I have this string :
*Field1=0(1936-S),Field13=0(2),Field2=0(),Field4=0(19.01.17),Field3=0(),Field5700=0(),Field5400=0(KS),Field14=0(21)*
And I need store n-th value of FieldN so I tried this:
(?<=Field[0-9]=0){4}?.*?(?=,)
This give me (1936-S) when I put N=4, I expect 19.01.17.
Can you help me please?
You may use
^\*(?:Field[0-9]+=0\([^)]*\),){3}Field[0-9]+=0\(\K[^)]*
(demo) or a capturing group based pattern:
^\*(?:Field[0-9]+=0\([^)]*\),){3}Field[0-9]+=0\(([^)]*)
See the regex demo
Details
^ - start of a string
\* - a * char
(?:Field[0-9]+=0\([^)]*\),){3} - 3 sequences of:
Field - a Field substring
[0-9]+ - 1 or more digits
=0\( - a =0( substring
[^)]* - any 0+ chars other than )
\), - a ), substring
Field[0-9]+=0\( - a Field subtring followed with 1+ digits, and then =0( substring
([^)]*) - Group 1: any 0+ chars other than ).
Related
I have seen in manuals how to use grep to match either a pattern or an exact string. However, I cannot figure out how to do both at the same time. I have a latex file where I want to find the following pattern:
\caption[SOME WORDS]
and replace it with:
\caption[\textit{SOME WORDS}]
I have tried with:
texfile <- sub('\\caption[','\\caption[\\textit ', texfile, fixed=TRUE)
but I do not know how to tell grep that there should be some text after the square bracket, and then a closed square bracket.
You can use
texfile <- "\\caption[SOME WORDS]" ## -> \caption[\textit{SOME WORDS}]
texfile <-gsub('(\\\\caption\\[)([^][]*)]','\\1\\\\textit{\\2}]', texfile)
cat(texfile)
## -> \caption[\textit{SOME WORDS}]
See the R demo online.
Details:
(\\caption\[) - Group 1 (\1 in the replacement pattern): a \caption[ string
([^][]*) - Group 2 (\2 in the replacement pattern): any zero or more chars other than [ and ]
] - a ] char.
Another solution based on a PCRE regex:
gsub('\\Q\\caption[\\E\\K([^][]*)]','\\\\textit{\\1}]', texfile, perl=TRUE)
See this R demo online. Details:
\Q - start "quoting", i.e. treating the patterns to the right as literal text
\caption[ - a literal fixed string
\E - stop quoting the pattern
\K - omit text matched so far
([^][]*) - Group 1 (\1): any zero or more non-bracket chars
] - a ] char.
My sample data is:
c("2\tNO PEMJNUM\t 2\tALTOGETHER HOW MANY JOBS\t216 - 217",
"1\tREFERENCE PERSON 2\tSPOUSE 3\tCHILD 4\tOTHER RELATIVE (PRIMARY FAMILY & UNREL) PRFAMTYP\t2\tFAMILY TYPE RECODE\t155 - 156",
"5\tUNABLE TO WORK PUBUS1\t 2\tLAST WEEK DID YOU DO ANY\t184 - 185",
"2\tNO PEIO1COW\t 2\tINDIVIDUAL CLASS OF WORKER CODE\t432 - 433"
For each line, I'm looking to extract (they are variable names):
Line 1: "PEMJNUM"
Line 2: "PRFAMTYP"
Line 3: "PUBUS1"
Line 4: "PEIO1COW"
My initial goal was to gsub remove the characters to the left and right of each variable name to leave just the variable names, but I was only able to grab everything to the right of the variable name and had issues with grabbing characters to the left. (as shown here https://regexr.com/67r6j).
Not sure if there's a better way to do this!
You can use sub in the following way:
x <- c("2\tNO PEMJNUM\t 2\tALTOGETHER HOW MANY JOBS\t216 - 217",
"1\tREFERENCE PERSON 2\tSPOUSE 3\tCHILD 4\tOTHER RELATIVE (PRIMARY FAMILY & UNREL) PRFAMTYP\t2\tFAMILY TYPE RECODE\t155 - 156",
"5\tUNABLE TO WORK PUBUS1\t 2\tLAST WEEK DID YOU DO ANY\t184 - 185",
"2\tNO PEIO1COW\t 2\tINDIVIDUAL CLASS OF WORKER CODE\t432 - 433")
sub("^(?:.*\\b)?(\\w+)\\s*\\b2\\b.*", "\\1", x, perl=TRUE)
# => [1] "PEMJNUM" "PRFAMTYP" "PUBUS1" "PEIO1COW"
See the online regex demo and the R demo.
Details:
^ - start of string
(?:.*\b)? - an optional non-capturing group that matches any zero or more chars (other than line break chars since I use perl=TRUE, if you need to match line breaks, too, add (?s) at the pattern start) as many as possible, and then a word boundary position
(\w+) - Group 1 (\1): one or more word chars
\s* - zero or more whitespaces
\b - a word boundary
2 - a 2 digit
\b - a word boundary
.* - the rest of the line/string.
If there are always whitespaces before 2, the regex can be written as "^(?:.*\\b)?(\\w+)\\s+2\\b.*".
How do I match the year such that it is general for the following examples.
a <- '"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}'
b <- 'Þegar það gerist (1998/I) (TV)'
I have tried the following, but did not have the biggest success.
gsub('.+\\(([0-9]+.+\\)).?$', '\\1', a)
What I thought it did was to go until it finds a (, then it would make a group of numbers, then any character until it meets a ). And if there are several matches, I want to extract the first group.
Any suggestions to where I go wrong? I have been doing this in R.
You could use
library(stringr)
strings <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)')
years <- str_match(strings, "\\((\\d+(?: B\\.C\\.)?)")[,2]
years
# [1] "1953" "1998"
The expression here is
\( # (
(\d+ # capture 1+ digits
(?: B\.C\.)? # B.C. eventually
)
Note that backslashes need to be escaped in R.
Your pattern contains .+ parts that match 1 or more chars as many as possible, and at best your pattern could grab last 4 digit chunks from the incoming strings.
You may use
^.*?\((\d{4})(?:/[^)]*)?\).*
Replace with \1 to only keep the 4 digit number. See the regex demo.
Details
^ - start of string
.*? - any 0+ chars as few as possible
\( - a (
(\d{4}) - Group 1: four digits
(?: - start of an optional non-capturing group
/ - a /
[^)]* - any 0+ chars other than )
)? - end of the group
\) - a ) (OPTIONAL, MAY BE OMITTED)
.* - the rest of the string.
See the R demo:
a <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)', 'Johannes Passion, BWV. 245 (1725 Version) (1996) (V)')
sub("^.*?\\((\\d{4})(?:/[^)]*)?\\).*", "\\1", a)
# => [1] "1953" "1998" "1996"
Another base R solution is to match the 4 digits after (:
regmatches(a, regexpr("\\(\\K\\d{4}(?=(?:/[^)]*)?\\))", a, perl=TRUE))
# => [1] "1953" "1998" "1996"
The \(\K\d{4} pattern matches ( and then drops it due to \K match reset operator and then a (?=(?:/[^)]*)?\\)) lookahead ensures there is an optional / + 0+ chars other than ) and then a ). Note that regexpr extracts the first match only.
I want to replace the last number in a string using regex and gsub
S <- "abcd2efghi2.txt"
The last number and the position of the last number can vary.
So I've tried the regex
?<=[\d+])\b
gsub("?<=[\d+])\b", "", S)
but that doesn't seem to work
Appreciate any help.
You can achieve that with a default TRE engine using the following regex:
\d+(\D*)$
Replace with the \1 backreference.
Details
\d+ - 1 or more digits
(\D*) - Capturing group 1: any 0+ non-digit symbols
$ - end of string
\1 - a backreference to the Group 1 value (so as to restore the text matched and consumed with the (\D*) subpattern).
See the regex demo.
R code demo:
sub("\\d+(\\D*)$", "\\1", S)
## => [1] "abcd2efghi.txt"
You could use this regex:
\d+(?=\D*$)
It matches a sequence of digits when everything that follows consists of non-digits (\D) until the end of the string ($).
I have this vector:
jvm<-c("test - PROD_DB_APP_185b#SERVER01" ,"uat - PROD_DB_APP_SYS[1]#SERVER2")
I need to extract text until "[" or if there is no "[", then until the "#" character.
result should be
PROD_DB_APP_185b
PROD_DB_APP_SYS
I've tried something like this:
str_match(jvm, ".*\\-([^\\.]*)([.*)|(#.*)")
not working, any ides?
A sub solution with base R:
jvm<-c("test - PROD_DB_APP_185b#SERVER01" ,"uat - PROD_DB_APP_SYS[1]#SERVER2")
sub("^.*?\\s+-\\s+([^#[]+).*", "\\1", jvm)
See the online R demo
Details:
^ - start of string
.*? - any 0+ chars as few as possible
\\s+-\\s+ - a hyphen enclosed with 1 or more whitespaces
([^#[]+) - capturing group 1 matching any 1 or more chars other than #
and [
.* - any 0+ chars, up to the end of string.
Or a stringr solution with str_extract:
str_extract(jvm, "(?<=-\\s)[^#\\[]+")
See the regex demo
Details:
(?<=-\\s) - a positive lookbehind that matches an empty string that is preceded with a - and a whitespace immediately to the left of the current location
[^#\\[]+ - 1 or more chars other than # and [.