This is not true.
On 01.10.2012 21:52, bangbig wrote: > I think it's better to understand how the RowSimilarityJob gets the result. > For two items, > itemA, 0, 0, a1, a2, a3, 0 > itemB, 0, b1, b2, b3, 0 , 0 > when computing, it just uses the blue parts of the vectors. > the cosine similarity thus is (a1*b2 + a2*b3)/(sqrt(a1*a1 + a2* a2) * > sqrt(b2*b2 + b3*b3)) > 1) if itemA and itemB have just one common word, the result is 1; > 2) if the values of the vectors are almost the same, the value would also be > nearly 1; > and for the two cases above, I think you can consider to use association > rules to consider the problem. > > At 2012-10-01 20:53:16,yamo93 <[email protected]> wrote: >> It seems that RowSimilarityJob does not have the same weakness, but i >> also use CosineSimilarity. Why ? >> >> On 10/01/2012 12:37 PM, Sean Owen wrote: >>> Yes, this is one of the weaknesses of this particular flavor of this >>> particular similarity metric. The more sparse, the worse the problem >>> is in general. There are some band-aid solutions like applying some >>> kind of weight against similarities based on small intersection size. >>> Or you can pretend that missing values are 0 (PreferenceInferrer), >>> which can introduce its own problems, or perhaps some mean value. >>> >>> On Mon, Oct 1, 2012 at 11:32 AM, yamo93 <[email protected]> wrote: >>>> Thanks for replying. >>>> >>>> So, documents with only one word in common have more chance to be similar >>>> than documents with more words in common, right ? >>>> >>>> >>>> >>>> On 10/01/2012 11:28 AM, Sean Owen wrote: >>>>> Similar items, right? You should look at the vectors that have 1.0 >>>>> similarity and see if they are in fact collinear. This is still by far >>>>> the most likely explanation. Remember that the vector similarity is >>>>> computed over elements that exist in both vectors only. They just have >>>>> to have 2 identical values for this to happen. >>>>> >>>>> On Mon, Oct 1, 2012 at 10:25 AM, yamo93 <[email protected]> wrote: >>>>>> For each item, i have 10 recommended items with a value of 1.0. >>>>>> It sounds like a bug somewhere. >>>>>> >>>>>> >>>>>> On 10/01/2012 11:06 AM, Sean Owen wrote: >>>>>>> It's possible this is correct. 1.0 is the maximum similarity and >>>>>>> occurs when two vector are just a scalar multiple of each other (0 >>>>>>> angle between them). It's possible there are several of these, and so >>>>>>> their 1.0 similarities dominate the result. >>>>>>> >>>>>>> On Mon, Oct 1, 2012 at 10:03 AM, yamo93 <[email protected]> wrote: >>>>>>>> I saw something strange : all recommended items, returned by >>>>>>>> mostSimilarItems(), have a value of 1.0. >>>>>>>> Is it normal ? >>>> >> >
