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.

In a paper recently published in the Condor, titled *Evaluating time-removal models for estimating availability of boreal birds during point-count surveys: sample size requirements and model complexity*, we assessed different ways of controlling for point-count duration in bird counts using data from the Boreal Avian Modelling Project. As the title indicates, the paper describes a cost-benefit analysis to make recommendations about when to use different types of the removal model. The paper is open access, so feel free to read the whole paper here).

- Fitting removal models with the detect R package
- Shiny slider examples with the intrval R package
- Phylogeny and species traits predict bird detectability
- PVA: Publication Viability Analysis, round 3
- The progress bar just got a lot cheaper

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