I have two possible endings for my string. The first with no numbers:
http://www.something.com/test.html
the second with numbers (up to two digits)
http://www.something.com/test-1.html
http://www.something.com/test-2.html
http://www.something.com/test-3.html
http://www.something.com/test-4.html
http://www.something.com/test-15.html
I need to strip the .html from the first case and -1.html (or whatever number) from the second. The idea is to make the two string comparable to find duplicates.
I think the following should manage the second case
gsub("-[0-9]|[1-9][0-9].html", "", string)
but is it possible to have a function to manage both cases?
You can perhaps use something like this:
(-[0-9]+)?\\.html
Note that it's safer to escape the dot because an unescaped dot will match any character.
regex101 demo
Related
Sample input:
___file___name___2000___ed2___1___2___3
DIFFERENT+FILENAME+(2000)+1+2+3+ed10
Desired output (eg, all letters and 4-digit numbers and literal 'ed' followed immediately by a digit of arbitrary length:
file name 2000 ed2
DIFFERENT FILENAME 2000 ed10
I am using:
[A-Za-z]+|[\d]{4}|ed\d+ which only returns:
file name 2000 ed
DIFFERENT FILENAME 2000 ed
I see that there is a related Q+A here:Regular Expression to match specific string followed by number?
eg using ed[0-9]* would match ed#, but unsure why it does not match in the above.
As written, your regex is correct. Remember, however, that regex tries to match its statements from left to right. Your ed\d+ is never going to match, because the ed was already consumed by your [A-Za-z] alternative. Reorder your regex and it'll work just fine:
ed\d+|[a-zA-Z]+|\d{4}
Demo
Nick's answer is right, but because in-order matching can be a less readable "gotcha", the best (order-insensitive) ways to do this kind of search are 1) with specified delimiters, and 2) by making each search term unique.
Jan's answer handles #1 well. But you would have to specify each specific delimiter, including its length (e.g. ___). It sounds like you may have some unusual delimiters, so this may not be ideal.
For #2, then, you can make each search term unique. (That is, you want the thing matching "file" and "name" to be distinct from the thing matching "2000", and both to be distinct from the thing matching "ed2".)
One way to do this is [A-Za-z]+(?![0-9a-zA-Z])|[\d]{4}|ed\d+. This is saying that for the first type of search term, you want an alphabet string which is followed by a non-alphanumeric character. This keeps it distinct from the third search term, which is an alphabet string followed by some digit(s). This also allows you to specify any range of delimiters inside of that negative lookbehind.
demo
You might very well use (just grab the first capturing group):
(?:^|___|[+(]) # delimiter before
([a-zA-Z0-9]{2,}) # the actual content
(?=$|___|[+)]) # delimiter afterwards
See a demo on regex101.com
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
Given a text "article_utf8" i want to remove a list of words:
remove = "el|la|de|que|y|a|en|un|ser|se|no|haber|..."
regex = re.compile(r'\b('+remove+r')\b', flags=re.IGNORECASE)
article_out = regex.sub("", article_utf8)
however this is incorrectly removing some words and parts of words for example:
1- aseguro becomes seguro
2- sería becomes í
3- coma becomes com
4- miercoles becomes 'ercoles'
Technically parts of a word can match a regexp. To solve this you would have to make sure that whatever sequence of letters your regexp matches is a single word and not part of it.
One way would be to make the regexp contain leading and trailing spaces, but words could also be separated with periods or commas so you would have to take those into account too if you want to catch all instances.
Alternatively, you can try splitting the list first into words using the built-in split method (https://docs.python.org/2/library/stdtypes.html#str.split). Then I would check each word in the resulting list, remove the ones I don't want and rejoin the strings. This method, however doesn't even need regexps so it's probably not what you intended despite being simple and practical.
After much testing, the following will remove the small words in a natural language string, without removing them from parts of other words:
regex = re.compile(r'[\s]?\b('+remove+')[\b\s\.\,]', flags=re.IGNORECASE)
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'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!