Let's say I have the string x <- "AbC" and I want to put an ampersand in between each letter. I would have assumed I could just do gsub("([a-zA-Z])([a-zA-Z])", "\\1 & \\2", x), but that produces "A & bC". Why doesn't gsub recognize the second set of letters that match the regex? It's not like gsub only replaces the first match found. If I have x <- "AbC DE" and run the same command, I get "A & bC D & E".
What am I missing in terms of how gsub is doing it's replacement? I would have expected outputs of "A & b & C" and "A & b & C D & E" from the two inputs above.
Because if a character present in one match, regex engine won't match the same character again. That is, it won't do overlapping matches.. Use lookaround to overcome this..
gsub("([a-zA-Z])(?=[a-zA-Z])", "\\1 & ", x, perl=T)
DEMO
Related
I have strings like this:
a <- "this string has an even number of words"
b <- "this string doesn't have an even number of words"
I want to replace every other space with a new line. So the output would look like this...
myfunc(a)
# "this string\nhas an\neven number\nof words"
myfunc(b)
# "this string\ndoesn't have\nan even\nnumber of\nwords"
I've accomplished this by doing a strsplit, paste-ing a newline on even numbered words, then paste(a, collapse=" ") them back together into one string. Is there a regular expression to use with gsub that can accomplish this?
#Jota suggested a simple and concise way:
myfunc = function(x) gsub("( \\S+) ", "\\1\n", x) # Jota's
myfunc2 = function(x) gsub("([^ ]+ [^ ]+) ", "\\1\n", x) # my idea
lapply(list(a,b), myfunc)
[[1]]
[1] "this string\nhas an\neven number\nof words"
[[2]]
[1] "this string\ndoesn't have\nan even\nnumber of\nwords"
How it works. The idea of "([^ ]+ [^ ]+) " regex is (1) "find two sequences of words/nonspaces with a space between them and a space after them" and (2) "replace the trailing space with a newline".
#Jota's "( \\S+) " is trickier -- it finds any word with a space before and after it and then replaces the trailing space with a newline. This works because the first word that is caught by this is the second word of the string; and the next word caught by it is not the third (since we have already "consumed"/looked at the space in front of the third word when handling the second word), but rather the fourth; and so on.
Oh, and some basic regex stuff.
[^xyz] means any single char except the chars x, y, and z.
\\s is a space, while \\S is anything but a space
x+ means x one or more times
(x) "captures" x, allowing for reference in the replacement, like \\1
I hope this isn't a duplicated, I didn't find an answer and I need help from regexp wizards.
I have a string and I would like to replace the second space found in it by a \n, but I don't know how to use indices (this way) in a regular expression :
For example :
# I have :
"a b c d e f"
# I want :
> "a b/nc d e f"
Also I would like to know how I can "repeat" this replacement: each two occurences of space replace by \n.
For example :
"a b c d e f"
> "a b\nc d\ne f"
(\\S+\\s+\\S+)\\s+
You can use this and replace by \1\n or $1\n.See demo.
https://regex101.com/r/yG7zB9/29
I try to extract passages like 44.11.36.00-1 (precisely, nn.nn.nn.nn-n, where n stands for any number from 0-9) from text in R.
I want to extract passages if they are "sticked" to non-number marks:
44.11.36.00-1 extracted from nsfghstighsl44.11.36.00-1vsdfgh is OK
44.11.36.00-1 extracted from fa0044.11.36.00-1000 is NOT
I have read that str_extract_all is not working with Lookbehind and Lookahead expressions, so I sadly came back to grep, but cannot deal with it:
> pattern1 <- "(?<![0-9]{1})[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}-[0-9]{1}(?![0-9]{1})"
> grep(pattern1, "dyj44.11.36.00-1aregjspotgji 44113600-1 agdtklj441136001 ", perl=TRUE, value = TRUE)
[1] "dyj44.11.36.00-1aregjspotgji 44113600-1 agdtklj441136001 "
which is not the result I expected.
I thought that:
(?<![0-9]{1}) means "match expression which is not preceeded by a number"
[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}-[0-9]{1} stands for the expression I seek for
(?![0-9]{1}) means "match expression which is not followed by a number"
You don't actually need lookahead or lookbehind with this approach. Just parenthesize the portion you want extracted:
library(gsubfn)
x <- c("nsfghstighsl44.11.36.00-1vsdfgh", "fa0044.11.36.00-1000") # test data
pat <- "(^|\\D)(\\d{2}[.]\\d{2}[.]\\d{2}[.]\\d{2}-\\d)(\\D|$)"
strapply(x, pat, ~ ..2, simplify = c)
## "44.11.36.00-1"
Note that ~ ..2 is short for the function function(...) ..2 which means grab the match to the second parenthesized portion in the regular expression. We could also have written function(x, y, z) y or x + y + z ~ y .
Note: The question seems to say that a non-numeric must come directly before and after the string but based on comments that have since disappeared it appears that what was really wanted was that the string be either at the beginning or just after a non-number and must either be at the end or folowed by a non-number. The answer has been so modified.
AS #Roland said in his comment, you need to use regmatches instead of grep
> s <- "nsfghstighsl44.11.36.00-1vsdfgh"
> m <- gregexpr("(?<![0-9]{1})[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}\\.[0-9]{2}-[0-9]{1}(?![0-9]{1})", s, perl=TRUE)
> regmatches(s, m)
[1] "44.11.36.00-1"
A reduced one,
> x <- c('nsfghstighsl44.11.36.00-1vsdfgh', 'fa0044.11.36.00-1000')
> m <- gregexpr("(?<!\\d)\\d{2}\\.\\d{2}\\.\\d{2}\\.\\d{2}-\\d(?!\\d)", x, perl=TRUE)
> regmatches(x, m)
[1] "44.11.36.00-1"
let's define the following string s:
s <- "$ A; B; C;"
I need to translate s into:
"$ A; $B; $C;"
the semicolon is the separator. However, $ is only one of 3 special characters which can appear in the string. The data frame m holds all 3 special characters:
m <- data.frame(sp = c("$", "%", "&"))
I first used strsplit to split the string using the semicolon as the separator
> strsplit(s, ";")
[[1]]
[1] "$ A" " B" " C"
I think the next step would be to use grep or match to check if the first string contains any of the 3 special characters defined in data frame m. If so, maybe use gsub to insert the matched special character into the remaining sub strings. Then simple use paste with collapse = "" to merge the substrings together again. Does that make sense?
Cheers
What about something like this:
getmeout = gsub("[$|%|& ]", "", unlist(strsplit(s, ";")))
whatspecial = unique(gsub("[^$|%|&]", "", s))
whatspecial
# [1] "$"
getmeout
# [1] "A" "B" "C"
paste0(whatspecial, getmeout, sep=";", collapse="")
# [1] "$A;$B;$C;"
Here is one method:
library(stringr)
separator <- '; '
# extract the first part
first.part <- str_split(s, separator)[[1]][1]
first.part
# [1] "$ A"
# try to identify your special character
special <- m$sp[str_detect(first.part, as.character(m$sp))]
special
# [1] $
# Levels: $ & %
# make sure you only matched one of them
stopifnot(length(special) == 1)
# search and replace
gsub(separator, paste(separator, special, sep=""), s)
# [1] "$ A; $B; $C;"
Let me know if I missed some of your assumptions.
Back-referencing turns it into a one-liner:
s <- c( "$ A; B; C;", "& A; B; C;", "% A; B; C;" )
ms = c("$", "%", "&")
s <- gsub( paste0("([", paste(ms,collapse="") ,"]) ([A-Z]); ([A-Z]); ([A-Z]);") , "\\1 \\2; \\1 \\3; \\1 \\4" , s)
> s
[1] "$ A; $ B; $ C" "& A; & B; & C" "% A; % B; % C"
You can then make the regular expression appropriately generic (match more than one space, more than one alphanumeric character, etc.) if you need to.
I needed to retain the words enclosed in brackets and delete the others in the following string.
(a(b(c)d)(e)f)
So what I expected would be (((c))(e)).
To delete a, b, d, f, I tried the 'not followed by' regex.
str <- "(a(b(c)d)(e)f)"
gsub("([a-z]+)(?!\\))", "", str) #(sub. anything that isn't followed by a ")" )
The message shows my regex in invalid. As I can see, the brackets in the second part of the regex "(?!\))" don't match properly. As for my editor, the first "(" matches with the immediately following ")", which is not meant to be a closure bracket (the one to its right is). I could make out just this error from my regex. Can you please tell me what actually is wrong? Is there any other way to do this?
In two steps, and using positive lookaheads:
str1 <- gsub("\\([a-z](?=\\()", "\\(", str, perl=TRUE)
str1
# [1] "(((c)d)(e)f)"
str2 <- gsub("\\)[a-z](?=\\))", "\\)", str1, perl=TRUE)
str2
# [1] "(((c))(e))"
Edit: it turns out you can even do it in one:
gsub("([\\(\\)])[a-z](?=\\1)", "\\1", str, perl=TRUE)
# [1] "(((c))(e))"
I agree with #Dason's comment:
st <- "(a(b(c)d)(e)f)"
while(grepl("\\([a-z]+\\(",st)) {
st <- sub("\\([a-z]+(\\(.+\\))[a-z]+\\)","\\1",st)
}
> st
[1] "(c)(e)"
Written on my iPad :-)