I have a problem with R's performance: here is my Script.

The problem is that I need to use it in a base with ~6m of clients (with one linear model per customer) and it's taking to long to process.

Can anyone help me to improve the performance of my script? I think that the problem is in the data.frame rbind function.

  • $\begingroup$ Have you looked at biglm? $\endgroup$ Sep 20, 2018 at 13:36
  • $\begingroup$ On line 26 you are using a "for" loop...In general you should avoid this in R as it slows things down. Try to vectorise your function instead. $\endgroup$
    – bstrain
    Sep 20, 2018 at 14:19
  • $\begingroup$ this is probably also very inefficient: results <- bind_rows(results, row) . You could preallocate a results vector with i elements $\endgroup$
    – knb
    Sep 21, 2018 at 21:35
  • 1
    $\begingroup$ You want to fit 6million regression models, one for each client? That's an trivially parallel problem and the data science approach would be to rent a few thousand Amazon EC2 systems for a short time. $\endgroup$
    – Spacedman
    Sep 26, 2018 at 16:33

1 Answer 1


There's a few problems in your script actually, and growing objects (whether using rbind() or bind_rows() ) seriously messes up your performance. So if you know how many rows you have, start by creating an object with that many rows.

Next, you rely on a data frame. If you just store the info in separate vectors and only combine them in the end, you remove again a ton of overhead. Take a look at following three mockup examples:

# The naive way
f1 <- function(n){
  out <- data.frame(x = numeric(), y = numeric())
  for(i in seq.int(n)){
    tmp <- data.frame(x = rnorm(1), y=rnorm(1))
    out <- bind_rows(out, tmp)
# Preallocating memory
f2 <- function(n){
  out <- data.frame(x = numeric(n), y = numeric(n))
  for(i in seq.int(n)){
    out$x[i] <- rnorm(1)
out$y[i] <- rnorm(1)
# Using vectors
f3 <- function(n){
  outx <- numeric(n)
  outy <- numeric(n)
  for(i in seq.int(n)){
    outx[i] <- rnorm(1)
    outy[i] <- rnorm(1)
  return(data.frame(x = outx, y = outy))

On my machine, I get the following timings:

  columns = c("test","elapsed","relative","replications")
##     test elapsed relative replications
## 1 f1(100)    2.16    27.00          100
## 2 f2(100)    0.34     4.25          100
## 3 f3(100)    0.08     1.00          100

Just changing this can seriously cut your running time already.

  • $\begingroup$ Thanks for your answer!! I'm going to try this solution. $\endgroup$ Oct 22, 2018 at 14:33

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