I am trying to extract the first group of non-whitespace characters that follows an Arabic string for each text in a set of about 2,100 total texts. Some of these texts contain the string, while others do not. This would be a very easy task, using str_extract from the stringr package, if the string were in English. However, for some reason this function doesn't work when using an Arabic string within the look-behind pattern:
library(stringr)
test_texts <- c(
"My text كلمة containing some Arabic",
"My text كلمة again containing some Arabic",
"My text that doesn't contain any Arabic"
)
str_extract(test_texts, "(?<=text )\\S+")
# [1] "كلمة" "كلمة" "that"
str_extract(test_texts, "(?<=containing )\\S+")
# [1] "some" "some" NA
str_extract(test_texts, "(?<=كلمة )\\S+") #returns NAs even though string is there
# [1] NA NA NA
Note that this works if I'm not using a look-behind pattern:
str_extract(test_texts, "كلمة \\S+")
# [1] "كلمة containing" "كلمة again" NA
Why does the Arabic mess things up only when using a look-behind pattern?
I am using R version 3.2.3, on OS X 10.11.3, and stringr version 1.0.0.
It seems there is some issue how str_extract processes the right-to-left text inside the positive lookbehind. As a workaround, you may use str_match with a regex having a capturing group around the subpattern capture the value you need:
> res <- str_match(test_texts, "كلمة +(\\S+)")
> res[,2]
[1] "containing" "again" NA
This solution allows matching the non-whitespace chunk even if there are more than 1 space after the Arabic word.
You can grep for non-ascii characters like this:
str_extract(test_texts, "[^\001-\177]+")
[1] "كلمة" "كلمة" NA
str_extract(test_texts, "(?<=[^\001-\177] )\\S+")
[1] "containing" "again" NA
And this seems to work... just adding brackets to what you had. This may not be sufficient either since the characters can be in any order if they are in brackets.
str_extract(test_texts, "(?<=[كلمة] )\\S+")
[1] "containing" "again" NA
Related
This is the name of a file that I have on R:
> lst.files[1]
[1] "clt_Amon_CanESM2_rcp45_185001-230012.nc"
What I need to do is capture just the part until the 4th underscore (including), so it would be something like this:
clt_Amon_CanESM2_rcp45_
How can I get this in R?
If you know you always have (at least) four underscores, then you could do something like this:
regmatches(lst, regexec(".*_.*_.*_.*_", lst.files[1]))[[1]]
# [1] "clt_Amon_CanESM2_rcp45_"
If potentially not always four, but no underscores in the second part, you could do something like this:
regmatches(lst, regexec(".*_", lst.files[1]))[[1]]
# [1] "clt_Amon_CanESM2_rcp45_"
This doesn't require any extra package, just base R.
Using the qdap package, you can do the following.
x <- "clt_Amon_CanESM2_rcp45_185001-230012.nc"
library(qdap)
beg2char(x, "_", 4, include = TRUE)
# [1] "clt_Amon_CanESM2_rcp45_"
We can also capture the repeating patterns as a group using sub. We match one more more characters from the beginning (^) of the string that is not an underscore ([^_]+) followed by an underscore (\\_) which is repeated 4 times ({4}), capture that as a group by wrapping with parentheses followed by zero or more characters (.*). We replace it with the capture group (\\1) to get the expected output.
sub('^(([^_]+\\_){4}).*', '\\1', str1)
#[1] "clt_Amon_CanESM2_rcp45_"
data
str1 <- "clt_Amon_CanESM2_rcp45_185001-230012.nc"
I am trying to replace some text in a character vector using regex in R where, if there is a set of letters inside a bracket, the bracket content is to erplace the whole thing. So, given the input:
tst <- c("85", "86 (TBA)", "87 (LAST)")
my desired output would be equivalent to c("85", "TBA", "LAST")
I tried gsub("\\(([[:alpha:]])\\)", "\\1", tst) but it didn't replace anything. What do I need to correct in my regular expression here?
I think you want
gsub(".*\\(([[:alpha:]]+)\\)", "\\1", tst)
# [1] "85" "TBA" "LAST"
Your first expression was trying to match exactly one alpha character rather than one-or-more. I also added the ".*" to capture the beginning part of the string so it gets replaced as well, otherwise, it would be left untouched.
gsub("(?=.*\\([^)]*\\)).*\\(([^)]*)\\)", "\\1", tst, perl=TRUE)
## [1] "85" "TBA" "LAST"
You can try this.See demo.Replace by \1.
https://regex101.com/r/sH8aR8/38
The following would work. Note that white-spaces within the brackets may be problematic
A<-sapply(strsplit(tst," "),tail,1)
B<-gsub("\\(|\\)", "", A)
I like the purely regex answers better. I'm showing a solution using the qdapRegex package that I maintain as the result is pretty speedy and easy to remember and generalize. It pulls out the strings that are in parenthesis and then replaces any NA (no bracket) with the original value. Note that the result is a list and you'd need to use unlist to match your desired output.
library(qdpRegex)
m <- rm_round(tst, extract=TRUE)
m[is.na(m)] <- tst[is.na(m)]
## [[1]]
## [1] "85"
##
## [[2]]
## [1] "TBA"
##
## [[3]]
## [1] "LAST"
I am trying something I thought would be easy. I'm looking for a single regex solution (though others are welcomed for completeness). I want to split on n occurrences of a delimiter.
Here is some data:
x <- "I like_to see_how_too"
pat <- "_"
Desired outcome
Say I want to split on first occurrence of _:
[1] "I like" "to see_how_too"
Say I want to split on second occurrence of _:
[1] "I like_to see" "how_too"
Ideally, if the solution is a regex one liner generalizable to nth occurrence; the solution will use strsplit with a single regex.
Here's a solution that doesn't fit my parameters of single regex that works with strsplit
x <- "I like_to see_how_too"
y <- "_"
n <- 1
loc <- gregexpr("_", x)[[1]][n]
c(substr(x, 1, loc-1), substr(x, loc + 1, nchar(x)))
Here is another solution using the gsubfn package and some regex-fu. To change the nth occurrence of the delimiter, you can simply swap the number that is placed inside of the range quantifier — {n}.
library(gsubfn)
x <- 'I like_to see_how_too'
strapply(x, '((?:[^_]*_){1})(.*)', c, simplify =~ sub('_$', '', x))
# [1] "I like" "to see_how_too"
If you would like the nth occurrence to be user defined, you could use the following:
n <- 2
re <- paste0('((?:[^_]*_){',n,'})(.*)')
strapply(x, re, c, simplify =~ sub('_$', '', x))
# [1] "I like_to see" "how_too"
Non-Solution
Since R is using PCRE, you can use \K to remove everything that matches the pattern before \K from the main match result.
Below is the regex to split the string at the 3rd _
^[^_]*(?:_[^_]*){2}\K_
If you want to split at the nth occurrence of _, just change 2 to (n - 1).
Demo on regex101
That was the plan. However, strsplit seems to think differently.
Actual execution
Demo on ideone.com
x <- "I like_to see_how_too but_it_seems to_be_impossible"
strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
It still fails to work on a stronger assertion \A
strsplit(x, "\\A[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
Explanation?
This behavior hints at the fact that strsplit find the first match, do a substring to extract the first token and the remainder part, and find the next match in the remainder part.
This removes all the states from the previous matches, and leaves us with a clean state when it tries to match the regex on the remainder. This makes the task of stopping the strsplit function at first match and achieving the task at the same time impossible. There is not even a parameter in strsplit to limit the number of splits.
Rather than split you do match to get your split strings.
Try this regex:
^((?:[^_]*_){1}[^_]*)_(.*)$
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
RegEx Demo
Update: It seems R also supports PCRE and in that case you can do split as well using this PCRE regex:
^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
(*FAIL) behaves like a failing negative assertion and is a synonym for (?!)
(*SKIP) defines a point beyond which the regex engine is not allowed to backtrack when the subpattern fails later
(*SKIP)(*FAIL) together provide a nice alternative of restriction that you cannot have a variable length lookbehind in above regex.
RegEx Demo2
x <- "I like_to see_how_too"
strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## > strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like" "to see" "how" "too"
## > strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like_to see" "how_too"
This uses gsubfn to to preprocess the input string so that strsplit can handle it. The main advantage is that one can specify a vector of numbers, k, indicating which underscores to split on.
It replaces the occurrences of underscore defined by k by a double underscore and then splits on double underscore. In this example we split at the 2nd and 4th underscore:
library(gsubfn)
k <- c(2, 4) # split at 2nd and 4th _
p <- proto(fun = function(., x) if (count %in% k) "__" else "_")
strsplit(gsubfn("_", p, "aa_bb_cc_dd_ee_ff"), "__")
giving:
[[1]]
[1] "aa_bb" "cc_dd" "ee_ff"
If empty fields are allowed then use any other character sequence not in the string, e.g. "\01" in place of the double underscore.
See section 4 of the gusbfn vignette for more info on using gusbfn with proto objects to retain state between matches.
I want to extract the tags (twitter handles) from tweets.
tweet <- "#me bla bla bla bla #2_him some text #me_"
The following only extracts part of some substrings due to the punctuation in some tags
regmatches(tweet, gregexpr("#[[:alnum:]]*", tweet))[[1]]
[1] "#me" "#2" "#me"
I don't know what regular expression would return the entire string (#tag).
Thanks!
If you want to match all non-spaces, just use the corresponding regular expression
regmatches(tweet, gregexpr("#[^ ]*", tweet))[[1]]
# [1] "#me" "#2_him" "#me_"
You can use the following. \S will match any non-white space character. As well, you want to use the + quantifier instead of * otherwise you will end up matching the # character by itself if one did exist in the string.
> regmatches(tweet, gregexpr("#\\S+", tweet))[[1]]
# [1] "#me" "#2_him" "#me_"
Instead of [[:alnum:]]* use \w* because _ does not comes under alphanumeric character list(ie, [[:alnum:]] matches alphanumeric[A-Za-z0-9] characters. ) but it comes under word character ([A-Za-z0-9_]) list.
> regmatches(tweet, gregexpr("#\\w*", tweet))[[1]]
[1] "#me" "#2_him" "#me_"
The qdapRegex package has a function specifically designed for this task rm_tag:
library(qdapRegex)
rm_tag(tweet, extract=TRUE)
## [[1]]
## [1] "#me" "#2_him" "#me_"
I try to use stringr package to extract part of a string, which is between two particular patterns.
For example, I have:
my.string <- "nanaqwertybaba"
left.border <- "nana"
right.border <- "baba"
and by the use of str_extract(string, pattern) function (where pattern is defined by a POSIX regular expression) I would like to receive:
"qwerty"
Solutions from Google did not work.
In base R you can use gsub. The parentheses in the pattern create numbered capturing groups. Here we select the second group in the replacement, i.e. the group between the borders. The . matches any character. The * means that there is zero or more of the preceeding element
gsub(pattern = "(.*nana)(.*)(baba.*)",
replacement = "\\2",
x = "xxxnanaRisnicebabayyy")
# "Risnice"
I do not know whether and how this is possible with functions provided by stringr but you can also use base regexpr and substring:
pattern <- paste0("(?<=", left.border, ")[a-z]+(?=", right.border, ")")
# "(?<=nana)[a-z]+(?=baba)"
rx <- regexpr(pattern, text=my.string, perl=TRUE)
# [1] 5
# attr(,"match.length")
# [1] 6
substring(my.string, rx, rx+attr(rx, "match.length")-1)
# [1] "qwerty"
I would use str_match from stringr: "str_match extracts capture groups formed by
() from the first match. It returns a character matrix with one column for the complete match and one column for each group." ref
str_match(my.string, paste(left.border, '(.+)', right.border, sep=''))[,2]
The code above creates a regular expression with paste concatenating the capture group (.+) that captures 1 or more characters, with left and right borders (no spaces between strings).
A single match is assumed. So, [,2] selects the second column from the matrix returned by str_match.
You can use the package unglue:
library(unglue)
my.string <- "nanaqwertybaba"
unglue_vec(my.string, "nana{res}baba")
#> [1] "qwerty"