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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

1 2 3 | 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

1 2 3 4 | 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,

1 2 3 4 5 | 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

1 2 3 4 5 6 7 8 | 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

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

note that it is also possible to use

1 | 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

1 2 3 4 5 6 7 | 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

1 2 3 4 | 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

1 2 3 | 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

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

The create on your side the following objects

1 2 3 4 | 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

1 2 3 4 | 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

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

and do the same with all providers

1 | for(j in 2:m) sumV = sumV+ function2(Xlist[[j]],ylist[[j]], Q2list[[j]]) |

1 2 3 4 | solve(R2) %*% sumV [,1] X1 -17.579095 X2 3.932409 |

which is what we were expecting…

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