There are many posts like this and I have found a few solutions but there are not perfect.
One of them:
"aabbhahahahahahahahahahahasetsetset".replace(/[^\w\s]|(.+)\1+/gi, '$1')
The results is:
abhahahahahahaset
I want to get result:
abhaset
How to do this ?
.+ is greedy. It takes as much as it can. That is half of the has so that \1 can match the second half. Making the repetition ungreedy should do the trick:
/[^\w\s]|(.+?)\1+/gi
By the way, the i doesn't change anything here.
To get rid of nested repetitions (e.g. transform aaBBaaBB into aB (via aaBB or aBaB)) simply run the replacement multiple times until the result does not change any more.
var pattern = /[^\w\s]|(.+?)\1+/g;
var output = "aaBBaaBB";
var input;
do
{
input = output;
output = input.replace(pattern, "$1");
} while (input != output)
I admit the naming of output is a bit awkward for the first repetition, but you know... the two most difficult problems in computer science are cache invalidation, naming things and off-by-one errors.
.+ will match the maximum amount possible, so hahahaha satisfies (.+)\1 with haha and haha. You want to match the minimum amount possible, so use a reluctant quantifier.
"aabbhahahahahahahahahahahasetsetset".replace(/[^\w\s]|(.+?)\1+/gi, '$1')
http://jsfiddle.net/HQRDg/
Related
I want to be clear I'm not looking for solutions. I'm really trying to understand what is being done. With that said all pointers and recommendations are welcomed. I am woking through freecodecamp.com task of Check for Palindromes. Below is the description.
Return true if the given string is a palindrome. Otherwise, return
false.
A palindrome is a word or sentence that's spelled the same way both
forward and backward, ignoring punctuation, case, and spacing.
Note You'll need to remove all non-alphanumeric characters
(punctuation, spaces and symbols) and turn everything lower case in
order to check for palindromes.
We'll pass strings with varying formats, such as "racecar", "RaceCar",
and "race CAR" among others.
We'll also pass strings with special symbols, such as "2A3*3a2", "2A3
3a2", and "2_A3*3#A2".
This is what I have for code right now again I'm working through this and using chrome dev tools to figure out what works and what doesn't.
function palindrome(str) {
// Good luck!
str = str.toLowerCase();
//str = str.replace(/\D\S/i);
str = str.replace(/\D\s/g, "");
for (var i = str.length -1; i >= 0; i--)
str += str[i];
}
palindrome("eye");
What I do not understand is when the below code is run in dev tools the "e" is missing.
str = str.replace(/\D\s/g, "");
"raccar"
So my question is what part of the regex am I miss understanding? From my understand the regex should only be getting rid of spaces and integers.
/\D\s/g is replacing any character not a digit, followed by a space with "".
So, in race car, the Regex matches "e " and replaces it with "", making the string raccar
For digit, you need to use \d. I think using an OR would get you what you want. So, you may try something like /\d|\s/g to get a digit or a space.
Hope this helps in some way in your understanding!
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.
I've got some string that contain invisible characters, but they are in somewhat predictable places. Typically the surround the piece of text I want to extract, and then after the 2nd occurrence I want to keep the rest of the text.
I can't seem to figure out how to both key off of the invisible characters, and exclude them from my result. To match invisibles I've been using this regex: /\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F/ which does seem to work.
Here's an example: [invisibles]Keep as match 1[invisibles]Keep as match 2
Here's what I've been using so far without success:
/([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)(.+)([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)/(.+)
I've got the capture groups in there, but it's bee a while since I've had to use regex's in this way, so I know I'm missing something important. I was hoping to just make the invisible matches non-capturing groups, but it seems that JavaScript does not support this.
Something like this seems like what you want. The second regex you have pretty much works, but the / is in totally the wrong place. Perhaps you weren't properly reading out the group data.
var s = "\x0EKeep as match 1\x0EKeep as match 2";
var r = /[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)/;
var match = s.match(r);
var part1 = match[1];
var part2 = match[2];
I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}
By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.
If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.
Friends,
I'm new to both Javascript and Regular Expressions and hope you can help!
Within a Javascript function I need to check to see if a comma(,) appears 1 or more times. If it does then there should be one or more numbers either side of it.
e.g.
1,000.00 is ok
1,000,00 is ok
,000.00 is not ok
1,,000.00 is not ok
If these conditions are met I want the comma to be removed so 1,000.00 becomes 1000.00
What I have tried so is:
var x = '1,000.00';
var regex = new RegExp("[0-9]+,[0-9]+", "g");
var y = x.replace(regex,"");
alert(y);
When run the alert shows ".00" Which is not what I was expecting or want!
Thanks in advance for any help provided.
strong text
Edit
strong text
Thanks all for the input so far and the 3 answers given. Unfortunately I don't think I explained my question well enough.
What I am trying to achieve is:
If there is a comma in the text and there are one or more numbers either side of it then remove the comma but leave the rest of the string as is.
If there is a comma in the text and there is not at least one number either side of it then do nothing.
So using my examples from above:
1,000.00 becomes 1000.00
1,000,00 becomes 100000
,000.00 is left as ,000.00
1,,000.00 is left as 1,,000.00
Apologies for the confusion!
Your regex isn't going to be very flexible with higher orders than 1000 and it has a problem with inputs which don't have the comma. More problematically you're also matching and replacing the part of the data you're interested in!
Better to have a regex which matches the forms which are a problem and remove them.
The following matches (in order) commas at the beginning of the input, at the end of the input, preceded by a number of non digits, or followed by a number of non digits.
var y = x.replace(/^,|,$|[^0-9]+,|,[^0-9]+/g,'');
As an aside, all of this is much easier if you happen to be able to do lookbehind but almost every JS implementation doesn't.
Edit based on question update:
Ok, I won't attempt to understand why your rules are as they are, but the regex gets simpler to solve it:
var y = x.replace(/(\d),(\d)/g, '$1$2');
I would use something like the following:
^[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)$
[0-9]{1,3}: 1 to 3 digits
(,[0-9]{3})*: [Optional] More digit triplets seperated by a comma
(\.[0-9]+): [Optional] Dot + more digits
If this regex matches, you know that your number is valid. Just replace all commas with the empty string afterwards.
It seems to me you have three error conditions
",1000"
"1000,"
"1,,000"
If any one of these is true then you should reject the field, If they are all false then you can strip the commas in the normal way and move on. This can be a simple alternation:
^,|,,|,$
I would just remove anything except digits and the decimal separator ([^0-9.]) and send the output through parseFloat():
var y = parseFloat(x.replace(/[^0-9.]+/g, ""));
// invalid cases:
// - standalone comma at the beginning of the string
// - comma next to another comma
// - standalone comma at the end of the string
var i,
inputs = ['1,000.00', '1,000,00', ',000.00', '1,,000.00'],
invalid_cases = /(^,)|(,,)|(,$)/;
for (i = 0; i < inputs.length; i++) {
if (inputs[i].match(invalid_cases) === null) {
// wipe out everything but decimal and dot
inputs[i] = inputs[i].replace(/[^\d.]+/g, '');
}
}
console.log(inputs); // ["1000.00", "100000", ",000.00", "1,,000.00"]