I am trying to get the number of open brackets in a character string in R. I am using the str_count function from the stringr package
s<- "(hi),(bye),(hi)"
str_count(s,"(")
Error in stri_count_regex(string, pattern, opts_regex = attr(pattern,
: ` Incorrectly nested parentheses in regexp pattern.
(U_REGEX_MISMATCHED_PAREN)
I am hoping to get 3 for this example
( is a special character. You need to escape it:
str_count(s,"\\(")
# [1] 3
Alternatively, given that you're using stringr, you can use the coll function:
str_count(s,coll("("))
# [1] 3
You could also use gregexpr along with length in base R:
sum(gregexpr("(", s, fixed=TRUE)[[1]] > 0)
[1] 3
gregexpr takes in a character vector and returns a list with the starting positions of each match. I added fixed=TRUE in order to match literals.length will not work because gregexpr returns -1 when a subexpression is not found.
If you have a character vector of length greater than one, you would need to feed the result to sapply:
# new example
s<- c("(hi),(bye),(hi)", "this (that) other", "what")
sapply((gregexpr("(", s, fixed=TRUE)), function(i) sum(i > 0))
[1] 3 1 0
If you want to do it in base R you can split into a vector of individual characters and count the "(" directly (without representing it as a regular expression):
> s<- "(hi),(bye),(hi)"
> chars <- unlist(strsplit(s,""))
> length(chars[chars == "("])
[1] 3
Related
Suppose I have list of string "S[+229]EC[+57]VDSTDNSSK[+229]PSSEPTSHVAR" and need to get a vector of string that contains only numbers with bracket like eg. [+229][+57].
Is there a convenient way in R to do this?
Using base R, then try it with
> unlist(regmatches(s,gregexpr("\\[\\+\\d+\\]",s)))
[1] "[+229]" "[+57]" "[+229]"
Or you can use
> gsub(".*?(\\[.*\\]).*","\\1",gsub("\\].*?\\[","] | [",s))
[1] "[+229] | [+57] | [+229]"
We can use str_extract_all from stringr
stringr::str_extract_all(x, "\\[\\+\\d+\\]")[[1]]
#[1] "[+229]" "[+57]" "[+229]"
Wrap it in unique if you need only unique values.
Similarly, in base R using regmatches and gregexpr
regmatches(x, gregexpr("\\[\\+\\d+\\]", x))[[1]]
data
x <- "S[+229]EC[+57]VDSTDNSSK[+229]PSSEPTSHVAR"
Seems like you want to remove the alphabetical characters, so
gsub("[[:alpha:]]", "", x)
where [:alpha:] is the class of alphabetical (lower-case and upper-case) characters, [[:alpha:]] says 'match any single alphabetical character', and gsub() says substitute, globally, any alphabetical character with the empty string "". This seems better than trying to match bracketed numbers, which requires figuring out which characters need to be escaped with a (double!) \\.
If the intention is to return the unique bracketed numbers, then the approach is to extract the matches (rather than remove the unwanted characters). Instead of using gsub() to substitute matches to a regular expression with another value, I'll use gregexpr() to identify the matches, and regmatches() to extract the matches. Since numbers always occur in [], I'll simplify the regular expression to match one or more (+) characters from the collection +[:digit:].
> xx <- regmatches(x, gregexpr("[+[:digit:]]+", x))
> xx
[[1]]
[1] "+229" "+57" "+229"
xx is a list of length equal to the length of x. I'll write a function that, for any element of this list, makes the values unique, surrounds the values with [ and ], and concatenates them
fun <- function(x)
paste0("[", unique(x), "]", collapse = "")
This needs to be applied to each element of the list, and simplified to a vector, a task for sapply().
> sapply(xx, fun)
[1] "[+229][+57]"
A minor improvement is to use vapply(), so that the result is robust (always returning a character vector with length equal to x) to zero-length inputs
> x = character()
> xx <- regmatches(x, gregexpr("[+[:digit:]]+", x))
> sapply(xx, fun) # Hey, this returns a list :(
list()
> vapply(xx, fun, "character") # vapply() deals with 0-length inputs
character(0)
I am trying to count the number of dots in a character string.
I have tried to use str_count but it gives me the number of letters of the string instead.
ex_str <- "This.is.a.string"
str_count(ex_str, '.')
nchar(ex_str)
. is a special regex symbol, so you need to escape it:
str_count(ex_str, '\\.')
# [1] 3
Using just base R you could do:
nchar(gsub("[^.]", "", ex_str))
Using stringi:
stri_count_fixed(ex_str, '.')
Another base R solution could be:
length(grepRaw(".", ex_str, fixed = TRUE, all = TRUE))
[1] 3
You may also use the base function gregexpr:
sum(gregexpr(".", ex_str, fixed=TRUE)[[1]] > 0)
[1] 3
You can use stringr::str_count with a fixed(...) argument to avoid treating it as a regular expression:
str_count(ex_str, fixed('.'))
See the online R demo:
library(stringr)
ex_str <- "This.is.a.string"
str_count(ex_str, fixed('.'))
## => [1] 3
If I have a sentence separated by spaces
s<-("C java","C++ java")
grep("C",s)
gives output as
[1] [2]
while I only require
[1]
How to do that? ( I have used c\++ to identify c++ separately but matching with C gives [1] and [2] both as the output)
If we want to match 1 only, then we can use the start (^) and end ($) of the string to denote that there are no characters after or before 'C'
grep("^C$",s)
#[1] 1
data
s<- c("C","C++","java")
s<-c("C","C++","java")
which(s %in% "C")
grep() gives a positive result for any match within a string
Let's say I want to change the string X0_Y1_Z2 into X0_Y1_Z1, i.e. to decrease the last number by one. I tried it by the following statement in R, which doesn't work:
sub("(\\S+_\\S+_)\\S(\\d)", paste0("\\1", as.numeric("\\2")-1), "X0_Y1_Z2", perl=T)
How can I do it?
If you always have the string in this same format, and you only have 1 last digit to decrement, use a simple substring:
> paste0(substring(s, 1, nchar(s)-1), as.numeric(substring(s, nchar(s))) - 1)
> [1] "X0_Y1_Z1"
In order to match the last digit chunk in a string, use [0-9]+$ regex. To increase the value, use gsubfn package. See an example code:
> library(gsubfn)
> s <- "X0_Y1_Z2"
> gsubfn('[0-9]+$', ~ as.numeric(x)-1, s)
[1] "X0_Y1_Z1"
If you need to validate the string the way you did, use more groups and the anchors ^ and $ will require the whole string to match the pattern (a "full string match"):
> p <- "^(\\S+_\\S+_\\S)(\\d+)$"
> gsubfn(p, function(x1,x2) paste0(x1, as.numeric(x2)-1), s)
[1] "X0_Y1_Z1"
rquote <- "R's internals are irrefutably intriguing"
chars <- strsplit(rquote, split = "")[[1]]
in the above code we need to find the number of r's(R and r) in rquote
You could use substrings.
## find position of first 'u'
u1 <- regexpr("u", rquote, fixed = TRUE)
## get count of all 'r' or 'R' before 'u1'
lengths(gregexpr("r", substr(rquote, 1, u1), ignore.case = TRUE))
# [1] 5
This follows what you ask for in the title of the post. If you want the count of all the "r", case insensitive, then simplify the above to
lengths(gregexpr("r", rquote, ignore.case = TRUE))
# [1] 6
Then there's always stringi
library(stringi)
## count before first 'u'
stri_count_regex(stri_sub(rquote, 1, stri_locate_first_regex(rquote, "u")[,1]), "r|R")
# [1] 5
## count all R or r
stri_count_regex(rquote, "r|R")
# [1] 6
To get the number of R's before the first u, you need to make an intermediate step. (You probably don't need to. I'm sure akrun knows some incredibly cool regular expression to get the job done, but it won't be as easy to understand as this).
rquote <- "R's internals are irrefutably intriguing"
before_u <- gsub("u[[:print:]]+$", "", rquote)
length(stringr::str_extract_all(before_u, "(R|r)")[[1]])
You may try this,
> length(str_extract_all(rquote, '[Rr]')[[1]])
[1] 6
To get the count of all r's before the first u
> length(str_extract_all(rquote, perl('u.*(*SKIP)(*F)|[Rr]'))[[1]])
[1] 5
EDIT: Just saw before the first u. In that case, we can get the position of the first 'u' from either which or match.
Then use grepl in the 'chars' up to the position (ind) to find the logical index of 'R' with ignore.case=TRUE and use sum using the strsplit output from the OP's code.
ind <- which(chars=='u')[1]
Or
ind <- match('u', chars)
sum(grepl('r', chars[seq(ind)], ignore.case=TRUE))
#[1] 5
Or we can use two gsubs on the original string ('rquote'). First one removes the characters starting with u until the end of the string (u.$) and the second matches all characters except R, r ([^Rr]) and replace it with ''. We can use nchar to get count of the characters remaining.
nchar(gsub('[^Rr]', '', sub('u.*$', '', rquote)))
#[1] 5
Or if we want to count the 'r' in the entire string, gregexpr to get the position of matching characters from the original string ('rquote') and get the length
length(gregexpr('[rR]', rquote)[[1]])
#[1] 6