I have a string
text = "Math\n \n \n 600 rubles / 45 min."
text2 = "Math\n \n \n in a group"
And I want to replace\n \n \n with " " only if digits are following.
As a result, I want to have:
"Math 600 rubles / 45 min."
"Math\n \n \n in a group"
I tried gsub("\n \n \n [\\d]", " ", text), but it replaces the first digit too.
You may use a pattern that will match 3 occurrences of \n followed with 6+ spaces and then capture the digit and replace with a backreference to the Group 1:
gsub("(?:\n {6,}){3}(\\d)", " \\1", text)
See the R demo
Details
(?:\n {6,}){3} - 3 consecutive occurrences of:
\n - a newline
{6,} - 6 or more spaces
(\\d) - Group 1 (referred to with \1 from the replacement pattern): any digit.
I came up with the following pattern:
gsub("\\n[[:blank:]]*\\n[[:blank:]]*\\n[[:blank:]]*(\\d+)", " \\1", text)
This pattern matches three newlines, in succession, ending with a number. It allows for an arbitrary and unfixed amount of whitespace between each newline. This makes the match flexible, and helps to avoid a misfire from counting spaces incorrectly (or new incoming data not behaving as you expect).
The main problems I see with your current call to gsub is that you are using fixed width spaces in between newlines. Also, [\\d] is never used in the replacement. Hence, you are consuming that number but it won't show up the replacement.
Demo
text =c("Math\n \n \n 600 rubles / 45 min.","Math\n \n \n in a group")
gsub('((\n\\s+){1,})(?=\\d)',' ',text,perl=T)
#[1] "Math 600 rubles / 45 min." "Math\n \n \n in a group"
Related
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.*".
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.
Using stringr i tried to detect a € sign at the end of a string as follows:
str_detect("my text €", "€\\b") # FALSE
Why is this not working? It is working in the following cases:
str_detect("my text a", "a\\b") # TRUE - letter instead of €
grepl("€\\b", "2009in €") # TRUE - base R solution
But it also fails in perl mode:
grepl("€\\b", "2009in €", perl=TRUE) # FALSE
So what is wrong about the €\\b-regex? The regex €$ is working in all cases...
When you use base R regex functions without perl=TRUE, TRE regex flavor is used.
It appears that TRE word boundary:
When used after a non-word character matches the end of string position, and
When used before a non-word character matches the start of string position.
See the R tests:
> gsub("\\b\\)", "HERE", ") 2009in )")
[1] "HERE 2009in )"
> gsub("\\)\\b", "HERE", ") 2009in )")
[1] ") 2009in HERE"
>
This is not a common behavior of a word boundary in PCRE and ICU regex flavors where a word boundary before a non-word character only matches when the character is preceded with a word char, excluding the start of string position (and when used after a non-word character requires a word character to appear right after the word boundary):
There are three different positions that qualify as word boundaries:
- Before the first character in the string, if the first character is a word character.
- After the last character in the string, if the last character is a word character.
- Between two characters in the string, where one is a word character and the other is not a word character.
\b
is equivalent to
(?:(?<!\w)(?=\w)|(?<=\w)(?!\w))
which is to say it matches
between a word char and a non-word char,
between a word char and the start of the string, and
between a word char and the end of the string.
€ is a symbol, and symbols aren't word characters.
$ uniprops €
U+20AC <€> \N{EURO SIGN}
\pS \p{Sc}
All Any Assigned Common Zyyy Currency_Symbol Sc Currency_Symbols S Gr_Base Grapheme_Base Graph X_POSIX_Graph GrBase Print X_POSIX_Print Symbol Unicode
If your language supports look-behinds and look-aheads, you could use the following to find a boundary between a space and non-space (treating the start and end as a space).
(?:(?<!\S)(?=\S)|(?<=\S)(?!\S))
How to remove words with dashes (as prefixes or suffixes) from such string:
x <- "word -o -dod -3 -33 dp-pd -d- --- 140 -- s- S- SS- s3- 3e- 33- 3- s SS avf-ada"
And obtain:
word dp-pd 140 s SS avf-ada
Occasionally, standalone dashes also can be removed.
I've found a solution thanks to regex101: (\s-\S+)|(\S+-\s)
I suggest using
x <- "word -o -dod -3 -33 dp-pd -d- --- 140 -- s- S- SS- s3- 3e- 33- 3- s SS avf-ada -"
trimws(gsub("(?:\\S+-\\B|\\B-\\S+|\\B-\\B)\\s*", "", x, perl=TRUE))
See the regex demo and an R demo.
Details:
(?:\S+-\B|\B-\S+|\B-\B) - either of the two alternatives:
\S+-\B - 1+ chars other than whitespace, - and a non-word boundary, that is, the - must be either at the end of string or before a non-word char
| - or
\B-\S+ - a non-word boundary, that is, the - should only be matched if preceded with a non-word char or start of string, then a hyphen and 1+ chars other than whitespace
\B-\B - any - enclosed with non-word boundaries (at the end/start of string or between non-word chars)
\s* - 0+ whitespaces.
The perl=TRUE needs to be used because of the non-word boundary that does not work correctly with a TRE regex version.
I have a string format that I would like to select from a character vector. The form is
123 123 1234
where the two spaces can also be a hyphen. i.e. 3 digits followed by space or hyphen, followed by 3 digits, followed by space or hyphen, followed by 4 digits
I am trying to do this by the following:
grep("^([0-9]{3}[ -.])([0-9]{3}[ -.])([0-9]{4}$)",mytext)
however this yields:
integer(0)
What am I doing wrong?
Your string has a whitespace at the end, so you can either consider that white space, like so:
grep("^([0-9]{3}[ -.])([0-9]{3}[ -.])([0-9]{4} $)",mytext)
Or remove the end of line assertion "$", like so:
grep("^([0-9]{3}[ -.])([0-9]{3}[ -.])([0-9]{4})",mytext)
Also, as pointed out by Wiktor Stribiżew, the character class [ -.] will match any character in the range between " " and ".". To match "-","." and " " you have to escape the "-" or put it at the end of the class. Like [ \-.] or [ .-]