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Finding matches using regular expression

How to find the length of consecutive zeros that is surrounded by ones at both ends of a binary?
For example, in 10010001 the 2 matches are 1001 and 10001
1001 the length of zeroes is 2
10001 the length of zeroes is 3
I used match which returned only the last one i.e. 10001.
'1010001'.match(/1(0+)1$/g)

You need lookahead assertions here:
console.log('1010001'.match(/10+(?=1)/g).map(function(x) {
return x.length - 1;
}));

First, replace all '1' by '11', then remove $ - from your regular expression.
console.log('10010001'.replace(/1/g, '11').match(/1(0+)1/g));

In regular expressions, $ is a special character matching the end of the string (MDN).
But that's only half the problem. String#match captured the trailing 1 in the first group and could not create a second overlapping group for '10001'. Try using RegExp#exec instead. The regular expression is stateful, and in this case, you'll want to move the last index back one for each match you find.
var re = /10+1/g;
var str = '10010001';
var matches = [];
var result;
while ((result = re.exec(str)) !== null) {
matches.push(result[0]);
re.lastIndex--;
}
console.log(matches);

Instead of using Regex, I would keep it simple.
Split the string on '1's
Get the length of each section
Remove the first and last section length because they aren't surrounded by '1's
Find the highest section length
const maxNumberOfZeros = Math.max(...'000000101000100000'.split('1').map(str => str.length).slice(1, -1));
console.log(maxNumberOfZeros);

function solution(N) {
let s = (N >>> 0).toString(2).split('');
let max = 0;
//1 1 0 0 1 0 0 0 1 0
if(s.length > 2){
let lastDigit = s[s.length - 1];
let firstDigit = s[0];
while(lastDigit == '0'){
s.pop();
if(!s.length || s.length == 1) return 0;
lastDigit = s[s.length - 1];
}
while(firstDigit == '0'){
s.shift();
if(!s.length || s.length == 1) return 0;
firstDigit = s[0];
}
let x = s.join('').split('1').filter(i => i !== '').sort().reverse();
return x.length ? x[0].length : 0;
}
return 0;
}

The $ within RegExp matches end of string.
You can use Array.prototype.reduce() and logic to evaluate 1 and 0 as a Boolean to determine if the sequence matches the required pattern
let nums = ["1000000001010001", "0000000101", "100000000"];
let countZerosBetweenOnes = str =>
(res => ([...str].reduce((a, b, idx, arr) =>
(!!+a && !+b && arr.find((n, i) => i > idx && !!+n) && res.push(+a)
, !+a && !+b && res.length && ++res[res.length - 1], b))
, res.sort((a, b) => a - b).pop() || 0))([]);
nums.forEach(n => console.log(countZerosBetweenOnes(n)));

Check how many times a char appears in a string

Simply trying to find how many times a given character appears in a string but I can't solve it any other way then this simple for-loop. Is there a method that would solve this quicker or more eloquently other than using Regex?
function countCharacter(str, char) {
var count = 0;
for(var i = 0; i < str.length; i++){
if(str.charAt(i) === char)
count++;
}
return count;
}

There are many possible ways are available in the market.
I am adding a few of them.
Method 1:
str = "The man is as good as his word"
str.split('a')
output: (4) ["The m", "n is ", "s good ", "s his word"]
str.split('a').length - 1
output: 3
Method 2:
str = "The man is as good as his word"
str.split('').map( function(char,i)
{ if(char === 'a')
return i;
}
).filter(Boolean)
Output: (3) [5, 11, 19]
str.split('').map( function(char,i)
{ if(char === 'a')
return i;
}
).filter(Boolean).length
ouput: 3
Edit: As per comment we can also make use of filter().
str.split('').filter(function(char, i){
if(char == 'a'){
return i;
}
})
output: (3) ["a", "a", "a"]
str.split('').filter(function(char, i){
if(char == 'a'){
return i;
}
}).length
output: 3

----edited by adding more cases from answers----
there are several ways, you can use split/for/regex/reduce/indexOf like this:
function countCharacter_reduce(str, ch) {
return Array.prototype.reduce.call(str, (prev, cur) => cur === ch && ++prev && prev, 0);
}
function countCharacter_split(str, ch) {
return str.split(ch).length - 1;
}
function countCharacter_for(str, ch) {
for (var count = 0, ii = 0; ii < str.length; ii++) {
if (str[ii] === ch)
count++;
}
return count;
}
function countCharacter_regex(str, ch) {
return str.length - str.replace(new RegExp(ch, 'g'), '').length;
}
function countCharacter_indexOf(str, char) {
var start = 0;
var count = 0;
while ((start = str.indexOf(char, start) + 1) !== 0) {
count++;
}
return count;
}
performance of them by running 1,000,000 times on counting '/' in a string.
-- case1: running 1000000 times on ( 'this/is/a/path/with/extension', '/' )
countCharacter_reduce: 2389.880ms
countCharacter_regex: 496.251ms
countCharacter_split: 114.709ms
countCharacter_for: 152.012ms
countCharacter_indexOf: 90.330ms
-- case2: running 1000000 times on ( '////////////', '/' )
countCharacter_reduce: 1138.051ms
countCharacter_regex: 619.886ms
countCharacter_split: 121.945ms
countCharacter_for: 74.542ms
countCharacter_indexOf: 204.242ms
Conclusion ('>' means 'has better performance'):
for|split|indexOf > regex > reduce.
furthermore,
if the string contains more searching characters (like case2),
for>split>indexOf,
otherwise (like case1)
indexOf > split > for.
BTW: you can change the for indexOf method to fit multi-characters search (example is single character)

using reduce:
function countCharacter(str, char) {
return str.split('').reduce((a, x) => x === char ? ++a : a, 0);
}

I guess this involves regex which you wanted to avoid but it's pretty quick:
function countCharacter(str, char) {
return str.length - str.replace(new RegExp(char,"g"),"").length;
}
You can also try the str.split(char).length-1 approach suggested by Jaromanda.
Or, go all out with some fun recursion (pass 0 to startingFrom):
function countCharacter(str, char, startingFrom) {
var idx = str.indexOf(char, startingFrom);
return idx == -1 ? 0 : 1 + countCharacter(str, char, idx + 1);
}
You can get rid of the annoying extra argument at the cost of some efficiency:
function countCharacter(str, char) {
var idx = str.indexOf(char);
return idx == -1 ? 0 : 1 + countCharacter(str.substr(idx+1), char);
}
And here's a version optimized for speed (this is about 3 times faster on my browser than your original and much faster than the regex versions, according to jsperf):
function countCharacter(str, char) {
var start = 0;
var count = 0;
while((start = str.indexOf(char, start)+1) !== 0) {
count++;
}
return count;
}
Note that the indexOf approaches will generally be substantially faster than manually iterating through the string. See jsperf

Here you go. One line code
"hello".match(new RegExp('l','g')).length
replace 'l' with any char here, new RegExp('l','g').
that is
str.match(new RegExp(char,'g')).length

Have you thought of using the split() method? How about this -
function myFunction(str, char) {
return string.split(char).length - 1
}
Let me know what you think of this solution.

How to split a string based on commas except inside parenthesis

I have a string that contains a create query, and I'm trying to split it based on commas. Unfortunately, some of the lines have commas in them surrounded by parenthesis.
Example:
dbo.Display_Test1.Column,
CASE WHEN CHARINDEX(' To ', Test3) > 0 AND CHARINDEX('XVR', Test3) = 0 THEN LEFT(Test3, (CHARINDEX(' To ', Test3) - 12)) ELSE Test3 END AS Test3,
dbo.Display_Test2.Column,
ISNULL((CASE WHEN [2-Display-Test4].[Total Number] > 0 AND [1-Display-Test5].SumOfNumber = 0 THEN 0 ELSE (([2-Display-Test4].[Total Number] * 1000) / [1-Display-Test5].SumOfNumber) END), 0) AS Test6,
I want to split my string based on commas that aren't inside of (possibly multiple) parenthesis. For reference, my example should be split where the line breaks are (although my string doesn't have the line breaks in it).
I've tried a number of different solutions, but none work quite right:
/(?:\(*[^()]*\)|[^,])+/g works on lines 1,3, and 4, but fails on line 2. It breaks up the line into multiple matches.
/((?:[^,(]+|(\((?:[^()]+)|$1\)))+)/g works on lines 1,2, and 3, but fails on line 4. It also breaks up the line into multiple matches.
I can't quite seem to make it work. Any help is appreciated.

One solution possible : ( note : we remove the separations comas )
var str = "dbo.Display_Test1.Column, CASE WHEN CHARINDEX(' To ', Test3) > 0 AND CHARINDEX('XVR', Test3) = 0 THEN LEFT(Test3, (CHARINDEX(' To ', Test3) - 12)) ELSE Test3 END AS Test3, dbo.Display_Test2.Column, ISNULL((CASE WHEN [2-Display-Test4].[Total Number] > 0 AND [1-Display-Test5].SumOfNumber = 0 THEN 0 ELSE (([2-Display-Test4].[Total Number] * 1000) / [1-Display-Test5].SumOfNumber) END), 0) AS Test6,";
// if the string don't end by a coma , we add it.
var strArr = str.replace(/,*\s*$/ , ',').split('');
var res;
res = strArr.reduce(function( trans , charValue ){
if(charValue === '(') {
trans.deep++;
}
if(charValue === ')') {
trans.deep--;
}
if( trans.deep === 0){
if(charValue===',') {
trans.arr.push( trans.str);
trans.str = '';
}else{
trans.str += charValue;
}
}else{
trans.str += charValue;
}
return trans;
}, { arr : [] , str : '' ,deep : 0});
document.write('<pre>' + JSON.stringify(res.arr , null , ' ') + '</pre>');
Edit :
As suggested by Thriggle in the comments i've added the case where the string don't end by a coma.
changed variables name .

Of course if we want to work with unlimited nesting of parentheses then it would be impossible to avoid a loop. Still for some bounded nesting we do can write a regular expression capturing it.
For up to 1 parentheses level it could be
/((?:[^(),]|\([^()]*\))+)/g
For up to 2 parentheses level (probably it's your case)
/((?:[^(),]|\((?:[^()]|\([^()]*\))*\))+)/g
And so on, recursively substituting [^()]* with (?:[^()]|\([^()]*\))*

Regex won't get you where you want and still allow arbitrary complexity.
One approach is to iterate through each character in the string, keeping track of when you hit open and close parentheses, and only splitting on commas when the numbers of open and close parentheses cancel each other out (so you know you're not within a parenthetical statement).
function splitQuery(input){
var arr = [];
var lastStart = 0;
var open = 0;
for(var i = 0, len = input.length; i < len; i++){
var curr = input[i];
if(curr === "("){open += 1;}
else if(curr === ")"){ open = open < 1 ? 0 :open -=1;}
else if(curr === ","){
if(open === 0){
arr.push(input.substring(lastStart,i));
lastStart = i+1;
}
}else if(i+1 === len){
arr.push(input.substring(lastStart,i+1));
}
}
return arr;
}
Just be aware that this approach can be expensive (from a performance perspective) when dealing with especially large strings.
Here's a working example:
var query = "dbo.Display_Test1.Column, CASE WHEN CHARINDEX(' To ', Test3) > 0 AND CHARINDEX('XVR', Test3) = 0 THEN LEFT(Test3, (CHARINDEX(' To ', Test3) - 12)) ELSE Test3 END AS Test3, dbo.Display_Test2.Column, ISNULL((CASE WHEN [2-Display-Test4].[Total Number] > 0 AND [1-Display-Test5].SumOfNumber = 0 THEN 0 ELSE (([2-Display-Test4].[Total Number] * 1000) / [1-Display-Test5].SumOfNumber) END), 0) AS Test6,";
var split = splitQuery(query);
var output = document.getElementById("output")
for(var el in split){
output.insertAdjacentHTML("beforeend",el + ": "+ split[el]+"<hr/>");
}
function splitQuery(input){
var arr = [];
var lastStart = 0;
var open = 0;
for(var i = 0, len = input.length; i < len; i++){
var curr = input[i];
if(curr === "("){open += 1;}
else if(curr === ")"){ open = open < 1 ? 0 :open -=1;}
else if(curr === ","){
if(open === 0){
arr.push(input.substring(lastStart,i));
lastStart = i+1;
}
}else if(i+1 === len){
arr.push(input.substring(lastStart,i+1));
}
}
return arr;
}
<div id="output"></div>

Inserting into a number string

Have the function DashInsert(num) insert dashes ('-') between each two odd numbers in num. For example: if num is 454793 the output should be 4547-9-3. Don't count zero as an odd number.
Here is my code (not working). When I run it, I get the same response as an infinite loop where I have to kill the page but I can't see why. I know there are ways to do this by keeping it as a string but now I'm wondering why my way isn't working. Thanks...
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
}
}
num = num.join("");
return num;
}

Using num.splice you are inserting new entries into the array, therefor increasing its length – and that makes the value of i “running behind” the increasing length of the array, so the break condition is never met.
And apart from that, on the next iteration after inserting a -, num[i-1] will be that - character, and therefor you are practically trying to check if '-' % 2 != 0 … that makes little sense as well.
So, when you insert a - into the array, you have to increase i by one as well – that will a) account for the length of the array having increased by one, and also it will check the next digit after the - on the next iteration:
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
i++; // <- this is the IMPORTANT part!
}
}
num = num.join("");
return num;
}
alert(DashInsert("454793"));
http://jsfiddle.net/37wA9/

Once you insert a dash -, the if statement is checking this '-'%2 != 0 which is always true and thus inserts another dash, ad infinitum.
Here's one way to do it with replace using a regex and function:
function DashInsert(n) {
var f = function(m,i,s) { return m&s[i+1]&1 ? m+'-' : m; };
return String(n).replace(/\d/g,f);
}
DashInsert(454793) // "4547-9-3"

When you are adding a dash, this dash will be processed as a number on the next iteration. You need to forward one step.
function DashInsert(num) {
var num = num.split("");
for (var i = 1; i < num.length; i++) {
if ((num[i - 1] % 2 != 0) && (num[i] % 2 != 0)) {
num.splice(i, 0, "-");
i++; // This is the only thing that needs changing
}
}
num = num.join("");
return num;
}

It's because there are cases when you use the % operator on dash '-' itself, e.g. right after you splice a dash into the array.
You can correct this behavior by using a clone array.
function DashInsert(num) {
num = num.split("");
var clone = num.slice(0);
var offset = 0;
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
clone.splice(i + offset, 0, "-");
offset++;
}
}
return clone.join("");
}
alert(DashInsert("45739"));
Output: 45-7-3-9
Demo: http://jsfiddle.net/262Bf/

To complement the great answers already given, I would like to share an alternative implementation, that doesn't modify arrays in-place:
function DashInsert(num) {
var characters = num.split("");
var numbers = characters.map(function(chr) {
return parseInt(chr, 10);
});
var withDashes = numbers.reduce(function(result, current) {
var lastNumber = result[result.length - 1];
if(lastNumber == null || current % 2 === 0 || lastNumber % 2 === 0) {
return result.concat(current);
} else {
return result.concat("-", current);
}
}, []);
return withDashes.join("");
}
It's longer, but IMHO reveals the intention better, and avoids the original issue.

How to sort strings in JavaScript numerically

I would like to sort an array of strings (in JavaScript) such that groups of digits within the strings are compared as integers not strings. I am not worried about signed or floating point numbers.
For example, the result should be ["a1b3","a9b2","a10b2","a10b11"] not ["a1b3","a10b11","a10b2","a9b2"]
The easiest way to do this seems to be splitting each string on boundaries around groups of digits. Is there a pattern I can pass to String.split to split on character boundaries without removing any characters?
"abc11def22ghi".split(/?/) = ["abc","11","def","22","ghi"];
Or is there another way to compare strings that does not involve splitting them up, perhaps by padding all groups of digits with leading zeros so they are the same length?
"aa1bb" => "aa00000001bb", "aa10bb" => "aa00000010bb"
I am working with arbitrary strings, not strings that have a specific arrangement of digit groups.
I like the /(\d+)/ one liner from Gaby to split the array. How backwards compatible is that?
The solutions that parse the strings once in a way that can be used to rebuild the originals are much more efficient that this compare function. None of the answers handle some strings starting with digits and others not, but that would be easy enough to remedy and was not explicit in the original question.
["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"].sort(function (inA, inB) {
var result = 0;
var a, b, pattern = /(\d+)/;
var as = inA.split(pattern);
var bs = inB.split(pattern);
var index, count = as.length;
if (('' === as[0]) === ('' === bs[0])) {
if (count > bs.length)
count = bs.length;
for (index = 0; index < count && 0 === result; ++index) {
a = as[index]; b = bs[index];
if (index & 1) {
result = a - b;
} else {
result = !(a < b) ? (a > b) ? 1 : 0 : -1;
}
}
if (0 === result)
result = as.length - bs.length;
} else {
result = !(inA < inB) ? (inA > inB) ? 1 : 0 : -1;
}
return result;
}).toString();
Result: "!!,9,9.5,10,a3,a3b,a3b3,a3b20,a3b100,a20,a100,~~"

Another variant is to use an instance of Intl.Collator with the numeric option:
var array = ["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"];
var collator = new Intl.Collator([], {numeric: true});
array.sort((a, b) => collator.compare(a, b));
console.log(array);

I think this does what you want
function sortArray(arr) {
var tempArr = [], n;
for (var i in arr) {
tempArr[i] = arr[i].match(/([^0-9]+)|([0-9]+)/g);
for (var j in tempArr[i]) {
if( ! isNaN(n = parseInt(tempArr[i][j])) ){
tempArr[i][j] = n;
}
}
}
tempArr.sort(function (x, y) {
for (var i in x) {
if (y.length < i || x[i] < y[i]) {
return -1; // x is longer
}
if (x[i] > y[i]) {
return 1;
}
}
return 0;
});
for (var i in tempArr) {
arr[i] = tempArr[i].join('');
}
return arr;
}
alert(
sortArray(["a1b3", "a10b11", "a10b2", "a9b2"]).join(",")
);

Assuming you want to just do a numeric sort by the digits in each array entry (ignoring the non-digits), you can use this:
function sortByDigits(array) {
var re = /\D/g;
array.sort(function(a, b) {
return(parseInt(a.replace(re, ""), 10) - parseInt(b.replace(re, ""), 10));
});
return(array);
}
It uses a custom sort function that removes the digits and converts to a number each time it's asked to do a comparison. You can see it work here: http://jsfiddle.net/jfriend00/t87m2/.

Use this compare function for sorting...
function compareLists(a, b) {
var alist = a.split(/(\d+)/), // Split text on change from anything
// to digit and digit to anything
blist = b.split(/(\d+)/); // Split text on change from anything
// to digit and digit to anything
alist.slice(-1) == '' ? alist.pop() : null; // Remove the last element if empty
blist.slice(-1) == '' ? blist.pop() : null; // Remove the last element if empty
for (var i = 0, len = alist.length; i < len; i++) {
if (alist[i] != blist[i]){ // Find the first non-equal part
if (alist[i].match(/\d/)) // If numeric
{
return +alist[i] - +blist[i]; // Compare as number
} else {
return alist[i].localeCompare(blist[i]); // Compare as string
}
}
}
return true;
}
Syntax
var data = ["a1b3", "a10b11", "b10b2", "a9b2", "a1b20", "a1c4"];
data.sort(compareLists);
alert(data);
There is a demo at http://jsfiddle.net/h9Rqr/7/.

Here's a more complete solution that sorts according to both letters and numbers in the strings
function sort(list) {
var i, l, mi, ml, x;
// copy the original array
list = list.slice(0);
// split the strings, converting numeric (integer) parts to integers
// and leaving letters as strings
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].match(/(\d+|[a-z]+)/g);
for( mi = 0, ml = list[i].length; mi < ml ; mi++ ) {
x = parseInt(list[i][mi], 10);
list[i][mi] = !!x || x === 0 ? x : list[i][mi];
}
}
// sort deeply, without comparing integers as strings
list = list.sort(function(a, b) {
var i = 0, l = a.length, res = 0;
while( res === 0 && i < l) {
if( a[i] !== b[i] ) {
res = a[i] < b[i] ? -1 : 1;
break;
}
// If you want to ignore the letters, and only sort by numbers
// use this instead:
//
// if( typeof a[i] === "number" && a[i] !== b[i] ) {
// res = a[i] < b[i] ? -1 : 1;
// break;
// }
i++;
}
return res;
});
// glue it together again
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].join("");
}
return list;
}

I needed a way to take a mixed string and create a string that could be sorted elsewhere, so that numbers sorted numerically and letters alphabetically. Based on answers above I created the following, which pads out all numbers in a way I can understand, wherever they appear in the string.
function padAllNumbers(strIn) {
// Used to create mixed strings that sort numerically as well as non-numerically
var patternDigits = /(\d+)/g; // This recognises digit/non-digit boundaries
var astrIn = strIn.split( patternDigits ); // we create an array of alternating digit/non-digit groups
var result = "";
for (var i=0;i<astrIn.length; i++) {
if (astrIn[i] != "") { // first and last elements can be "" and we don't want these padded out
if (isNaN(astrIn[i])) {
result += astrIn[i];
} else {
result += padOneNumberString("000000000",astrIn[i]);
}
}
}
return result;
}
function padOneNumberString(pad,strNum,left) {
// Pad out a string at left (or right)
if (typeof strNum === "undefined") return pad;
if (typeof left === "undefined") left = true;
var padLen = pad.length - (""+ strNum).length;
var padding = pad.substr(0,padLen);
return left? padding + strNum : strNum + padding;
}

Sorting occurs from left to right unless you create a custom algorithm. Letters or digits are compared digits first and then letters.
However, what you want to accomplish as per your own example (a1, a9, a10) won’t ever happen. That would require you knowing the data beforehand and splitting the string in every possible way before applying the sorting.
One final alternative would be:
a) break each and every string from left to right whenever there is a change from letter to digit and vice versa; &
b) then start the sorting on those groups from right-to-left. That will be a very demanding algorithm. Can be done!
Finally, if you are the generator of the original "text", you should consider NORMALIZING the output where a1 a9 a10 could be outputted as a01 a09 a10. This way you could have full control of the final version of the algorithm.