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My previous post was explaining how mathematically it was possible to parallelize computation to estimate the parameters of a linear regression. More speficially, we have a matrix $\mathbf{X}$ which is $n\times k$ matrix and $\mathbf{y}$ a $n$-dimensional vector, and we want to compute $\widehat{\mathbf{\beta}}=[\mathbf{X}^T\mathbf{X}]^{-1}\mathbf{X}^T\mathbf{y}$ by spliting the job. Instead of using the $n$ observations, we’ve seen that it was to possible to compute “something” using the first $n_1$ rows, then the next $n_2$ rows, etc. Then, finally, we “aggregate” the $m$ objects created to get our overall estimate.

## Parallelizing on multiple cores

Let us see how it works from a computational point of view, to run each computation on a different core of the machine. Each core will see a slave, computing what we’ve seen in the previous post. Here, the data we use are

y = cars$dist X = data.frame(1,cars$speed)
k = ncol(X)

On my laptop, I have three cores, so we will split it in $m=3$ chunks

library(parallel)
library(pbapply)
ncl = detectCores()-1
cl = makeCluster(ncl)

This is more or less what we will do: we have our dataset, and we split the jobs, We can then create lists containing elements that will be sent to each core, as Ewen suggested,

chunk = function(x,n) split(x, cut(seq_along(x), n, labels = FALSE))
a_parcourir = chunk(seq_len(nrow(X)), ncl)
for(i in 1:length(a_parcourir)) a_parcourir[[i]] = rep(i, length(a_parcourir[[i]]))
Xlist = split(X, unlist(a_parcourir))
ylist = split(y, unlist(a_parcourir))

It is also possible to simplify the QR functions we will use

compute_qr = function(x){
list(Q=qr.Q(qr(as.matrix(x))),R=qr.R(qr(as.matrix(x))))
}
get_Vlist = function(j){
Q3 = QR1[[j]]$Q %*% Q2list[[j]] t(Q3) %*% ylist[[j]] } clusterExport(cl, c("compute_qr", "get_Vlist"), envir=environment()) Then, we can run our functions on each core. The first one is  QR1 = parLapply(cl=cl,Xlist, compute_qr) note that it is also possible to use  QR1 = pblapply(Xlist, compute_qr, cl=cl) which will include a progress bar (that can be nice when the database is rather large). Then use  R1 = pblapply(QR1, function(x) x$R, cl=cl) %>% do.call("rbind", .)
Q1 = qr.Q(qr(as.matrix(R1)))
R2 = qr.R(qr(as.matrix(R1)))
Q2list = split.data.frame(Q1, rep(1:ncl, each=k))
clusterExport(cl, c("QR1", "Q2list", "ylist"), envir=environment())
Vlist = pblapply(1:length(QR1), get_Vlist, cl=cl)
sumV = Reduce('+', Vlist)

and finally the ouput is

solve(R2) %*% sumV
[,1]
X1 -17.579095
X2   3.932409

which is what we were expecting…

## Using multiple sources

In practice, it might also happen that various “servers” have the data, but we cannot get a copy. But it is possible to run some functions on their server, and get some output, that we can use afterwards. Datasets are supposed to be available somewhere. We can send a request, and get a matrix. Then we we aggregate all of them, and send another request. That’s what we will do here. Provider $j$ should run $f_1(\mathbf{X})$ on his part of the data, that function will return $R^{(1)}_j$. More precisely, to the first provider, send

function1 = function(subX){
return(qr.R(qr(as.matrix(subX))))}
R1 = function1(Xlist[])

and actually, send that function to all providers, and aggregate the output

for(j in 2:m) R1 = rbind(R1,function1(Xlist[[j]]))

The create on your side the following objects

Q1 = qr.Q(qr(as.matrix(R1)))
R2 = qr.R(qr(as.matrix(R1)))
Q2list=list()
for(j in 1:m) Q2list[[j]] = Q1[(j-1)*k+1:k,]

Finally, contact one last time the providers, and send one of your objects

function2=function(subX,suby,Q){
Q1=qr.Q(qr(as.matrix(subX)))
Q2=Q
return(t(Q1%*%Q2) %*% suby)}

Provider $j$ should then run $f_2(\mathbf{X},\mathbf{y},Q_j^{(2)})$ on his part of the data, using also $Q_j^{(2)}$ as argument (that we obtained on own side) and that function will return $(\mathbf{Q}^{(2)}_j\mathbf{Q}^{(1)}_j)^{T}_j\mathbf{y}_j$. For instance, ask the first provider to run

sumV = function2(Xlist[],ylist[], Q2list[])

and do the same with all providers

for(j in 2:m) sumV = sumV+ function2(Xlist[[j]],ylist[[j]], Q2list[[j]])
solve(R2) %*% sumV
[,1]
X1 -17.579095
X2   3.932409

which is what we were expecting…