I'd like to find the longest repeating string within a string, implemented in JavaScript and using a regular-expression based approach.
I have an PHP implementation that, when directly ported to JavaScript, doesn't work.
The PHP implementation is taken from an answer to the question "Find longest repeating strings?":
preg_match_all('/(?=((.+)(?:.*?\2)+))/s', $input, $matches, PREG_SET_ORDER);
This will populate $matches[0][X] (where X is the length of $matches[0]) with the longest repeating substring to be found in $input. I have tested this with many input strings and found am confident the output is correct.
The closest direct port in JavaScript is:
var matches = /(?=((.+)(?:.*?\2)+))/.exec(input);
This doesn't give correct results
input Excepted result matches[0][X]
======================================================
inputinput input input
7inputinput input input
inputinput7 input input
7inputinput7 input 7
XXinputinputYY input XX
I'm not familiar enough with regular expressions to understand what the regular expression used here is doing.
There are certainly algorithms I could implement to find the longest repeating substring. Before I attempt to do that, I'm hoping a different regular expression will produce the correct results in JavaScript.
Can the above regular expression be modified such that the expected output is returned in JavaScript? I accept that this may not be possible in a one-liner.
Javascript matches only return the first match -- you have to loop in order to find multiple results. A little testing shows this gets the expected results:
function maxRepeat(input) {
var reg = /(?=((.+)(?:.*?\2)+))/g;
var sub = ""; //somewhere to stick temp results
var maxstr = ""; // our maximum length repeated string
reg.lastIndex = 0; // because reg previously existed, we may need to reset this
sub = reg.exec(input); // find the first repeated string
while (!(sub == null)){
if ((!(sub == null)) && (sub[2].length > maxstr.length)){
maxstr = sub[2];
}
sub = reg.exec(input);
reg.lastIndex++; // start searching from the next position
}
return maxstr;
}
// I'm logging to console for convenience
console.log(maxRepeat("aabcd")); //aa
console.log(maxRepeat("inputinput")); //input
console.log(maxRepeat("7inputinput")); //input
console.log(maxRepeat("inputinput7")); //input
console.log(maxRepeat("7inputinput7")); //input
console.log(maxRepeat("xxabcdyy")); //x
console.log(maxRepeat("XXinputinputYY")); //input
Note that for "xxabcdyy" you only get "x" back, as it returns the first string of maximum length.
It seems JS regexes are a bit weird. I don't have a complete answer, but here's what I found.
Although I thought they did the same thing re.exec() and "string".match(re) behave differently. Exec seems to only return the first match it finds, whereas match seems to return all of them (using /g in both cases).
On the other hand, exec seems to work correctly with ?= in the regex whereas match returns all empty strings. Removing the ?= leaves us with
re = /((.+)(?:.*?\2)+)/g
Using that
"XXinputinputYY".match(re);
returns
["XX", "inputinput", "YY"]
whereas
re.exec("XXinputinputYY");
returns
["XX", "XX", "X"]
So at least with match you get inputinput as one of your values. Obviously, this neither pulls out the longest, nor removes the redundancy, but maybe it helps nonetheless.
One other thing, I tested in firebug's console which threw an error about not supporting $1, so maybe there's something in the $ vars worth looking at.
Related
now I have two strings,
var str1 = "A10B1C101D11";
var str2 = "A1B22C101D110E1";
What I intend to do is to tell the difference between them, the result will look like
A10B1C101D11
A10 B22 C101 D110E1
It follows the same pattern, one character and a number. And if the character doesn't exist or the number is different between them, I will say they are different, and highlight the different part. Can regular expression do it or any other good solution? thanks in advance!
Let me start by stating that regexp might not be the best tool for this. As the strings have a simple format that you are aware of it will be faster and safer to parse the strings into tokens and then compare the tokens.
However you can do this with Regexp, although in javascript you are hampered by the lack of lookbehind.
The way to do this is to use negative lookahead to prevent matches that are included in the other string. However since javascript does not support lookbehind you might need to go search from both directions.
We do this by concatenating the strings, with a delimiter that we can test for.
If using '|' as a delimiter the regexp becomes;
/(\D\d*)(?=(?:\||\D.*\|))(?!.*\|(.*\d)?\1(\D|$))/g
To find the tokens in the second string that are not present in the first you do;
var bothstring=str2.concat("|",str1);
var re=/(\D\d*)(?=(?:\||\D.*\|))(?!.*\|(.*\d)?\1(\D|$))/g;
var match=re.exec(bothstring);
Subsequent calls to re.exec will return later matches. So you can iterate over them as in the following example;
while (match!=null){
alert("\""+match+"\" At position "+match.index);
match=re.exec(t);
}
As stated this gives tokens in str2 that are different in str1. To get the tokens in str1 that are different use the same code but change the order of str1 and str2 when you concatenate the strings.
The above code might not be safe if dealing with potentially dirty input. In particular it might misbehave if feed a string like "A100|A100", the first A100 will not be considered as having a missing object because the regexp is not aware that the source is supposed to be two different strings. If this is a potential issue then search for occurences of the delimiting character.
You call break the string into an array
var aStr1 = str1.split('');
var aStr2 = str2.split('');
Then check which one has more characters, and save the smaller number
var totalCharacters;
if(aStr1.length > aStr2.length) {
totalCharacters = aStr2.length
} else {
totalCharacters = aStr1.length
}
And loop comparing both
var diff = [];
for(var i = 0; i<totalCharacters; i++) {
if(aStr1[i] != aStr2[i]) {
diff.push(aStr1[i]); // or something else
}
}
At the very end you can concat those last characters from the bigger String (since they obviously are different from the other one).
Does it helps you?
I am trying to extract the first character after the last underscore in a string with an unknown number of '_' in the string but in my case there will always be one, because I added it in another step of the process.
What I tried is this. I also tried the regex by itself to extract from the name, but my result was empty.
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var string = match(/[^_]*$/)[1]
string.charAt(0)
So the final desired result is 'D'. If the RegEx can only get me what is behind the last '_' that is fine because I know I can use the charAt like currently shown. However, if the regex can do the whole thing, even better.
If you know there will always be at least one underscore you can do this:
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var firstCharAfterUnderscore = s.charAt(s.lastIndexOf("_") + 1);
// OR, with regex
var firstCharAfterUnderscore = s.match(/_([^_])[^_]*$/)[1]
With the regex, you can extract just the one letter by using parentheses to capture that part of the match. But I think the .lastIndexOf() version is easier to read.
Either way if there's a possibility of no underscores in the input you'd need to add some additional logic.
I'm fairly sure after spending the night trying to find an answer that this isn't possible, and I've developed a work around - but, if someone knows of a better method, I would love to hear it...
I've gone through a lot of iterations on the code, and the following is just a line of thought really. At some point I was using the global flag, I believe, in order for match() to work, and I can't remember if it was necessary now or not.
var str = "#abc#def#ghi&jkl";
var regex = /^(?:#([a-z]+))?(?:&([a-z]+))?$/;
The idea here, in this simplified code, is the optional group 1, of which there is an unspecified amount, will match #abc, #def and #ghi. It will only capture the alpha characters of which there will be one or more. Group 2 is the same, except matches on & symbol. It should also be anchored to the start and end of the string.
I want to be able to back reference all matches of both groups, ie:
result = str.match(regex);
alert(result[1]); //abc,def,ghi
alert(result[1][0]); //abc
alert(result[1][1]); //def
alert(result[1][2]); //ghi
alert(result[2]); //jkl
My mate says this works fine for him in .net, unfortunately I simply can't get it to work - only the last matched of any group is returned in the back reference, as can be seen in the following:
(additionally, making either group optional makes a mess, as does setting global flag)
var str = "#abc#def#ghi&jkl";
var regex = /(?:#([a-z]+))(?:&([a-z]+))/;
var result = str.match(regex);
alert(result[1]); //ghi
alert(result[1][0]); //g
alert(result[2]); //jkl
The following is the solution I arrived at, capturing the whole portion in question, and creating the array myself:
var str = "#abc#def#ghi&jkl";
var regex = /^([#a-z]+)?(?:&([a-z]+))?$/;
var result = regex.exec(str);
alert(result[1]); //#abc#def#ghi
alert(result[2]); //jkl
var result1 = result[1].toString();
result[1] = result1.split('#')
alert(result[1][1]); //abc
alert(result[1][2]); //def
alert(result[1][3]); //ghi
alert(result[2]); //jkl
That's simply not how .match() works in JavaScript. The returned array is an array of simple strings. There's no "nesting" of capture groups; you just count the ( symbols from left to right.
The first string (at index [0]) is always the overall matched string. Then come the capture groups, one string (or null) per array element.
You can, as you've done, rearrange the result array to your heart's content. It's just an array.
edit — oh, and the reason your result[1][0] was "g" is that array indexing notation applied to a string gets you the individual characters of the string.
I need to do a lot of regex things in javascript but am having some issues with the syntax and I can't seem to find a definitive resource on this.. for some reason when I do:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test)
it shows
"afskfsd33j, fskfsd33"
I'm not sure why its giving this output of original and the matched string, I am wondering how I can get it to just give the match (essentially extracting the part I want from the original string)
Thanks for any advice
match returns an array.
The default string representation of an array in JavaScript is the elements of the array separated by commas. In this case the desired result is in the second element of the array:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test[1]);
Each group defined by parenthesis () is captured during processing and each captured group content is pushed into result array in same order as groups within pattern starts. See more on http://www.regular-expressions.info/brackets.html and http://www.regular-expressions.info/refcapture.html (choose right language to see supported features)
var source = "afskfsd33j"
var result = source.match(/a(.*)j/);
result: ["afskfsd33j", "fskfsd33"]
The reason why you received this exact result is following:
First value in array is the first found string which confirms the entire pattern. So it should definitely start with "a" followed by any number of any characters and ends with first "j" char after starting "a".
Second value in array is captured group defined by parenthesis. In your case group contain entire pattern match without content defined outside parenthesis, so exactly "fskfsd33".
If you want to get rid of second value in array you may define pattern like this:
/a(?:.*)j/
where "?:" means that group of chars which match the content in parenthesis will not be part of resulting array.
Other options might be in this simple case to write pattern without any group because it is not necessary to use group at all:
/a.*j/
If you want to just check whether source text matches the pattern and does not care about which text it found than you may try:
var result = /a.*j/.test(source);
The result should return then only true|false values. For more info see http://www.javascriptkit.com/javatutors/re3.shtml
I think your problem is that the match method is returning an array. The 0th item in the array is the original string, the 1st thru nth items correspond to the 1st through nth matched parenthesised items. Your "alert()" call is showing the entire array.
Just get rid of the parenthesis and that will give you an array with one element and:
Change this line
var test = tesst.match(/a(.*)j/);
To this
var test = tesst.match(/a.*j/);
If you add parenthesis the match() function will find two match for you one for whole expression and one for the expression inside the parenthesis
Also according to developer.mozilla.org docs :
If you only want the first match found, you might want to use
RegExp.exec() instead.
You can use the below code:
RegExp(/a.*j/).exec("afskfsd33j")
I've just had the same problem.
You only get the text twice in your result if you include a match group (in brackets) and the 'g' (global) modifier.
The first item always is the first result, normally OK when using match(reg) on a short string, however when using a construct like:
while ((result = reg.exec(string)) !== null){
console.log(result);
}
the results are a little different.
Try the following code:
var regEx = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
var result = sample_string.match(regEx);
console.log(JSON.stringify(result));
// ["1 cat","2 fish"]
var reg = new RegExp('[0-9]+ (cat|fish)','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null) {
console.dir(JSON.stringify(result))
};
// '["1 cat","cat"]'
// '["2 fish","fish"]'
var reg = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null){
console.dir(JSON.stringify(result))
};
// '["1 cat","1 cat","cat"]'
// '["2 fish","2 fish","fish"]'
(tested on recent V8 - Chrome, Node.js)
The best answer is currently a comment which I can't upvote, so credit to #Mic.
I have a RegExp like the following simplified example:
var exp = /he|hell/;
When I run it on a string it will give me the first match, fx:
var str = "hello world";
var match = exp.exec(str);
// match contains ["he"];
I want the first and longest possible match,
and by that i mean sorted by index, then length.
Since the expression is combined from an array of RegExp's, I am looking for a way to find the longest match without having to rewrite the regular expression.
Is that even possible?
If it isn't, I am looking for a way to easily analyze the expression, and arrange it in the proper order. But I can't figure out how since the expressions could be a lot more complex, fx:
var exp = /h..|hel*/
How about /hell|he/ ?
All regex implementations I know of will (try to) match characters/patterns from left to right and terminate whenever they find an over-all match.
In other words: if you want to make sure you get the longest possible match, you'll need to try all your patterns (separately), store all matches and then get the longest match from all possible matches.
You can do it. It's explained here:
http://www.regular-expressions.info/alternation.html
(In summary, change the operand order or group with question mark the second part of the search.)
You cannot do "longest match" (or anything involving counting, minus look-aheads) with regular expressions.
Your best bet is to find all matches, and simply compare the lengths in the program.
I don't know if this is what you're looking for (Considering this question is almost 8 years old...), but here's my grain of salt:
(Switching the he for hell will perform the search based on the biggest first)
var exp = /hell|he/;
var str = "hello world";
var match = exp.exec(str);
if(match)
{
match.sort(function(a, b){return b.length - a.length;});
console.log(match[0]);
}
Where match[0] is going to be the longest of all the strings matched.