My string pattern is as follows:
1233 fox street, omaha NE ,69131-7233
Jeffrey Jones, 666 Church Street, Omaha NE ,69131-72339
Betty Davis, LLC, 334 Aloha Blvd., Fort Collins CO ,84444-00333
,1233 Decker street, omaha NE ,69131-7233
I need to separate the above string into four variables: name, address, city_state, zipcode.
Since the pattern has three to four commas, I am starting at the right to separate the field into multiple fields.
rubular.com says the pattern ("(,\\d.........)$"))) or the pattern ",\d.........$" will match the zipcode at the end of the string.
regex101.com, finds neither of the above patterns comes up with a match.
When I try to separate with:
#need to load pkg:tidyr for the `separate`
function
library(tidyr)
separate(street_add, c("street_add2", "zip", sep= ("(,\d.........)$")))
or with:
separate(street_add, c("street_add2", "zip", sep= (",\d.........$")))
In both scenarios, R splits at the first comma in the string.
How do I split the string into segments?
Thank you.
Use
sep=",(?=[^,]*$)"
See regex proof.
EXPLANATION
--------------------------------------------------------------------------------
, ','
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
[^,]* any character except: ',' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
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.*".
This question already has answers here:
Regex - Split String on Comma, Skip Anything Between Balanced Parentheses
(2 answers)
Closed 1 year ago.
I have the following string:
Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews
I want to replace the semicolons that are not in parenthesis to commas. There can be any number of brackets and any number of semicolons within the brackets and the result should look like this:
Almonds , Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt), Cashews
This is my current code:
x<- Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews
gsub(";(?![^(]*\\))",",",x,perl=TRUE)
[1] "Almonds , Roasted Peanuts (Peanuts, Canola Oil (Antioxidants (319; 320)); Salt), Cashews "
The problem I am facing is if there's a nested () inside a bigger bracket, the regex I have will replace the semicolon to comma.
Can I please get some help on regex that will solve the problem? Thank you in advance.
The pattern ;(?![^(]*\)) means matching a semicolon, and assert that what is to the right is not a ) without a ( in between.
That assertion will be true for a nested opening parenthesis, and will still match the ;
You could use a recursive pattern to match nested parenthesis to match what you don't want to change, and then use a SKIP FAIL approach.
Then you can match the semicolons and replace them with a comma.
[^;]*(\((?>[^()]+|(?1))*\))(*SKIP)(*F)|;
In parts, the pattern matches
[^;]* Match 0+ times any char except ;
( Capture group 1
\( Match the opening (
(?> Atomic group
[^()]+ Match 1+ times any char except ( and )
| Or
(?1) Recurse the whole first sub pattern (group 1)
)* Close the atomic group and optionally repeat
\) Match the closing )
) Close group 1
(*SKIP)(*F) Skip what is matched
| Or
; Match a semicolon
See a regex demo and an R demo.
x <- c("Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews",
"Peanuts (32.5%); Macadamia Nuts (14%; PPPG(AHA)); Hazelnuts (9%); nuts(98%)")
gsub("[^;]*(\\((?>[^()]+|(?1))*\\))(*SKIP)(*F)|;",",",x,perl=TRUE)
Output
[1] "Almonds , Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt), Cashews"
[2] "Peanuts (32.5%), Macadamia Nuts (14%; PPPG(AHA)), Hazelnuts (9%), nuts(98%)"
I want to change one of the _ to another character, for example to -, the reason is there are problems reading in these filenames. I want a to become like b. So I want to change the second last underscore(_), how to specify this in an efficient way?
gsub("_", "-"), it must also be specified to a certain location.
a <- c("2018-01-09_B2_HILIC_POS_123_-14b_090.mzML", "2018-01-09_B2_HILIC_POS_243_-12a_026.mzML", "2020-01-09_B2_HILIC_POS_415_893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV7001248356_040.mzML")
b <- c("2018-01-09_B2_HILIC_POS_123--14b_090.mzML", "2018-01-09_B2_HILIC_POS_243--12a_026.mzML", "2020-01-09_B2_HILIC_POS_415-893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV4004365711_040.mzML")
Here is one base R option using sub :
sub('(.*)(_)(.*_.*)$', '\\1-\\3', a)
#[1] "2018-01-09_B2_HILIC_POS_123--14b_090.mzML"
#[2] "2018-01-09_B2_HILIC_POS_243--12a_026.mzML"
#[3] "2020-01-09_B2_HILIC_POS_415-893a_059.mzML"
#[4] "2020-01-18_B3_HILIC_POS-LV7001248356_040.mzML"
Here we divide data into 3 groups -
The 1st group is everything until second last underscore which is captured using (.*) and used as a backreference (\\1).
The 2nd group is second last underscore which us replaced with -.
The 3rd one is everything after second last underscore which is captured using (.*_.*) and used as a backreference (\\3).
Use
sub("_(?=[^_]*_[^_]*$)", "-", a, perl=TRUE)
See regex proof.
Explanation
--------------------------------------------------------------------------------
_ '_'
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
[^_]* any character except: '_' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
_ '_'
--------------------------------------------------------------------------------
[^_]* any character except: '_' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
See R proof:
a <- c("2018-01-09_B2_HILIC_POS_123_-14b_090.mzML", "2018-01-09_B2_HILIC_POS_243_-12a_026.mzML", "2020-01-09_B2_HILIC_POS_415_893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV7001248356_040.mzML")
sub("_(?=[^_]*_[^_]*$)", "-", a, perl=TRUE)
Results:
[1] "2018-01-09_B2_HILIC_POS_123--14b_090.mzML"
[2] "2018-01-09_B2_HILIC_POS_243--12a_026.mzML"
[3] "2020-01-09_B2_HILIC_POS_415-893a_059.mzML"
[4] "2020-01-18_B3_HILIC_POS-LV7001248356_040.mzML"
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))