I'm having a problem with regex in R. Maybe I've just been looking at it too long. I've got strings of the form
'thing1 - thing2'
'thingA - thingB'
where the first is separated from the second by a space, a dash, and another space. The first thing is a combination of letters, digits, slashes, and periods; the second can be the same, or not exist (in which case there is also no separating dash). I want to use regmatches with gregexpr to find patterns matching the first and second parts. That's something like
regmatches(
'thing1 - thing2',
gregexpr('^(\\w|\\s|\\.|/)+(\\s-\\s){0,1}', 'thing1 - thing2', perl=T)
)
Fine and well. But sometimes thing1 is tricky, with a dash with no spaces (eg 10-43), or it can be the exact string Blue - MC, which obviously messes up the "separate by \\s-\\s" rule. And I just can't seem to get the regex right! I tried
regmatches(
c('10-43', 'Blue - MC'),
gregexpr(
'^\\w(\\w|\\s|\\.|/\\S-\\S)+\\s-\\s{0,1}|^Blue\\s-\\sMC',
c('10-43', 'Blue - MC'), perl=T
)
)
and I get c('10', 'Blue'). Help? Thanks!
and
I know you said you want to use gregexpr and regmatches, but why not strsplit since all you're doing is splitting the strings that will "always be separated by a space-dash-space"?
Per your comment, you can split at space-dash-space, but keep Blue - MC by simply removing Blue - MC from the list before applying the split. Then you can add it back in afterward.
> things <- c('thing1 - thing2', 'thingA - thingB', 'thingC', 'Blue - MC')
> w <- which(things == 'Blue - MC')
> ( s <- c(strsplit(things[-w], " - ", fixed = TRUE), things[w]) )
#[[1]]
#[1] "thing1" "thing2"
#[[2]]
#[1] "thingA" "thingB"
#[[3]]
#[1] "thingC"
#[[4]]
#[1] "Blue - MC"
Then if you only want the first of each of those,
> sapply(s, "[", 1)
#[1] "thing1" "thingA" "thingC" "Blue - MC"
When I want to capture parts of a message, I like to use the regcapturedmatches.R helper function. I would use it like this
v <- c("thing1 - thing2", "thingalone","Blue-MC","1 - 2")
m <- gregexpr('^(.*?)(?:\\s-\\s(.*))?$', v, perl=T)
regmatches(v, m)
do.call(rbind, regcapturedmatches(v,m))
That returns
[,1] [,2]
[1,] "thing1" "thing2"
[2,] "thingalone" ""
[3,] "Blue-MC" ""
[4,] "1" "2"
Which I believe satisfies your expectations.
Related
This question already has answers here:
Regular Expression to get a string between parentheses in Javascript
(10 answers)
Closed 7 years ago.
Suppose I have a string like "A B C (123-456-789)", I'm wondering what's the best way to retrieve "123-456-789" from it.
strsplit("A B C (123-456-789)", "\\(")
[[1]]
[1] "A B C" "123-456-789)"
If we want to extract the digits with - between the braces, one option is str_extract. If there are multiple patterns within a string, use str_extract_all
library(stringr)
str_extract(str1, '(?<=\\()[0-9-]+(?=\\))')
#[1] "123-456-789"
str_extract_all(str2, '(?<=\\()[0-9-]+(?=\\))')
In the above codes, we are using regex lookarounds to extract the numbers and the -. The positive lookbehind (?<=\\()[0-9-]+ matches numbers along with - ([0-9-]+) in (123-456-789 and not in 123-456-789. Similarly the lookahead ('[0-9-]+(?=\)') matches numbers along with - in 123-456-789) and not in 123-456-798. Taken together it matches all the cases that satisfy both the conditions (123-456-789) and extract those in between the lookarounds and not with cases like (123-456-789 or 123-456-789)
With strsplit you can specify the split as [()]. We keep the () inside the square brackets to [] to treat it as characters or else we have to escape the parentheses ('\\(|\\)').
strsplit(str1, '[()]')[[1]][2]
#[1] "123-456-789"
If there are multiple substrings to extract from a string, we could loop with lapply and extract the numeric split parts with grep
lapply(strsplit(str2, '[()]'), function(x) grep('\\d', x, value=TRUE))
Or we can use stri_split from stringi which has the option to remove the empty strings as well (omit_empty=TRUE).
library(stringi)
stri_split_regex(str1, '[()A-Z ]', omit_empty=TRUE)[[1]]
#[1] "123-456-789"
stri_split_regex(str2, '[()A-Z ]', omit_empty=TRUE)
Another option is rm_round from qdapRegex if we are interested in extracting the contents inside the brackets.
library(qdapRegex)
rm_round(str1, extract=TRUE)[[1]]
#[1] "123-456-789"
rm_round(str2, extract=TRUE)
data
str1 <- "A B C (123-456-789)"
str2 <- c("A B C (123-425-478) A", "ABC(123-423-428)",
"(123-423-498) ABCDD",
"(123-432-423)", "ABC (123-423-389) GR (124-233-848) AK")
or with sub from base R:
sub("[^(]+\\(([^)]+)\\).*", "\\1", "A B C (123-456-789)")
#[1] "123-456-789"
Explanation:
[^(]+ : matches anything except an opening bracket
\\( : matches an opening bracket, which is just before what you want
([^)]+) : matches the pattern you want to capture (which is then retrieved in replacement="\\1"), which is anything except a closing bracket
\\).* matches a closing bracket followed by anything, 0 or more times
Another option with look-ahead and look-behind
sub(".*(?<=\\()(.+)(?=\\)).*", "\\1", "A B C (123-456-789)", perl=TRUE)
#[1] "123-456-789"
The capture groups in sub will target your desired output:
sub('.*\\((.*)\\).*', '\\1', str1)
[1] "123-456-789"
Extra check to make sure I pass #akrun's extended example:
sub('.*\\((.*)\\).*', '\\1', str2)
[1] "123-425-478" "123-423-428" "123-423-498" "123-432-423" "124-233-848"
You may try these gsub functions.
> gsub("[^\\d-]", "", x, perl=T)
[1] "123-456-789"
> gsub(".*\\(|\\)", "", x)
[1] "123-456-789"
> gsub("[^0-9-]", "", x)
[1] "123-456-789"
Few more...
> gsub("[0-9-](*SKIP)(*F)|.", "", x, perl=T)
[1] "123-456-789"
> gsub("(?:(?![0-9-]).)*", "", x, perl=T)
[1] "123-456-789"
Try this also:
k<-"A B C (123-456-789)"
regmatches(k,gregexpr("*.(\\d+).*",k))[[1]]
[1] "(123-456-789)"
With suggestion from #Arun:
regmatches(k, gregexpr('(?<=\\()[^A-Z ]+(?=\\))', k, perl=TRUE))[[1]]
With suggestion from #akrun:
regmatches(k, gregexpr('[0-9-]+', k))[[1]]
I need to match any 'r' that is preceded by two different vowels. For example, 'our' or 'pear' would be matching but 'bar' or 'aar' wouldn't. I did manage to match for the two different vowels, but I still can't make that the condition (...) of lookbehind for the ensuing 'r'. Neither (?<=...)r nor ...\\Kr yields any results. Any ideas?
x <- c('([aeiou])(?!\\1)(?=(?1))')
y <- c('our','pear','bar','aar')
y[grepl(paste0(x,collapse=''),y,perl=T)]
## [1] "our" "pear"`
These two solutions seem to work:
the why not way:
x <- '(?<=a[eiou]|e[aiou]|i[aeou]|o[aeiu]|u[aeio])r'
y[grepl(x, y, perl=T)]
the \K way:
x <- '([aeiou])(?!\\1)[aeiou]\\Kr'
y[grepl(x, y, perl=T)]
The why not way variant (may be more efficient because it searches the "r" before):
x <- 'r(?<=a[eiou]r|e[aiou]r|i[aeou]r|o[aeiu]r|u[aeio]r)'
or to quickly exclude "r" not preceded by two vowels (without to test the whole alternation)
x <- 'r(?<=[aeiou][aeiou]r)(?<=a[eiou]r|e[aiou]r|i[aeou]r|o[aeiu]r|u[aeio]r)'
As HamZa points out in the comments using skip and fail verbs is one way to do what we want. Basically we tell it to ignore cases where we have two identical vowels followed by "r"
# The following is the beginning of the regex and isn't just R code
# the ([aeiou]) captures the first vowel, the \\1 references what we captured
# so this gives us the same vowel two times in a row
# which we then follow with an "r"
# Then we tell it to skip/fail for this
([aeiou])\\1r(*SKIP)(*FAIL)
Now we told it to skip those cases so now we tell it "or cases where we have two vowels followed by an 'r'" and since we already eliminated the cases where those two vowels are the same this will get us what we want.
|[aeiou]{2}r
Putting it together we end up with
y <- c('our','pear','bar','aar', "aa", "ae", "are", "aeer", "ssseiras")
grep("([aeiou])\\1r(*SKIP)(*FAIL)|[aeiou]{2}r", y, perl = TRUE, value = TRUE)
#[1] "our" "pear" "sseiras"
Here is a less than elegant solution:
y[grepl("[aeiou]{2}r", y, perl=T) & !grepl("(.)\\1r", y, perl=T)]
Probably has some corner case failures where the first set matches at different location than the second set (will have to think about that), but something to get you started.
Another one through negative lookahead assertion.
> y <- c('our','pear','bar','aar', "aa", "ae", "are", "aeer", "ssseiras")
> grep("(?!(?:aa|ee|ii|oo|uu)r)[aeiou][aeiou]r", y, perl=TRUE, value=TRUE)
[1] "our" "pear" "ssseiras"
> grep("(?!aa|ee|ii|oo|uu)[aeiou][aeiou]r", y, perl=TRUE, value=TRUE)
[1] "our" "pear" "ssseiras"
(?!aa|ee|ii|oo|uu) asserts that the first two chars in the match won't be aa or ee or .... or uu. So this [aeiou][aeiou] would match any two vowels other but it wouldn't be repeated . That's why we set the condition at first. r matches the r which follows the vowels.
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.
My timestamp is in the form
0992006 09:00
I need to remove the leading zeros to get this form:
992006 9:00
Here's the code I'm using now, which doesn't remove leading zeros:
prediction$TIMESTAMP <- as.character(format(prediction$TIMESTAMP, '%j%Y %H:%M'))
Simplest way is to create your own boundary that asserts either the start of the string or a space precedes.
gsub('(^| )0+', '\\1', '0992006 09:00')
# [1] "992006 9:00"
You could do the same making the replacement exempt using a trick. \K resets the starting point of the reported match and any previously consumed characters are no longer included.
gsub('(^| )\\K0+', '', '0992006 09:00', perl=T)
# [1] "992006 9:00"
Or you could use sub and match until the second set of leading zeros.
sub('^0+([0-9]+ )0+', '\\1', '0992006 09:00')
# [1] "992006 9:00"
And to cover all possibilities, if you know that you will ever have a format like 0992006 00:00, simply remove the + quantifier from zero in the regular expression so it only removes the first leading zero.
str1 <- "0992006 09:00"
gsub("(?<=^| )0+", "", str1, perl=TRUE)
#[1] "992006 9:00"
For situations like below, it could be:
str2 <- "0992006 00:00"
gsub("(?<=^| )0", "", str2, perl=TRUE)
#[1] "992006 0:00"
Explanation
Here the idea is to use look behind (?<=^| )0+ to match 0s
if it occurs either at the beginning of the string
(?<=^
or |
if it follows after a space )0+
and replace those matched 0s by "" in the second part of the gsub argument.
In the second string, the hour and minutes are all 0's. So, using the first code would result in:
gsub("(?<=^| )0+", "", str2, perl=TRUE)
#[1] "992006 :00"
Here, it is unclear what the OP would accept as a result. So, I thought, instead of removing the whole 0s before the :, it would be better if one 0 was left. So, I replaced the multiple 0+ code to just one 0 and replace that by "".
Here's another option using a lookbehind
gsub("(^0)|(?<=\\s)0", "", "0992006 09:00", perl = TRUE)
## [1] "992006 9:00"
With sub:
sub("^[0]+", "", prediction$TIMESTAMP)
[1] "992006 09:00"
You can also use stringr without a regular expression, by using the substrings.
> library(stringr)
> str_c(str_sub(word(x, 1:2), 2), collapse = " ")
# [1] "992006 9:00"
Some more Perl regexes,
> gsub("(?<!:)\\b0+", "", "0992006 09:00", perl=T)
[1] "992006 9:00"
> gsub("(?<![\\d:])0+", "", "0992006 09:00", perl=T)
[1] "992006 9:00"
I have a vector a strings with a number of spaces in. I would like to split this into two vectors split by the final space. For example:
vec <- c('This is one', 'And another', 'And one more again')
Should become
vec1 = c('This is', 'And', 'And one more again')
vec2 = c('one', 'another', 'again')
Is there a quick and easy way to do this? I have done similar things before using gsub and regex, and have managed to get the second vector using the following
vec2 <- gsub(".* ", "", vec)
But can't work out how to get vec1.
Thanks in advance
Here is one way using a lookahead assertion:
do.call(rbind, strsplit(vec, ' (?=[^ ]+$)', perl=TRUE))
# [,1] [,2]
# [1,] "This is" "one"
# [2,] "And" "another"
# [3,] "And one more" "again"