I'm trying to create a pattern that would identify a money in a string. My expression so far is:
(\d{1,3}[\.,\s]{0,2})*\d{3}[\.,\s]{0,2}\d{0,2}[\s]{0,2}[zl|zł|zlotych|złotych|pln|PLN]{0,1}
and my main problem is with the last part: [zl|zł|zlotych|złotych|pln|PLN], which should find one of the national notations for money value (sth like $ or usd or dollars) but I'm doing it wrong, since it also matches something like '108.1 z'. Is it possible to change the last part, so that it would match only expressions that contain the whole expressions like 'zl', 'pln' and so on, and not single letters?
Yes, don't use [], which defines a character class, but instead use () to group your words.
(\d{1,3}[\.,\s]{0,2})*\d{3}[\.,\s]{0,2}\d{0,2}[\s]{0,2}(zl|zł|zlotych|złotych|pln|PLN)?
As you had it written, [zl|zł|zlotych|złotych|pln|PLN], means "match any of the characters contained in the []", or the equivalent of: [zl|łotychpnPLN] (duplicates removed)
If you don't want the money symbol captured, then start the group with ?:, i.e.:
(\d{1,3}[\.,\s]{0,2})*\d{3}[\.,\s]{0,2}\d{0,2}[\s]{0,2}(?:zl|zł|zlotych|złotych|pln|PLN)?
Use parentheses (which delimit groups) rather than square brackets (which delimit character classes) around that last group.
As a matter of style, use ? instead of {0,1}.
(\d{1,3}[\.,\s]{0,2})*\d{3}[\.,\s]{0,2}\d{0,2}[\s]{0,2}(zl|zł|zlotych|złotych|pln|PLN)?
You have a few problems here. First off, inside [] characters are taken as literals, so the first two [] blocks should be [.,\s].
Next (as the other answers say), the last [] block needs to be a group, not a character class, so replace the [] with ().
Finally, at the end you can replace {0, 1} with ?. It won't make a difference, but it's neater.
The regex should look like this:
(\d{1,3}[.,\s]{0,2})*\d{3}[.,\s]{0,2}\d{0,2}[\s]{0,2}(zl|zł|zlotych|złotych|pln|PLN)?
For the future, for regex questions it's really helpful if you post a typical input string and desired match along with your question!
Related
I am trying to catch strings around the acronym ADJ. The strings look like this:
·NOM·JJ·ADJ+CASE_DEF_GEN
·NOM·JJ·ADJ+CASE_DEF_ACC
·NOM·JJ·ADJ+CASE_INDEF_GEN
·NOM·DT+JJ·DET+ADJ+NSUFF_FEM_SG+CASE_DEF_GEN
·NOM·JJ·ADJ+CASE_INDEF_GEN
·NOM·JJ·ADJ+NSUFF_FEM_SG+CASE_INDEF_GEN
·NOM·DT+JJ·DET+ADJ+NSUFF_FEM_SG+CASE_DEF_ACC
So far I have this:
/[A-Z·\+#_]*?[·\+]ADJ[·\+][A-Z_·\+#]*?/g
But it only matches from the beginning of the strings until "ADJ+" ·NOM·DT+JJ·DET+ADJ+.
Since the rest of the strings after ADJ have the same composition of the beginning of the strings before ADJ, I thought this /[A-Z·\+#_]*?[·\+]/g should work, but it doesn't.
How do I get it to match the rest of the string?
My guess is that you want to make sure if you have an ADJ in the string, which if so, maybe we could simplify our expression to something similar to:
([A-Z·+#_]*)\bADJ\b([A-Z·+#_]*)
The expression is explained on the top right panel of this demo, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs step by step, if you like.
That *? quantifier after the +ADJ+ phrase is satisfied with the empty string right after it, since the ? makes the quantifier before it match "the minimum number of times possible" and for * that is zero times.
So drop the ?, which also has no purpose for the rest of the line
perl -wE'$_=q(-XADJX-JJ+ADJ-REST-);
($before, $after) = /(.*?)[+\-]ADJ[+\-](.*)/;
say for $before,$after'
Removing the ? at the end would match the whole strings,
/[A-Z·\+#_]*?[·\+]ADJ[·\+][A-Z_·\+#]*/g
I am not entirely sure why you needed a ? in a *.
I am re-phrasing my question to clear confusions!
I want to match if a string has certain letters for this I use the character class:
[ACD]
and it works perfectly!
but I want to match if the string has those letter(s) 2 or more times either repeated or 2 separate letters
For example:
[AKL] should match:
ABCVL
AAGHF
KKUI
AKL
But the above should not match the following:
ABCD
KHID
LOVE
because those are there but only once!
that's why I was trying to use:
[ACD]{2,}
But it's not working, probably it's not the right Regex.. can somebody a Regex guru can help me solve this puzzle?
Thanks
PS: I will use it on MYSQL - a differnt approach can also welcome! but I like to use regex for smarter and shorter query!
To ensure that a string contains at least two occurencies in a set of letters (lets say A K L as in your example), you can write something like this:
[AKL].*[AKL]
Since the MySQL regex engine is a DFA, there is no need to use a negated character class like [^AKL] in place of the dot to avoid backtracking, or a lazy quantifier that is not supported at all.
example:
SELECT 'KKUI' REGEXP '[AKL].*[AKL]';
will return 1
You can follow this link that speaks on the particular subject of the LIKE and the REGEXP features in MySQL.
If I understood you correctly, this is quite simple:
[A-Z].*?[A-Z]
This looks for your something in your set, [A-Z], and then lazily matches characters until it (potentially) comes across the set, [A-Z], again.
As #Enigmadan pointed out, a lazy match is not necessary here: [A-Z].*[A-Z]
The expression you are using searches for characters between 2 and unlimited times with these characters ACDFGHIJKMNOPQRSTUVWXZ.
However, your RegEx expression is excluding Y (UVWXZ])) therefore Z cannot be found since it is not surrounded by another character in your expression and the same principle applies to B ([ACD) also excluded in you RegEx expression. For example Z and A would match in an expression like ZABCDEFGHIJKLMNOPQRSTUVWXYZA
If those were not excluded on purpose probably better can be to use ranges like [A-Z]
If you want 2 or more of a match on [AKL], then you may use just [AKL] and may have match >= 2.
I am not good at SQL regex, but may be something like this?
check (dbo.RegexMatch( ['ABCVL'], '[AKL]' ) >= 2)
To put it in simple English, use [AKL] as your regex, and check the match on the string to be greater than 2. Here's how I would do in Java:
private boolean search2orMore(String string) {
Matcher matcher = Pattern.compile("[ACD]").matcher(string);
int counter = 0;
while (matcher.find())
{
counter++;
}
return (counter >= 2);
}
You can't use [ACD]{2,} because it always wants to match 2 or more of each characters and will fail if you have 2 or more matching single characters.
your question is not very clear, but here is my trial pattern
\b(\S*[AKL]\S*[AKL]\S*)\b
Demo
pretty sure this should work in any case
(?<l>[^AKL\n]*[AKL]+[^AKL\n]*[AKL]+[^AKL\n]*)[\n\r]
replace AKL for letters you need can be done very easily dynamicly tell me if you need it
Is this what you are looking for?
".*(.*[AKL].*){2,}.*" (without quotes)
It matches if there are at least two occurences of your charactes sorrounded by anything.
It is .NET regex, but should be same for anything else
Edit
Overall, MySQL regular expression support is pretty weak.
If you only need to match your capture group a minimum of two times, then you can simply use:
select * from ... where ... regexp('([ACD].*){2,}') #could be `2,` or just `2`
If you need to match your capture group more than two times, then just change the number:
select * from ... where ... regexp('([ACD].*){3}')
#This number should match the number of matches you need
If you needed a minimum of 7 matches and you were using your previous capture group [ACDF-KM-XZ]
e.g.
select * from ... where ... regexp('([ACDF-KM-XZ].*){7,}')
Response before edit:
Your regex is trying to find at least two characters from the set[ACDFGHIJKMNOPQRSTUVWXZ].
([ACDFGHIJKMNOPQRSTUVWXZ]){2,}
The reason A and Z are not being matched in your example string (ABCDEFGHIJKLMNOPQRSTUVWXYZ) is because you are looking for two or more characters that are together that match your set. A is a single character followed by a character that does not match your set. Thus, A is not matched.
Similarly, Z is a single character preceded by a character that does not match your set. Thus, Z is not matched.
The bolded characters below do not match your set
ABCDEFGHIJKLMNOPQRSTUVWXYZ
If you were to do a global search in the string, only the italicized characters would be matched:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
I want to remove all numbers from a paragraph except from some words.
My attempt is using a negative look-ahead:
gsub('(?!ami.12.0|allo.12)[[:digit:]]+','',
c('0.12','1245','ami.12.0 00','allo.12 1'),perl=TRUE)
But this doesn't work. I get this:
"." "" "ami.. " "allo."
Or my expected output is:
"." "" 'ami.12.0','allo.12'
You can't really use a negative lookahead here, since it will still replace when the cursor is at some point after ami.
What you can do is put back some matches:
(ami.12.0|allo.12)|[[:digit:]]+
gsub('(ami.12.0|allo.12)|[[:digit:]]+',"\\1",
c('0.12','1245','ami.12.0 00','allo.12 1'),perl=TRUE)
I kept the . since I'm not 100% sure what you have, but keep in mind that . is a wildcard and will match any character (except newlines) unless you escape it.
Your regex is actually finding every digit sequence that is not the start of "ami.12.0" or "allo.12". So for example, in your third string, it gets to the 12 in ami.12.0 and looks ahead to see if that 12 is the start of either of the two ignored strings. It is not, so it continues with replacing it. It would be best to generalize this, but in your specific case, you can probably achieve this by instead doing a negative lookbehind for any prefixes of the words (that can be followed by digit sequences) that you want to skip. So, you would use something like this:
gsub('(?<!ami\\.|ami\\.12\\.|allo\\.)[[:digit:]]+','',
c('0.12','1245','ami.12.0 00','allo.12 1'),perl=TRUE)
I have the following list of strings:
name <- c("hsa-miR-555p","hsa-miR-519b-3p","hsa-let-7a")
What I want to do is for each of the above strings
replace the text after second delimiter (-) with "zzz".
Yielding:
hsa-miR-zzz
hsa-miR-zzz
hsa-let-zzz
What's the way to do it?
Might as well use something like:
gsub("^((?:[^-]*-){2}).*", "\\1zzz", name)
(?:[^-]*-) is a non-capturing group which consists of several non-dash characters followed by a single dash character and the {2} just after means this group occurs twice only. Then, match everything else for the replacement. Note I used an anchor just in case to avoid unintended substitutions.
Perhaps something like this:
> gsub("([A-Za-z]+-)([A-Za-z]+-)(.*)", "\\1\\2zzz", name)
[1] "hsa-miR-zzz" "hsa-miR-zzz" "hsa-let-zzz"
There are actually several ways to approach this, depending on how "regular" your expressions actually are. For example, do they all start with "hsa-"? What are the options for the "middle" group? Might there be more than three dashes?
I need a regular expression that can be used to find the Nth entry in a comma-separated list.
For example, say this list looks like this:
abc,def,4322,mail#mailinator.com,3321,alpha-beta,43
...and I wanted to find the value of the 7th entry (alpha-beta).
My first thought would not be to use a regular expression, but to use something that splits the string into an array on the comma, but since you asked for a regex.
most regexes allow you to specify a minimum or maximum match, so something like this would probably work.
/(?:[^\,]*,){5}([^,]*)/
This is intended to match any number of character that are not a comma followed by a comma six times exactly (?:[^,]*,){5} - the ?: says to not capture - and then to match and capture any number of characters that are not a comma ([^,]+). You want to use the first capture group.
Let me know if you need more info.
EDIT: I edited the above to not capture the first part of the string. This regex works in C# and Ruby.
You could use something like:
([^,]*,){$m}([^,]*),
As a starting point. (Replace $m with the value of (n-1).) The content would be in capture group 2. This doesn't handle things like lists of size n, but that's just a matter of making the appropriate modifications for your situation.
#list = split /,/ => $string;
$it = $list[6];
or just
$it = (split /,/ => $string)[6];
Beats writing a pattern with a {6} in it every time.