Problem: I want to match those strings which contains two digits.Their position is random and a digit should match 2 times.
Example for better understanding my question:
3abc3
a22de
b7abc7a
For these strings it must match.If a string contains two digits but they are different then it shouldn't match.
Example:
3abcd2 not supposed to match
3abc3 -> supposed to match
I tried using {n}, but it not helps, because it thinks the two number follows each other.
You can use this grep:
grep -E '([0-9]).*\1' file
3abc3
a22de
b7abc7a
About this Regex:
([0-9]) # match and capture any digit in group #1
.* # match 0 or more of any character in between
\1 # using back-reference \1, make sure we have same digit as in group #1
Related
there are 4 strings as shown below
ABC_FIXED_20220720_VALUEABC.csv
ABC_FIXED_20220720_VALUEABCQUERY_answer.csv
ABC_FIXED_20220720_VALUEDEF.csv
ABC_FIXED_20220720_VALUEDEFQUERY_answer.csv
Two strings are considered as matched based on a matching substring value (VALUEABC, VALUEDEF in the above shown strings). Thus I am looking to match first 2 (having VALUEABC) and then next 2 (having VALUEDEF). The matched strings are identified based on the same value returned for one regex group.
What I tried so far
ABC.*[0-9]{8}_(.*[^QUERY_answer])(?:QUERY_answer)?.csv
This returns regex group-1 (from (.*[^QUERY_answer])) value "VALUEABC" for first 2 strings and "VALUEDEF" for next 2 strings and thus desired matching achieved.
But the problem with above regex is that as soon as the value ends with any of the characters of "QUERY_answer", the regex doesn't match any value for the grouping. For instance, the below 2 strings doesn't match at all as the VALUESTU ends with "U" here :
ABC_FIXED_20220720_VALUESTU.csv
ABC_FIXED_20220720_VALUESTUQUERY_answer.csv
I tried to use Negative Lookahead:
ABC.*[0-9]{8}_(.*(?!QUERY_answer))(?:QUERY_answer)?.csv
but in this case the grouping-1 value is returned as "VALUESTU" for first string and "VALUESTUQUERY_answer" for second string, thus effectively making the 2 strings unmatched.
Any way to achieve the desired matching?
With your shown samples please try following regex.
^ABC_[^_]*_[0-9]+_(.*?)(?:QUERY_answer)?\.csv$
OR to match exact 8 digits try:
^ABC_[^_]*_[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv$
Here is the online demo for above regex.
Explanation: Adding detailed explanation for above regex.
^ABC_[^_]*_ ##Matching from starting of value ABC followed by _ till next occurrence of _.
[0-9]+_ ##Matching continuous occurrences of digits followed by _ here.
(.*?) ##Creating one and only capturing group using lazy match which is opposite of greedy match.
(?:QUERY_answer)? ##In a non-capturing group matching QUERY_answer and keeping it optional.
\.csv$ ##Matching dot literal csv at the end of the value.
You need
ABC.*[0-9]{8}_(.*?)(?:QUERY_answer)?\.csv
See the regex demo.
Note
.*[^QUERY_answer] matches any zero or more chars other than line break chars as many as possible, and then any one char other than Q, U, E, etc., i.e. any char in the negated character class. This is replaced with .*?, to match any zero or more chars other than line break chars as few as possible.
(?:QUERY_answer)? - the group is made non-capturing to reduce grouping complexity.
\.csv - the . is escaped to match a literal dot.
I need to find the first set of 5 numbers in a text like this :
;SUPER U CHARLY SUR MARNE;;;rte de Pavant CHARLY SUR MARNE Picardie 02310;Charly-sur-Marne;;;02310;;;;;;;;;;;;;;
I need to find the first 02310 only.
My regex but it found all set of 5 numbers :
([^\d]|^)\d{5}([^\d]|$)
To match the first 5-digit number you may use
^.*?\K(?<!\d)\d{5}(?!\d)
See the regex demo. As you want to remove the match, simply keep the Replace With field blank. The ^ matches the start of a line, .*? matches any 0+ chars other than line break chars, as few as possible, and \K operator drops the text matched so far. Then, (?<!\d)\d{5}(?!\d) matches 5 digits not enclosed with other digits.
Another variation includes a capturing group/backreference:
Find What: ^(.*?)(?<!\d)\d{5}(?!\d)
Replace With: $1
See this regex demo.
Here, instead of dropping the found text before the number, (.*?) is captured into Group 1 and $1 in the replacement pattern puts it back.
I would've use
(^(?:(?!\d{5}).)+)(\d{5})(?!\d)
It finds fragment from beginning of the string till end of first 5-digit number, but in case of replacement you can use $1 or $2 to substitute corresponding part. For example replacement $1<$2> will surround number by < and >.
To find the first 5 digits in the text, you could also match not a digit \D* or 1-4 digits followed by matching 5 digits:
^(?=.*\b\d{5}\b)(?:\D*|\d{1,4})*\K\d{5}(?!\d)
^ Start of string
(?=.*\b\d{5}\b) Assert that there are 5 consecutive digits between word boundaries
(?:\D*|\d{1,4})* Repeat matching 0+ times not a digit or 1-4 digits
\K\d{5} Forget what was matched, then match 5 digits
(?!\d) Assert what followed is not a digit
Regex demo
I have patterns like
FQC19515_TCELL001_20190319_165944.pdf
FQC19515_TBNK001_20190319_165944.pdf
I can match word TCELL and TBNK with this RegEX
^(\D+)-(\d+)-(\d+)([A-Z1-9]+)?.*
But if I have patterns like
FLW194640_T20NK022_20190323_131348.pdf
FLW194228_C1920_SOME_DEBRIS_REMOVED.pdf
the above regex returns
T2 and C192 instead of T20NK and C1920 respectively
Is there a general regex that matches Nzeros out side of these word boundaries?
Let's consider all 4 examples of your input:
FQC19515_TCELL001_20190319_165944.pdf
FQC19515_TBNK001_20190319_165944.pdf
FLW194640_T20NK022_20190323_131348.pdf
FLW194228_C1920_SOME_DEBRIS_REMOVED.pdf
The first group, between start of line and the first "_" (e.g. FQC19515 in row 1)
consists of:
a non-empty sequence of letters,
a non-empty sequence of digits.
So the regex matching it, including the start of line anchor and a capturing group is:
^([A-Z]+\d+)
You used \D instead of [A-Z] but I think that [A-Z] is
more specific, as it matches only letters an not e.g. "_".
The next source char is _, so the regex can also include _.
A now the more diificult part: The second group to be captured has
actually 2 variants:
a sequence of letters and a sequence of digits (after that there is
a "_"),
a sequence of letters, a sequence of digits and another sequence of
letters (after that there are digits that you want to omit).
So the most intuitive way is to define 2 alternatives, each with
a respective positive lookahead:
alternative 1: [A-Z]+\d+(?=_),
alternative 2: [A-Z]+\d+[A-Z]+(?=\d).
But there is a bit shorter way. Notice that both alternatives start
from [A-Z]+\d+.
So we can put this fragment at the first place and only the rest
include as a non-capturing group ((?:...)), with 2 alternatives.
All the above should be surrounded with a capturing group:
([A-Z]+\d+(?:(?=_)|[A-Z]+(?=\d)))
So the whole regex can be:
^([A-Z]+\d+)_([A-Z]+\d+(?:(?=_)|[A-Z]+(?=\d)))
with m option ("^" matches also the start of each line).
For a working example see https://regex101.com/r/GDdt10/1
Your regex: ^(\D+)-(\d+) is wrong as after a sequence of non-digits
(\D+) you specified a minus which doesn't occur in your source.
Also the second minus does not correspond to your input.
Edit
To match all your strings, I modified slightly the previous regex.
The changes are limited to the matching group No 2 (after _):
Alternative No 1: [A-Z]{2,}+(?=\d) - two or more letters, after them
there is a digit, to be omitted. It will match TCELL and TBNK.
Alternative No 2: [A-Z]+\d+(?:(?=_)|[A-Z]+(?=\d)) - the previous
content of this group. It will match two remaining cases.
So the whole regex is:
^([A-Z]+\d+)_([A-Z]{2,}+(?=\d)|[A-Z]+\d+(?:(?=_)|[A-Z]+(?=\d)))
For a working example see https://regex101.com/r/GDdt10/2
As far as I understand, you could use:
^[A-Z]+\d+_\K[A-Z0-9]{5}
Explanation:
^ # beginning of line
[A-Z]+ # 1 or more capitals
\d+_ # 1 or more digit and 1 underscore
\K # forget all we have seen until this position
[A-Z0-9]{5} # 5 capitals or digits
Demo
For example, this is the regular expression
([a]{2,3})
This is the string
aaaa // 1 match "(aaa)a" but I want "(aa)(aa)"
aaaaa // 2 match "(aaa)(aa)"
aaaaaa // 2 match "(aaa)(aaa)"
However, if I change the regular expression
([a]{2,3}?)
Then the results are
aaaa // 2 match "(aa)(aa)"
aaaaa // 2 match "(aa)(aa)a" but I want "(aaa)(aa)"
aaaaaa // 3 match "(aa)(aa)(aa)" but I want "(aaa)(aaa)"
My question is that is it possible to use as few groups as possible to match as long string as possible?
How about something like this:
(a{3}(?!a(?:[^a]|$))|a{2})
This looks for either the character a three times (not followed by a single a and a different character) or the character a two times.
Breakdown:
( # Start of the capturing group.
a{3} # Matches the character 'a' exactly three times.
(?! # Start of a negative Lookahead.
a # Matches the character 'a' literally.
(?: # Start of the non-capturing group.
[^a] # Matches any character except for 'a'.
| # Alternation (OR).
$ # Asserts position at the end of the line/string.
) # End of the non-capturing group.
) # End of the negative Lookahead.
| # Alternation (OR).
a{2} # Matches the character 'a' exactly two times.
) # End of the capturing group.
Here's a demo.
Note that if you don't need the capturing group, you can actually use the whole match instead by converting the capturing group into a non-capturing one:
(?:a{3}(?!a(?:[^a]|$))|a{2})
Which would look like this.
Try this Regex:
^(?:(a{3})*|(a{2,3})*)$
Click for Demo
Explanation:
^ - asserts the start of the line
(?:(a{3})*|(a{2,3})*) - a non-capturing group containing 2 sub-sequences separated by OR operator
(a{3})* - The first subsequence tries to match 3 occurrences of a. The * at the end allows this subsequence to match 0 or 3 or 6 or 9.... occurrences of a before the end of the line
| - OR
(a{2,3})* - matches 2 to 3 occurrences of a, as many as possible. The * at the end would repeat it 0+ times before the end of the line
-$ - asserts the end of the line
Try this short regex:
a{2,3}(?!a([^a]|$))
Demo
How it's made:
I started with this simple regex: a{2}a?. It looks for 2 consecutive a's that may be followed by another a. If the 2 a's are followed by another a, it matches all three a's.
This worked for most cases:
However, it failed in cases like:
So now, I knew I had to modify my regex in such a way that it would match the third a only if the third a is not followed by a([^a]|$). So now, my regex looked like a{2}a?(?!a([^a]|$)), and it worked for all cases. Then I just simplified it to a{2,3}(?!a([^a]|$)).
That's it.
EDIT
If you want the capturing behavior, then add parenthesis around the regex, like:
(a{2,3}(?!a([^a]|$)))
I have a dataset with repeating pattern in the middle:
YM10a15b5c27
and
YM1b5c17
How can I get what is between "YM" and the last two numbers?
I'm using this but is getting one number in the end and should not.
/([A-Z]+)([0-9a-z]+)([0-9]+)/
Capture exactly two characters in the last group:
/([A-Z]+)([0-9a-z]+)([0-9]{2})/
You should use:
/^(?:([a-z]+))([0-9a-z]+)(?=\1)/
^ matches the start of the sentence. This is really important, because if your code is aaaa1234aaaa, then without the ^, it would also match the aaaa of the end.
(?:([a-z]+)) is a non-capturing group which takes any letter from 'a' to 'z' as group 1
(?=\1) tells the regex to match the text as long as it is followed by the same code at the starting.
All you have to do is extract the code by group(2)
An example is shown here.
Solution
If you want to match these strings as whole words, use \b(([a-z])\2)([0-9a-z]+)(\1)\b. If you need to match them as separate strings, use ^(([a-z])\2)([0-9a-z]+)(\1)$.
Explanation
\b - a word boundary (or if ^ is used, start of string)
(([a-z])\2) - Group 1: any lowercase ASCII letter, exactly two occurrences (aa, bb, etc.)
([0-9a-z]+) - Group 3: 1 or more digits or lowercase ASCII letters
(\1) - Group 4: the same text as stored in Group 1
\b - a word boundary (or if $ is used, end of string).