Convert for loops into foreach loops - r

Question

I want to make the code below more efficient by using the foreach package. I tried it for a very long time but I don't manage to get the same result as when using the for-loops. I would like to use a nested foreach-loop including parallelization... And as output I would like to have two matrices with dim [R,b1] I would be very grateful for some suggestions!!
n <- c(100, 300, 500)
R <- 100
b0 <- 110
b1 <- seq(0.01, 0.1, length.out = 100)
## all combinations of n and b1
grid <- expand.grid(n, b1)
names(grid) <- c("n", "b1")
calcPower <- function( R, b0, grid) {
cl <- makeCluster(3)
registerDoParallel(cl)
## n and b1 coefficients
n <- grid$n
b1 <- grid$b1
## ensures reproducibility
set.seed(2020)
x <- runif(n, 18, 80)
x.dich <- factor( ifelse( x < median( x), 0, 1))
## enables to store two outputs
solution <- list()
## .options.RNG ensures reproducibility
res <- foreach(i = 1:R, .combine = rbind, .inorder = TRUE, .options.RNG = 666) %dorng% {
p.val <- list()
p.val.d <- list()
for( j in seq_along(b1)) {
y <- b0 + b1[j] * x + rnorm(n, 0, sd = 10)
mod.lm <- lm( y ~ x)
mod.lm.d <- lm( y ~ x.dich)
p.val <- c( p.val, ifelse( summary(mod.lm)$coef[2,4] <= 0.05, 1, 0))
p.val.d <- c( p.val.d, ifelse( summary(mod.lm.d)$coef[2,4] <= 0.05, 1, 0))
}
solution[[1]] <- p.val
solution[[2]] <- p.val.d
return(solution)
}
dp.val <- matrix( unlist(res[,1], use.names = FALSE), R, length(b1), byrow = TRUE)
dp.val.d <- matrix( unlist(res[,2], use.names = FALSE), R, length(b1), byrow = TRUE)
stopCluster(cl)
df <- data.frame(
effectS = b1,
power = apply( dp.val, 2, function(x){ mean(x) * 100}),
power.d = apply( dp.val.d, 2, function(x){ mean(x) * 100}),
n = factor(n))
return(df)
}
## simulation for different n
tmp <- with(grid,
by( grid, n,
calcPower, R = R, b0 = b0))
## combines the 3 results
df.power <- rbind(tmp[[1]], tmp[[2]], tmp[[3]])

Answer 1

I created a foreach loop in following code. There had to be some changes made. It is a lot easier to return a list then a matrix in foreach, since it's combined with rbind. Especially when you want to return multiple ones. My solution here is to save everything in a list and afterwards transform it into a matrix of length 100.
Note: there is one mistake in your code. summary( mod.lm.d)$coef[2,4] does not exist. I changed it to [2]. Adjust to your needing
solution <- list()
df2<-foreach(i = 1:R, .combine = rbind, .inorder=TRUE) %dopar%{
set.seed(i)
p.val <- list()
p.val.d <- list()
counter <- list()
for( j in seq_along(b1)){
x <- sort( runif(n, 18, 80))
x.dich <- factor( ifelse( x < median(x), 0, 1))
y <- b0 + b1[j] * x + rnorm( n, 0, sd = 10)
mod.lm <- lm( y ~ x)
mod.lm.d <- lm( y ~ x.dich)
p.val <- c(p.val, ifelse( summary( mod.lm)$coef[2] <= 0.05, 1, 0))
p.val.d <- c(p.val.d, ifelse( summary( mod.lm.d)$coef[2] <= 0.05, 1, 0))
counter <- c(counter, j)
}
solution[[1]] <- p.val
solution[[2]] <- p.val.d
solution[[3]] <- counter
return(solution)
}
dp.val <- unlist(df2[,1], use.names = FALSE)
dp.val.d <- unlist(df2[,2], use.names = FALSE)
dp.val.matr <- matrix(dp.val, R, length(b1))
dp.val.d.matr <- matrix(dp.val.d, R, length(b1))
stopCluster(cl)
for your comment:
A foreach does work with a normal for loop. Minimal reproducible example:
df<-foreach(i = 1:R, .combine = cbind, .inorder=TRUE) %dopar%{
x <- list()
for(j in 1:3){
x <- c(x,j)
}
return(x)
}