**Consistently Infrequent » R**, and kindly contributed to R-bloggers)

Earlier this year I came across a rather interesting page about optimisation in R from rwiki. The goal was to find the most efficient code to produce strings which follow the pattern below given a single integer input `n`

:

# n = 4 [1] "i001.002" "i001.003" "i001.004" "i002.003" "i002.004" "i003.004" # n = 5 [1] "i001.002" "i001.003" "i001.004" "i001.005" "i002.003" "i002.004" "i002.005" "i003.004" [9] "i003.005" "i004.005" # n = 6 [1] "i001.002" "i001.003" "i001.004" "i001.005" "i001.006" "i002.003" "i002.004" "i002.005" [9] "i002.006" "i003.004" "i003.005" "i003.006" "i004.005" "i004.006" "i005.006"

From this we can see that the general pattern for `n`

is:

# pseudo code FOR j = 1 to n-1 FOR k = j+1 to n PRINT "i" + format(j, digits = 3) + "." + format(k, digits = 3) k = k + 1 END FOR j = j + 1 END FOR # this produces (n-1)*n/2 strings.

It is rather heart warming to go though that rwiki page and see how we can sequentially optimise the algorithm in R to more efficiently produce the desired string sequence. I learned quite a lot from this page about R and how fun these types of challenges can be!

Looking at the tenth solution, it achieves its speed by recognising that there are a total of n unique strings (e.g. “001″, “002″) to the pattern. These can be assigned to a character vector which we subsequently call by working out the sequence of indices we require and passing them to the character vector:

generateIndex10 <- function(n) { s <- sprintf("%03d", seq_len(n)); paste( "i", rep(s[1:(n-1)], (n-1):1), ".", unlist(lapply(2:n, function(k) s[k:n]), use.names=FALSE), sep="" ) } generateIndex10(7) [1] "i001.002" "i001.003" "i001.004" "i001.005" "i001.006" "i001.007" "i002.003" "i002.004" [9] "i002.005" "i002.006" "i002.007" "i003.004" "i003.005" "i003.006" "i003.007" "i004.005" [17] "i004.006" "i004.007" "i005.006" "i005.007" "i006.007"

Playing around with the solution above, I noticed that a speed up would be possible if the following were implemented:

`rep`

on a string vector is slower than simply using`rep.int`

to work out the indices first and then passing those into the character vector.- Initialise the vectors to the correct size. This seems to produce a speed up though I’m not sure why it would in a vectorised solution. I could see the benefit if I were creating the vectors using an explicit R loop and concatenating the results but that is not the case here.
- The functions could be compiled into bytecodes using the new compiler package (NB given that tenth solution is from 2007, this option would not have been available back then).
- This problem could be set up to run in parallel. I didn’t do this because I’m still within running distance of being an R noob

Given the first three points above, this is the solution I came up with:

generateIndex11 <- function(n) { # initialise vectors len <- (n-1)*n/2 s <- vector(mode = "character", length = n) ind.1 <- vector(mode = "integer", length = len) ind.2 <- vector(mode = "integer", length = len) # set up strings s <- sprintf("%03d", seq_len(n)) # calculate indices ind.1 <- rep.int(1:(n-1), (n-1):1) ind.2 <- unlist(lapply(2:n, ":", n), use.names = FALSE) # paste strings together return(paste("i", s[ind.1], ".", s[ind.2], sep = "")) }

But is it faster? We can benchmark it and see!

# load packages library(compiler) library(rbenchmark) # compile functions compiled_generateIndex10 <- cmpfun(generateIndex10) compiled_generateIndex11 <- cmpfun(generateIndex11) # set size n = 5000 # run benchmark benchmark(generateIndex10(n), compiled_generateIndex10(n), generateIndex11(n), compiled_generateIndex11(n), columns = c("test", "replications", "elapsed", "relative"), order = "relative", replications = 10) ###--- TIMINGS ---### test replications elapsed relative 4 compiled_generateIndex11(n) 10 51.46 1.000000 3 generateIndex11(n) 10 51.67 1.004081 2 compiled_generateIndex10(n) 10 54.35 1.056160 1 generateIndex10(n) 10 61.20 1.189273

So it seems that there is a speed up between `generateIndex10`

and `generateIndex11`

. If we call the benchmark for different values of n and plot using the ggplot2 pack we get the following:

library(ggplot2) get_bm <- function(n) { gc() df = benchmark(generateIndex10(n), compiled_generateIndex10(n), generateIndex11(n), compiled_generateIndex11(n), columns = c("test", "elapsed"), order = "relative", replications = 10) df$n = n return(df) } df.list <- lapply(seq(1000, 9000, by = 250), get_bm) df <- do.call("rbind", df.list) ggplot(df, aes(x=n, y=elapsed, colour=test, shape=test)) + geom_point() + stat_smooth() + xlab("N") + ylab("Time (seconds)")

Which produces the graph below. I don’t know what’s happening at the extreme values but I suspect that it has something to do with memory because on my 8GB machine, the RAM was mostly at 99% for the largest values of n. I also like how stable the compiled versions of the functions are:

In conclusion, whist this compiled eleventh solution is faster for larger values of `n`

, for values below `1000`

the tenth solution is sometimes faster.

Writing this post, I noticed something and actually came up with an even faster twelfth solution. When I get time in the next couple of days I’ll make a post about it and also submit it to rwiki.

Thanks for reading, this was my first proper R blog post, constructive comments are most welcome! Also, please kindly bare in mind that I am not an R expert but rather just trying to improve my knowledge of the language

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