getting confused on
what the 2 represents, is it the number of columns or number of rows?
--
sp
On Mon, May 9, 2016 at 7:44 PM, Maniteja Nandana <
maniteja.modesty...@gmail.com> wrote:
>
> On 9 May 2016 5:24 pm, "Shishir Pandey" <shishir...@gmail.com> wrote:
> >
> > This is what I am having trouble understanding. What does each dimension
> of the vector represent? I am thinking of it as follows:
> >
> > [label_1, label_2, ..., label_N]
> >
> > a characteristic vector would be something like [1, 1, 0, ..., 1, 0, 0]
> >
> > This represents weather label_i is present in the set or not? In that
> case the answer would be different. A 0 is the two sets would represent
> that the label is not present in either of the sets and hence the union
> would be smaller than the dimension of the vector.
> >
> > --
> > sp
> >
>
> Sorry if I am misunderstanding here but in case you are referring to multi
> label classification here, something like an array of [[0,0],[0,1]] would
> be the value of predicted outputs, but an array of [0,1,3,2] would
> represent multi class output and is used in the example in discussion.
> Regarding the size of union, in multiclass and binary, intersection size
> is the number of times predicted class is same by total number of outputs.
> Here 0 need not mean the absence of class.
> But in multi label 0 means absence of label and in this case jaccard
> similarity is calculated for each output and weighted mean is calculated.
>
> In following, the first output has only one label in ground truth while
> two in prediction. While in the second example has the first output which
> has only one label in both ground truth and prediction.
>
> y_true = [[0, 1], [1, 1]]
> y_pred = [[1, 1], [1, 1]]
> jaccard_similarity_score
> 0.75 #(0.5 + 1)/2
>
> y_true = [[0, 1], [1, 1]]
> y_pred = [[0, 1], [1, 1]]
> jaccard_similarity_score
> 1#(1+1)/2
>
> Hope it helps.
>
> Regards,
> Maniteja.
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