I am checking if a string contains any special characters. This is what I have, and its not working,
if(grepl('^\\[:punct:]', val))
So if anybody can tell me what I am missing, that will be helpful.
Special characters
~ ` ! ## $ % ^ & * | : ; , ." |
As #thelatemail pointed out in the comments you can use:
grepl('[^[:punct:]]', val)
which will result in TRUE or FALSE for each value in your vector. You can add sum() to the beginning of the statement to get the total number of these cases.
You can also use:
grepl('[^[:alnum:]]', val)
which will check for any value that is not a letter or a number.
Related
I have a list of years like this:
2018-
2001–2020
1999-
2005-
I would like to create a regex to match the year with these criteria:
xxxx- matches xxxx
yyyy-nnnn matches nnnn
Can you please help me?
I've tried [[:digit:]]{4}$, or alternatively [[:digit:]]{4}-$, but they only partially work.
To get the last year in the "range," established by - character, the cleanest way
my $year = (split /-/, $range)[-1];
If there isn't anything after the last delimiter then the last returned element by split is what is before it, so the last element in its return list (obtained with index -1) is either the second given year -- as in 2001-2020 -- or the only one, as in other examples. This performs no checking of input.
With a regex, one way is to seek the last number in the string
my ($year) = $range =~ /([0-9]+)[^0-9]*$/;
where if you use [0-9]{4} then there is a small additional measure of checking.
The POSIX character class [[:digit:]] and its negation [[:^digit:]] (or \P{PosixDigit}) can be used instead if desired, but note that these match all manner of Unicode "digit characters," just like \d and \D do (a few hundred), on top of the ascii [0-9] (unless /a modifier is used).
A full test program, for both
use warnings;
use strict;
use feature 'say';
my #ranges = qw(2018- 2001-2020 1999- 2005-);
foreach my $range (#ranges) {
my $year = (split /-/, $range)[-1];
# Or, using regex
# my ($year) = $range =~ /([0-9]+)[^0-9]*$/;
say $year;
}
Prints as desired.
We can capture the 4 digits as group, followed by a - at the end ($) of the string and replace with the backreference (\\1) of the captured group
sub(".*(\\d{4})-?$", "\\1", str1)
#[1] "2018" "2020" "1999" "2005"
data
str1 <- c("2018-", "2001-2020", "1999-", "2005-")
You can split the text on "-" and get the last number.
x <- c("2018-", "2001-2020", "1999-", "2005-")
sapply(strsplit(str1, '-', fixed = TRUE), tail, 1)
#[1] "2018" "2020" "1999" "2005"
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
Regular Expression To exclude sub-string name(job corps)
Includes at least 1 upper case letter, 1 lower case letter, 1 number and 1 symbol except "#"
I have written something like below :
^((?!job corps).)(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[!#$%^&*]).*$
I tested with the above regular expression, not working for special character.
can anyone guide on this..
If I understand well your requirements, you can use this pattern:
^(?![^a-z]*$|[^A-Z]*$|[^0-9]*$|[^!#$%^&*]*$|.*?job corps)[^#]*$
If you only want to allow characters from [a-zA-Z0-9^#$%&*] changes the pattern to:
^(?![^a-z]*$|[^A-Z]*$|[^0-9]*$|[^!#$%^&*]*$|.*?job corps)[a-zA-Z0-9^#$%&*]*$
details:
^ # start of the string
(?! # not followed by any of these cases
[^a-z]*$ # non lowercase letters until the end
|
[^A-Z]*$ # non uppercase letters until the end
|
[^0-9]*$
|
[^!#$%^&*]*$
|
.*?job corps # any characters and "job corps"
)
[^#]* # characters that are not a #
$ # end of the string
demo
Note: you can write the range #$%& like #-& to win a character.
stribizhev, your answer is correct
^(?!.job corps)(?=.[0-9])(?=.[a-z])(?=.[A-Z])(?=.[!#$%^&])(?!.#).$
can verify the expression in following url:
http://www.freeformatter.com/regex-tester.html
I am trying to use Pyparsing to identify a keyword which is not beginning with $ So for the following input:
$abc = 5 # is not a valid one
abc123 = 10 # is valid one
abc$ = 23 # is a valid one
I tried the following
var = Word(printables, excludeChars='$')
var.parseString('$abc')
But this doesn't allow any $ in var. How can I specify all printable characters other than $ in the first character position? Any help will be appreciated.
Thanks
Abhijit
You can use the method I used to define "all characters except X" before I added the excludeChars parameter to the Word class:
NOT_DOLLAR_SIGN = ''.join(c for c in printables if c != '$')
keyword_not_starting_with_dollar = Word(NOT_DOLLAR_SIGN, printables)
This should be a bit more efficient than building up with a Combine and a NotAny. But this will match almost anything, integers, words, valid identifiers, invalid identifiers, so I'm skeptical of the value of this kind of expression in your parser.
I'm working to grab two different elements in a string.
The string look like this,
str <- c('a_abc', 'b_abc', 'abc', 'z_zxy', 'x_zxy', 'zxy')
I have tried with the different options in ?grep, but I can't get it right, 'm doing something like this,
grep('[_abc]:[_zxy]',str, value = TRUE)
and what I would like is,
[1] "a_abc" "b_abc" "z_zxy" "x_zxy"
any help would be appreciated.
Use normal parentheses (, not the square brackets [
grep('_(abc|zxy)',str, value = TRUE)
[1] "a_abc" "b_abc" "z_zxy" "x_zxy"
To make the grep a bit more flexible, you could do something like:
grep('_.{3}$',str, value = TRUE)
Which will match an underscore _ followed by any character . three times {3} followed immediately by the end of the string $
this should work: grep('_abc|_zxy', str, value=T)
X|Y matches when either X matches or Y matches
In this case just doing:
str[grep("_",str)]
will work... is it more complicated in your specific case?