I want to check if a character can be safely converted to a numeric by using a regex.
However, I don't see my error. Example:
stringr::str_detect("4.", pattern = "-{0,1}[0-9]+(.[0-9]+){0,1}")
This produces a TRUE. My intention was to specifiy that whenever a . follows the first sequence of numbers, there must be at least one other number, therefore (.[0-9]+){0,1}.
What's wrong here?
Note:
(.[0-9]+){0,1} is an optional pattern because {0,1} (=?) makes the .[0-9]+ pattern sequence match one or zero times. So, yes, one or more digits ([0-9]+) must follow any char other than line break chars (matched with an unescaped .), but this pattern is optional, and thus you cannot require anything with it.
. is unescaped, so it matches any char other than line break chars. Escape it to match a literal dot
Your regex is not anchored, and can match partial substrings in a longer string. Use ^ and $ to make the pattern match the whole string.
So, consider using
stringr::str_detect("4.", pattern = "^-?[0-9]+(?:\\.[0-9]+)?$")
where
^ - start of string
-? - an optional - char
[0-9]+ - one or more digits
(?:\.[0-9]+)? - a non-capturing group matching an optional sequence of a . and then one or more digits
$ - end of string.
I'm creating a small example to be put into mutate(). Not sure why this doesn't work.
> str_extract("rs1234-<b>C</b>","^rs*\\d$")
[1] NA
I'd be great if you can point to my misunderstanding of the language instead of merely providing a solution. I expect to get "rs1234".
The ^rs*\d$ regex matches
^ - start of string
rs* - r and zero or more occurrences of s char
\d - a digit
$ - end of string.
So, your pattern matches strings like rsssss1, r3, etc.
You need
str_extract("rs1234-<b>C</b>", "^rs\\d+")
where ^rs\d+ matches rs at the start of string and then one or more digits. See this regex demo.
But if I just want the substring in between "rs" and the last number. What should I do?
You would use rs.*\d:
str_extract("rs1234-<b>C</b>", "rs.*\\d")
where rs.*\d matches rs, then any zero or more chars other than line break chars as many as possible and then a digit.
NOTE: If you need to match line endings, too, you need to prepend the last pattern with (?s) inline DOTALL modifier.
See this regex demo.
I have the following QRegExpValidator
QRegExpValidator doubleValidator = new QRegExpValidator(QRegExp("[-+]?[0-9]*[\\.,]?[0-9]+([eE][-+]?[0-9]+)?"));
It's supposed to be a Double numbers validator that accepts numbers, only one "e" sign, one comma OR dot and one + or - sign at the beggining of the string or after the "e" sign. It works for every case, except that it allows the string to start with a comma or dot. I tried to use [^\\.,] and variations and they did in fact work, but in this case, it would also allow to put two +- signs.
How can I make this to work?
The [-+]?[0-9]*[.,]?[0-9]+([eE][-+]?[0-9]+)? pattern allows the , or . at the start because [-+]? and [0-9]* can match empty strings due to the ? (one or zero occurrences) and * (zero or more occurrences) quantifiers, and then [.,] matches a single occurrence of . or ,. Besides, if the method you are using does not anchor the pattern by default, you also need ^ and $ anchors around the pattern.
I suggest fixing that with
"^[-+]?[0-9]+([.,][0-9]+)?([eE][-+]?[0-9]+)?$"
^ ^^^^^^^^^^^^^^ ^
Note you do not need to escape the dot inside a character class, [.] always matches a dot char only.
The [0-9]+([.,][0-9]+)? matches 1+ digits and then an optional sequence of a . or , followed with 1+ digits.
I have a following string. I tried to remove all the strings before the last space but it seems I can't achieve it.
I tried to follow this post
Use gsub remove all string before first white space in R
str <- c("Veni vidi vici")
gsub("\\s*","\\1",str)
"Venividivici"
What I want to have is only "vici" string left after removing everything before the last space.
Your gsub("\\s*","\\1",str) code replaces each occurrence of 0 or more whitespaces with a reference to the capturing group #1 value (which is an empty string since you have not specified any capturing group in the pattern).
You want to match up to the last whitespace:
sub(".*\\s", "", str)
If you do not want to get a blank result in case your string has trailing whitespace, trim the string first:
sub(".*\\s", "", trimws(str))
Or, use a handy stri_extract_last_regex from stringi package with a simple \S+ pattern (matching 1 or more non-whitespace chars):
library(stringi)
stri_extract_last_regex(str, "\\S+")
# => [1] "vici"
Note that .* matches any 0+ chars as many as possible (since * is a greedy quantifier and . in a TRE pattern matches any char including line break chars), and grabs the whole string at first. Then, backtracking starts since the regex engine needs to match a whitespace with \s. Yielding character by character from the end of the string, the regex engine stumbles on the last whitespace and calls it a day returning the match that is removed afterwards.
See the R demo and a regex demo online:
str <- c("Veni vidi vici")
gsub(".*\\s", "", str)
## => [1] "vici"
Also, you may want to see how backtracking works in the regex debugger:
Those red arrows show backtracking steps.
I'm a regular expression newbie and I can't quite figure out how to write a single regular expression that would "match" any duplicate consecutive words such as:
Paris in the the spring.
Not that that is related.
Why are you laughing? Are my my regular expressions THAT bad??
Is there a single regular expression that will match ALL of the bold strings above?
Try this regular expression:
\b(\w+)\s+\1\b
Here \b is a word boundary and \1 references the captured match of the first group.
Regex101 example here
I believe this regex handles more situations:
/(\b\S+\b)\s+\b\1\b/
A good selection of test strings can be found here: http://callumacrae.github.com/regex-tuesday/challenge1.html
The below expression should work correctly to find any number of duplicated words. The matching can be case insensitive.
String regex = "\\b(\\w+)(\\s+\\1\\b)+";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0), m.group(1));
}
Sample Input : Goodbye goodbye GooDbYe
Sample Output : Goodbye
Explanation:
The regex expression:
\b : Start of a word boundary
\w+ : Any number of word characters
(\s+\1\b)* : Any number of space followed by word which matches the previous word and ends the word boundary. Whole thing wrapped in * helps to find more than one repetitions.
Grouping :
m.group(0) : Shall contain the matched group in above case Goodbye goodbye GooDbYe
m.group(1) : Shall contain the first word of the matched pattern in above case Goodbye
Replace method shall replace all consecutive matched words with the first instance of the word.
Try this with below RE
\b start of word word boundary
\W+ any word character
\1 same word matched already
\b end of word
()* Repeating again
public static void main(String[] args) {
String regex = "\\b(\\w+)(\\b\\W+\\b\\1\\b)*";// "/* Write a RegEx matching repeated words here. */";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE/* Insert the correct Pattern flag here.*/);
Scanner in = new Scanner(System.in);
int numSentences = Integer.parseInt(in.nextLine());
while (numSentences-- > 0) {
String input = in.nextLine();
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0),m.group(1));
}
// Prints the modified sentence.
System.out.println(input);
}
in.close();
}
Regex to Strip 2+ duplicate words (consecutive/non-consecutive words)
Try this regex that can catch 2 or more duplicate words and only leave behind one single word. And the duplicate words need not even be consecutive.
/\b(\w+)\b(?=.*?\b\1\b)/ig
Here, \b is used for Word Boundary, ?= is used for positive lookahead, and \1 is used for back-referencing.
Example
Source
The widely-used PCRE library can handle such situations (you won't achieve the the same with POSIX-compliant regex engines, though):
(\b\w+\b)\W+\1
Here is one that catches multiple words multiple times:
(\b\w+\b)(\s+\1)+
No. That is an irregular grammar. There may be engine-/language-specific regular expressions that you can use, but there is no universal regular expression that can do that.
This is the regex I use to remove duplicate phrases in my twitch bot:
(\S+\s*)\1{2,}
(\S+\s*) looks for any string of characters that isn't whitespace, followed whitespace.
\1{2,} then looks for more than 2 instances of that phrase in the string to match. If there are 3 phrases that are identical, it matches.
Since some developers are coming to this page in search of a solution which not only eliminates duplicate consecutive non-whitespace substrings, but triplicates and beyond, I'll show the adapted pattern.
Pattern: /(\b\S+)(?:\s+\1\b)+/ (Pattern Demo)
Replace: $1 (replaces the fullstring match with capture group #1)
This pattern greedily matches a "whole" non-whitespace substring, then requires one or more copies of the matched substring which may be delimited by one or more whitespace characters (space, tab, newline, etc).
Specifically:
\b (word boundary) characters are vital to ensure partial words are not matched.
The second parenthetical is a non-capturing group, because this variable width substring does not need to be captured -- only matched/absorbed.
the + (one or more quantifier) on the non-capturing group is more appropriate than * because * will "bother" the regex engine to capture and replace singleton occurrences -- this is wasteful pattern design.
*note if you are dealing with sentences or input strings with punctuation, then the pattern will need to be further refined.
The example in Javascript: The Good Parts can be adapted to do this:
var doubled_words = /([A-Za-z\u00C0-\u1FFF\u2800-\uFFFD]+)\s+\1(?:\s|$)/gi;
\b uses \w for word boundaries, where \w is equivalent to [0-9A-Z_a-z]. If you don't mind that limitation, the accepted answer is fine.
This expression (inspired from Mike, above) seems to catch all duplicates, triplicates, etc, including the ones at the end of the string, which most of the others don't:
/(^|\s+)(\S+)(($|\s+)\2)+/g, "$1$2")
I know the question asked to match duplicates only, but a triplicate is just 2 duplicates next to each other :)
First, I put (^|\s+) to make sure it starts with a full word, otherwise "child's steak" would go to "child'steak" (the "s"'s would match). Then, it matches all full words ((\b\S+\b)), followed by an end of string ($) or a number of spaces (\s+), the whole repeated more than once.
I tried it like this and it worked well:
var s = "here here here here is ahi-ahi ahi-ahi ahi-ahi joe's joe's joe's joe's joe's the result result result";
print( s.replace( /(\b\S+\b)(($|\s+)\1)+/g, "$1"))
--> here is ahi-ahi joe's the result
Try this regular expression it fits for all repeated words cases:
\b(\w+)\s+\1(?:\s+\1)*\b
I think another solution would be to use named capture groups and backreferences like this:
.* (?<mytoken>\w+)\s+\k<mytoken> .*/
OR
.*(?<mytoken>\w{3,}).+\k<mytoken>.*/
Kotlin:
val regex = Regex(""".* (?<myToken>\w+)\s+\k<myToken> .*""")
val input = "This is a test test data"
val result = regex.find(input)
println(result!!.groups["myToken"]!!.value)
Java:
var pattern = Pattern.compile(".* (?<myToken>\\w+)\\s+\\k<myToken> .*");
var matcher = pattern.matcher("This is a test test data");
var isFound = matcher.find();
var result = matcher.group("myToken");
System.out.println(result);
JavaScript:
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/;
const input = "This is a test test data";
const result = regex.exec(input);
console.log(result.groups.myToken);
// OR
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/g;
const input = "This is a test test data";
const result = [...input.matchAll(regex)];
console.log(result[0].groups.myToken);
All the above detect the test as the duplicate word.
Tested with Kotlin 1.7.0-Beta, Java 11, Chrome and Firefox 100.
You can use this pattern:
\b(\w+)(?:\W+\1\b)+
This pattern can be used to match all duplicated word groups in sentences. :)
Here is a sample util function written in java 17, which replaces all duplications with the first occurrence:
public String removeDuplicates(String input) {
var regex = "\\b(\\w+)(?:\\W+\\1\\b)+";
var pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
var matcher = pattern.matcher(input);
while (matcher.find()) {
input = input.replaceAll(matcher.group(), matcher.group(1));
}
return input;
}
As far as I can see, none of these would match:
London in the
the winter (with the winter on a new line )
Although matching duplicates on the same line is fairly straightforward,
I haven't been able to come up with a solution for the situation in which they
stretch over two lines. ( with Perl )
To find duplicate words that have no leading or trailing non whitespace character(s) other than a word character(s), you can use whitespace boundaries on the left and on the right making use of lookarounds.
The pattern will have a match in:
Paris in the the spring.
Not that that is related.
The pattern will not have a match in:
This is $word word
(?<!\S)(\w+)\s+\1(?!\S)
Explanation
(?<!\S) Negative lookbehind, assert not a non whitespace char to the left of the current location
(\w+) Capture group 1, match 1 or more word characters
\s+ Match 1 or more whitespace characters (note that this can also match a newline)
\1 Backreference to match the same as in group 1
(?!\S) Negative lookahead, assert not a non whitespace char to the right of the current location
See a regex101 demo.
To find 2 or more duplicate words:
(?<!\S)(\w+)(?:\s+\1)+(?!\S)
This part of the pattern (?:\s+\1)+ uses a non capture group to repeat 1 or more times matching 1 or more whitespace characters followed by the backreference to match the same as in group 1.
See a regex101 demo.
Alternatives without using lookarounds
You could also make use of a leading and trailing alternation matching either a whitespace char or assert the start/end of the string.
Then use a capture group 1 for the value that you want to get, and use a second capture group with a backreference \2 to match the repeated word.
Matching 2 duplicate words:
(?:\s|^)((\w+)\s+\2)(?:\s|$)
See a regex101 demo.
Matching 2 or more duplicate words:
(?:\s|^)((\w+)(?:\s+\2)+)(?:\s|$)
See a regex101 demo.
Use this in case you want case-insensitive checking for duplicate words.
(?i)\\b(\\w+)\\s+\\1\\b