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.
Related
The use case is I want to compare a query string of characters to an array of words, and return the matches. A match is when a word contains all the characters in the query string, order doesn't matter, repeated characters are okay. Regex seems like it may be too powerful (a sledgehammer where only a hammer is needed). I've written a solution that compares the characters by looping through them and using indexOf, but it seems consistently slower. (http://jsperf.com/indexof-vs-regex-inside-a-loop/10) Is Regex the fastest option for this type of operation? Are there ways to make my alternate solution faster?
var query = "word",
words = ['word', 'wwoorrddss', 'words', 'argument', 'sass', 'sword', 'carp', 'drowns'],
reStoredMatches = [],
indexOfMatches = [];
function match(word, query) {
var len = word.length,
charMatches = [],
charMatch,
char;
while (len--) {
char = word[len];
charMatch = query.indexOf(char);
if (charMatch !== -1) {
charMatches.push(char);
}
}
return charMatches.length === query.length;
}
function linearIndexOf(words, query) {
var wordsLen = words.length,
wordMatch,
word;
while (wordsLen--) {
word = words[wordsLen];
wordMatch = match(word, query);
if (wordMatch) {
indexOfMatches.push(word);
}
}
}
function linearRegexStored(words, query) {
var wordsLen = words.length,
re = new RegExp('[' + query + ']', 'g'),
match,
word;
while (wordsLen--) {
word = words[wordsLen];
match = word.match(re);
if (match !== null) {
if (match.length >= query.length) {
reStoredMatches.push(word);
}
}
}
}
Note that your regex is wrong, that's most certainly why it goes so fast.
Right now, if your query is "word" (as in your example), the regex is going to be:
/[word]/g
This means look for one of the characters: 'w', 'o', 'r', or 'd'. If one matches, then match() returns true. Done. Definitively a lot faster than the most certainly more correct indexOf(). (i.e. in case of a simple match() call the 'g' flag is ignored since if any one thing matches, the function returns true.)
Also, you mention the idea/concept of any number of characters, I suppose as shown here:
'word', 'wwoorrddss'
The indexOf() will definitively not catch that properly if you really mean "any number" for each and every character. Because you should match an infinite number of cases. Something like this as a regex:
/w+o+r+d+s+/g
That you will certainly have a hard time to write the right code in plain JavaScript rather than use a regex. However, either way, that's going to be somewhat slow.
From the comment below, all the letters of the word are required, in order to do that, you have to have 3! tests (3 factorial) for a 3 letter word:
/(a.*b.*c)|(a.*c.*b)|(b.*a.*c)|(b.*c.*a)|(c.*a.*b)|(c.*b.*a)/
Obviously, a factorial is going to very quickly grow your number of possibilities and blow away your memory in a super long regex (although you can simplify if a word has the same letter multiple times, you do not have to test that letter more than once).
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
...
That's probably why your properly written test in plain JavaScript is much slower.
Also, in your case you should write the words nearly as done in Scrabble dictionaries: all letters once in alphabetical order (Scrabble keeps duplicates). So the word "word" would be "dorw". And as you shown in your example, the word "wwoorrddss" would be "dorsw". You can have some backend tool to generate your table of words (so you still write them as "word" and "words", and your tool massage those and convert them to "dorw" and "dorsw".) Then you can sort the letters of the words you are testing in alphabetical order and the result is that you do not need a silly factorial for the regex, you can simply do this:
/d.*o.*r.*w/
And that will match any word that includes the word "word" such as "password".
One easy way to sort the letters will be to split your word in an array of letters, and then sort the array. You may still get duplicates, it will depend on the sort capabilities. (I don't think that the default JavaScript sort will remove duplicates automatically.)
One more detail, if you test is supposed to be case insensitive, then you want to transform your strings to lowercase before running the test. So something like:
query = query.toLowerCase();
early on in your top function.
You are trying to speed up the algorithm "chars in word are a subset of the chars of query." You can short circuit this check and avoid some assignments (that are more readable but not strictly needed). Try the following version of match
function match(word, query) {
var len = word.length;
while (len--) {
if (query.indexOf(word[len]) === -1) { // found a missing char
return false;
}
}
return true; // couldn't find any missing chars
}
This gives a 4-5X improvement
Depending on the application you could try presorting words and presorting each word in words as another optimization.
The regexp match algorithm constructs a finite state automaton and makes its decisions on the current state and character read from left to right. This involves reading each character once and make a decision.
For static strings (to look a fixed string on a couple of text) you have better algorithms, like Knuth-Morris that allow you to go faster than one character at a time, but you must understand that this algorithm is not for matching regular expressions, just plain strings.
if you are interested in Knuth-Morris (there are several other algorithms) just have a round in wikipedia. http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm
A good thing you can do is to investigate if you regexp match routines do it with an DFA or a NDFA, as NDFAs occupy less memory and are easier to compute, but DFAs do it faster, but with some compilation penalties and more memory occupied.
Knuth-Morris algorithm also needs to compile the string into an automaton before working, so perhaps it doesn't apply to your problem if you are using it just to find one word in some string.
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'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.
have a regex problem :(
what i would like to do is to find out the contents between two or more numbers.
var string = "90+*-+80-+/*70"
im trying to edit the symbols in between so it only shows up the last symbol and not the ones before it. so trying to get the above variable to be turned into 90+80*70. although this is just an example i have no idea how to do this. the length of the numbers, how many "sets" of numbers and the length of the symbols in between could be anything.
many thanks,
Steve,
The trick is in matching '90+-+' and '80-+/' seperately, and selecting only the number and the last constant.
The expression for finding the a number followed by 1 or more non-numbers would be
\d+[^\d]+
To select the number and the last non-number, add parens:
(\d+)[^\d]*([^\d])
Finally add a /g to repeat the procedure for each match, and replace it with the 2 matched groups for each match:
js> '90+*-+80-+/*70'.replace(/(\d+)[^\d]*([^\d])/g, '$1$2');
90+80*70
js>
Or you can use lookahead assertion and simply remove all non-numerical characters which are not last: "90+*-+80-+/*70".replace(/[^0-9]+(?=[^0-9])/g,'');
You can use a regular expression to match the non-digits and a callback function to process the match and decide what to replace:
var test = "90+*-+80-+/*70";
var out = test.replace(/[^\d]+/g, function(str) {
return(str.substr(-1));
})
alert(out);
See it work here: http://jsfiddle.net/jfriend00/Tncya/
This works by using a regular expression to match sequences of non-digits and then replacing that sequence of non-digits with the last character in the matched sequence.
i would use this tutorial, first, then review this for javascript-specific regex questions.
This should do it -
var string = "90+*-+80-+/*70"
var result = '';
var arr = string.split(/(\d+)/)
for (i = 0; i < arr.length; i++) {
if (!isNaN(arr[i])) result = result + arr[i];
else result = result + arr[i].slice(arr[i].length - 1, arr[i].length);
}
alert(result);
Working demo - http://jsfiddle.net/ipr101/SA2pR/
Similar to #Arnout Engelen
var string = "90+*-+80-+/*70";
string = string.replace(/(\d+)[^\d]*([^\d])(?=\d+)/g, '$1$2');
This was my first thinking of how the RegEx should perform, it also looks ahead to make sure the non-digit pattern is followed by another digit, which is what the question asked for (between two numbers)
Similar to #jfriend00
var string = "90+*-+80-+/*70";
string = string.replace( /(\d+?)([^\d]+?)(?=\d+)/g
, function(){
return arguments[1] + arguments[2].substr(-1);
});
Instead of only matching on non-digits, it matches on non-digits between two numbers, which is what the question asked
Why would this be any better?
If your equation was embedded in a paragraph or string of text. Like:
This is a test where I want to clean up something like 90+*-+80-+/*70 and don't want to scrap the whole paragraph.
Result (Expected) :
This is a test where I want to clean up something like 90+80*70 and don't want to scrap the whole paragraph.
Why would this not be any better?
There is more pattern matching, which makes it theoretically slower (negligible)
It would fail if your paragraph had embedded numbers. Like:
This is a paragraph where Sally bought 4 eggs from the supermarket, but only 3 of them made it back in one piece.
Result (Unexpected):
This is a paragraph where Sally bought 4 3 of them made it back in one piece.
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.