Hello list, I wonder if anyone might be able to help me troubleshoot an attempt at porting some simple Python code to R.
The function below is supposed to take a matrix containing item ratings from various users and, given a vector containing at least 1 rating and 1 missing value, employ a 'weighted slope one' algorithm to predict the missing values. The algorithm itself is fairly simple and described here:[1] and more formally here: [3]. But comparing my answers to those generated by the Python implementation here [2], I obtain different answers. Unfortunately I don't know enough Python to tell where I might be going wrong. Any suggestions? [1] http://en.wikipedia.org/wiki/Slope_One#Slope_one_collaborative_filtering_for_rated_resources [2] http://www.serpentine.com/wordpress/wp-content/uploads/2006/12/slope_one.py.txt [3] http://www.daniel-lemire.com/fr/documents/publications/lemiremaclachlan_sdm05.pdf # 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])) })) }) # estimates 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 1, using example from [1] john=c(item1=5, item2=3, item3=2) mark=c(item1=3, item2=4, item3=NA) tr.set1=rbind(john,mark) lucy1=c(item1=NA, item2=2, item3=5) pred(tr.set1,lucy1) ## item1=4.33 --> correct # test 2, using example from [2] 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} [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
