Parallelizing Linear Regression or Using Multiple Sources

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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

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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

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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,

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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

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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

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  QR1 = parLapply(cl=cl,Xlist, compute_qr)

note that it is also possible to use

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  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

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  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

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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

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function1 = function(subX){
return(qr.R(qr(as.matrix(subX))))}
R1 = function1(Xlist[[1]])

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

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

The create on your side the following objects

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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

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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

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sumV = function2(Xlist[[1]],ylist[[1]], Q2list[[1]])

and do the same with all providers

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

which is what we were expecting…

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