I'm trying to come up with an example where positive look-around works but
non-capture groups won't work, to further understand their usages. The examples I"m coming up with all work with non-capture groups as well, so I feel like I"m not fully grasping the usage of positive look around.
Here is a string, (taken from a SO example) that uses positive look ahead in the answer. The user wanted to grab the second column value, only if the value of the
first column started with ABC, and the last column had the value 'active'.
string ='''ABC1 1.1.1.1 20151118 active
ABC2 2.2.2.2 20151118 inactive
xxx x.x.x.x xxxxxxxx active'''
The solution given used 'positive look ahead' but I noticed that I could use non-caputure groups to arrive at the same answer.
So, I'm having trouble coming up with an example where positive look-around works, non-capturing group doesn't work.
pattern =re.compile('ABC\w\s+(\S+)\s+(?=\S+\s+active)') #solution
pattern =re.compile('ABC\w\s+(\S+)\s+(?:\S+\s+active)') #solution w/out lookaround
If anyone would be kind enough to provide an example, I would be grateful.
Thanks.
The fundamental difference is the fact, that non-capturing groups still consume the part of the string they match, thus moving the cursor forward.
One example where this makes a fundamental difference is when you try to match certain strings, that are surrounded by certain boundaries and these boundaries can overlap. Sample task:
Match all as from a given string, that are surrounded by bs - the given string is bababaca. There should be two matches, at positions 2 and 4.
Using lookarounds this is rather easy, you can use b(a)(?=b) or (?<=b)a(?=b) and match them. But (?:b)a(?:b) won't work - the first match will also consume the b at position 3, that is needed as boundary for the second match. (note: the non-capturing group isn't actually needed here)
Another rather prominent sample are password validations - check that the password contains uppercase, lowercase letters, numbers, whatever - you can use a bunch of alternations to match these - but lookaheads come in way easier:
(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])(?=.*[!?.])
vs
(?:.*[a-z].*[A-Z].*[0-9].*[!?.])|(?:.*[A-Z][a-z].*[0-9].*[!?.])|(?:.*[0-9].*[a-z].*[A-Z].*[!?.])|(?:.*[!?.].*[a-z].*[A-Z].*[0-9])|(?:.*[A-Z][a-z].*[!?.].*[0-9])|...
Related
I am writing a string.Format-like method. In order to do this, I am adopting Regex to determine commands and parameters: e.g. Format(#"\m{0,1,2}", byteArr0, byteArr1, byteArr2)
For the first Regex, return 2 groups:
'\m'
'{0,1,2}'
Another Regex takes the value of '{0,1,2}' and has 3 matches:
0
1
2
These values are the indexes corresponding to the byteArr params.
This command structure is likely to grow so I'm really trying to figure this out and learn enough to be able to modify the Regex for future requirements.I would think that a single Regex would do all of the above but there is value in having 2 separate Regex(es/ices ???) expressions.
Any way, to get the first group '\m' the Regex is:
"(\\)(\w{1,1})" // I want the '{0,1,2}' group also
To get the integer matches '{0,1,2}' I was trying:
"(?<=\{)([^}]*)(?=\})"
I am having difficulty in achieving: (1) 2 groups on the first expression and (2) 3 matches on the integers within the braces delimited by a comma in the second expression.
Your first regex (\\)(\w{1,1}) can be greatly simplified.
You don't want to capture the \ separately to the m so no need to wrap them in their own sets of parenthesis.
\w{1,1} is the same as just \w.
So we have \\\w to match the first part \m.
Now to deal with the second part, really we can ignore everything other than the 0,1,2 in the example since there are no numbers elsewhere so you'd just use: \d+ and iterate through the matches.
But lets assume the example could actually be \9{1,2,3}.
Now \d+ would match the 9 so to avoid this we could use [{,](\d+)[,}]. This says capture a number that has either a , or { on the left of it and a , or } on the right.
You're right in saying that we can match the whole string with a single regex, something like this would do it:
(\\\w){((\d+),?)+}
However the problem with this is when you examine the contents of the capture groups afterwards, the last number caught by the (\d+) overwrites all the other values that were caught in there. So you'd be left with group 1: \m and group 2: 2 for your example.
With that in mind I recommend using 2 regexs:
For the 1st part: \\\w
For the numbers: I'd forget about the [{,](\d+)[,}] (and the many other ways you could do it), the cleanest way might just be to grab whatever is inside the {...} and then match with a simple \d+.
So to do this first use (\\\w)\{([^/}]+)\} to grab the \m into group 1 and the 1,2,3 into group 2, then just use \d+ on that.
FYI, your (?<=\{)([^}]*)(?=\}) works fine, but you can't but anything before the lookbehind i.e. the \\\w. In the vast majority of cases where a lookbehind can be used, you can do what you want by just using capture groups and ignoring everything else :
My regex \{([^/}]+)\} is pretty much the same as you (?<=\{)([^}]*)(?=\}) except rather than looking ahead and looking behind for the { and } I just leave them outside the capture groups that are going to be used.
Consider the following Regexes...
(^.*?)(?={.*})
\d+
Good Luck!
Considering
NN = number/digit
x = any single letter
I want to match these patterns:
1. NN
2. NNx
3. NN.NN
4. NN.NNx
5. NN.NN.NN
6. NN.NN.NNx
Example that needs to be match:
1. 20
2. 20a
3. 20.20
4. 20.20a
5. 20.20.20
6. 20.20.20a
Right now I am trying to use this regex:
\b\d+\.?\d+\.?\d+?[a-z]?\b
But if fails.
Any help would be greatly appreciate, thanks! XD
EDIT:
I am matching this:
<fn:footnote fr="10.23.20a"> (Just a sample)
Now I have a regex that will extract the '10.23.20a'
Now I will check if this value will be valid, the 6 examples above will be the only string that will be accepted.
This examples are invalid:
1. 20.a
2. 20a.20.20
3. etc.
Many thanks for your help men! :D
You always have \d+, which is one or more digits. So you require at least three digits. Try grouping the digits with their periods:
^\d+(?:[.]\d+){0,2}[a-z]?$
The ?: is just an optimization (and a good practice) that suppresses capturing. [.] and \. are completely interchangeable, but I prefer the readability of the former. Choose whatever you like best.
If you actually want to capture the numbers and the letter, there two options:
^(?<first>\d+)(?:[.](?<second>\d+))?(?:[.](?<third>\d+))?(?<letter>[a-z])?$
Note that the important point is to group a period and the digits together and make them optional together. You could as well use unnamed groups, it doesn't really matter. However, if you use my version, you can now access the parts through (for instance)
match.Groups["first"].Value
where match is a Match object returned by Regex.Match or Regex.Matches for example.
Alternatively, you can use .NET's feature of capturing multiple values with one group:
^(?<d>\d+)(?:[.](?<d>\d+){0,2}(?<letter>[a-z])?$
Now match.Groups["d"].Captures will contain a list of all captured numbers (1 to 3). And match.Groups["letter"].Value will still contain the letter if it was there.
Try this
^\d+(?:(?:\.\d+)*[a-z]?)$
I have a list of post codes which should be excluded from my shipping methods.
Suppose I have to exclude Scilly Isles, Isle of Man and few others.
For the above 2 areas valid post codes are IM1-IM9, IM86, IM87, IM89. And if it is IM25 or IM85 it is invalid.
I have writtent following expression. But it is returning even it is IM25 or IM 85.
var regex = new Regex("(PO3[0-9]|PO4[0-1]|GY[1-9]|JE[1-5]|IM[1-9]|TR[1-9])");
If I am passing IM85, to my expression it should return false. for IM1-IM9,, IM86, IM87, IM89 it should return true.
Same with TR post codes also. TR1-TR27 is a valid post code. If I give TR28, it should return false.
I am using '|' to seperate multiple patterns. Is that the right way of including multiple patterns in 1 expression.
What do you expect? What should be matched and what not? And please give an example of the string you want to test.
If you match your pattern against "IM25" it will match because you do allow IM[1-9] in your pattern, so you get a valid partial match. If you want to avoid that (I am not sure what you want to achieve) and want to allow really only a single digit after the first letters, use a "word boundary" \b and specify exactly what you want to allow, something like this:
(PO3[0-9]|PO4[0-1]|GY[1-9]|JE[1-5]|IM([1-9]|8[6-9])|TR([1-9]|2[0-7]))\b
See it here on Regexr
this would allow for the "IM" part also 6-9 as a second digit when there is a 8 before.
Update
It is still not clear what the context of your task is. I assume you have a list of valid Postcodes, probably it would be better, you extract the post code or only the first part of it (for that you can eventually use a regex) and check if it is in the list or not.
The actual validation is on the wikipedia site... Google has the answers ;) http://en.wikipedia.org/wiki/Postcodes_in_the_United_Kingdom#Validation
(GIR 0AA)|(((A[BL]|B[ABDFHLNRSTX]?|C[ABFHMORTVW]|D[ADEGHLNTY]|E[HNX]?|F[KY]|G[LUY]?|H[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|L[ADELNSU]?|M[EKL]?|N[EGNPRW]?|O[LX]|P[AEHLOR]|R[GHM]|S[AEGKLMNOPRSTY]?|T[ADFNQRSW]|UB|W[ADFNRSV]|YO|ZE)[1-9]?[0-9]|((E|N|NW|SE|SW|W)1|EC[1-4]|WC[12])[A-HJKMNPR-Y]|(SW|W)([2-9]|[1-9][0-9])|EC[1-9][0-9]) [0-9][ABD-HJLNP-UW-Z]{2})
I still think you need more clarification. As a huge Regex guy, I would like to point out that multi-digit ranges should try to be put into the code side, not the Regex side, just for your sanity. But I personally like to play with Regex in this way. Regex reads one character at a time, so it only recognizes zero through nine. Not ten, not twenty eight. If you want to allow the following:
28 through 347
Then it becomes pretty complicated.
To put it into words, you want to allow:
If Two Digits, allow 2-9 for the first digit, and:
If the first digit is a Two, then allow 8/9 for the second digit,
ElseIf the first Digit is 3-9, then allow 0-9 for the second digit
Elseif Three Digits, allow 1-3 for the first Digit, and:
If the first digit is a Three, then allow 0-4 for the second digit, and:
If the second digit is a Four, then allow 0-7 for the third digit,
ElseIf the second digit is 0-3, then allow 0-9 for the third digit.
ElseIf the first digit is 1/2, then allow 0-9 for both the Second and Third digits.
Then with that, you can write a proper Regex like so, which searches for a word boundary or non-Digit surrounding a 2-pair or 3-pair. With this type of Problem-Solving, you should be able to figure out your Regex issue. Otherwise, let us know more about EXACTLY What you want to Match and NOT Match:
(\b|\D)((2[89]|[3-9][0-9])(\b|\D)|(3(4[0-7]|[0-3][0-9])|[12][0-9][0-9])(\b|\D))
I have changed my approach.
Instead of going for a regular expression which is becoming more complex, I am saving all the excluded outward codes of UK post codes.
And if any post code contains the particular outward code, excluding the post code from the list.
Outward codes are in this format
XX-YYY
XXX-YYY
XXXX-YYY
In all above formats, X represents outward code of an UK postcode.
I am trying to become better at regular expressions. I am having a hard time trying to understand what does (?> expression ) means. Where can I find more info on non-backtacking subexpressoins? The description of THIS link says:
Greedy subexpression, also known as a non-backtracking subexpression.
This is matched only once and then does not participate in
backtracking.
this other link: http://msdn.microsoft.com/en-us/library/bs2twtah(v=vs.71).aspx has also a definition of non-backtracking subexpression but I still am having a hard time understanding what it means plus I cannot think of an example where I will use (?>exp)
As always, regular-expressions.info is a good place to start.
Use an atomic group if you want to make sure that whatever has once been matched will stay part of the match.
For example, to match a number of "words" that may or may not be separated by spaces, then followed by a colon, a user tried the regex:
(?:[A-Za-z0-9_.&,-]+\s*)+:
When there was a match, everything was fine. But when there wasn't, his PC would become non-responsive with 100% CPU load because of catastrophic backtracking because the regex engine would vainly try to find a matching combination of words that would allow the following colon to match. Which was of course impossible.
By using an atomic group, this could have been prevented:
(?>[A-Za-z0-9_.&,-]+\s*)+:
Now whatever has been matched stays matched - no backtracking and therefore fast failing times.
Another good example I recently came across:
If you want to match all numbers that are not followed by an ASCII letter, you might want to use the regex \d+(?![A-Za-z]). However, this will fail with inputs like 123a because the regex engine will happily return the match 12 by backtracking until the following character is no longer a letter. If you use (?>\d+)(?![A-Za-z]), this won't happen. (Of course, \d+(?![\dA-Za-z]) would also work)
The Regex Tutorial has a page on it here: http://www.regular-expressions.info/atomic.html
Basically what it does is discards backtracking information, meaning that a(?>bc|b)c matches abcc but not abc.
The reason it doesn't match the second string is because it finds a match with bc, and discards backtracking information about the bc|b alternation. It essentially forgets the |b part of it. Therefore, there is no c after the bc, and the match fails.
The most useful method of using atomic groups, as they are called, is to optimize slow regexes. You can find more information on the aforementioned page.
Read up on possessive quantifiers [a-z]*+ make the backtracking engine remember only the previous step that matched not all of the previous steps that matched.
This is useful when a lot of acceptable steps are probable and they will eat up memory if each step is stored for any possible backtracking regression.
Possessive quantifiers are a shorthand for atomic groups.
I'm currently writing a library where I wish to allow the user to be able to specify spreadsheet cell(s) under four possible alternatives:
A single cell: "A1";
Multiple contiguous cells: "A1:B10"
Multiple separate cells: "A1,B6,I60,AA2"
A mix of 2 and 3: "B2:B12,C13:C18,D4,E11000"
Then, to validate whether the input respects these formats, I intended to use a regular expression to match against. I have consulted this article on Wikipedia:
Regular Expression (Wikipedia)
And I also found this related SO question:
regex matching alpha character followed by 4 alphanumerics.
Based on the information provided within the above-linked articles, I would try with this Regex:
Default Readonly Property Cells(ByVal cellsAddresses As String) As ReadOnlyDictionary(Of String, ICell)
Get
Dim validAddresses As Regex = New Regex("A-Za-z0-9:,A-Za-z0-9")
If (Not validAddresses.IsMatch(cellsAddresses)) then _
Throw New FormatException("cellsAddresses")
// Proceed with getting the cells from the Interop here...
End Get
End Property
Questions
1. Is my regular expression correct? If not, please help me understand what expression I could use.
2. What exception is more likely to be the more meaningful between a FormatException and an InvalidExpressionException? I hesitate here, since it is related to the format under which the property expect the cells to be input, aside, I'm using an (regular) expression to match against.
Thank you kindly for your help and support! =)
I would try this one:
[A-Za-z]+[0-9]+([:,][A-Za-z]+[0-9]+)*
Explanation:
Between [] is a possible group of characters for a single position
[A-Za-z] means characters (letters) from 'A' to 'Z' and from 'a' to 'z'
[0-9] means characters (digits) from 0 to 9
A "+" appended to a part of a regex means: repeat that one or more times
A "*" means: repeat the previous part zero or more times.
( ) can be used to define a group
So [A-Za-z]+[0-9]+ matches one or more letters followed by one or more digits for a single cell-address.
Then that same block is repeated zero or more times, with a ',' or ':' separating the addresses.
Assuming that the column for the spreadsheet is any 1- or 2-letter value and the row is any positive number, a more complex but tighter answer still would be:
^[A-Z]{1,2}[1-9]\d*(:[A-Z]{1,2}[1-9]\d*)?(,[A-Z]{1,2}[1-9]\d*(:[A-Z]{1,2}[1-9]\d*)?)*$
"[A-Z]{1,2}[1-9]\d*" is the expression for a single cell reference. If you replace "[A-Z]{1,2}[1-9]\d*" in the above with then the complex expression becomes
^<cell>(:<cell>)?(,<cell>(:<cell>*)?)*$
which more clearly shows that it is a cell or a range followed by one or more "cell or range" entries with commas in between.
The row and column indicators could be further refined to give a tighter still, yet more complex expression. I suspect that the above could be simplified with look-ahead or look-behind assertions, but I admit those are not (yet) my strong suit.
I'd go with this one, I think:
(([A-Z]+[1-9]\d*:)?[A-Z]+[1-9]\d*,)*([A-Z]+[1-9]\d*:)?[A-Z]+[1-9]\d*
This only allows capital letters as the prefix. If you want case insensitivity, use RegexOptions.IgnoreCase.
You could simplify this by replacing [A-Z]+[1-9]\d* with plain old [A-Z]\d+, but that will only allow a one-letter prefix, and it also allows stuff like A0 and B01. Up to you.
EDIT:
Having thought hard about DocMax's mention of lookarounds, and using Hans Kesting's answer as inspiration, it occurs to me that this should work:
^[A-Z]+\d+((,|(?<!:\w*):)[A-Z]+\d+)*$
Or if you want something really twisted:
^([A-Z]+\d+(,|$|(?<!:\w*):))*(?<!,|:)
As in the previous example, replace \d+ with [1-9]\d* if you want to prevent leading zeros.
The idea behind the ,|(?<!\w*:): is that if a group is delimited by a comma, you want to let it through; but if it's a colon, it's only allowed if the previous delimiter wasn't a colon. The (,|$|...) version is madness, but it allows you to do it all with only one [A-Z]+\d+ block.
However! Even though this is shorter, and I'll admit I feel a teeny bit clever about it, I pity the poor fellow who has to come along and maintain it six months from now. It's fun from a code-golf standpoint, but I think it's best for practical purposes to go with the earlier version, which is a lot easier to read.
i think your regex is incorrect, try (([A-Za-z0-9]*)[:,]?)*
Edit : to correct the bug pointed out by Baud : (([A-Za-z0-9]*)[:,]?)*([A-Za-z0-9]+)
and finally - best version : (([A-Za-z]+[0-9]+)[:,]?)*([A-Za-z]+[0-9]+)
// ah ok this wont work probably... but to answer 1. - no i dont think your regex is correct
( ) form a group
[ ] form a charclass (you can use A-Z a-d 0-9 etc or just single characters)
? means 1 or 0
* means 0 or any
id suggest reading http://www.regular-expressions.info/reference.html .
thats where i learned regexes some time ago ;)
and for building expressions i use Rad Software Regular Expression Designer
Let's build this step by step.
If you are following an Excel addressing format, to match a single-cell entry in your CSL, you would use the regular expression:
[A-Z]{1,2}[1-9]\d*
This matches the following in sequence:
Any character in A to Z once or twice
Any digit in 1 to 9
Any digit zero or more times
The digit expression will prevent inputting a cell address with leading zeros.
To build the expression that allows for a cell address pair, repeat the expression preceded by a colon as optional.
[A-Z]{1,2}[1-9]\d*(:[A-Z]{1,2}[1-9]\d*)?
Now allow for repeating the pattern preceded by a comma zero or more times and add start and end string delimiters.
^[A-Z]{1,2}[1-9]\d*(:[A-Z]{1,2}[1-9]\d*)?(,[A-Z]{1,2}[1-9]\d*(:[A-Z]{1,2}[1-9]\d*)?)*$
Kind of long and obnoxious, I admit, but after trying enough variants, I can't find a way of shortening it.
Hope this is helpful.