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How can I split a string but keep the delimiters in javascript?

I have searched online but it really doesnt make much sense.
I am trying to split a string and keep the delimiters too.
For example, I have this string:
var str = '45612+54721/121*124.2';
And, I would like to split it based on the operations & keep the operations.
So the output must look like this
[45612], [+], [54721], [/], [121], [*], [124.2];

If you use strings’ split with a regex, any captured groups become part of the resulting array.
var str = '45612+54721/121*124.2';
var tokens = str.split(/([+/*])/);
console.log(tokens);

How to split a string by a character not directly preceded by a character of the same type?

Let's say I have a string: "We.need..to...split.asap". What I would like to do is to split the string by the delimiter ., but I only wish to split by the first . and include any recurring .s in the succeeding token.
Expected output:
["We", "need", ".to", "..split", "asap"]
In other languages, I know that this is possible with a look-behind /(?<!\.)\./ but Javascript unfortunately does not support such a feature.
I am curious to see your answers to this question. Perhaps there is a clever use of look-aheads that presently evades me?
I was considering reversing the string, then re-reversing the tokens, but that seems like too much work for what I am after... plus controversy: How do you reverse a string in place in JavaScript?
Thanks for the help!

Here's a variation of the answer by guest271314 that handles more than two consecutive delimiters:
var text = "We.need.to...split.asap";
var re = /(\.*[^.]+)\./;
var items = text.split(re).filter(function(val) { return val.length > 0; });
It uses the detail that if the split expression includes a capture group, the captured items are included in the returned array. These capture groups are actually the only thing we are interested in; the tokens are all empty strings, which we filter out.
EDIT: Unfortunately there's perhaps one slight bug with this. If the text to be split starts with a delimiter, that will be included in the first token. If that's an issue, it can be remedied with:
var re = /(?:^|(\.*[^.]+))\./;
var items = text.split(re).filter(function(val) { return !!val; });
(I think this regex is ugly and would welcome an improvement.)

You can do this without any lookaheads:
var subject = "We.need.to....split.asap";
var regex = /\.?(\.*[^.]+)/g;
var matches, output = [];
while(matches = regex.exec(subject)) {
output.push(matches[1]);
}
document.write(JSON.stringify(output));
It seemed like it'd work in one line, as it did on https://regex101.com/r/cO1dP3/1, but had to be expanded in the code above because the /g option by default prevents capturing groups from returning with .match (i.e. the correct data was in the capturing groups, but we couldn't immediately access them without doing the above).
See: JavaScript Regex Global Match Groups
An alternative solution with the original one liner (plus one line) is:
document.write(JSON.stringify(
"We.need.to....split.asap".match(/\.?(\.*[^.]+)/g)
.map(function(s) { return s.replace(/^\./, ''); })
));
Take your pick!

Note: This answer can't handle more than 2 consecutive delimiters, since it was written according to the example in the revision 1 of the question, which was not very clear about such cases.
var text = "We.need.to..split.asap";
// split "." if followed by "."
var res = text.split(/\.(?=\.)/).map(function(val, key) {
// if `val[0]` does not begin with "." split "."
// else split "." if not followed by "."
return val[0] !== "." ? val.split(/\./) : val.split(/\.(?!.*\.)/)
});
// concat arrays `res[0]` , `res[1]`
res = res[0].concat(res[1]);
document.write(JSON.stringify(res));

Faster way match characters between strings than Regex?

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.

Regex one-liner for splitting string at nth character where n is a variable length

I've found a few similar questions, but none of them are clean one-liners, which I feel should be possible. I want to split a string at the last instance of specific character (in my case .).
var img = $('body').attr('data-bg-img-url'); // the string http://sub.foo.com/img/my-img.jpg
var finalChar = img.split( img.split(/[.]+/).length-1 ); // returns int 3 in above string example
var dynamicRegex = '/[.$`finalChar`]/';
I know I'm breaking some rules here, wondering if someone smarter than me knows the correct way to put that together and compress it?
EDIT - The end goal here is to split and store http://sub.foo.com/img/my-img and .jpg as separate strings.

In regex, .* is greedy, meaning it will match as much as possible. Therefore, if you want to match up to the last ., you could do:
/^.*\./
And from the looks, you are trying to get the file extension, so you would want to add capture:
var result = /^.*\.(.*)$/.exec( str );
var extension = result[1];
And for both parts:
var result = /^(.*)\.(.*)$/.exec( str );
var path = result[1];
var extension = result[2];

You can use the lastIndexOf() method on the period and then use the substring method to obtain the first and second string. The split() method is better used in a foreach scenario where you want to split at all instances. Substring is preferable for these types of cases where you are breaking at a single instance of the string.

Find longest repeating substring in JavaScript using regular expressions

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.