I am not much familiar in regular expression, I wanted to do the following comparison by using regular expression.
Source word is : Hello124
In a list, I have following strings
Hello12
Hello
Hel
Hel123
Her
the output I want is ( Hello12, Hello, Hel ). i.e from source sting, I will reduce last char one by one and find the match in the list. Please let me know, Is that possible to use regular expression to optimize this functionality?
I am using C++ with stl::tr1 library.
You could try this:
^H(?:e(?:l(?:l(?:o(?:1(?:24?)?)?)?)?)?)?$
But in most languages it would be easier just to evaluate query.StartsWith(word) for each word.
Of course, you can solve this problem by using regular expressions, for example using the following: h|he|hel|hell|hello|hello1|hello12|hello124.
However, this is not very nice and an overkill. As far as I know, every language supporting regex also supports querying for substrings (you may want to look here if you find yours).
Related
I'm working on a project and I want to remove text between two parentheses in a string.
Example:
std::string str = "I want to remove (this)."
How would I go about doing that?
I've searched google and stackoverflow an haven't found anything.
I'd use a regular expression for that. Check out the link I provided. As for the expression to use the following expression
(\()(?:[^\)\\]*(?:\\.)?)*\)
That guy worked for me.
Conditionally replace regex matches in string
Do not get regular and common expressions confused. This is not like the more common expression of :-) or :-O or >:( All-though effective These expressions are mutually exclusive expressions that not many languages understand but are more commonly used.
Let's say I've got such regex (python notation) r'^namespace/(\w+)/([0-9]+)/', is there a way to reverse this regex and find a string fulfilling it?
By reversing I don't mean manual constructing 'namespace/' + 'a_1' + '/' + '1', but systematic way to reverse any regular expression consisting of some special characters. So that for every regex I can generate (any) string fulfilling it.
The only thing that comes to my mind is to parse the given regex with some other regexs, but it does not seem acceptable solution. Although I expect the whole operation to have huge complexity, I still look for at least a bit more sophisticated way to do it.
The only thing that comes to my mind is to parse the given regex with some other regexs, but it does not seem acceptable solution
You don't need to parse the regex with regexes, but yes you will need to parse it. When you have an AST of the regular expression, you can easily traverse that and build a possible match in linear time (for plain regular expression, nothing too fancy like lookaround).
Check Enumerating Regular Languages for an example code and continuative links.
Regular expression:
/Hello .*, what's up?/i
String which may contain any number of wildcard characters (%):
"% world, what's up?" (matches)
"Hello world, %?" (matches)
"Hello %, what's up?" (matches)
"Hey world, what's up?" (no match)
"Hello %, blabla." (no match)
I have thought of a solution myself, but I'd like to see what you are able to come up with (considering performance is a high priority). A requirement is the ability to use any regular expression; I only used .* in the example, but any valid regular expression should work.
A little automata theory might help you here. You say
this is a simplified version of matching a regular expression with a regular expression[1]
Actually, that does not seem to be the case. Instead of matching the text of a regular expression, you want to find regular expressions that can match the same string as a given regular expression.
Luckily, this problem is solvable :-) To see whether such a string exists, you would need to compute the union of the two regular languages and test whether the result is not the empty language. This might be a non-trivial problem and solving it efficiently [enough] may be hard, but standard algorithms for this do already exist. Basically you would need to translate the expression into a NFA, that one into a DFA which you then can union.
[1]: Indeed, the wildcard strings you're using in the question build some kind of regular language, and can be translated to corresponding regular expressions
Not sure that I fully understand your question, but if you're looking for performance, avoid regular expressions. Instead you can split the string on %. Then, take a look at the first and last matches:
// Anything before % should match at start of the string
targetString.indexOf(splits[0]) === 0;
// Anything after % should match at the end of the string
targetString.indexOf(splits[1]) + splits[1].length === targetString.length;
If you can use % multiple times within the string, then the first and last splits should follow the above rules. Anything else just needs to be in the string, and .indexOf is how you can check that.
I came to realize that this is impossible with a regular language, and therefore the only solution to this problem is to replace the wildcard symbol % with .* and then match two regular expressions with each other. This can however not be done by traditional regular expressions, look at this SO-question and it's answers for details.
Or perhaps you should edit the underlying Regular Expression engine for supporting wildcard based strings. Anyone being able to answer this question by extending the default implementation will be accepted as answer to this question ;-)
I'm curious why this doesn't work, and need to know why/how to work around it; I'm trying to detect whether some input is a question, I'm pretty sure string.match is what I need, but:
print(string.match("how much wood?", "(how|who|what|where|why|when).*\\?"))
returns nil. I'm pretty sure Lua's string.match uses regular expressions to find matches in a string, as I've used wildcards (.) before with success, but maybe I don't understand all the mechanics? Does Lua require special delimiters in its string functions? I've tested my regular expression here, so if Lua used regular regular expressions, it seems like the above code would return "how much wood?".
Can any of you tell me what I'm doing wrong, what I mean to do, or point me to a good reference where I can get comprehensive information about how Lua's string manipulation functions utilize regular expressions?
Lua doesn't use regex. Lua uses Patterns, which look similar but match different input.
.* will also consume the last ? of the input, so it fails on \\?. The question mark should be excluded. Special characters are escaped with %.
"how[^?]*%?"
As Omri Barel said, there's no alternation operator. You probably need to use multiple patterns, one for each alternative word at the beginning of the sentence. Or you could use a library that supports regex like expressions.
According to the manual, patterns don't support alternation.
So while "how.*" works, "(how|what).*" doesnt.
And kapep is right about the question mark being swallowed by the .*.
There's a related question: Lua pattern matching vs. regular expressions.
As they have already answered before, it is because the patterns in lua are different from the Regex in other languages, but if you have not yet managed to get a good pattern that does all the work, you can try this simple function:
local function capture_answer(text)
local text = text:lower()
local pattern = '([how]?[who]?[what]?[where]?[why]?[when]?[would]?.+%?)'
for capture in string.gmatch(text, pattern) do
return capture
end
end
print(capture_answer("how much wood?"))
Output: how much wood?
That function will also help you if you want to find a question in a larger text string
Ex.
print(capture_answer("Who is the best football player in the world?\nWho are your best friends?\nWho is that strange guy over there?\nWhy do we need a nanny?\nWhy are they always late?\nWhy does he complain all the time?\nHow do you cook lasagna?\nHow does he know the answer?\nHow can I learn English quickly?"))
Output:
who is the best football player in the world?
who are your best friends?
who is that strange guy over there?
why do we need a nanny?
why are they always late?
why does he complain all the time?
how do you cook lasagna?
how does he know the answer?
how can i learn english quickly?
Given a regular expression, how can I list all possible matches?
For example: AB[CD]1234, I want it to return a list like:
ABC1234
ABD1234
I searched the web, but couldn't find anything.
Exrex can do this:
$ python exrex.py 'AB[CD]1234'
ABC1234
ABD1234
The reason you haven't found anything is probably because this is a problem of serious complexity given the amount of combinations certain expressions would allow. Some regular expressions could even allow infite matches:
Consider following expressions:
AB[A-Z0-9]{1,10}1234
AB.*1234
I think your best bet would be to create an algorithm yourself based on a small subset of allowed patterns. In your specific case, I would suggest to use a more naive approach than a regular expression.
For some simple regular expressions like the one you provided (AB[CD]1234), there is a limited set of matches. But for other expressions (AB[CD]*1234) the number of possible matches are not limited.
One method for locating all the posibilities, is to detect where in the regular expression there are choices. For each possible choice generate a new regular expression based on the original regular expression and the current choice. This new regular expression is now a bit simpler than the original one.
For an expression like "A[BC][DE]F", the method will proceed as follows
getAllMatches("A[BC][DE]F")
= getAllMatches("AB[DE]F") + getAllMatches("AC[DE]F")
= getAllMatches("ABDF") + getAllMatches("ABEF")
+ getAllMatches("ACDF")+ getAllMatches("ACEF")
= "ABDF" + "ABEF" + "ACDF" + "ACEF"
It's possible to write an algorithm to do this but it will only work for regular expressions that have a finite set of possible matches. Your regexes would be limited to using:
Optional: ?
Characters: . \d \D
Sets: like [1a-c]
Negated sets: [^2-9d-z]
Alternations: |
Positive lookarounds
So your regexes could NOT use:
Repeaters: * +
Word patterns: \w \W
Negative lookarounds
Some zero-width assertions: ^ $
And there are some others (word boundaries, lazy & greedy quantifiers) I'm not sure about yet.
As for the algorithm itself, another user posted a link to this answer which describes how to create it.
Well you could convert the regular expression into an equivalent finite state machine (is relatively simple and can be done algorithmly) and then recursively folow every possible path through that fsm, outputting the followed paths through the machine. It's neither very hard nor computer intensive per output (you will normally get a HUGE amount of output however). You should however take care to disallow potentielly infinite passes (like .*). This can be done by having a maximum allowed path length, after which the tracing is aborted
A regular expression is intended to do nothing more than match to a pattern, that being said, the regular expression will never 'list' anything, only match. If you want to get a list of all matches I believe you will need to do it on your own.
Impossible.
Really.
Consider look ahead assertions. And what about .*, how will you generate all possible strings that match that regex?
It may be possible to find some code to list all possible matches for something as simple as you are doing. But most regular expressions you would not even want to attempt listing all possible matches.
For example AB.*1234 would be AB followed by absolutely anything and then 1234.
I'm not entirely sure this is even possible, but if it were, it would be so cpu/time intensive for many situations that it would not be useful.
For instance, try to make a list of all matches for A.*Z
There are sites that help with building a good regular expression though:
http://www.fileformat.info/tool/regex.htm
http://www.regular-expressions.info/javascriptexample.html
http://www.regextester.com/