I'm really putting time into learning regex and I'm playing with different toy scenarios. One setup I can't get to work is to grab from the beginning of a string to n occurrence of a character where n > 1.
Here I can grab from the beginning of the string to the first underscore but I can't generalize this to the second or third underscore.
x <- c("a_b_c_d", "1_2_3_4", "<_?_._:")
gsub("_.*$", "", x)
Here's what I'm trying to achieve with regex. (`sub`/`gsub`):
## > sapply(lapply(strsplit(x, "_"), "[", 1:2), paste, collapse="_")
## [1] "a_b" "1_2" "<_?"
#or
## > sapply(lapply(strsplit(x, "_"), "[", 1:3), paste, collapse="_")
## [1] "a_b_c" "1_2_3" "<_?_."
Related post: regex from first character to the end of the string
Here's a start. To make this safe for general use, you'll need it to properly escape regular expressions' special characters:
x <- c("a_b_c_d", "1_2_3_4", "<_?_._:", "", "abcd", "____abcd")
matchToNth <- function(char, n) {
others <- paste0("[^", char, "]*") ## matches "[^_]*" if char is "_"
mainPat <- paste0(c(rep(c(others, char), n-1), others), collapse="")
paste0("(^", mainPat, ")", "(.*$)")
}
gsub(matchToNth("_", 2), "\\1", x)
# [1] "a_b" "1_2" "<_?" "" "abcd" "_"
gsub(matchToNth("_", 3), "\\1", x)
# [1] "a_b_c" "1_2_3" "<_?_." "" "abcd" "__"
How about:
gsub('^(.+_.+?).*$', '\\1', x)
# [1] "a_b" "1_2" "<_?"
Alternatively you can use {} to indicate the number of repeats...
sub('((.+_){1}.+?).*$', '\\1', x) # {0} will give "a", {1} - "a_b", {2} - "a_b_c" and so on
So you don't have to repeat yourself if you wanted to match the nth one...
second underscore in perl style regex:
/^(.?_.?_)/
and third:
/^(.*?_.*?_.*?_)/
Maybe something like this
x
## [1] "a_b_c_d" "1_2_3_4" "<_?_._:"
gsub("(.*)_", "\\1", regmatches(x, regexpr("([^_]*_){1}", x)))
## [1] "a" "1" "<"
gsub("(.*)_", "\\1", regmatches(x, regexpr("([^_]*_){2}", x)))
## [1] "a_b" "1_2" "<_?"
gsub("(.*)_", "\\1", regmatches(x, regexpr("([^_]*_){3}", x)))
## [1] "a_b_c" "1_2_3" "<_?_."
Using Justin's approach this was what I devised:
beg2char <- function(text, char = " ", noc = 1, include = FALSE) {
inc <- ifelse(include, char, "?")
specchar <- c(".", "|", "(", ")", "[", "{", "^", "$", "*", "+", "?")
if(char %in% specchar) {
char <- paste0("\\", char)
}
ins <- paste(rep(paste0(char, ".+"), noc - 1), collapse="")
rep <- paste0("^(.+", ins, inc, ").*$")
gsub(rep, "\\1", text)
}
x <- c("a_b_c_d", "1_2_3_4", "<_?_._:")
beg2char(x, "_", 1)
beg2char(x, "_", 2)
beg2char(x, "_", 3)
beg2char(x, "_", 4)
beg2char(x, "_", 3, include=TRUE)
Related
I have "elephant_giraffe_lion" and "monkey_tiger" strings.
The condition here is if there are two or more delimiters, I want to split at the second delimiter and if there is only one delimiter, I want to split at that delimiter. So the results I want to get in this example are "elephant_giraffe" and "monkey".
mystring<-c("elephant_giraffe_lion", "monkey_tiger")
result
"elephant_giraffe" "monkey"
You can anchor your split to the end of the string using $,
unlist(strsplit(mystring, "_[a-z]+$"))
# [1] "elephant_giraffe" "monkey"
Edit
The above only matches the last "_", not accounting for cases where there are more than two "_". For the more general case, you could try
mystring<-c("elephant_giraffe_lion", "monkey_tiger", "dogs", "foo_bar_baz_bap")
tmp <- gsub("([^_]+_[^_]+).*", "\\1", mystring)
tmp[tmp==mystring] <- sapply(strsplit(tmp[tmp==mystring], "_"), `[[`, 1)
tmp
# [1] "elephant_giraffe" "monkey" "dogs" "foo_bar"
You could also use gsubfn, to process the match with a function
library(gsubfn)
f <- function(x,y) if (y==x) strsplit(y, "_")[[1]][[1]] else y
gsubfn("([^_]+_[^_]+).*", f, mystring, backref=1)
# [1] "elephant_giraffe" "monkey" "dogs" "foo_bar"
As I posted an answer on your other related question, a base R solution:
x <- c('elephant_giraffe_lion', 'monkey_tiger', 'foo_bar_baz_bap')
sub('^(?|([^_]*_[^_]*)_.*|([^_]*)_[^_]*)$', '\\1', x, perl=TRUE)
# [1] "elephant_giraffe" "monkey" "foo_bar"
I would like to capture the first match, and return NA if there is no match.
regexpr("a+", c("abc", "def", "cba a", "aa"), perl=TRUE)
# [1] 1 -1 3 1
# attr(,"match.length")
# [1] 1 -1 1 2
x <- c("abc", "def", "cba a", "aa")
m <- regexpr("a+", x, perl=TRUE)
regmatches(x, m)
# [1] "a" "a" "aa"
So I expected "a", NA, "a", "aa"
Staying with regexpr:
r <- regexpr("a+", x)
out <- rep(NA,length(x))
out[r!=-1] <- regmatches(x, r)
out
#[1] "a" NA "a" "aa"
use regexec instead, since it returns a list which will allow you to catch the character(0)'s before unlisting
R <- regmatches(x, regexec("a+", x))
unlist({R[sapply(R, length)==0] <- NA; R})
# [1] "a" NA "a" "aa"
In R 3.3.0, it is possible to pull out both the matches and the non-matched results using the invert=NA argument. From the help file, it says
if invert is NA, regmatches extracts both non-matched and matched substrings, always starting and ending with a non-match (empty if the match occurred at the beginning or the end, respectively).
The output is a list, typically, in most cases of interest, (matching a single pattern), regmatches with this argument will return a list with elements of either length 3 or 1. 1 is the case of where no matches are found and 3 is the case with a match.
myMatch <- regmatches(x, m, invert=NA)
myMatch
[[1]]
[1] "" "a" "bc"
[[2]]
[1] "def"
[[3]]
[1] "cb" "a" " a"
[[4]]
[1] "" "aa" ""
So to extract what you want (with "" in place of NA), you can use sapply as follows:
myVec <- sapply(myMatch, function(x) {if(length(x) == 1) "" else x[2]})
myVec
[1] "a" "" "a" "aa"
At this point, if you really want NA instead of "", you can use
is.na(myVec) <- nchar(myVec) == 0L
myVec
[1] "a" NA "a" "aa"
Some revisions:
Note that you can collapse the last two lines into a single line:
myVec <- sapply(myMatch, function(x) {if(length(x) == 1) NA_character_ else x[2]})
The default data type of NA is logical, so using it will result in additional data conversions. Using the character version NA_character_, avoids this.
An even slicker extraction method for the final line is to use [:
sapply(myMatch, `[`, 2)
[1] "a" NA "a" "aa"
So you can do the whole thing in a fairly readable single line:
sapply(regmatches(x, m, invert=NA), `[`, 2)
Using more or less the same construction as yours -
chars <- c("abc", "def", "cba a", "aa")
chars[
regexpr("a+", chars, perl=TRUE) > 0
][1] #abc
chars[
regexpr("q", chars, perl=TRUE) > 0
][1] #NA
#vector[
# find all indices where regexpr returned positive value i.e., match was found
#][return the first element of the above subset]
Edit - Seems like I misunderstood the question. But since two people have found this useful I shall let it stay.
You can use stringr::str_extract(string, pattern). It will return NA if there is no matches. It has simpler function interface than regmatches() as well.
string<-c(" this is a string ")
Is it possible to trim-off the white spaces on both the sides of the string (or just one side as required) and replace it with a desired character, such as this, in R? The number of white spaces differ on each side of the string and have to be retained on replacement.
"~~~~~~~this is a string~~"
This seems like an inefficient way of doing it, but maybe you should be looking in the direction of gregexpr and regmatches instead of gsub:
x <- " this is a string "
pattern <- "^ +?\\b|\\b? +$"
startstop <- gsub(" ", "~", regmatches(x, gregexpr(pattern, x))[[1]])
text <- paste(regmatches(x, gregexpr(pattern, x), invert=TRUE)[[1]], collapse="")
paste0(startstop[1], text, startstop[2])
# [1] "~~~~this is a string~~"
And, for fun, as a function, and a "vectorized" function:
## The function
replaceEnds <- function(string) {
pattern <- "^ +?\\b|\\b? +$"
startstop <- gsub(" ", "~", regmatches(string, gregexpr(pattern, string))[[1]])
text <- paste(regmatches(string, gregexpr(pattern, string), invert = TRUE)[[1]],
collapse = "")
paste0(startstop[1], text, startstop[2])
}
## use Vectorize here if you want to apply over a vector
vReplaceEnds <- Vectorize(replaceEnds)
Some sample data:
myStrings <- c(" Four at the start, 2 at the end ",
" three at the start, one at the end ")
vReplaceEnds(myStrings)
# Four at the start, 2 at the end three at the start, one at the end
# "~~~~Four at the start, 2 at the end~~" "~~~three at the start, one at the end~"
Use gsub:
gsub(" ", "~", " this is a string ")
[1] "~~~~this~is~a~string~~"
This function uses regular expressions to replace (i.e. sub), all occurrences of a pattern inside a string.
In your case, you have to express the pattern in a special way:
gsub("(^ *)|( *$)", "~~~", " this is a string ")
[1] "~~~this is a string~~~"
The pattern means:
(^ *): Find one or more spaces at the start of the string
( *$): Find one or more spaces at the end of the string
`|: The OR operator
Now you can use this approach to tackle your problem of replacing each space with a new character:
txt <- " this is a string "
foo <- function(x, new="~"){
lead <- gsub("(^ *).*", "\\1", x)
last <- gsub(".*?( *$)", "\\1", x)
mid <- gsub("(^ *)|( *$)", "", x)
paste0(
gsub(" ", new, lead),
mid,
gsub(" ", new, last)
)
}
> foo(" this is a string ")
[1] "~~~~this is a string~~"
> foo(" And another one ")
[1] "~And another one~~~~~~~~"
For more, see ?gsub or ?regexp.
Or using a more complex pattern matching and gsub...
gsub("\\s(?!\\b)|(?<=\\s)\\s(?=\\b)", "~", " this is a string " , perl = TRUE )
#[1] "~~~~this is a string~~"
Or with #AnandaMahto's data:
gsub("\\s(?!\\b)|(?<=\\s)\\s(?=\\b)", "~", myStrings , perl = TRUE )
#[1] "~~~~Four at the start, 2 at the end~~"
#[2] "~~~three at the start, one at the end~"
Explanation
This uses the positive and negative lookahead and look behind assertions:
\\s(?!\\b) - match a space, \\s not followed by a word boundary, (?!\\b). This would work by itself for everything except the last space before the first word, i.e. by itself we would get
"~~~~ this is a string~~". So we need another pattern...
(?<=\\s)\\s(?=\\b) - match a space, \\s that is preceded by another space, (?<=\\s) and is followed by a word boundary, (?=\\b).
And it is gsub so it tries to make the maximal number of matches that it can.
The below code works so long as before and after strings have no characters that are special to a regex:
before <- 'Name of your Manager (note "self" if you are the Manager)' #parentheses cause problem in regex
after <- 'CURRENT FOCUS'
pattern <- paste0(c('(?<=', before, ').*?(?=', after, ')'), collapse='')
ex <- regmatches(x, gregexpr(pattern, x, perl=TRUE))
Does R have a function to escape strings to be used in regexes?
In Perl, there is http://perldoc.perl.org/functions/quotemeta.html for doing exactly that. If the doc is correct when it says
Returns the value of EXPR with all the ASCII non-"word" characters backslashed. (That is, all ASCII characters not matching /[A-Za-z_0-9]/ will be preceded by a backslash in the returned string, regardless of any locale settings.)
then you can achieve the same by doing:
quotemeta <- function(x) gsub("([^A-Za-z_0-9])", "\\\\\\1", x)
And your pattern should be:
pattern <- paste0(c('(?<=', quotemeta(before), ').*?(?=', quotemeta(after), ')'),
collapse='')
Quick sanity check:
a <- "he'l(lo)"
grepl(a, a)
# [1] FALSE
grepl(quotemeta(a), a)
# [1] TRUE
Use \Q...\E to surround the verbatim subpatterns:
# test data
before <- "A."
after <- ".Z"
x <- c("A.xyz.Z", "ABxyzYZ")
pattern <- sprintf('(?<=\\Q%s\\E).*?(?=\\Q%s\\E)', before, after)
which gives:
> gregexpr(pattern, x, perl = TRUE) > 0
[1] TRUE FALSE
dnagirl, such a function exists and is glob2rx
a <- "he'l(lo)"
tt <- glob2rx(a)
# [1] "^he'l\\(lo)$"
before <- 'Name of your Manager (note "self" if you are the Manager)'
tt <- glob2rx(before)
# [1] "^Name of your Manager \\(note \"self\" if you are the Manager)$"
You can just remove the "^" and "$" from the strings by doing:
substr(tt, 2, nchar(tt)-1)
# [1] "he'l\\(lo)"
Is it possible in R to search for a regex in a vector as if all the elements are a collapsed single element? If we collapse all the elements into one to do this, it becomes impossible to put them back to their element-wise form after the search.
here is a vector.
vector<-c("I", "met", "a", "cow")
now, the search word is "meta" (elements 2 and 3 collapsed).
Let's say my task is to merge the two elements across which the search string lies.
So what I expect is this:
vector = "I", "meta", "cow"
Is it possible to do this? Please help.
If you'd like something that matches "meta" but not "taco", this will do the trick:
myFun <- function(vector, word) {
D <- "UnLiKeLyStRiNg"
## Construct a string on which you'll perform regex-search
xx <- paste0(paste0(D, vector, collapse=""), D)
## Construct the regex pattern
start <- paste0("(?<=", D, ")")
mid <- paste0(strsplit(word, "")[[1]], collapse=paste0("(", D, ")?"))
end <- paste0("(?=", D, ")")
pat <- paste0(start, mid, end)
## Use it
strsplit(gsub(pat, word, xx, perl=TRUE), D)[[1]][-1]
}
vector <- c("I", "met", "a", "cow")
myFun(vector, "meta")
# [1] "I" "meta" "cow"
myFun(vector, "taco")
# [1] "I" "met" "a" "cow"
myFun(vector, "Imet")
# [1] "Imet" "a" "cow"
myFun(vector, "Ime")
# [1] "I" "met" "a" "cow"
If only complete elements should merged, you could try this approach:
mergeRegExpr <- function(x, pattern) {
str <- paste(x, sep="", collapse="")
## find starting position of each word
wordStart <- head(cumsum(c(1, nchar(x))), -1)
## look for pattern
rx <- regexpr(pattern=pattern, text=str, fixed=TRUE)
## pos of matching pattern == rx+nchar(pattern)-1
rxEnd <- rx+attr(rx, "match.length")-1
## which vector elements doesn't match pattern
sel <- wordStart < rx | wordStart > rxEnd
## insert merged elements
return(append(x[sel], paste(x[!sel], collapse=""), rx-1))
}
vector <- c("I", "met", "a", "cow")
mergeRegExpr(vector, "meta")
# "I" "meta" "cow"
mergeRegExpr(vector, "acow")
# "I" "met" "acow"
mergeRegExpr(vector, "Imeta")
# "Imeta" "cow"
## partial matching doesn't work
mergeRegExpr(vector, "taco")
# "I" "metacow"
Building on Carl Witthoft's comment, my solution was not with regex, but with basic matching:
# A slightly longer vector
v = c("I", "met", "a", "cow", "today",
"You", "met", "a", "cow", "today")
# Create the combinations of each pair
temp1 = sapply(1:(length(v)-1),
function(x) paste0(v[x], v[x+1]))
# Grab the index of the desired search term
temp2 = which(temp1 %in% "meta")
# The following also works.
# Don't know what's faster/better.
# temp2 = grep("meta", temp1)
# Do some manual substitution and deletion
v[temp2] <- "meta"
v <- v[-(temp2+1)]
I don't think this is an ideal situation at all though.