Weighted slope one algorithm? (porting from Python to R)

prediction, python, r, recommendation-engine

Solution

I used the same reference (Bryan O'Sullivan's python code) to write an R version of Slope One a while back. I'm pasting the code below in case it helps.

predict <- function(userprefs, data.freqs, data.diffs) {
    seen <- names(userprefs)

    preds <- sweep(data.diffs[ , seen, drop=FALSE], 2, userprefs, '+') 
    preds <- preds * data.freqs[ , seen]
    preds <- apply(preds, 1, sum)

    freqs <- apply(data.freqs[ , seen, drop=FALSE], 1, sum)

    unseen <- setdiff(names(preds), seen)
    result <- preds[unseen] / freqs[unseen]
    return(result[is.finite(result)])
}

update <- function(userdata, freqs, diffs) {
    for (ratings in userdata) {
        items <- names(ratings)
        n <- length(ratings)

        ratdiff <- rep(ratings, n) - rep(ratings, rep(n, n))
        diffs[items, items] <- diffs[items, items] + ratdiff

        freqs[items, items] <- freqs[items, items] + 1
    }
    diffs <- diffs / freqs
    return(list(freqs=freqs, diffs=diffs))
}


userdata <- list(alice=c(squid=1.0, cuttlefish=0.5, octopus=0.2),
                 bob=c(squid=1.0, octopus=0.5, nautilus=0.2),
                 carole=c(squid=0.2, octopus=1.0, cuttlefish=0.4, nautilus=0.4),
                 dave=c(cuttlefish=0.9, octopus=0.4, nautilus=0.5))

items <- c('squid', 'cuttlefish', 'nautilus', 'octopus')
n.items <- length(items)
freqs <- diffs <- matrix(0, nrow=n.items, ncol=n.items, dimnames=list(items, items))

result <- update(userdata, freqs, diffs)
print(result$freqs)
print(result$diffs)

userprefs <- c(squid=.4)
predresult <- predict(userprefs, result$freqs, result$diffs)
print(predresult)

Problem

I was reading about the Weighted slope one algorithm ( and more formally here (PDF)) which is supposed to take item ratings from different users and, given a user vector containing at least 1 rating and 1 missing value, predict the missing ratings. I found a Python implementation of the algorithm, but I'm having a hard time porting it to R (which I'm more comfortable with). Below is my attempt. Any suggestions on how to make it work? Thanks in advance, folks. ``` # take a 'training' set, tr.set and a vector with some missing ratings, d pred=function(tr.set,d) { tr.set=rbind(tr.set,d) n.items=ncol(tr.set) # tally frequencies to use as weights freqs=sapply(1:n.items, function(i) { unlist(lapply(1:n.items, function(j) { sum(!(i==j)&!is.na(tr.set[,i])&!is.na(tr.set[,j])) })) }) # estimate product-by-product mean differences in ratings diffs=array(NA, dim=c(n.items,n.items)) diffs=sapply(1:n.items, function(i) { unlist(lapply(1:n.items, function(j) { diffs[j,i]=mean(tr.set[,i]-tr.set[,j],na.rm=T) })) }) # create an output vector with NAs for all the items the user has already rated pred.out=as.numeric(is.na(d)) pred.out[!is.na(d)]=NA a=which(!is.na(pred.out)) b=which(is.na(pred.out)) # calculated the weighted slope one estimate pred.out[a]=sapply(a, function(i) { sum(unlist(lapply(b,function (j) { sum((d[j]+diffs[j,i])*freqs[j,i])/rowSums(freqs)[i] }))) }) names(pred.out)=colnames(tr.set) return(pred.out) } # end function # test, using example from [3] alice=c(squid=1.0, octopus=0.2, cuttlefish=0.5, nautilus=NA) bob=c(squid=1.0, octopus=0.5, cuttlefish=NA, nautilus=0.2) carole=c(squid=0.2, octopus=1.0, cuttlefish=0.4, nautilus=0.4) dave=c(squid=NA, octopus=0.4, cuttlefish=0.9, nautilus=0.5) tr.set2=rbind(alice,bob,carole,dave) lucy2=c(squid=0.4, octopus=NA, cuttlefish=NA, nautilus=NA) pred(tr.set2,lucy2) # not correct # correct(?): {'nautilus': 0.10, 'octopus': 0.23, 'cuttlefish': 0.25} ```

Original source