Related
I want to retrieve the sequence of consecutive blanks in a string. Like :
mystring="lalalal lalalal lalala lalalala "
retrieve_sequence_of_consecutive_blanks(mystring)
[1] 2 1 3 1
Actually, I got a solution, with this
sequence_of_blanks=function(vectors_of_strings){
tokens=strsplit(x = mystring,split = "",fixed = TRUE)
sequence=lapply(X = tokens,FUN = rle)
resultats=lapply(sequence, function(item){
resultats=item$lengths[which(item$values==" ")]
})
}
My question is about performance, do you think if there is better way to do it? What about a regex solution? What about a python solution?
You may match all space chunks and get their lengths, e.g.
library(stringr)
nchar(unlist(str_extract_all(mystring, " +")))
Or the base R equivalent:
nchar(unlist(regmatches(mystring, gregexpr(" +", mystring))))
Both yield
[1] 2 1 3 1
In Python, you may use
[x.count(" ") for x in re.findall(" +", mystring)]
See the Python demo
If you plan to count any whitespace, replace the literal space with \s. Tweak as per your further requirements.
You could use
myrle <- rle(charToRaw(mystring) == charToRaw(" "))
myrle$lengths[myrle$values]
which is a bit faster:
microbenchmark::microbenchmark(
OP = sequence_of_blanks(mystring),
akrun = tabulate(cumsum(c(TRUE, diff(str_locate_all(mystring, " ")[[1]][,2]) !=1))),
wiktor = nchar(unlist(str_extract_all(mystring, " +"))),
# charToRaw(mystring) == charToRaw(" "),
fprive = { myrle <- rle(charToRaw(mystring) == charToRaw(" ")); myrle$lengths[myrle$values] }
)
Unit: microseconds
expr min lq mean median uq max neval
OP 32.826 37.680 42.97734 42.3940 46.3405 115.239 100
akrun 40.718 44.874 48.40903 48.4360 50.7050 78.991 100
wiktor 24.166 29.753 34.73199 35.0955 36.7370 129.626 100
fprive 23.258 25.877 29.50010 28.6000 31.6730 43.721 100
If you really need performance, designing some Rcpp function for your particular use giving as arguments charToRaw(mystring) and charToRaw(" ") would improve performance.
If you want a bit more of performance using simple base R:
length_seq_blanks <- function(string) {
x <- nchar(unlist(strsplit(string, "[a-z]+")))
x[x > 0]
}
length_seq_blanks(mystring)
[1] 2 1 3 1
Benchmark
microbenchmark::microbenchmark(
snoram = {
length_seq_blanks <- function(string) {
x <- nchar(unlist(strsplit(string, "[a-z]+")))
x[x > 0]
}
length_seq_blanks(mystring)
},
fprive = {
myrle <- rle(charToRaw(mystring) == charToRaw(" "))
myrle$lengths[myrle$values]
},
unit = "relative"
)
Unit: relative
expr min lq mean median uq max neval
snoram 1.000000 1.000000 1.000000 1.000000 1.000000 1.00000 100
fprive 1.866597 1.818247 1.734015 1.684211 1.634093 1.20812 100
Is there a more efficient way to count the most frequently appearing digit in a string? My R code below calls gsub() 10 times for each string; and I have gazillions of strings to process.
> txt = 'wow:011 test 234567, abc=8951111111111aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
> max(vapply(0:9, function(i) nchar(gsub(paste0('[^',i,']'), '', txt)), integer(1L)))
[1] 12
I don't care about the digit itself. I just want the count of the most frequent one.
I would prefer to use R's core packages, unless some external package offers a significant outperformance. I use x64 R version 3.4.1 (2017-06-30) on Windows 10.
UPDATE:
Here is the (apples-to-apples) performance comparison of excellent suggestions below.
> microbenchmark(
+ original = max(vapply(0:9, function(i) nchar(gsub(paste0('[^',i,']'), '', s)), integer(1L))),
+ strsplit = max(table(unlist(strsplit(gsub("\\D+", "", s), "")))),
+ gregexpr = max(vapply(0:9, function(d) sum(unlist(gregexpr(d, s)) > 0), integer(1L))),
+ stringi = max(vapply(0:9, function(x) stri_count_fixed(s, x), integer(1L))),
+ raw=max(vapply(0x30:0x39, function(x) sum(charToRaw(s)==x), integer(1L))),
+ tabulate = max(tabulate(as.integer(charToRaw(paste('a',s))))[48:57]),
+ times=1000L)
Unit: microseconds
expr min lq mean median uq max neval
original 476.172 536.9770 567.86559 554.8600 580.0530 8054.805 1000
strsplit 366.071 422.3660 448.69815 445.3810 469.6410 798.389 1000
gregexpr 302.622 345.2325 423.08347 360.3170 378.0455 9082.416 1000
stringi 112.589 135.2940 149.82411 144.6245 155.1990 3910.770 1000
raw 58.161 71.5340 83.57614 77.1330 82.1090 6249.642 1000
tabulate 18.039 29.8575 35.20816 36.3890 40.7430 72.779 1000
Why the weird calculation?
This odd formula helps identify some plainly-looking fake identifiers entered by the user. For example, some non-creative users (I'm a guilty one as well) fill out same digits for their phone numbers. Frequently, in data analysis, it would be better to have no phone number at all than a fake phone number that changes from one dataset to another. Naturally, if there is a check-digit, it would be an additional easy validation.
max(table(unlist(strsplit(gsub("\\D+", "", txt), ""))))
#OR
max(sapply(0:9, function(d) sum(unlist(gregexpr(d, txt)) > 0)))
#[1] 12
Or if you do care about the digit
with(rle(sort(unlist(strsplit(gsub("\\D+", "", txt), "")))),
setNames(c(max(lengths)), values[which.max(lengths)]))
# 1
#12
library(microbenchmark)
set.seed(42)
t = paste(sample(c(letters, 0:9), 1e5, TRUE), collapse = "")
microbenchmark(original = max(sapply(0:9, function(i) nchar(gsub(paste0('[^',i,']'), '', t)))),
strsplit = max(table(unlist(strsplit(gsub("\\D+", "", t), "")))),
gregexpr = max(sapply(0:9, function(d) sum(unlist(gregexpr(d, t)) > 0))))
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# original 215.371764 220.862807 233.368696 228.757529 239.809292 308.94393 100 c
# strsplit 11.224226 11.856327 12.956749 12.320586 12.893789 30.61072 100 b
# gregexpr 7.542871 7.958818 8.680391 8.302971 8.728735 13.79921 100 a
Using charToRaw to count digits in string:
# To count only digits in string, filter out ASCii codes for numbers from 0 to 9 which is 48 to 57 according to https://ascii.cl/
# You need to add na.rm = TRUE in case some of your strings contain only one digit
txt = 'wow:011 test 234567, abc=8951111111111aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
max(tabulate(as.integer(charToRaw(txt)))[48:57], na.rm = TRUE)
#[1] 12
txt='22222222222'
max(tabulate(as.integer(charToRaw(txt)))[48:57], na.rm = TRUE)
#[1] 11
#Andrew already did benchmarking test which proves that using charToRaw is fastest approach to count digits in string.
If you do not care about the digit and just want to count most frequent character/digit then you just remove filtering ASCII codes [48:57].
txt = 'wow:011 test 234567, abc=8951111111111aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa'
max(tabulate(as.integer(charToRaw(txt))))
#[1] 32
txt='22222222222'
max(tabulate(as.integer(charToRaw(txt))))
#[1] 11
Building on Santosh's approach, this is significantly faster than the other options...
max(tabulate(as.integer(charToRaw(txt)))[48:57]) #48:57 picks out ASCII digits
library(microbenchmark)
set.seed(42)
t = paste(sample(c(letters, 0:9), 1e5, TRUE), collapse = "")
microbenchmark(original = max(sapply(0:9, function(i) nchar(gsub(paste0('[^',i,']'), '', t)))),
strsplit = max(table(unlist(strsplit(gsub("\\D+", "", t), "")))),
gregexpr = max(sapply(0:9, function(d) sum(unlist(gregexpr(d, t)) > 0))),
tabulate = max(tabulate(as.integer(charToRaw(t)))[48:57]))
Unit: milliseconds
expr min lq mean median uq max neval
original 807.947235 860.112901 1169.744733 935.169003 1154.057709 3513.1401 100
strsplit 34.100444 36.453163 55.457896 42.881400 58.208820 390.1453 100
gregexpr 27.205510 29.333569 42.616817 33.146572 49.840566 246.9001 100
tabulate 1.189702 1.208321 2.150022 1.226319 1.297068 37.4300 100
I have a character vector of stock tickers where the ticker name is concatenated to the country in which that ticker is based in the following form: country_name/ticker_name. I am trying to split each string and delete everything from the '/' back, returning a character vector of only the ticker names. Here is an example vector:
sample_string <- c('US/SPY', 'US/AOL', 'US/MTC', 'US/PHA', 'US/PZI',
'US/AOL', 'US/BRCM')
My initial thought would be to use the stringr library. I don't have really any experience with that package, but here is what I was trying:
library(stringr)
split_string <- str_split(sample_string, '/')
But I was unsure how to return only the second element of each list as a single vector.
How would I do this over a large character vector (~105 million entries)?
Some benchmark here including all the methods suggested by #David Arenburg, and another method using str_extract from stringr package.
sample_string <- rep(sample_string, 1000000)
library(data.table); library(stringr)
s1 <- function() sub(".*/(.*)", "\\1", sample_string)
s2 <- function() sub(".*/", "", sample_string)
s3 <- function() str_extract(sample_string, "(?<=/)(.*)")
s4 <- function() tstrsplit(sample_string, "/", fixed = TRUE)[[2]]
length(sample_string)
# [1] 7000000
identical(s1(), s2())
# [1] TRUE
identical(s1(), s3())
# [1] TRUE
identical(s1(), s4())
# [1] TRUE
microbenchmark::microbenchmark(s1(), s2(), s3(), s4(), times = 5)
# Unit: seconds
# expr min lq mean median uq max neval
# s1() 3.916555 3.917370 4.046708 3.923246 3.925184 4.551184 5
# s2() 3.584694 3.593755 3.726922 3.610284 3.646449 4.199426 5
# s3() 3.051398 3.062237 3.354410 3.138080 3.722347 3.797985 5
# s4() 1.908283 1.964223 2.349522 2.117521 2.760612 2.996971 5
The tstrsplit method is the fastest.
Update:
Add another method from #Frank, this comparison is not strictly accurate which depends on the actual data, if there is a lot of duplicated cases as the sample_string is produced above, the advantage is quite obvious:
s5 <- function() setDT(list(sample_string))[, v := tstrsplit(V1, "/", fixed = TRUE)[[2]], by=V1]$v
identical(s1(), s5())
# [1] TRUE
microbenchmark::microbenchmark(s1(), s2(), s3(), s4(), s5(), times = 5)
# Unit: milliseconds
# expr min lq mean median uq max neval
# s1() 3905.97703 3913.264 3922.8540 3913.4035 3932.2680 3949.3575 5
# s2() 3568.63504 3576.755 3713.7230 3660.5570 3740.8252 4021.8426 5
# s3() 3029.66877 3032.898 3061.0584 3052.6937 3086.9714 3103.0604 5
# s4() 1322.42430 1679.475 1985.5440 1801.9054 1857.8056 3266.1101 5
# s5() 82.71379 101.899 177.8306 121.6682 209.0579 373.8141 5
Some helpful notes about your question: Firstly, there is a str_split_fixed function in the stringrpackage which does what you want it to do by calling lapply.
library(data.table); library(stringr)
sample_string <- c('US/SPY', 'US/AOL', 'US/MTC', 'US/PHA', 'US/PZI',
'US/AOL', 'US/BRCM')
sample_string <- rep(sample_string, 1e5)
split_string <- str_split_fixed(sample_string, '/', 2)[,2]
It works by calling stringi::stri_split_fixed and is not dissimilar to
do.call("c", lapply(str_split(sample_string, '/'),"[[",2))
Secondly, another way to think about extracting each second element of the list is by doing exactly what tstrsplit is doing internally.
transpose(strsplit(sample_string, "/", fixed = T))[[2]]
On a total side note, the above should be marginally faster than calling tstrsplit. This of course, is probably not worth typing at length but it helps to know what the function does.
library(data.table); library(stringr)
s4 <- function() tstrsplit(sample_string, "/", fixed = TRUE)[[2]]
s5 <- function() transpose(strsplit(sample_string, "/", fixed = T))[[2]]
identical(s4(), s5())
microbenchmark::microbenchmark(s4(), s5(), times = 20)
microbenchmark::microbenchmark(s4(), s5(), times = 20)
Unit: milliseconds
expr min lq mean median uq max neval
s4() 161.0744 193.3611 255.8136 234.9945 271.6811 434.7992 20
s5() 140.8569 176.5600 233.3570 194.1676 251.7921 420.3431 20
Regarding this second method, in short, transposing this list of length 7 million, each with 2 elements will convert your result to a list of length 2, each with 7 million elements. You are then extracting the second element of this list.
I have a column containing values of 3 strings separated by semicolons. I need to just extract the part of the string which comes before the first semicolon.
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
What I want is: Get the first part of the string (till the first semicolon).
Desired output : SNSR_RMIN_PSX150Y_CSH
I tried gsub without success.
You could try sub
sub(';.*$','', Type)
#[1] "SNSR_RMIN_PSX150Y_CSH"
It will match the pattern i.e. first occurence of ; to the end of the string and replace with ''
Or use
library(stringi)
stri_extract(Type, regex='[^;]*')
#[1] "SNSR_RMIN_PSX150Y_CSH"
The stringi package works very fast here:
stri_extract_first_regex(Type, "^[^;]+")
## [1] "SNSR_RMIN_PSX150Y_CSH"
I benchmarked on the 3 main approaches here:
Unit: milliseconds
expr min lq mean median uq max neval
SAPPLY() 254.88442 267.79469 294.12715 277.4518 325.91576 419.6435 100
SUB() 182.64996 186.26583 192.99277 188.6128 197.17154 237.9886 100
STRINGI() 89.45826 91.05954 94.11195 91.9424 94.58421 124.4689 100
Here's the code for the Benchmarks:
library(stringi)
SAPPLY <- function() sapply(strsplit(Type, ";"), "[[", 1)
SUB <- function() sub(';.*$','', Type)
STRINGI <- function() stri_extract_first_regex(Type, "^[^;]+")
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
Type <- rep(Type, 100000)
library(microbenchmark)
microbenchmark(
SAPPLY(),
SUB(),
STRINGI(),
times=100L)
you can also use strsplit
strsplit(Type, ";")[[1]][1]
[1] "SNSR_RMIN_PSX150Y_CSH"
When performance is important you can use substr in combination with regexpr from base.
substr(Type, 1, regexpr(";", Type, fixed=TRUE)-1)
#[1] "SNSR_RMIN_PSX150Y_CSH"
Timings: (Reusing the part from #tyler-rinker)
library(stringi)
SAPPLY <- function() sapply(strsplit(Type, ";"), "[[", 1)
SUB <- function() sub(';.*$','', Type)
SUB2 <- function() sub(';.*','', Type)
SUB3 <- function() sub('([^;]*).*','\\1', Type)
STRINGI <- function() stri_extract_first_regex(Type, "^[^;]+")
STRINGI2 <- function() stri_extract_first_regex(Type, "[^;]*")
SUBSTRREG <- function() substr(Type, 1, regexpr(";", Type)-1)
SUBSTRREG2 <- function() substr(Type, 1, regexpr(";", Type, fixed=TRUE)-1)
SUBSTRREG3 <- function() substr(Type, 1, regexpr(";", Type, fixed=TRUE, useBytes = TRUE)-1)
Type <- c("SNSR_RMIN_PSX150Y_CSH;SP_12;I0.00V50HX0HY3000")
Type <- rep(Type, 100000)
library(microbenchmark)
microbenchmark(SAPPLY(), SUB(), SUB2(), SUB3(), STRINGI()
, STRINGI2(), SUBSTRREG(), SUBSTRREG2(), SUBSTRREG3())
#Unit: milliseconds
# expr min lq mean median uq max neval
# SAPPLY() 382.23750 395.92841 412.82508 410.05236 427.58816 460.28508 100
# SUB() 111.92120 114.28939 116.41950 115.57371 118.15573 123.92400 100
# SUB2() 94.27831 96.50462 98.14741 97.38199 99.15260 119.51090 100
# SUB3() 167.77139 172.51271 175.07144 173.83121 176.27710 190.97815 100
# STRINGI() 38.27645 39.33428 39.94134 39.71842 40.50182 42.55838 100
# STRINGI2() 38.16736 39.19250 40.14904 39.63929 40.37686 56.03174 100
# SUBSTRREG() 45.04828 46.39867 47.13018 46.85465 47.71985 51.07955 100
# SUBSTRREG2() 10.67439 11.02963 11.29290 11.12222 11.43964 13.64643 100
# SUBSTRREG3() 10.74220 10.95139 11.39466 11.06632 11.46908 27.72654 100
I'd like to read only the first character from each line of a text file, ignoring the rest.
Here's an example file:
x <- c(
"Afklgjsdf;bosfu09[45y94hn9igf",
"Basfgsdbsfgn",
"Cajvw58723895yubjsdw409t809t80",
"Djakfl09w50968509",
"E3434t"
)
writeLines(x, "test.txt")
I can solve the problem by reading everything with readLines and using substring to get the first character:
lines <- readLines("test.txt")
substring(lines, 1, 1)
## [1] "A" "B" "C" "D" "E"
This seems inefficient though. Is there a way to persuade R to only read the first characters, rather than having to discard them?
I suspect that there ought to be some incantation using scan, but I can't find it. An alternative might be low level file manipulation (maybe with seek).
Since performance is only relevant for larger files, here's a bigger test file for benchmarking with:
set.seed(2015)
nch <- sample(1:100, 1e4, replace = TRUE)
x2 <- vapply(
nch,
function(nch)
{
paste0(
sample(letters, nch, replace = TRUE),
collapse = ""
)
},
character(1)
)
writeLines(x2, "bigtest.txt")
Update: It seems that you can't avoid scanning the whole file. The best speed gains seem to be using a faster alternative to readLines (Richard Scriven's stringi::stri_read_lines solution and Josh O'Brien's data.table::fread solution), or to treat the file as binary (Martin Morgan's readBin solution).
If you allow/have access to Unix command-line tools you can use
scan(pipe("cut -c 1 test.txt"), what="", quiet=TRUE)
Obviously less portable but probably very fast.
Using #RichieCotton's benchmarking code with the OP's suggested "bigtest.txt" file:
expr min lq mean median uq
RC readLines 14.797830 17.083849 19.261917 18.103020 20.007341
RS read.fwf 125.113935 133.259220 148.122596 138.024203 150.528754
BB scan pipe cut 6.277267 7.027964 7.686314 7.337207 8.004137
RC readChar 1163.126377 1219.982117 1324.576432 1278.417578 1368.321464
RS scan 13.927765 14.752597 16.634288 15.274470 16.992124
data.table::fread() seems to beat all of the solutions so far proposed, and has the great virtue of running comparably fast on both Windows and *NIX machines:
library(data.table)
substring(fread("bigtest.txt", sep="\n", header=FALSE)[[1]], 1, 1)
Here are microbenchmark timings on a Linux box (actually a dual-boot laptop, booted up as Ubuntu):
Unit: milliseconds
expr min lq mean median uq max neval
RC readLines 15.830318 16.617075 18.294723 17.116666 18.959381 27.54451 100
JOB fread 5.532777 6.013432 7.225067 6.292191 7.727054 12.79815 100
RS read.fwf 111.099578 113.803053 118.844635 116.501270 123.987873 141.14975 100
BB scan pipe cut 6.583634 8.290366 9.925221 10.115399 11.013237 15.63060 100
RC readChar 1347.017408 1407.878731 1453.580001 1450.693865 1491.764668 1583.92091 100
And here are timings from the same laptop booted up as a Windows machine (with the command-line tool cut supplied by Rtools):
Unit: milliseconds
expr min lq mean median uq max neval cld
RC readLines 26.653266 27.493167 33.13860 28.057552 33.208309 61.72567 100 b
JOB fread 4.964205 5.343063 6.71591 5.538246 6.027024 13.54647 100 a
RS read.fwf 213.951792 217.749833 229.31050 220.793649 237.400166 287.03953 100 c
BB scan pipe cut 180.963117 263.469528 278.04720 276.138088 280.227259 387.87889 100 d
RC readChar 1505.263964 1572.132785 1646.88564 1622.410703 1688.809031 2149.10773 100 e
Figure out the file size, read it in as a single binary blob, find the offsets of the characters of interest (don't count the last '\n', at the end of the file!), and coerce to final form
f0 <- function() {
sz <- file.info("bigtest.txt")$size
what <- charToRaw("\n")
x = readBin("bigtest.txt", raw(), sz)
idx = which(x == what)
rawToChar(x[c(1L, idx[-length(idx)] + 1L)], multiple=TRUE)
}
The data.table solution (was I think the fastest so far -- need to include the first line as part of the data!)
library(data.table)
f1 <- function()
substring(fread("bigtest.txt", header=FALSE)[[1]], 1, 1)
and in comparison
> identical(f0(), f1())
[1] TRUE
> library(microbenchmark)
> microbenchmark(f0(), f1())
Unit: milliseconds
expr min lq mean median uq max neval
f0() 5.144873 5.515219 5.571327 5.547899 5.623171 5.897335 100
f1() 9.153364 9.470571 9.994560 10.162012 10.350990 11.047261 100
Still wasteful, since the entire file is read in to memory before mostly being discarded.
01/04/2015 Edited to bring the better solution to the top.
Update 2 Changing the scan() method to run on an open connection instead of opening and closing on every iteration allows to read line-by-line and eliminates the looping. The timing improved quite a bit.
## scan() on open connection
conn <- file("bigtest.txt", "rt")
substr(scan(conn, what = "", sep = "\n", quiet = TRUE), 1, 1)
close(conn)
I also discovered the stri_read_lines() function in the stringi package, Its help file says it's experimental at the moment, but it is very fast.
## stringi::stri_read_lines()
library(stringi)
stri_sub(stri_read_lines("bigtest.txt"), 1, 1)
Here are the timings for these two methods.
## timings
library(microbenchmark)
microbenchmark(
scan = {
conn <- file("bigtest.txt", "rt")
substr(scan(conn, what = "", sep = "\n", quiet = TRUE), 1, 1)
close(conn)
},
stringi = {
stri_sub(stri_read_lines("bigtest.txt"), 1, 1)
}
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# scan 50.00170 50.10403 50.55055 50.18245 50.56112 54.64646 100
# stringi 13.67069 13.74270 14.20861 13.77733 13.86348 18.31421 100
Original [slower] answer :
You could try read.fwf() (fixed width file), setting the width to a single 1 to capture the first character on each line.
read.fwf("test.txt", 1, stringsAsFactors = FALSE)[[1L]]
# [1] "A" "B" "C" "D" "E"
Not fully tested of course, but works for the test file and is a nice function for getting substrings without having to read the entire file.
Update 1 : read.fwf() is not very efficient, calling scan() and read.table() internally. We can skip the middle-men and try scan() directly.
lines <- count.fields("test.txt") ## length is num of lines in file
skip <- seq_along(lines) - 1 ## set up the 'skip' arg for scan()
read <- function(n) {
ch <- scan("test.txt", what = "", nlines = 1L, skip = n, quiet=TRUE)
substr(ch, 1, 1)
}
vapply(skip, read, character(1L))
# [1] "A" "B" "C" "D" "E"
version$platform
# [1] "x86_64-pc-linux-gnu"
Benchmarks for each answer, under Windows.
library(microbenchmark)
microbenchmark(
"RC readLines" = {
lines <- readLines("test.txt")
substring(lines, 1, 1)
},
"RS read.fwf" = read.fwf("test.txt", 1, stringsAsFactors = FALSE)$V1,
"BB scan pipe cut" = scan(pipe("cut -c 1 test.txt"),what=character()),
"RC readChar" = {
con <- file("test.txt", "r")
x <- readChar(con, 1)
while(length(ch <- readChar(con, 1)) > 0)
{
if(ch == "\n")
{
x <- c(x, readChar(con, 1))
}
}
close(con)
}
)
## Unit: microseconds
## expr min lq mean median uq
## RC readLines 561.598 712.876 830.6969 753.929 884.8865
## RS read.fwf 5079.010 6429.225 6772.2883 6837.697 7153.3905
## BB scan pipe cut 308195.548 309941.510 313476.6015 310304.412 310772.0005
## RC readChar 1238.963 1549.320 1929.4165 1612.952 1740.8300
## max neval
## 2156.896 100
## 8421.090 100
## 510185.114 100
## 26437.370 100
And on the bigger dataset:
## Unit: milliseconds
## expr min lq mean median uq max neval
## RC readLines 52.212563 84.496008 96.48517 103.319789 104.124623 158.086020 20
## RS read.fwf 391.371514 660.029853 703.51134 766.867222 777.795180 799.670185 20
## BB scan pipe cut 283.442150 482.062337 516.70913 562.416766 564.680194 567.089973 20
## RC readChar 2819.343753 4338.041708 4500.98579 4743.174825 4921.148501 5089.594928 20
## RS scan 2.088749 3.643816 4.16159 4.651449 4.731706 5.375819 20
I don't find it very informative to benchmark operations in the order of micro or milliseconds. But I understand that in some cases it can't be avoided. In those cases, still, I find it essential to test data of different (increasing sizes) to get a rough measure of how well the method scales..
Here's my run on #MartinMorgan's tests using f0() and f1() on 1e4, 1e5 and 1e6 rows and here are the results:
1e4
# Unit: milliseconds
# expr min lq mean median uq max neval
# f0() 4.226333 7.738857 15.47984 8.398608 8.972871 89.87805 100
# f1() 8.854873 9.204724 10.48078 9.471424 10.143601 84.33003 100
1e5
# Unit: milliseconds
# expr min lq mean median uq max neval
# f0() 71.66205 176.57649 174.9545 184.0191 187.7107 307.0470 100
# f1() 95.60237 98.82307 104.3605 100.8267 107.9830 205.8728 100
1e6
# Unit: seconds
# expr min lq mean median uq max neval
# f0() 1.443471 1.537343 1.561025 1.553624 1.558947 1.729900 10
# f1() 1.089555 1.092633 1.101437 1.095997 1.102649 1.140505 10
identical(f0(), f1()) returned TRUE on all the tests.
Update:
1e7
I also ran on 1e7 rows.
f1() (data.table) ran in 9.7 seconds, where as f0() ran in 7.8 seconds the first time, and 9.4 and 6.6s the second time.
However, f1() resulted in no noticeable change in memory while reading the entire 0.479GB file, whereas, f0() resulted in a spike of 2.4GB.
Another observation:
set.seed(2015)
x2 <- vapply(
1:1e5,
function(i)
{
paste0(
sample(letters, 100L, replace = TRUE),
collapse = "_"
)
},
character(1)
)
# 10 million rows, with 200 characters each
writeLines(unlist(lapply(1:100, function(x) x2)), "bigtest.txt")
## readBin() results in a 2 billion row vector
system.time(f0()) ## explodes on memory
Because the readBin() step results in a 2 billion length vector (~1.9GB to read the file), and which(x == what) step takes ~4.5+GB (= ~6.5GB in total) at which point I stopped the process.
fread() takes ~23 seconds in this case.
HTH