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
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
I've got some problems deleting duplicate elements in a string.
My data look similar to this:
idvisit path
1 1,16,23,59
2 2,14,14,19
3 5,19,23,19
4 10,10
5 23,23,27,29,23
I have a column containing an unique ID and a column containing a path for web page navigation.
The right column contains some cases, where pages just were reloaded and the page were tracked twice or even more.
The pages are separated with commas and are saved as factors.
My problem is, that I don't want to have multiple pages in a row, so the data should look like this.
idvisit path
1 1,16,23,59
2 2,14,19
3 5,19,23,19
4 10
5 23,27,29,23
The multiple pages next to each other should be removed. I know how to delete a specific multiple number using regexpressions, but I have about 20.000 different pages and can't do this for all of them.
Does anyone have a solution or a hint, for my problem?
Thanks
Sebastian
We can use tidyverse. Use the separate_rows to split the 'path' variable by the delimiter (,) to convert to a long format, then grouped by 'idvisit', we paste the run-length-encoding values
library(tidyverse)
separate_rows(df1, path) %>%
group_by(idvisit) %>%
summarise(path = paste(rle(path)$values, collapse=","))
# A tibble: 5 × 2
# idvisit path
# <int> <chr>
#1 1 1,16,23,59
#2 2 2,14,19
#3 3 5,19,23,19
#4 4 10
#5 5 23,27,29,23
Or a base R option is
df1$path <- sapply(strsplit(df1$path, ","), function(x) paste(rle(x)$values, collapse=","))
NOTE: If the 'path' column is factor class, convert to character before passing as argument to strsplit i.e. strsplit(as.character(df1$path), ",")
Using stringr package, with function: str_replace_all, I think it gets what you want using the following regular expression: ([0-9]+),\\1and then replace it with \\1 (we need to scape the \ special character):
library(stringr)
> str_replace_all("5,19,23,19", "([0-9]+),\\1", "\\1")
[1] "5,19,23,19"
> str_replace_all("10,10", "([0-9]+),\\1", "\\1")
[1] "10"
> str_replace_all("2,14,14,19", "([0-9]+),\\1", "\\1")
[1] "2,14,19"
You can use it in a array form: x <- c("5,19,23,19", "10,10", "2,14,14,19") then:
str_replace_all(x, "([0-9]+),\\1", "\\1")
[1] "5,19,23,19" "10" "2,14,19"
or using sapply:
result <- sapply(x, function(x) str_replace_all(x, "([0-9]+),\\1", "\\1"))
Then:
> result
5,19,23,19 10,10 2,14,14,19
"5,19,23,19" "10" "2,14,19"
Notes:
The first line is the attribute information:
> str(result)
Named chr [1:3] "5,19,23,19" "10" "2,14,19"
- attr(*, "names")= chr [1:3] "5,19,23,19" "10,10" "2,14,14,19"
If you don't want to see them (it does not affect the result), just do:
attributes(result) <- NULL
Then,
> result
[1] "5,19,23,19" "10" "2,14,19"
Explanation about the regular expression used: ([0-9]+),\\1
([0-9]+): Starts with a group 1 delimited by () and finds any digit (at least one)
,: Then comes a punctuation sign: , (we can include spaces here, but the original example only uses this character as delimiter)
\\1: Then comes an identical string to the group 1, i.e.: the repeated number. If that doesn't happen, then the pattern doesn't match.
Then if the pattern matches, it replaces it, with the value of the variable \\1, i.e. the first time the number appears in the pattern matched.
How to handle more than one duplicated number, for example 2,14,14,14,19?:
Just use this regular expression instead: ([0-9]+)(,\\1)+, then it matches when at least there is one repetition of the delimiter (right) and the number. You can try other possibilities using this regex101.com (in MHO it more user friendly than other online regular expression checkers).
I hope this would work for you, it is a flexible solution, you just need to adapt it with the pattern you need.
I want to extract alphanumeric characters from a partiular sentence in R.
I have tried the following:
aa=grep("[:alnum:]","abc")
.This should return integer(0),but it returns 1,which should not be the case as "abc" is not an alphanumeric.
What am I missing here?
Essentially I am looking for a function,that only searches for characters that are combinations of both alphabets and numbers,example:"ABC-0112","PCS12SCH"
Thanks in advance for your help.
[[:alnum:]] matches alphabets or digits. To match the string which contains the both then you should use,
x <- c("ABC", "ABc12", "--A-1", "abc--", "89=A")
grep("(.*[[:alpha:]].*[[:digit:]]|.*[[:digit:]].*[[:alpha:]])", x)
# [1] 2 3 5
or
which(grepl("[[:alpha:]]", x) & grepl("[[:digit:]]", x))
# [1] 2 3 5
I want to get indices of all occurences of character elements in some word. Assume these character elements I look for are: l, e, a, z.
I tried the following regex in grep function and tens of its modifications, but I keep receiving not what I want.
grep("/([leazoscnz]{1})/", "ylaf", value = F)
gives me
numeric(0)
where I would like:
[1] 2 3
To use grep work with individual characters of a string, you first need to split the string into separate character vectors. You can use strsplit for this:
strsplit("ylaf", split="")[[1]]
[1] "y" "l" "a" "f"
Next you need to simplify your regular expression, and try the grep again:
strsplit("ylaf", split="")[[1]]
grep("[leazoscnz]", strsplit("ylaf", split="")[[1]])
[1] 2 3
But it is easier to use gregexpr:
gregexpr("[leazoscnz]", "ylaf")
[[1]]
[1] 2 3
attr(,"match.length")
[1] 1 1
attr(,"useBytes")
[1] TRUE