# 23 Measuring performance

## 23.1 Profiling

**Q**: Profile the following function with`torture = TRUE`

. What is surprising? Read the source code of`rm()`

to figure out what’s going on.`f <- function(n = 1e5) { x <- rep(1, n) rm(x) }`

**A**:

## 23.2 Microbenchmarking

**Q**: Instead of using`bench::mark()`

, you could use the built-in function`system.time()`

. But`system.time()`

is much less precise, so you’ll need to repeat each operation many times with a loop, and then divide to find the average time of each operation, as in the code below.`n <- 1e6 system.time(for (i in 1:n) sqrt(x)) / n system.time(for (i in 1:n) x ^ 0.5) / n`

How do the estimates from

`system.time()`

compare to those from`bench::mark()`

? Why are they different?**A**: (TODO: Last part of the quesiton: Why are the results different?)`n <- 1e6 x <- runif(100) bench_res <- bench::mark(sqrt(x), x ^ 0.5) time_sqrt <- system.time(for (i in 1:n) sqrt(x)) / n time_powerhalf <- system.time(for (i in 1:n) x ^ 0.5) / n # compare results for sqrt(x) time_sqrt[["elapsed"]] #> [1] 1.03e-06 as.numeric(bench_res$median)[[1]] #> [1] 6.49e-07 # compare results for x ^ sqrt(0.5) time_powerhalf[["elapsed"]] #> [1] 9.65e-06 as.numeric(bench_res$median)[[2]] #> [1] 9.61e-06`

- both approaches get the order of magnitude right, but the average method is a little faster.
- this is surprising to me, because this approach also get’s the (small) overhead of the
`for`

-loop `system.time`

-approach is only able to return the average execution time, which is not optimal for the skewed distribution of execution times.

**Q**: Here are two other ways to compute the square root of a vector. Which do you think will be fastest? Which will be slowest? Use microbenchmarking to test your answers.`x ^ (1 / 2) exp(log(x) / 2)`

**A**: We’ll use the “bench”-package to estimate the relative execution time of these expression, with the fastest expression standardized to 1.`x <- runif(100) bench::mark(sqrt(x), # (1) x ^ 0.5, # (2) x ^ (1 / 2), # (3) exp(log(x) / 2), # (4) relative = TRUE) %>% dplyr::select(expression, median) %>% dplyr::arrange(median) #> # A tibble: 4 x 2 #> expression median #> <bch:expr> <dbl> #> 1 sqrt(x) 1 #> 2 exp(log(x)/2) 9.20 #> 3 x^0.5 15.0 #> 4 x^(1/2) 15.5`

As supposed,

`exp(log(x)/2)`

needs the longest time to calculate the square root of`x`

.

## 23.3 Old exercises

**Q**: Instead of using`microbenchmark()`

, you could use the built-in function`system.time()`

. But`system.time()`

is much less precise, so you’ll need to repeat each operation many times with a loop, and then divide to find the average time of each operation, as in the code below.`n <- 1:1e6 system.time(for (i in n) sqrt(x)) / length(n) system.time(for (i in n) x ^ 0.5) / length(n)`

How do the estimates from

`system.time()`

compare to those from`microbenchmark()`

? Why are they different?**Q**: Here are two other ways to compute the square root of a vector. Which do you think will be fastest? Which will be slowest? Use microbenchmarking to test your answers.`x ^ (1 / 2) exp(log(x) / 2)`

**A**: The second one looks more complex, but you never know…unless you test it.`x <- runif(100) microbenchmark::microbenchmark( sqrt(x), x ^ 0.5, x ^ (1 / 2), exp(log(x) / 2) ) #> Unit: nanoseconds #> expr min lq mean median uq max neval #> sqrt(x) 650 954 1227 1172 1436 2795 100 #> x^0.5 9277 9616 10563 9998 10326 39721 100 #> x^(1/2) 9462 9862 10217 10076 10362 16313 100 #> exp(log(x)/2) 5661 5914 6390 6234 6436 17884 100`

**Q**: Use microbenchmarking to rank the basic arithmetic operators (`+`

,`-`

,`*`

,`/`

, and`^`

) in terms of their speed. Visualise the results. Compare the speed of arithmetic on integers vs. doubles.**A**: Since I am on a Windows system, where these short execution times are hard to measure, I just ran the following code on a linux and paste the results here:`mb_integer <- microbenchmark::microbenchmark( 1L + 1L, 1L - 1L, 1L * 1L, 1L / 1L, 1L ^ 1L, times = 1000000, control = list(order = "random", warmup = 20000)) mb_double <- microbenchmark::microbenchmark( 1 + 1, 1 - 1, 1 * 1, 1 / 1, 1 ^ 1, times = 1000000, control = list(order = "random", warmup = 20000)) mb_integer # and got the following output: # Unit: nanoseconds # expr min lq mean median uq max neval # 1L + 1L 50 66 96.45262 69 73 7006051 1e+06 # 1L - 1L 52 69 88.76438 71 76 587594 1e+06 # 1L * 1L 51 68 88.51854 70 75 582521 1e+06 # 1L/1L 50 65 94.40669 68 74 7241972 1e+06 # 1L^1L 67 77 102.96209 84 92 574519 1e+06 mb_double # Unit: nanoseconds # expr min lq mean median uq max neval # 1 + 1 48 66 92.44331 69 75 7217242 1e+06 # 1 - 1 50 66 88.13654 68 77 625462 1e+06 # 1 * 1 48 66 135.88379 70 77 42974915 1e+06 # 1/1 48 65 87.11615 69 77 659032 1e+06 # 1^1 79 92 127.07686 103 135 641524 1e+06`

To visualise and compare the results, we make some short spaghetties:

`mb_median <- data.frame(operator = c("+", "-", "*", "/", "^"), int = c(69, 71, 70, 68, 84), # same as mb_integer$median dbl = c(69, 68, 70, 69, 103), # same as mb_double$median stringsAsFactors = FALSE) mb_median <- tidyr::gather(mb_median, type, time, int, dbl) mb_median <- dplyr::mutate(mb_median, type = factor(type, levels = c("int", "dbl"))) library(ggplot2) #> Registered S3 methods overwritten by 'ggplot2': #> method from #> [.quosures rlang #> c.quosures rlang #> print.quosures rlang ggplot(mb_median, aes(x = type, y = time, group = operator, color = operator)) + geom_point(show.legend = FALSE) + geom_line(show.legend = FALSE, size = 1.5) + geom_label(aes(label = operator), show.legend = FALSE) + theme_minimal() + ylab("time in nanoseconds") + theme(axis.title.x = element_blank(), axis.title.y = element_text(size = 14), axis.text.x = element_text(size = 14), axis.text.y = element_text(size = 10)) + scale_y_continuous(breaks = seq(0, max(mb_median$time), 10))`

**Q**: You can change the units in which the microbenchmark results are expressed with the`unit`

parameter. Use`unit = "eps"`

to show the number of evaluations needed to take 1 second. Repeat the benchmarks above with the eps unit. How does this change your intuition for performance?

**Q**:`scan()`

has the most arguments (21) of any base function. About how much time does it take to make 21 promises each time scan is called? Given a simple input (e.g.,`scan(text = "1 2 3", quiet = T)`

) what proportion of the total run time is due to creating those promises?**A**: According to the textbook every extra argument slows the function down by approximately 20 nanoseconds, which I can’t reproduce on my system:`f5 <- function(a = 1, b = 2, c = 4, d = 4, e = 5) NULL f6 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6) NULL f7 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6, g = 7) NULL f8 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6, g = 7, h = 8) NULL microbenchmark::microbenchmark(f5(), f6(), f7(), f8(), times = 10000) #> Unit: nanoseconds #> expr min lq mean median uq max neval #> f5() 309 327 549 335 392 609674 10000 #> f6() 348 365 613 374 435 593795 10000 #> f7() 386 406 693 415 538 597765 10000 #> f8() 415 432 775 440 610 1040266 10000`

However, for now we just assume that 20 nanoseconds are correct and in kind of doubt, we recommend to benchmark this value individually. With this assumption we calculate

`21 * 20 = 420`

nanoseconds of extra time for each call of`scan()`

.For a percentage, we first benchmark a simple call of

`scan()`

:`(mb_prom <- microbenchmark::microbenchmark( scan(text = "1 2 3", quiet = T), times = 100000, unit = "ns", control = list(warmup = 1000) )) #> Unit: nanoseconds #> expr min lq mean median uq max #> scan(text = "1 2 3", quiet = T) 26742 29556 39667 32079 36463 1.16e+08 #> neval #> 1e+05 mb_prom_median <- summary(mb_prom)$median`

This lets us calculate, that ~1.31% of the median run time are caused by the extra arguments.

**Q**: Read “Evaluating the Design of the R Language”. What other aspects of the R-language slow it down? Construct microbenchmarks to illustrate.**Q**: How does the performance of S3 method dispatch change with the length of the class vector? How does performance of S4 method dispatch change with number of superclasses? How about RC?**Q**: What is the cost of multiple inheritance and multiple dispatch on S4 method dispatch?**Q**: Why is the cost of name lookup less for functions in the base package?

**Q**: The performance characteristics of`squish_ife()`

,`squish_p()`

, and`squish_in_place()`

vary considerably with the size of`x`

. Explore the differences. Which sizes lead to the biggest and smallest differences?**Q**: Compare the performance costs of extracting an element from a list, a column from a matrix, and a column from a data frame. Do the same for rows.