Talk:Slope One

Messy article

Given several others below had the same reaction as I did, I decided to remove the image, and change the n² references. The discussion on algorithmic complexity is almost certainly wrong. Chebuctonian (talk) 03:40, 20 December 2007 (UTC)

This article does not cite relevant sources in its claims that Slope One is better than linear regression -- clearly the context for this matters. Please elaborate more on that and... and everything, specifically the "big picture". Also please cite sources. —The preceding unsigned comment was added by 68.100.224.150 (talk • contribs) 19:23, 13 Jul 2007 (UTC)

Slope one collaborative filtering for rated resources

Hence, given n items, to implement Slope One, all that is needed is to compute and store the average differences and the number of common ratings for each of the n2 pairs of items.

Is it more accurate to say it's actually n(n-1)/2 pairs (might still say it's o(n²) though)?
We don't need to store (item_i,item_i) and (item_i,item_j) = - (item_j,item_i).

Is that correct?

Sounds correct. —Preceding unsigned comment added by 24.37.15.142 (talk) 13:06, 2 September 2007 (UTC)

Algorithmic Complexity of Slope One

Updating the database when a user has already entered x ratings, and enters a new one, requires x time steps.

I don't understand why it's x time steps.
Adding a rating to one item, we need to update the difference average for every pair this item is part of.
So with m users and n items, would it be more like m(n-1) operations? Could you explain why the number of ratings for a given user matters to calculate the average differences of an item?

Answer: If a user didn't rate both i and j, then it contributes nothing to the prediction of ratings on item i from ratings on item j (or to the prediction of ratings on item j from ratings on item i). Hence, suppose that a user has a single rating entered for item i, and he rates something more, say item j, for a total of two rated items, then it contributes only to the prediction of item i from item j (or vice versa). If he rates a third item, say item k, then there is a contribution to the prediction of item j from item k (and vice versa) and to the prediction of item i from item k (and vice versa) and so on. —Preceding unsigned comment added by 24.37.15.142 (talk) 13:10, 2 September 2007 (UTC)

Image totally wrong

Okay, what the hell? This doesn't even add up: (4-3+2(2-3))/(1+2) = -1/3 not 0. And item 1 is rated more on average than item 2... --Tgr 02:04, 5 October 2007 (UTC)

It still doen't make sense, and it has been implemented. I wish I knew what was going on. I have same problem you said, doesn't even add up! --FeralOink (talk) 07:01, 29 December 2012 (UTC)

Indeed... my entire system crashed trying to process the given MySQL/PHP... 6 months of work gone just like that... and I failed my FYP to top it all. Thanks Slope One. —Preceding unsigned comment added by 121.7.129.3 (talk) 15:50, 20 July 2010 (UTC)

Bullshit.