I want to match every cases of "-", but not these ones:
[\d]-[A-Z]
[A-Z]-[\d]
I tried this pattern: ((?<![A-Z])-(?![0-9]))|((?<![0-9])-(?![A-Z])) but some results are incorrect like: "RUA VF-32 N"
Can anyone help me?
A simple approach is to use grep with your current logic and inverting the result, and then run another grep to only keep those items that have a hyphen in them:
x <- c("QUADRA 120 - ASA BRANCA","FAZENDA LAGE -RODOVIA RIO VERDE","C-15","99-B","A-A")
grep("-", grep("[A-Z]-\\d|\\d-[A-Z]", x, invert=TRUE, value=TRUE), value=TRUE, fixed=TRUE)
# => [1] "QUADRA 120 - ASA BRANCA" "FAZENDA LAGE -RODOVIA RIO VERDE"
# [3] "A-A"
Here, [A-Z]-\\d|\\d-[A-Z] matches a hyphen either in between an uppercase ASCII etter or a digit or betweena digit and an ASCII uppercase letter. If there is a match, the result is inverted due to invert=TRUE.
See the R demo.
To only match - in all contexts other than in between a letter and a digit, you may use the PCRE regex based on SKIP-FAIL technique like
> grep("(?:\\d-[A-Z]|[A-Z]-\\d)(*SKIP)(*F)|-", x, perl=TRUE)
[1] 1 2
See this regex demo
Details
(?:\d-[A-Z]|[A-Z]-\d) - a non-capturing group that matches either a digit, - and then uppercase ASCII letter, or an uppercase ASCII letter, - and a digit
(*SKIP)(*F) - omit the current match and proceed looking for the next match at the end of the "failed" match
| - or
- - a hyphen.
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.*".
For a text field, I would like to expose those that contain invalid characters. The list of invalid characters is unknown; I only know the list of accepted ones.
For example for French language, the accepted list is
A-z, 1-9, [punc::], space, àéèçè, hyphen, etc.
The list of invalid charactersis unknown, yet I want anything unusual to resurface, for example, I would want
This is an 2-piece à-la-carte dessert to pass when
'Ã this Øs an apple' pumps up as an anomalie
The 'not contain' notion in R does not behave as I would like, for example
grep("[^(abc)]",c("abcdef", "defabc", "apple") )
(those that does not contain 'abc') match all three while
grep("(abc)",c("abcdef", "defabc", "apple") )
behaves correctly and match only the first two. Am I missing something
How can we do that in R ? Also, how can we put hypen together in the list of accepted characters ?
[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+
The above regex matches any of the following (one or more times). Note that the parameter ignore.case=T used in the code below allows the following to also match uppercase variants of the letters.
a-z Any lowercase ASCII letter
1-9 Any digit in the range from 1 to 9 (excludes 0)
[:punct:] Any punctuation character
The space character
àâæçéèêëîïôœùûüÿ Any valid French character with a diacritic mark
- The hyphen character
See code in use here
x <- c("This is an 2-piece à-la-carte dessert", "Ã this Øs an apple")
gsub("[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+", "", x, ignore.case=T)
The code above replaces all valid characters with nothing. The result is all invalid characters that exist in the string. The following is the output:
[1] "" "ÃØ"
If by "expose the invalid characters" you mean delete the "accepted" ones, then a regex character class should be helpful. From the ?regex help page we can see that a hyphen is already part of the punctuation character vector;
[:punct:]
Punctuation characters:
! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { | } ~
So the code could be:
x <- 'Ã this Øs an apple'
gsub("[A-z1-9[:punct:] àéèçè]+", "", x)
#[1] "ÃØ"
Note that regex has a predefined, locale-specific "[:alpha:]" named character class that would probably be both safer and more compact than the expression "[A-zàéèçè]" especially since the post from ctwheels suggests that you missed a few. The ?regex page indicates that "[0-9A-Za-z]" might be both locale- and encoding-specific.
If by "expose" you instead meant "identify the postion within the string" then you could use the negation operator "^" within the character class formalism and apply gregexpr:
gregexpr("[^A-z1-9[:punct:] àéèçè]+", x)
[[1]]
[1] 1 8
attr(,"match.length")
[1] 1 1
Say I have a dataframe df in which a column df$strings contains strings like
[cat 00.04;09]
[cat 00.04;10]
and so on. I want to remove all characters between "[cat" and "]" to yield
[cat]
[cat]
I've tried this using gsub but it's not working and I'm not sure what I'm doing wrong:
gsub('cat*?\\]', '', df)
Note that cat*?\\] patten matches ca, then any 0+ t chars but as few as possible and then ].
You want to match any chars other than ] between [cat and ]:
gsub('\\[cat[^]]*\\]', '[cat]', df$strings)
Here,
\\[ - matches [
cat - matches cat
[^]]* - 0+ chars other than ] (note that ] inside the bracket expression should not be escaped when placed at the start - else, if you escape it, you will need to add perl=TRUE argument since PCRE regex engine can handle regex escapes inside bracket expressions (not the default TRE))
\\] - a ] (you do not even need to escape it, you may just use ]).
See the R demo:
x <- c("[cat 00.04;09]", "[cat 00.04;10]")
gsub('\\[cat[^]]*\\]', '[cat]', x)
## => [1] "[cat]" "[cat]"
If cat can be any word, use
gsub('\\[(\\w+)[^]]*\\]', '[\\1]', x)
where (\\w+) is a capturing group with ID=1 that matches 1 or more word chars, and \\1 in the replacement pattern is a replacement backreference that stands for the group value.
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.