I think the key point here is that these vectors should be logically
thought of as sparse vectors (not sure how they are represented in Mahout).
 If a value in the vector at some position i is empty, it is essentially
not part of the calculation.  And only positions, i, that have a value in
both vectors can be used as part of the calculation for the denominator.

On Wed, Aug 22, 2012 at 2:41 PM, Sean Owen <[email protected]> wrote:

> Depends a bit on what you mean in the example here -- are the 0 values
> observed values, or "null", a lack of an observed value?
>
> If they are really 0, then the implementation will calculate the
> values you listed. But I think you really mean the input is...
>
> v1=[1.0, 1.0, 1.0,   ]
> v2=[     , 1.0, 1.0,   ]
> v3=[1.0,   ,   ,   ]
> v4=[1.0,1.0,1.0,1.0]
> v5=[1.0,1.0,1.0,1.0]
>
> You can't assume the missing values are 0 in general. That may make
> sense in some cases, but, for example, if your values are ratings on a
> scale of 1 to 5 this amounts to assuming that all unrated items are
> completely hated. The results will be nonsense.
>
> (Really this isn't the right example to truly illustrate that, try a
> dummy data set pretending that these are 1- to 5-star movie ratings
> and I think you'll see the similarities that result from assuming
> they're 0 don't make intuitive sense.
>
> If you want this behavior, to assume null == 0, that's what the
> PreferenceInferrer is for. You can inject any default you want, the
> one that makes sense for the data set.
>
> On Wed, Aug 22, 2012 at 5:31 PM, Francis Kelly <[email protected]>
> wrote:
> > Thanks very much for you quick - I really appreciate it!
> >
> > My follow-up question is an attempt to better understand this choice
> > of implementation.
> >
> > To take a concrete example, let's suppose that we have a system with 4
> > users, so item vectors are 4 dimensional and we have the following 5
> > vectors (I realize this is completely pathological example, but bear
> > with me). We have:
> >
> > v1=[1.0, 1.0, 1.0, 0]
> > v2=[0.0, 1.0, 1.0, 0]
> > v3=[1.0, 0, 0, 0]
> > v4=[1.0,1.0,1.0,1.0]
> > v5=[1.0,1.0,1.0,1.0]
> >
> > As I understand it, then, in the definition of
> > UncenteredCosineSimilarity, the Cosine Similarity between all of the
> > above would be 1.0.
> >
> > Whereas in the "traditional" definition of Cosine Similarity, we'd
> > have the following correlation values:
> > cs(v1,v2)=0.816
> > cs(v1,v3)=0.577
> > cs(v1,v4)=0.866
> > cs(v4,v5)=1.0
> >
> > Assuming I'm correct to this point, could you elaborate a little bit
> > on the rationale behind this choice? It would seem to me that, for
> > example, v1 and v2 are "more similar" (with 2 ratings in common) than
> > v1 and v3 (with just 1 rating in common). But obviously, you've
> > thought of already, so I'm curious to understand what I'm missing
> > here. I'm guessing it has something to do with your comment that the
> > calculation "is only going to make sense if the data indeed has a mean
> > of zero by nature."
> >
> > Thanks for your time on this question and all of your efforts on
> > Mahout -- it's a great project.
> >
> > best,
> > Francis
> >
> > On Wed, Aug 22, 2012 at 5:11 PM, Sean Owen <[email protected]> wrote:
> >> The similarity is only defined over the dimensions where both series
> >> have a value, yes. So the denominator and numerator are equal in this
> >> case, giving a cosine of 1, which is right in the sense that in 1D
> >> space the cosine must be 1 or -1; two vectors can only point in
> >> exactly the same or exactly opposite directions. The calculation
> >> you're trying is equivalent to pretending that the dimensions with no
> >> value have value 0.0. That is only going to make sense if the data
> >> indeed has a mean of zero by nature.
> >>
> >> On Wed, Aug 22, 2012 at 12:27 PM, Francis Kelly <
> [email protected]> wrote:
> >>> I'm writing with a question about the UncenteredCosineSimilarity
> >>> metric in Mahout 0.7 (in the context of a
> >>> GenericItemBasedRecommender).
> >>>
> >>> I'm getting a correlation value that I don't understand and I'm hoping
> >>> that someone can explain it to me.
> >>>
> >>> When I step through the code with a debugger, I find that when I'm
> >>> comparing two items in AbstractItemSimilarity.com at lines 265-266, we
> >>> have:
> >>>
> >>> PreferenceArray xPrefs = dataModel.getPreferencesForItem(itemID1);
> >>> PreferenceArray yPrefs = dataModel.getPreferencesForItem(itemID2);
> >>>
> >>> Upon inspection, we see the following vectors:
> >>>
> >>> xPrefs=GenericItemPreferenceArray[itemID:6,{1=0.31,3=0.49,4=0.62}]
> >>> yPrefs=GenericItemPreferenceArray[itemID:7,{2=0.43,4=0.21,5=0.52}].
> >>>
> >>> My understanding of the Cosine Similarity metric is that we take the
> >>> dot product of the vectors and divide it by the product of the
> >>> vectors' lengths. Assuming that's the case, we should have a
> >>> denominator of 0.62 * 0.21 = 0.13 because the above vectors only
> >>> overlap for userid=4. For the denominator -- and this is where the
> >>> code is confusing me -- I would assume that we would have the product
> >>> of the first vector length (sqrt(0.31^2 + 0.49^2 + 0.62^2) = 0.84) and
> >>> the second (sqrt(0.43^2 + 0.21^2 + 0.52^2)= 0.70).
> >>>
> >>> The code, however, appears only to consider the places the vectors
> >>> overlap (in other words, userid=4) to compute the lengths. Thus, when
> >>> I find myself at line 332:
> >>>
> >>> result = computeResult(count, sumXY, sumX2, sumY2, sumXYdiff2);
> >>>
> >>> I find that sumX2 = 0.38 = 0.62^2 and sumY2 = 0.044 = 0.21^2. In other
> >>> words, sumX2 only considers the value for userid=4 and, sumY2 only
> >>> considers the value for userid=4 and not all values in each vector.
> >>>
> >>> And, indeed, following the code through the ultimate result it
> >>> produces is a correlation value of 1.0 for these vectors: 0.62*0.21 /
> >>> (sqrt(0.62^2)*sqrt(0.21^2)). I would have computed a correlation value
> >>> of 0.13/(0.84 * 0.70) = 0.21. If someone could explain the discrepancy
> >>> to me I'd be extremely grateful.
> >>>
> >>>
> >>> Thanks in advance,
> >>> Francis
>



-- 

Thanks,
John C

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