# 23 Measuring performance

## 23.1 Profiling

1. 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

1. 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:

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] 1.007e-06
as.numeric(bench_res$mean)[[1]] #> [1] 8.169481e-07 # compare results for x ^ sqrt(0.5) time_powerhalf[["elapsed"]] #> [1] 9.885e-06 as.numeric(bench_res$mean)[[2]]
#> [1] 1.057554e-05
• 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.
1. 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)
y <- 1/2
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
#>           <chr>     <dbl>
#> 1       sqrt(x)  1.000000
#> 2 exp(log(x)/2)  9.108887
#> 3         x^0.5 15.047374
#> 4       x^(1/2) 15.351634

At first, the result seems surprising. The call to exp(log(x)/2), with the most function calls is actually considerably faster than (2) and (3).

## 23.3 Old exercises

1. 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?

2. 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)  718  967.0  1141.70  1127.5  1252.5  2032   100
#>          x^0.5 9265 9654.0 10169.92  9823.0 10100.5 19160   100
#>        x^(1/2) 9506 9822.5 10584.82 10053.0 10281.5 37142   100
#>  exp(log(x)/2) 5627 5889.5  6470.31  6051.0  6217.0 35361   100
3. 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)
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$time), 10)) 4. Q: You can change the units in which the microbenchmark results are expressed with the unit parameter. Use unit = "eps" to show the number of evaluations needed to take 1 second. Repeat the benchmarks above with the eps unit. How does this change your intuition for performance? 1. Q: scan() has the most arguments (21) of any base function. About how much time does it take to make 21 promises each time scan is called? Given a simple input (e.g., scan(text = "1 2 3", quiet = T)) what proportion of the total run time is due to creating those promises? A: According to the textbook every extra argument slows the function down by approximately 20 nanoseconds, which I can’t reproduce on my system: f5 <- function(a = 1, b = 2, c = 4, d = 4, e = 5) NULL f6 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6) NULL f7 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6, g = 7) NULL f8 <- function(a = 1, b = 2, c = 4, d = 4, e = 5, f = 6, g = 7, h = 8) NULL microbenchmark::microbenchmark(f5(), f6(), f7(), f8(), times = 10000) #> Unit: nanoseconds #> expr min lq mean median uq max neval #> f5() 301 316 492.0463 324 364.5 626679 10000 #> f6() 335 352 548.3376 361 417.0 883407 10000 #> f7() 375 391 559.0169 400 473.0 560387 10000 #> f8() 403 426 961.0390 437 544.0 3339123 10000 However, for now we just assume that 20 nanoseconds are correct and in kind of doubt, we recommend to benchmark this value individually. With this assumption we calculate 21 * 20 = 420 nanoseconds of extra time for each call of scan(). For a percentage, we first benchmark a simple call of scan(): (mb_prom <- microbenchmark::microbenchmark( scan(text = "1 2 3", quiet = T), times = 100000, unit = "ns", control = list(warmup = 1000) )) #> Unit: nanoseconds #> expr min lq mean median uq #> scan(text = "1 2 3", quiet = T) 26263 28568 39776.87 31465 35585 #> max neval #> 85847317 1e+05 mb_prom_median <- summary(mb_prom)$median

This lets us calculate, that ~1.33% of the median run time are caused by the extra arguments.

2. Q: Read “Evaluating the Design of the R Language”. What other aspects of the R-language slow it down? Construct microbenchmarks to illustrate.

3. 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?

4. Q: What is the cost of multiple inheritance and multiple dispatch on S4 method dispatch?

5. Q: Why is the cost of name lookup less for functions in the base package?

1. 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?

2. 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.