As title, is there a way in PHP, with preg_match_all to catch all the repetitions of chars group?
For instante catch
hahahaha
jajajaj
hihihi
It's fine to catch repetition of any char, like abababab, acacacacac.
Also, is there a way to count the number of repetition?
The idea is to catch all this "forms" of smiling on social media.
I figured out that there are also other cases, such as misspelled instances like ahahhahaah (where you have two consecutive a or h). Any ideas?
How about this:
preg_match_all('/((?i)[a-z])((?i)[a-z])(\1\2)+/', $str, $m);
$matches = $m[0]; //$matches will contain an array of matches
A bit complicated, but it does work. To explain, the first subpattern (((?i)[a-z])) matches any character between a and z, no matter the case. The second subpattern (((?i)[a-z])) does the same thing. The third subpattern ((\1\2)+) matches one or more repetitions of the first two letters, in the same case as they were originally put. This regular expression also assumes that there's an even number of repetitions. If you don't want that, you can add \1? at the end, meaning that (as long as it contains one or more repetitions), it can end with the first character (for instance, hahah and ikikikik would both be valid, but not asa).
To retrieve the number of repetitions for a specific match, you can do:
$numb = strlen($matches[$index])/2 - 1; //-1 because the first two letters aren't repetitions
For the shortest repetition (e.g. ha gets repeated multiple times in hahahaha):
(.+?)\1+
See demo.
For the longest repetition (e.g. haha gets repeated in hahahaha):
(.+)\1+
Counting Repetitions
The non-regex solution is to compare the lengths of Group 1 (the repteated token) and the overall match.
With pure regex, in .NET, you could simply do (.+?)(\1)+ and look at the number of captures in the Group 1 CaptureCollection object.
In PHP, that's not possible, but there are some hacks. See, for instance, this question about matching a line number—it's the same technique. This is for "study purposes" only—you wouldn't want to use that in real life.
Related
I have a dictionary of 4 letter words. I want to write a regex to go through the dictionary and matches all words given a set of letters.
Suppose I pass in a,b,l,l. I want to find all words with exactly those letters.
I know I could do /[abl]{4}/ but that will also match words with 2 a's or 2 b's.
I feel like I need to do a negative look ahead. Something like:
[l|(ab)(?!\1)]{4}
The attempt here is that I want a word that starts with l or a or b and not followed by a or b.
First thing you need to anchor your pattern to describe where the string begins and ends:
for a whole string (^ start of the string, $ end of the string):
^[abl]{4}$
or to find words in a larger text, use word-boundaries (limit between a character from [A-Za-z0-9_] and something else):
\b[abl]{4}\b
Then you need to say that l must occur two times (or that a and b must occurs only one time, but it's more complicated):
for a whole string:
^(?=.*l.*l)[abl]{4}$
in a larger text:
\b(?=\w*l\w*l)[abl]{4}\b
To avoid two a or b, you can use an other lookahead:
for a whole string:
^(?=.*l.*l)(?=l*al*b|l*bl*a)[abl]{4}$
in a larger text:
\b(?=\w*l\w*l)(?=l*al*b|l*bl*a)[abl]{4}\b
About [l|(ab)(?!\1)]: in a character class, special regex characters or sequence of characters loose their special meaning and all characters are seen as literals. So [l|(ab)(?!\1)] is the same than [)(!|?1abl] for example. (Since \1 is an unknown escape sequence in a character class, the backslash is ignored.)
Note that with several constraints the pattern becomes quickly ugly. You should consider an other approach that consists to catch all words with \b[abl]{4}\b and to filter them in a second time (using count_chars for example).
$str ='abll labl ball aabl lblabla 1234';
$dict = 'abll';
$count = count_chars($dict);
$result = [];
if (preg_match_all('~\b[abl]{4}\b~', $str, $matches)) {
$result = array_filter($matches[0], function ($i) use ($count) {
return $count == count_chars($i);
});
}
print_r($result);
If you want specify letters dynamically and then generate regexp that will do all work - this will be a very expensive work.
Simple approach: you can generate simple regexp like /^[abl]{4}$/, get all words from dictionary that match him and then validate each word separately - check letters quantity.
More efficient approach: you can index your words in dictionary with sorted list of letters like this:
word: apple | index: aelpp
word: pale | index: aelp
And so on. To get all words from list of letters you simply should sort this letters and find exact match with "index" value.
Edit: So for 47 letters it would be
\b(?:((?(1)(?!))l1)|((?(2)(?!))l2)|...|((?(47)(?!))l47)){47}\b
Letters can be duplicates, say 4 a's and 15 r's (but no more), etc ...
( immune to permutations )
To match out of order items only once,
use a conditional to allow each item to match once,
but no more.
It's not complicated, and is immune to permutations.
Works every time !
\b(?:((?(1)(?!))a)|((?(2)(?!))b)|((?(3)(?!))l)|((?(4)(?!))l)){4}\b
Expanded
\b
(?:
( # (1)
(?(1)(?!))
a
)
|
( # (2)
(?(2)(?!))
b
)
|
( # (3)
(?(3)(?!))
l
)
|
( # (4)
(?(4)(?!))
l
)
){4}
\b
I am trying to retrieve matches from a comma separated list that is located inside parenthesis using regular expression. (I also retrieve the version number in the first capture group, though that's not important to this question)
What's worth noting is that the expression should ideally handle all possible cases, where the list could be empty or could have more than 3 entries = 0 or more matches in the second capture group.
The expression I have right now looks like this:
SomeText\/(.*)\s\(((,\s)?([\w\s\.]+))*\)
The string I am testing this on looks like this:
SomeText/1.0.4 (debug, OS X 10.11.2, Macbook Pro Retina)
Result of this is:
1. [6-11] `1.0.4`
2. [32-52] `, Macbook Pro Retina`
3. [32-34] `, `
4. [34-52] `Macbook Pro Retina`
The desired result would look like this:
1. [6-11] `1.0.4`
2. [32-52] `debug`
3. [32-34] `OS X 10.11.2`
4. [34-52] `Macbook Pro Retina`
According to the image above (as far as I can see), the expression should work on the test string. What is the cause of the weird results and how could I improve the expression?
I know there are other ways of solving this problem, but I would like to use a single regular expression if possible. Please don't suggest other options.
When dealing with a varying number of groups, regex ain't the best. Solve it in two steps.
First, break down the statement using a simple regex:
SomeText\/([\d.]*) \(([^)]*)\)
1. [9-14] `1.0.4`
2. [16-55] `debug, OS X 10.11.2, Macbook Pro Retina`
Then just explode the second result by ',' to get your groups.
Probably the \G anchor works best here for binding the match to an entry point. This regex is designed for input that is always similar to the sample that is provided in your question.
(?<=SomeText\/|\G(?!^))[(,]? *\K[^,)(]+
(?<=SomeText\/|\G) the lookbehind is the part where matches should be glued to
\G matches where the previous match ended (?!^) but don't match start
[(,]? *\ matches optional opening parenthesis or comma followed by any amount of space
\K resets beginning of the reported match
[^,)(]+ matches the wanted characters, that are none of ( ) ,
Demo at regex101 (grab matches of $0)
Another idea with use of capture groups.
SomeText\/([^(]*)\(|\G(?!^),? *([^,)]+)
This one without lookbehind is a bit more accurate (it also requires the opening parenthesis), of better performance (needs fewer steps) and probably easier to understand and maintain.
SomeText\/([^(]*)\( the entry anchor and version is captured here to $1
|\G(?!^),? *([^,)]+) or glued to previous match: capture to $2 one or more characters, that are not , ) preceded by optional space or comma.
Another demo at regex101
Actually, stribizhev was close:
(?:SomeText\/([^() ]*)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\))
Just had to make that one class expect at least one match
(?:SomeText\/([0-9.]+)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\)) is a little more clear as long as the version number is always numbers and periods.
I wanted to come up with something more elegant than this (though this does actually work):
SomeText\/(.*)\s\(([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?\)
Obviously, the
([^\,]+)?\,?\s?
is repeated 6 times.
(It can be repeated any number of times and it will work for any number of comma-separated items equal to or below that number of times).
I tried to shorten the long, repetitive list of ([^\,]+)?\,?\s? above to
(?:([^\,]+)\,?\s?)*
but it doesn't work and my knowledge of regex is not currently good enough to say why not.
This should solve your problem. Use the code you already have and add something like this. It will determine where commas are in your string and delete them.
Use trim() to delete white spaces at the start or the end.
$a = strpos($line, ",");
$line = trim(substr($line, 55-$a));
I hope, this helps you!
I found some partial help but cannot seem to fully accomplish what I need. I need to be able to do the following:
I need an regular expression to replace any 1 to 3 character words between two words that are longer than 3 characters with a match any expression:
For example:
walk to the beach ==> walk(.*)beach
If the 1 to 3 character word is not preceded by a word that's longer than 3 characters then I want to translate that 1 to 3 letter word to '<word> ?'
For example:
on the beach ==> on ?the ?beach
The simpler the rule the better (of course, if there's an alternative more complicated version that's more performant then I'll take that as well as I eventually anticipate heavy usage eventually).
This will be used in a PHP context most likely with preg_replace. Thus, if you can put it in that context then even better!
By the way so far I have got the following:
$string = preg_replace('/\s+/', '(.*)', $string);
$string = preg_replace('/\b(\w{1,3})(\.*)\b/', '${1} ?', $string);
but that results in:
walk to the beach ==> 'walk(.*)to ?beach'
which is not what I want. 'on the beach' seems to translate correctly.
I think you will need two replacements for that. Let's start with the first requirement:
$str = preg_replace('/(\w{4,})(?: \w{1,3})* (?=\w{4,})/', '$1(.*)', $str);
Of course, you need to replace those \w (which match letters, digits and underscores) with a character class of what you actually want to treat as a word character.
The second one is a bit tougher, because matches cannot overlap and lookbehinds cannot be of variable length. So we have to run this multiple times in a loop:
do
{
$str = preg_replace('/^\w{0,3}(?: \w{0,3})* (?!\?)/', '$0?', $str, -1, $count);
} while($count);
Here we match everything from the beginning of the string, as long as it's only up-to-3-letter words separated by spaces, plus one trailing space (only if it is not already followed by a ?). Then we put all of that back in place, and append a ?.
Update:
After all the talk in the comments, here is an updated solution.
After running the first line, we can assume that the only less-than-3-letter words left will be at the beginning or at the end of the string. All others will have been collapsed to (.*). Since you want to append all spaces between those with ?, you do not even need a loop (in fact these are the only spaces left):
$str = preg_replace('/ /', ' ?', $str);
(Do this right after my first line of code.)
This would give the following two results (in combination with the first line):
let us walk on the beach now go => let ?us ?walk(.*)beach ?now ?go
let us walk on the beach there now go => let ?us ?walk(.*)beach(.*)there ?now ?go
I have a string that contains 5 words. In the string one of the words is a Ham Radio Call Sign and can be anyone of the thousands of call signs in the US. In order to extract the Call Sign from the string I need to utilize the below pattern. The Call Sign I need to extract can be in any of the 5 positions in the string. The number is never the first character and the number is never the last character. The string is actually put together from an Array since it is originally read from a text file.
$string = $word[1] $word[2] $word[3] etc....
So the search can be either done on the whole string or each piece of the array.
Patterns:
1 Number and 3 Letters Example: AB4C A4BC
1 Number and 4 Letters Example: A4BCD
1 Number and 5 Letters Example: AB4CDE
I have tried everything I can think of and search till I cant search no more. I am sure I am over thinking this.
A two-step regular expression like this would do it:
$str = "hello A4AB there BC5AD";
$signs = array();
preg_match_all('/[A-Z][A-Z\d]{1,3}[A-Z]/', $str, $possible_signs);
foreach($possible_signs[0] as $possible_sign)
if (preg_match('/^\D+\d\D+$/', $possible_sign))
array_push($signs, $possible_sign);
print_r($signs); //Array ([0] => A4AB [1] => BC5AD)
Explanation
This is a regular expression approach, using two patterns. I don't think it could be done with one and still satisfy the exact requirements of the matching rules.
The first pattern enforces the following requirements:
substring starts and ends with a capital letter
substring contains only other capital letters or numbers between the first and last letter
substring is, overall, not more than 6 characters long
What I can't do in that same pattern, for complex REGEX reasons I won't go into (unless someone knows a way and can correct me), is enforce that only one number is contained.
#jeroen's answer does enforce this in a single pattern, but in turn does not enforce the correct length of the substring. Either way, we need a second pattern.
So after grabbing the initial matches, we loop over the results. We then apply each to a second pattern that enforces simply that there is only one number in the substring.
If so, we green-light the substring and it's added to the $signs array.
Hope this helps.
It depends on what the other words can contain, but you could use a regular expression like:
#\b[a-z]+\d[a-z]+\b#i
^ case insensitive
^^ a word boundary
^^^^^^ One or more letters
^^ One number
You can make it more restrictive by using {1,3} instead of + for the letters so that you have a sequence of 1 to 3 letters.
The complete expression would be something like:
$success = preg_match('#\b[a-z]+\d[a-z]+\b#i', $input_string, $matches);
where $matches[0] will contain the matched value, see the manual.
How can you match the following words by PHP, either by regex/globbing/...?
Examples
INNO, heppeh, isi, pekkep, dadad, mum
My attempt would be to make a regex which has 3 parts:
1st match match [a-zA-Z]*
[a-zA-Z]?
rotation of the 1st match // Problem here!
The part 3 is the problem, since I do not know how to rotate the match.
This suggests me that regex is not the best solution here, since it is too very inefficient for long words.
I think regex are a bad solution. I'd do something with the condition like: ($word == strrev($word)).
Regexs are not suitable for finding palindromes of an arbitrary length.
However, if you are trying to find all of the palindromes in a large set of text, you could use regex to find a list of things that might be palindromes, and then filter that list to find the words that actually are palindromes.
For example, you can use a regex to find all words such that the first X characters are the reverse of the last X characters (from some small fixed value of X, like 2 or 3), and then run a secondary filter against all the matches to see if the whole word is in fact a palindrome.
In PHP once you get the string you want to check (by regex or split or whatever) you can just:
if ($string == strrev($string)) // it's a palindrome!
i think this regexp can work
$re = '~([a-z])(.?|(?R))\1~';