A comment on my answer to this question which should give the desired result using strsplit does not, even though it seems to correctly match the first and last commas in a character vector. This can be proved using gregexpr and regmatches.
So why does strsplit split on each comma in this example, even though regmatches only returns two matches for the same regex?
# We would like to split on the first comma and
# the last comma (positions 4 and 13 in this string)
x <- "123,34,56,78,90"
# Splits on every comma. Must be wrong.
strsplit( x , '^\\w+\\K,|,(?=\\w+$)' , perl = TRUE )[[1]]
#[1] "123" "34" "56" "78" "90"
# Ok. Let's check the positions of matches for this regex
m <- gregexpr( '^\\w+\\K,|,(?=\\w+$)' , x , perl = TRUE )
# Matching positions are at
unlist(m)
[1] 4 13
# And extracting them...
regmatches( x , m )
[[1]]
[1] "," ","
Huh?! What is going on?
The theory of #Aprillion is exact, from R documentation:
The algorithm applied to each input string is
repeat {
if the string is empty
break.
if there is a match
add the string to the left of the match to the output.
remove the match and all to the left of it.
else
add the string to the output.
break.
}
In other words, at each iteration ^ will match the begining of a new string (without the precedent items.)
To simply illustrate this behavior:
> x <- "12345"
> strsplit( x , "^." , perl = TRUE )
[[1]]
[1] "" "" "" "" ""
Here, you can see the consequence of this behavior with a lookahead assertion as delimiter (Thanks to #JoshO'Brien for the link.)
Related
I'm trying to subdivide my metacharacter expression in my gsub() function. But it does not return anything found.
Task: I want to delete all sections of string that contain either .ST or -XST in my vector of strings.
As you can see below, using one expression works fine. But the | expression simply does not work. I'm following the metacharacter guide on https://www.stat.auckland.ac.nz/~paul/ItDT/HTML/node84.html
What can be the issue? And what caused this issue?
My data
> rownames(table.summary)[1:10]
[1] "AAK.ST" "ABB.ST" "ALFA.ST" "ALIV-SDB.ST" "AOI.ST" "ATCO-A.ST" "ATCO-B.ST" "AXFO.ST" "AXIS.ST" "AZN.ST"
> gsub(pattern = '[.](.*)$ | [-](.*)$', replacement = "", x = rownames(table.summary)[1:10])
[1] "AAK.ST" "ABB.ST" "ALFA.ST" "ALIV-SDB.ST" "AOI.ST" "ATCO-A.ST" "ATCO-B.ST" "AXFO.ST" "AXIS.ST" "AZN.ST"
> gsub(pattern = '[.](.*)$', replacement = "", x = rownames(table.summary)[1:10])
[1] "AAK" "ABB" "ALFA" "ALIV-SDB" "AOI" "ATCO-A" "ATCO-B" "AXFO" "AXIS" "AZN"
> gsub(pattern = '[-](.*)$', replacement = "", x = rownames(table.summary)[1:10])
[1] "AAK.ST" "ABB.ST" "ALFA.ST" "ALIV" "AOI.ST" "ATCO" "ATCO" "AXFO.ST" "AXIS.ST" "AZN.ST"
It seems you tested your regex with a flag like IgnorePatternWhitespace (VERBOSE, /x) that allows whitespace inside patterns for readability. You can use it with perl=T option:
d <- c("AAK.ST","ABB.ST","ALFA.ST","ALIV-SDB.ST","AOI.ST","ATCO-A.ST","ATCO-B.ST","AXFO.ST", "AXIS.ST","AZN.ST")
gsub('(?x)[.](.*)$ | [-](.*)$', '', d, perl=T)
## [1] "AAK" "ABB" "ALFA" "ALIV" "AOI" "ATCO" "ATCO" "AXFO" "AXIS" "AZN"
However, you really do not have to use that complex regex here.
If you plan to remove all substrings from ther first hyphen or dot up to the end, you may use the following regex:
[.-].*$
The character class [.-] will match the first . or - symbol and .* wil match all characters up to the end of the string ($).
See IDEONE demo:
d <- c("AAK.ST","ABB.ST","ALFA.ST","ALIV-SDB.ST","AOI.ST","ATCO-A.ST","ATCO-B.ST","AXFO.ST", "AXIS.ST","AZN.ST")
gsub("[.-].*$", "", d)
Result: [1] "AAK" "ABB" "ALFA" "ALIV" "AOI" "ATCO" "ATCO" "AXFO" "AXIS" "AZN"
This will find .ST or -XST at the end of the text and substitute it with empty characters string (effectively removing that part). Don't forget that gsub returns modified string, not modifies it in place. You won't see any change until you reassign return value back to some variable.
strings <- c("AAK.ST", "ABB.ST", "ALFA.ST", "ALIV-SDB.ST", "AOI.ST", "ATCO-A.ST", "ATCO-B.ST", "AXFO.ST", "AXIS.ST", "AZN.ST", "AAC-XST", "AAD-XSTV")
strings <- gsub('(\\.ST|-XST)$', '', strings)
Your regular expression ([.](.*)$ | [-](.*)$'), if not for unnecessary spaces, would remove everything from first dot (.) or dash (-) to end of text. This might be what you want, but not what you said you want.
Given the following vector of strings x
x <- c("hello", "foo_bar", "blah_blub_(bleep)", "blah_(xyz)", "xyz(_$_)")
I am looking for a regexp to extract everything before the optional parenthesis (and its content). So the final result for the above vector should be:
c("hello", "foo_bar", "blah_blub", "blah", "xyz")
I came up with the following regexp which, however, does not work (why?):
R> sub("^(.*)[_?\\(.*\\)]?$", \\1, x)
[1] "hello" "foo_bar" "blah_blub_(bleep)" "blah_(xyz)" "xyz(_$_)"
Any help is appreciated!
We can match the pattern of zero or more _ followed by ( followed by one more characters until the end of the string and replace it with ''.
sub('_*\\(.*$', '', x)
#[1] "hello" "foo_bar" "blah_blub" "blah" "xyz"
string = "ABC3JFD456"
Suppose I have the above string, and I wish to find what the first digit in the string is and store its value. In this case, I would want to store the value 3 (since it's the first-occuring digit in the string). grepl("\\d", string) only returns a logical value, but does not tell me anything about where or what the first digit is. Which regular expression should I use to find the value of the first digit?
Base R
regmatches(string, regexpr("\\d", string))
## [1] "3"
Or using stringi
library(stringi)
stri_extract_first(string, regex = "\\d")
## [1] "3"
Or using stringr
library(stringr)
str_extract(string, "\\d")
## [1] "3"
1) sub Try sub with the indicated regular expression which takes the shortest string until a digit, a digit and then everything following and replaces it with the digit:
sub(".*?(\\d).*", "\\1", string)
giving:
[1] "3"
This also works if string is a vector of strings.
2) strapplyc It would also be possible to use strapplyc from gsubfn in which case an even simpler regular expression could be used:
strapplyc(string, "\\d", simplify = TRUE)[1]
giving the same or use this which gives the same answer again but also works if string is a vector of strings:
sapply(strapplyc(string, "\\d"), "[[", 1)
Get the locations of the digits
tmp <- gregexpr("[0-9]", string)
iloc <- unlist(tmp)[1]
Extract the first digit
as.numeric(substr(string,iloc,iloc))
Using regexpr is simpler
tmp<-regexpr("[0-9]",string)
if(tmp[[1]]>=0) {
iloc <- tmp[1]
num <- as.numeric(substr(string,iloc,iloc))
}
Using rex may make this type of task a little simpler.
string = c("ABC3JFD456", "ARST4DS324")
re_matches(string,
rex(
capture(name = "first_number", digit)
)
)
#> first_number
#> 1 3
#> 2 4
> which( sapply( strsplit(string, ""), grepl, patt="[[:digit:]]"))[1]
[1] 4
Or
> gregexpr("[[:digit:]]", string)[[1]][1]
[1] 4
So:
> splstr[[1]][ which( sapply( splstr, grepl, patt="[[:digit:]]"))[1] ]
[1] "3"
Note that a full result from a gregexpr call is a list, hence the need to extract its first element with "[[":
> gregexpr("[[:digit:]]", string)
[[1]]
[1] 4 8 9 10
attr(,"match.length")
[1] 1 1 1 1
attr(,"useBytes")
[1] TRUE
A gsub solution that is based on replacing the substrings preceding and following the first digit with the empty string:
gsub("^\\D*(?=\\d)|(?<=\\d).*", "", string, perl = TRUE)
# [1] "3"
This question is very similar to Removing multiple spaces and trailing spaces using gsub, except that I'd like to apply it to commas instead of spaces.
For example, I'd like a function TrimCommas to turn x into y:
x <- c("a,b,c", ",a,b,,c", ",,,a,,,b,c,,,")
# y <- TrimCommas(x) # presumably
y <- c("a,b,c", "a,b,c", "a,b,c")
The solution for spaces was gsub("^ *|(?<= ) | *$", "", x, perl=T), so I'm hoping comparing the solution for this will help explain some regex fundamentals as well.
Isn't the solution pretty similar?
x <- c("a,b,c", ",a,b,,c", ",,,a,,,b,c,,,")
gsub("^,*|(?<=,),|,*$", "", x, perl=T)
# [1] "a,b,c" "a,b,c" "a,b,c"
There are three parts to the regex ^,*|(?<=,),|,*$:
^,* -- this matches 0 or more commas at the beginning of the string
(?<=,), -- this is a positive lookbehind to see if there a comma behind a comma, so it matches , in ,,
,*$ -- this matches 0 or more commas at the end of the string
As you can see all of the above are substituted with nothing.
You can make this generic to any character (" ", ",", etc.) with this function:
TrimMult <- function(x, char=" ") {
return(gsub(paste0("^", char, "*|(?<=", char, ")", char, "|", char, "*$"),
"", x, perl=T))
}
I am not familiar at all with regular expressions, and would like to do pattern matching and replacement in R.
I would like to replace the pattern #1, #2 in the vector: original = c("#1", "#2", "#10", "#11") with each value of the vector vec = c(1,2).
The result I am looking for is the following vector: c("1", "2", "#10", "#11")
I am not sure how to do that. I tried doing:
for(i in 1:2) {
pattern = paste("#", i, sep = "")
original = gsub(pattern, vec[i], original, fixed = TRUE)
}
but I get :
#> original
#[1] "1" "2" "10" "11"
instead of: "1" "2" "#10" "#11"
I would appreciate any help I can get! Thank you!
Specify that you are matching the entire string from start (^) to end ($).
Here, I've matched exactly the conditions you are looking at in this example, but I'm guessing you'll need to extend it:
> gsub("^#([1-2])$", "\\1", original)
[1] "1" "2" "#10" "#11"
So, that's basically, "from the start, look for a hash symbol followed by either the exact number one or two. The one or two should be just one digit (that's why we don't use * or + or something) and also ends the string. Oh, and capture that one or two because we want to 'backreference' it."
Another option using gsubfn:
library(gsubfn)
gsubfn("^#([1-2])$", I, original) ## Function substituting
[1] "1" "2" "#10" "#11"
Or if you want to explicitly use the values of your vector , using vec values:
gsubfn("^#[1-2]$", as.list(setNames(vec,c("#1", "#2"))), original)
Or formula notation equivalent to function notation:
gsubfn("^#([1-2])$", ~ x, original) ## formula substituting
Here's a slightly different take that uses zero width negative lookahead assertion (what a mouthful!). This is the (?!...) which matches # at the start of a string as long as it is not followed by whatever is in .... In this case two (or equivalently, more as long as they are contiguous) digits. It replaces them with nothing.
gsub( "^#(?![0-9]{2})" , "" , original , perl = TRUE )
[1] "1" "2" "#10" "#11"