I am trying to find a regular expression that will allow me to know if there is a dinucleotide(Two letters) that appears 2 times in a row in my sequence. I give you an example:
Let's suppose I have this sequence (The character ; is to make clear that I am talking about dinucleotides):
"AT;GC;TA;CC;AG;AG;CC;CA;TA;TA"
The result I expect is that it matches the pattern AGAG and TATA.
I have tried this already but it fails because it gives me any pair of dinucleotides, not the same pair :
([ATGC]{2}){2}
You will need to use backreferences.
Start with matching one pair:
[ATGC]{2}
will match any pair of two of the four letters.
You need to put that in capturing parentheses and refer to the contents of the parentheses with \1, like so:
([ATGC]{2});\1
Suppose the string were
"TA;TA;GC;TA;CC;AG;AG;CC;CA;TA;TA"
^^ ^^ ^^ ^^ ^^ ^^
If you wish to match "TA" twice (and "AG" once) you could apply #Andy's solution.
If you wish to match "TA" just once, no matter the number of instances of "TA;TA" in the string, you could match
([ATGC]{2});\1(?!.*\1;\1)
and retrieve the contents of capture group 1.
Demo
The expression can be broken down as follows.
([ATGC]{2}) # match two characters, each from the character class,
# and save to capture group 1
;\1 # match ';' followed by the content of capture group 1
(?! # begin a negative lookahead
.* # match zero or more characters
\1;\1 # match the content of capture group 1 followed by ';'
# followed by the content of capture group 1
) # end negative lookahead
Related
I want to select the same element with exact n occurrences.
Match letters that repeats exact 3 times in this String: "aaaaabbbcccccccccdddee"
this should return "bbb" and "ddd"
If I define what I should match like "b{3}" or "d{3}", this would be easier, but I want to match all elements
I've tried and the closest I came up is this regex: (.)\1{2}(?!\1)
Which returns "aaa", "bbb", "ccc", "ddd"
And I can't add negative lookbehind, because of "non-fixed width" (?<!\1)
One possibility is to use a regex that looks for a character which is not followed by itself (or beginning of line), followed by three identical characters, followed by another character which is not the same as the second three i.e.
(?:(.)(?!\1)|^)((.)\3{2})(?!\3)
Demo on regex101
The match is captured in group 2. The issue with this though is that it absorbs a character prior to the match, so cannot find adjacent matches: as shown in the demo, it only matches aaa, ccc and eee in aaabbbcccdddeee.
This issue can be resolved by making the entire regex a lookahead, a technique which allows for capturing overlapping matches as described in this question. So:
(?=(?:(.)(?!\1)|^)((.)\3{2})(?!\3))
Again, the match is captured in group 2.
Demo on regex101
You could match what you don't want to keep, which is 4 or more times the same character.
Then use an alternation to capture what you want to keep, which is 3 times the same character.
The desired matches are in capture group 2.
(.)\1{3,}|((.)\3\3)
(.) Capture group 1, match a single character
\1{3,} Repeat the same char in group 1, 3 or more times
| Or
( Capture group 2
(.)\3\3 Capture group 3, match a single character followed by 2 backreferences matching 2 times the same character as in group 3
) Close group 2
Regex demo
This gets sticky because you cannot put a back reference inside a negative character set, so we'll use a lookbehind followed by a negative lookahead like this:
(?<=(.))((?!\1).)\2\2(?!\2))
This says find a character but don't include it in the match. Then look ahead to be certain the next character is different. Next consume it into capture group 2 and be certain that the next two characters match it, and the one after does not match.
Unfortunately, this does not work on 3 characters at the beginning of the string. I had to add a whole alternation clause to handle that case. So the final regex is:
(?:(?<=(.))((?!\1).)\2\2(?!\2))|^(.)\3\3(?!\3)
This handles all cases.
EDIT
I found a way to handle matches at the beginning of the string:
(?:(?<=(.))|^)((?!\1).)\2\2(?!\2)
Much nicer and more compact, and does not require looking in capture groups to get the answer.
If your environment permits the use of (*SKIP)(*FAIL), you can manage to return a lean set of matches by consuming substrings of four or more consecutive duplicate characters then discard them. In the alternation, match the desired 3 consecutive duplicated characters.
PHP Code: (Demo)
$string = 'aaaaabbbcccccccccdddee';
var_export(
preg_match_all(
'/(?:(.)\1{3,}(*SKIP)(*F)|(.)\2{2})/',
$string,
$m
)
? $m[0]
: 'no matches'
);
Output:
array (
0 => 'bbb',
1 => 'ddd',
)
This technique uses no lookarounds and does not generate false positive matches in the matches array (which would otherwise need to be filtered out).
This pattern is efficient because it never needs to look backward and by consuming the 4 or more consecutive duplicates, it can rule-out long substrings quickly.
I want to expect some characters only if a prior regex matched. If not, no characters (empty string) is expected.
For instance, if after the first four characters appears a string out of the group (A10, B32, C56, D65) (kind of enumeration) then a "_" followed by a 3-digit number like 123 is expected. If no element of the mentioned group appears, no other string is expected.
My first attempt was this but the ELSE branch does not work:
^XXX_(?<DT>A12|B43|D14)(?(DT)(_\d{1,3})|)\.ZZZ$
XXX_A12_123.ZZZ --> match
XXX_A11.ZZZ --> match
XXX_A12_abc.ZZZ --> no match
XXX_A23_123.ZZZ --> no match
These are examples of filenames. If the filename contains a string of the mentioned group like A12 or C56, then I expect that this element if followed by an underscore followed by 1 to 3 digits. If the filename does not contain a string of that group (no character or a character sequence different from the strings in the group) then I don't want to see the underscore followed by 1 to 3 digits.
For instance, I could extend the regex to
^XXX_(?<DT>A12|B43|D14)_\d{5}(?(DT)(_\d{1,3})|)_someMoreChars\.ZZZ$
...and then I want these filenames to be valid:
XXX_A12_12345_123_wellDone.ZZZ
XXX_Q21_00000_wellDone.ZZZ
XXX_Q21_00000_456_wellDone.ZZZ
...but this is invalid:
XXX_A12_12345_wellDone.ZZZ
How can I make the ELSE branch of the conditional statement work?
In the end I intend to have two groups like
Group A: (A11, B32, D76, R33)
Group B: (A23, C56, H78, T99)
If an element of group A occurs in the filename then I expect to find _\d{1,3} in the filename.
If an element of group B occurs ion the filename then the _\d{1,3} shall be optional (it may or may not occur in the filename).
I ended up in this regex:
^XXX_(?:(?A12|B43|D14))?(?(DT)(_\d{5}_\d{1,3})|(?!(?&DT))(?!.*_\d{3}(?!\d))).*\.ZZZ$
^XXX_(?:(?<DT>A12|B43|D14))?_\d{5}(?(DT)(_\d{1,3})|(?!(?&DT))(?!.*_\d{3}(?!\d))).+\.ZZZ$
Since I have to use this regex in the OpenApi #Pattern annotation I have the problem that I get the error:
Conditionals are not supported in this regex dialect.
As #The fourth bird suggested alternation seems to do the trick:
XXX_((((A12|B43|D14)_\d{5}_\d{1,3}))|((?:(A10|B10|C20)((?:_\d{5}_\d{3})|(?:_\d{3}))))).*\.ZZZ$
The else branch is the part after the |, but if you also want to match the 2nd example, the if clause would not work as you have already matched one of A12|B43|D14
The named capture group is not optional, so the if clause will always be true.
What you can do instead is use an alternation to match either the numeration part followed by an underscore and 3 digits, or match an uppercase char and 2 digits.
^XXX_(?:(?<DT>A12|B43|D14)_\d{1,3}|[A-Z]\d{2})\.ZZZ$
Regex demo
If you want to make use of the if/else clause, you can make the named capture group optional, and then check if group 1 exists.
^XXX_(?<DT>A12|B43|D14)?(?(DT)_\d{1,3}|[A-Z]\d{2})\.ZZZ$
Regex demo
For the updated question:
^XXX_(?<DT>A12|B43|D14)?(?(DT)(?:_\d{5})?_\d{3}(?!\d)|(?!A12|B43|D14|[A-Z]\d{2}_\d{3}(?!\d))).*\.ZZZ$
The pattern matches:
^ Start of string
XXX_ Match literally
(?<DT>A12|B43|D14)?
(?(DT) If we have group DT
(?:_\d{5})? Optionally match _ and 5 digits
_\d{3}(?!\d) Match _ and 3 digits
| Or
(?! Negative lookahead, assert not to the right
A12|B43|D14| Match one of the alternatives, or
[A-Z]\d{2}_\d{3}(?!\d) Match 1 char A-Z, 2 digits _ 3 digits not followed by a digit
) Close lookahead
) Close if clause
.* Match the rest of the line
\.ZZZ Match . and ZZZ
$ End of string
Regex demo
A straight in poker is five cards in a row, for example 23456 or 89TJQ. With a "sorted" hand, the regex could be written as:
^(A2345|23456|34567|45678|56789|6789T|789TJ|89TJQ|9TJQK|TJQKA)$
It's a bit verbose but straightforward enough. However, would it be possible to generate a (sensible) regex if the hand was unordered? For example, if the hand was 52634 or JQ89T??
One possible way would be to use a ?=.*<item> lookahead (which would essentially be "unsorted"), for example:
^(?:
(?=.*A)(?=.*2)(?=.*3)(?=.*4)(?=.*5)
|(?=.*2)(?=.*3)(?=.*4)(?=.*5)(?=.*6)
|(?=.*3)(?=.*4)(?=.*5)(?=.*6)(?=.*7)
|(?=.*4)(?=.*5)(?=.*6)(?=.*7)(?=.*8)
|(?=.*5)(?=.*6)(?=.*7)(?=.*8)(?=.*9)
|(?=.*6)(?=.*7)(?=.*8)(?=.*9)(?=.*T)
|(?=.*7)(?=.*8)(?=.*9)(?=.*T)(?=.*J)
|(?=.*8)(?=.*9)(?=.*T)(?=.*J)(?=.*Q)
|(?=.*9)(?=.*T)(?=.*J)(?=.*Q)(?=.*K)
|(?=.*T)(?=.*J)(?=.*Q)(?=.*K)(?=.*A)
)
.{5}$
Are there other / better approaches to finding if a straight exists using regex only?
You can use the following regex:
See regex in use here
(?!.*(.).*\1)(?:[A2345]{5}|[23456]{5}|[34567]{5}|[45678]{5}|[56789]{5}|[6789T]{5}|[789TJ]{5}|[89TJQ]{5}|[9TJQK]{5}|[TJQKA]{5})
This works by first using a negative lookahead to ensure that the string doesn't contain any duplicates (?!.*(.).*\1). Then it matches 5 characters from any of the straight possibilities.
(?!.*(.).*\1)
#^^^ ^ negative lookahead ensuring what follows doesn't match
# ^^ match any character any number of times
# ^^^ capture a character into capture group #1
# ^^ match any character any number of times
# ^^ match the same text as most recently matched by the 1st capture group
Against JQQ89, it works as follows:
- .* matches J
- (.) captures Q
- .* matches nothing
- \1 tries to match Q (and succeeds)
- Negative lookahead has a match, so fail the match.
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]|$)))
I am using this regex:
((?:[a-z][a-z]+))_(\d+)_((?:[a-z][a-z]+)\d+)_(\d{13})
to match strings like this:
SH_6208069141055_BC000388_20110412101855
separating into 4 groups:
SH
6208069141055
BC000388
20110412101855
Question: How do I make the first group optional, so that the resulting group is a empty string?
I want to get 4 groups in every case, when possible.
Input string for this case: (no underline after the first group)
6208069141055_BC000388_20110412101855
Making a non-capturing, zero to more matching group, you must append ?.
(?: ..... )?
^ ^____ optional
|____ group
You can easily simplify your regex to be this:
(?:([a-z]{2,})_)?(\d+)_([a-z]{2,}\d+)_(\d+)$
^ ^^
|--------------||
| first group ||- quantifier for 0 or 1 time (essentially making it optional)
I'm not sure whether the input string without the first group will have the underscore or not, but you can use the above regex if it's the whole string.
regex101 demo
As you can see, the matched group 1 in the second match is empty and starts at matched group 2.