October 15, 2016 Code R pbapply progress bar plyr

As a testament to my obsession with progress bars in R, here is
a quick investigation about the overhead cost of
drawing a progress bar during computations in R.
I compared several approaches including
my **pbapply** and Hadley Wickham’s **plyr**.

Let’s compare the good old `lapply`

function from base R,
a custom-made variant called `lapply_pb`

that was
proposed here, `l_ply`

from the **plyr** package,
and finally `pblapply`

from the **pbapply** package:

```
library(pbapply)
library(plyr)
lapply_pb <- function(X, FUN, ...) {
env <- environment()
pb_Total <- length(X)
counter <- 0
pb <- txtProgressBar(min = 0, max = pb_Total, style = 3)
wrapper <- function(...){
curVal <- get("counter", envir = env)
assign("counter", curVal +1 ,envir = env)
setTxtProgressBar(get("pb", envir = env), curVal + 1)
FUN(...)
}
res <- lapply(X, wrapper, ...)
close(pb)
res
}
f <- function(n, type = "lapply", s = 0.1) {
i <- seq_len(n)
out <- switch(type,
"lapply" = system.time(lapply(i, function(i) Sys.sleep(s))),
"lapply_pb" = system.time(lapply_pb(i, function(i) Sys.sleep(s))),
"l_ply" = system.time(l_ply(i, function(i)
Sys.sleep(s), .progress="text")),
"pblapply" = system.time(pblapply(i, function(i) Sys.sleep(s))))
unname(out["elapsed"] - (n * s))
}
```

Use the function `f`

to run all four variants. The expected run time
is `n * s`

(number of iterations x sleep duration),
therefore we can calculate the overhead from the
return objects as elapsed minus expected. Let’s get some numbers
for a variety of `n`

values and replicated `B`

times
to smooth out the variation:

```
n <- c(10, 100, 1000)
s <- 0.01
B <- 10
x1 <- replicate(B, sapply(n, f, type = "lapply", s = s))
x2 <- replicate(B, sapply(n, f, type = "lapply_pb", s = s))
x3 <- replicate(B, sapply(n, f, type = "l_ply", s = s))
x4 <- replicate(B, sapply(n, f, type = "pblapply", s = s))
m <- cbind(
lapply = rowMeans(x1),
lapply_pb = rowMeans(x2),
l_ply = rowMeans(x3),
pblapply = rowMeans(x4))
op <- par(mfrow=c(1, 2))
matplot(n, m, type = "l", lty = 1, lwd = 3,
ylab = "Overhead (sec)", xlab = "# iterations")
legend("topleft", bty = "n", col = 1:4, lwd = 3, text.col = 1:4,
legend = colnames(m))
matplot(n, m / n, type = "l", lty = 1, lwd = 3,
ylab = "Overhead / # iterations (sec)", xlab = "# iterations")
par(op)
dev.off()
```

The plot tells us that the overhead increases linearly
with the number of iterations when using `lapply`

without progress bar.
All other three approaches show similar patterns to each other
and the overhead is constant: lines are
parallel above 100 iterations after an initial increase.
The per iteration overhead is decreasing, approaching
the `lapply`

line. Note that all the differences are tiny
and there is no practical consequence
for choosing one approach over the other in terms of processing times.
This is good news and another argument for using progress bar
because its usefulness far outweighs the minimal
(<2 seconds here for 1000 iterations)
overhead cost.

As always, suggestions and feature requests are welcome. Leave a comment or visit the GitHub repo.

I moved to Canada in 2008 to start a postdoctoral fellowship with Prof. Subhash Lele at the stats department of the University of Alberta. Subhash at the time just published a paper about a statistical technique called data cloning. Data cloning is a way to use Bayesian MCMC algorithms to do frequentist inference. Yes, you read that right.

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