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I try to understand item based collaborative filtering by studying the recommenderlab documentation. On page 7 the calculation of expected ratings for items unrated by the user is very nicely illustrated but I cannot reproduce the results.

For item $i_3$ (row three of the matrix) the bold matrix elements are the $k = 3$ other most similar components and $\mathcal{S}(i_3) = \{i_2,i_5,i_8 \}$. The estimated rating for the third product is $$\hat{r}_{u3} = \frac{0.8\cdot 0 +0.4 \cdot 4+0.5 \cdot 5}{0.8+0.4+0.5} = 2.41$$ but it is supposed to equal 4.6.

what am I doing wrong here?

Screenshot of Page 7 of the recommenderlab documentation

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My equation in the vignette of recomenderlab is unfortunately slightly incorrect since the weight is only the sum of the most similar components $S(i)$ for which we have user ratings (i.e., $r_{ai} \ne ?$).

The corrected equation is:

\begin{equation} \hat{r}_{ai} = \frac{1}{\sum_{j \in S(i)\cap \{l\,;\,r_{al} \ne ?\}}{s_{ij}}} \sum_{j \in S(i)\cap \{l\,;\,r_{al} \ne ?\}}{s_{ij} r_{aj}} \end{equation}

This gives: $$ \hat{r}_{a3} = \frac{0.4 \cdot 4+0.5 \cdot 5}{0.4+0.5} = 4.6 $$

Unfortunately, the figure was also quite misleading, so I changed it as well (basically $u_a$ is now where $r_a$ was and vice versa).

Hope I got it right this time! Thanks for spotting the mistake. I will update the vignette for the next version of the package.

-Michael

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  • $\begingroup$ I have a question regarding this calculation. I am trying to build an item based recommendation system. My question is what prediction do you assign if the intersection of the top k items and the user ratings are empty? $\endgroup$
    – RDizzl3
    Commented Sep 11, 2017 at 23:17
  • $\begingroup$ Then you just have no recommendation. In an actual implementation of a recommender system, you probably will resort to recommending popular items instead. $\endgroup$ Commented Sep 13, 2017 at 3:30

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