My string patterns looks like this:
UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870 and I am trying to extract everything after the second last +, i.e. 180101:0050+10870.
Thus far, I managed to address the second last block 180101:0050 with this expression (?<=\+)[^\+]+(?=\+[^\+]*$) but fail to include the last block including the last +. Here is my sample: regex101
The expression is meant for R and I still need to escape the characters later on. This format it just for testing purposes in Regex101.
We could capture group based on the occurrence of + from the end ($) of the string.
sub(".*\\+([^+]+\\+[^+]+$)", "\\1", str1)
#[1] "180101:0050+10870"
data
str1 <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
You may use
\+\K[^+]+\+[^+]*$
Or, if you would like to use it with stringr::str_extract:
(?<=\+)[^+]+\+[^+]*$
See the regex demo. Details:
\+ - a + char
\K - match reset operator
(?<=\+) - location right after a + symbol
[^+]+ - one or more chars other than +
\+ - a +
[^+]+ - one or more chars other than +
$ - end of string.
See R demo online:
x <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
regmatches(x, regexpr("\\+\\K[^+]+\\+[^+]*$", x, perl=TRUE))
## => [1] "180101:0050+10870"
library(stringr)
str_extract(x, "(?<=\\+)[^+]+\\+[^+]*$")
## => [1] "180101:0050+10870"
Another way you can do in this case:
library(stringr)
str_extract("UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870", "\\d+:\\d+\\+\\d+")
#"180101:0050+10870"
Related
I want to extract all substrings that begin with M and are terminated by a *
The string below as an example;
vec<-c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
Would ideally return;
MGMTPRLGLESLLE
MTPRLGLESLLE
I have tried the code below;
regmatches(vec, gregexpr('(?<=M).*?(?=\\*)', vec, perl=T))[[1]]
but this drops the first M and only returns the first string rather than all substrings within.
"GMTPRLGLESLLE"
You can use
(?=(M[^*]*)\*)
See the regex demo. Details:
(?= - start of a positive lookahead that matches a location that is immediately followed with:
(M[^*]*) - Group 1: M, zero or more chars other than a * char
\* - a * char
) - end of the lookahead.
See the R demo:
library(stringr)
vec <- c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
matches <- stringr::str_match_all(vec, "(?=(M[^*]*)\\*)")
unlist(lapply(matches, function(z) z[,2]))
## => [1] "MGMTPRLGLESLLE" "MTPRLGLESLLE"
If you prefer a base R solution:
vec <- c("SHVANSGYMGMTPRLGLESLLE*A*MIRVASQ")
matches <- regmatches(vec, gregexec("(?=(M[^*]*)\\*)", vec, perl=TRUE))
unlist(lapply(matches, tail, -1))
## => [1] "MGMTPRLGLESLLE" "MTPRLGLESLLE"
This could be done instead with a for loop on a char array converted from you string.
If you encounter a M you start concatenating chars to a new string until you encounter a *, when you do encounter a * you push the new string to an array of strings and start over from the first step until you reach the end of your loop.
It's not quite as interesting as using REGEX to do it, but it's failsafe.
It is not possible to use regular expressions here, because regular languages don't have memory states required for nested matches.
stringr::str_extract_all("abaca", "a[^a]*a") only gives you aba but not the sorrounding abaca.
The first M was dropped, because (?<=M) is a positive look behind which is by definition not part of the match, but just behind it.
How can I match all words starting with plan_ and not ending with template without using invert = TRUE? In the below example, I'd like to match only the second string. I tried with negative lookahead but it does not work, maybe because of greediness?
names <- c("plan_x_template", "plan_x")
grep("^plan.*(?!template)$",
names,
value = TRUE, perl = TRUE
)
#> [1] "plan_x_template" "plan_x"
I mean one can also solve the problem with two regex calls but I'd like to see how it works the other way :-)
is_plan <- grepl("^plan_", names)
is_template <- grepl("_template$", names)
names[is_plan & !is_template]
#> [1] "plan_x"
You may use
names <- c("plan_x_template", "plan_x")
grep("^plan(?!.*template)",
names,
value = TRUE, perl = TRUE
)
See the R online demo
The ^plan(?!.*template) pattern matches:
^ - a start of string
plan - a plan substring
(?!.*template) - a negative lookahead that fails the match if, immediately to the left of the current location, there are 0+ chars other than line break chars (since perl = TRUE is used and the pattern is processed with a PCRE engine, the . does not match all possible chars as opposed to the default grep TRE regex engine), as many as possible, followed with template substring.
NOTE: In case of multiline strings, you need to use a DOTALL modifier in the regex, "(?s)^plan(?!.*template)".
I have a vector of strings that look like this:
a - bc/def_g - A/mn/us/ww
opq - rs/ts_uf - BC/wx/yza
Abc - so/dhie7u - XYZ/En/xy/jkq - QWNE
I'd like to get the text after 2nd dash (-) but before first flash (/), i.e. the result should look like
A
BC
XYZ
What is the best way to do it (the vector has more than 500K rows.)
Thanks
Suppose your string is defined like this:
string <- c("a - bc/def_g - A/mn/us/ww",
"opq - rs/ts_uf - BC/wx/yza",
"Abc - so/dhie7u - XYZ/En/xy/jkq - QWNE")
Then you can use sub
> sub(".*\\-\\s+([A-Z]+)/.*", "\\1", string)
[1] "A" "BC" "XYZ"
See regex in use here
^[^-]*-[^-]*-\s*\K[^/]+
^ Assert position at the start of the line
[^-]* Match any character except - any number of times
- Match this literally
[^-]* Match any character except - any number of times
- Match this literally
\s* Match any number of whitespace characters
\K Resets the starting point of the pattern. Any previously consumed characters are no longer included in the final match
[^/]+ Match any character except / one or more times
Alternatively, as suggested by Jan in the comments below (I believe it has since been deleted) ^(?:\[^-\]*-){2}\s*\K\[^/\]+ may be used. It's shorter and easily scalable, but more adds steps.
See code in use here
x <- c("a - bc/def_g - A/mn/us/ww", "opq - rs/ts_uf - BC/wx/yza", "Abc - so/dhie7u - XYZ/En/xy/jkq - QWNE")
m <- regexpr("^[^-]*-[^-]*-\\s*\\K[^/]+", x, perl=T)
regmatches(x, m)
Result: [1] "A" "BC" "XYZ"
Lest's say I have a string:
test <- "(pop+corn)-bread+salt"
I want to replace the plus sign that is only between parenthesis by '|', so I get:
"(pop|corn)-bread+salt"
I tried:
gsub("([+])","\\|",test)
But it replaces all the plus signs of the string (obviously)
If you want to replace all + symbols that are inside parentheses (if there may be 1 or more), you can use any of the following solutions:
gsub("\\+(?=[^()]*\\))", "|", x, perl=TRUE)
See the regex demo. Here, the + is only matched when it is followed with any 0+ chars other than ( and ) (with [^()]*) and then a ). It is only good if the input is well-formed and there is no nested parentheses as it does not check if there was a starting (.
gsub("(?:\\G(?!^)|\\()[^()]*?\\K\\+", "|", x, perl=TRUE)
This is a safer solution since it starts matching + only if there was a starting (. See the regex demo. In this pattern, (?:\G(?!^)|\() matches the end of the previous match (\G(?!^)) or (|) a (, then [^()]*? matches any 0+ chars other than ( and ) chars, and then \K discards all the matched text and \+ matches a + that will be consumed and replaced. It still does not handle nested parentheses.
Also, see an online R demo for the above two solutions.
library(gsubfn)
s <- "(pop(+corn)+unicorn)-bread+salt+malt"
gsubfn("\\((?:[^()]++|(?R))*\\)", ~ gsub("+", "|", m, fixed=TRUE), s, perl=TRUE, backref=0)
## => [1] "(pop(|corn)|unicorn)-bread+salt+malt"
This solves the problem of matching nested parentheses, but requires the gsubfn package. See another regex demo. See this regex description here.
Note that in case you do not have to match nested parentheses, you may use "\\([^()]*\\)" regex with the gsubfn code above. \([^()]*\) regex matches (, then any zero or more chars other than ( and ) (replace with [^)]* to match )) and then a ).
We can try
sub("(\\([^+]+)\\+","\\1|", test)
#[1] "(pop|corn)-bread+salt"
It's a simple question, but I'm not good with Regex. (I tried many expressions without success)
I want to replace all the text (replace for nothing) after a pattern.
My pattern is something like this:
/canais/*/
My data is:
/canais/b3/conheca-o-pai-dos-indices-da-b3/
/canais/cpbs/cvm-abre-audiencia-publica-de-instruc
/canais/stocche-forbes/dividendo-controverso/
The desired result is:
/canais/b3/
/canais/cpbs/
/canais/stocche-forbes/
How can I do it with gsub?
Thanks
You may use the following sub:
x <- c("/canais/b3/conheca-o-pai-dos-indices-da-b3/","/canais/cpbs/cvm-abre-audiencia-publica-de-instruc","/canais/stocche-forbes/dividendo-controverso/")
sub("^(/canais/[^/]+/).*", "\\1", x)
See the online R demo
Details:
^ - start of string
(/canais/[^/]+/) - Group 1 (later referred to with \1) capturing:
/canais/ - a substring /canais/
[^/]+ - 1 or more chars other than /
/ - a slash
.* - any 0+ chars up to the end of string.