Develop Reference

How to create number suffixes such as "100K" without having to be repetitive?

I've been having a problem that when my auto clicker in my clicker game goes fast enough to get to 200 thousand, it starts to lag, and then it doesn't function properly, or as fast.
Is there a way to make 100 thousand turn into 100K, and 101 thousand turn into 101K without being repetitive?
I tried this with my original code, and realized putting up to 1000 suffixes into each function would be a little too hard:
if (number >= 100000) {
document.getElementById(ID).innerHTML = "100K"
}
if (number >= 101000) {
document.getElementById(ID).innerHTML = "101K"
}
and on and on.
I don't want multiple if statements!
This would work, but it would take up way too much space, and I know there is an easier way to it, but I just couldn't find it. Can anyone provide a way to do this?

Try separating the job of formatting your number into a different function.
SUFFIXES = 'KMBTqQsSOND' // or whatever you'd like them to be
function getSuffixedNumber(num) {
var power = Math.floor(Math.log10(num));
var index = Math.floor(power / 3);
num = Math.round(num / Math.pow(10, (index * 3))); // first 3 digits of the number
return num + (SUFFIXES[index - 1] || ''); // default to no suffix if we get an out of bounds index
}
You can call the function like this: var x = getSuffixedNumber(101000), the value of x will be "101K".

Adding space to fill array length

I am working with a javascript program that needs to be formatted a certain way. Basically, I need to have each section of information from an array be a set length, for example 12 characters long, and no more than that.
The problem I am running into comes when a value in the array is NOT 12 characters long. If I have a value that is less than the 12 characters the remaining character allotment needs to be filled with blank spaces.
The length of each section of information varies in size and is not always 12. How can I add X number of blank spaces, should the length not meet the maximum requirement, for each section?
This is where I am at with adding space:
str = str + new Array(str.length).join(' ');
I am pretty sure what I have above is wrong but I believe I am on the right track with the .join function. Any ideas?
EDIT: I was asked to show a wanted outcome. It is a bit complicated because this javascript is being run out of a web report tool and not out of something like Visual Studio so its not traditional JS.
The outcome expected should look something like:
Sample Image
So as shown above the data is in one line, cutting off longer strings of information or filling in blank spaces if its too short for the "column" to keep that nice even look.

try this code and leverage the wonders of the map function:
let say your array is:
var myArr = ["123456789012", "12345678901", "123"];
now just apply this function
myArr.map(function(item){ //evalueate each item inside the array
var strLength = item.length; //apply this function to each item
if (strLength < 12){
return item + ' '.repeat(12-item.length) //add the extra spaces as needed
} else {
return item; // return the item because it's length is 12 or +
}
})
What you are looking for is the ' '.repeat(x) - where x is the times you want to repeat the string you have set, it could be '*'.repeat(2) and you would get '**', if you want to understand more about it look at the docs

depending on which version of javascript, this might work:
if (str.length < 12) str += ' '.repeat(12 - str.length);

Not exactly sure how you're setup -- but something like the following will accept an array and return another array with all its values being 12 characters in length.
var array = ['Test', 'Testing', 'Tested', 'This is not a Test'];
var adjustedArray = correctLength(array, 12);
function correctLength(array, length) {
array.map(function(v, i) {
if (array[i].length < length) {
array[i] += Array((length+1) - array[i].length).join('_');
}
// might not need this if values are already no greater than 12
array[i] = array[i].substring(0, length);
});
return array;
}
console.log(adjustedArray);

compressing a string of 0's and 1's in js

Itroduction
I'm currently working on John Conway's Game of Life in js. I have the game working (view here) and i'm working on extra functionalities such as sharing your "grid / game" to your friends. To do this i'm extracting the value's of the grid (if the cell is alive or dead) into a long string of 0's and 1's.
This string has a variable length since the grid is not always the same size. for example:
grid 1 has a length and width of 30 => so the string's length is 900
grid 2 has a length and width of 50 => so the string's length is 2500
The problem
As you can see these string's of 0's and 1's are way too long to copy around and share.
However hard i try I don't seem to be able to come up with a code that would compress a string this long to a easy to handle one.
Any ideas on how to compress (and decompress) this?
I have considered simply writing down every possible grid option for the gird sizes 1x1 to 100x100 and giving them a key/reference to use as sharable code. Doing that by hand would be madness but maybe any of you has an idea on how to create an algorithm that can do this?
GitHub repository

In case it wasn't already obvious, the string you're trying to store looks like a binary string.
Counting systems
Binary is a number in base-2. This essentially means that there are two characters being used to keep count. Normally we are used to count with base-10 (decimal characters). In computer science the hexadecimal system (base-16) is also widely being used.
Since you're not storing the bits as bits but as bytes (use var a = 0b1100001; if you ever wish to store them like bits) the 'binary' you wish to store just takes as much space as any other random string with the same length.
Since you're using the binary system each position just has 2 possible values. When using the hexadecimal value a single position can hold up to 16 possible values. This is already a big improvement when it comes to storing the data compactly. As an example 0b11111111 and 0xff both represents the decimal number 255.
In your situation that'd shave 6 bytes of every 8 bytes you have to store. In the end you'd be stuck with a string just 1/4th of the length of the original string.
Javascript implementation
Essentially what we want to do is to interpret the string you store as binary and retrieve the hexadecimal value. Luckily JavaScript has built in functionality to achieve stuff like this:
var bin =
'1110101110100011' +
'0000101111100001' +
'1010010101011010' +
'0000110111011111' +
'1111111001010101' +
'0111000011100001' +
'1011010100110001' +
'0111111110010100' +
'0111110110100101' +
'0000111101100111' +
'1100001111011100' +
'0101011100001111' +
'0110011011001101' +
'1000110010001001' +
'1010100010000011' +
'0011110000000000';
var returnValue = '';
for (var i = 0; i < parseInt(bin.length / 8); i++) {
returnValue += parseInt(bin.substr(i*8, 8), 2).toString(16);
}
console.log(bin.length); // Will return 265
console.log(returnValue.length); // Will return 64
We're saying "parse this string and interpret it like a base-2 number and store it as a hexadecimal string".
Decoding is practically the same. Replace all occurrences of the number 8 in the example above with 2 and vice versa.
Please note
A prerequisite for this code to work correctly is that the binary length is dividable by 8. See the following example:
parseInt('00011110', 2).toString(16); // returns '1e'
parseInt('1e', 16).toString(2); // returns '11110'
// Technically both representations still have the same decimal value
When decoding you should add leading zeros until you have a full byte (8 bits).
In case the positions you have to store are not dividable by 8 you can, for example, add padding and add a number to the front of the output string to identify how much positions to strip.
Wait, there's more
To get even shorter strings you can build a lookup table with 265 characters in which you search for the character associated with the specific position. (This works because you're still storing the hexadecimal value as a string.) Sadly neither the ASCII nor the UTF-8 encodings are suited for this as there are blocks with values which have no characters defined.
It may look like:
// Go fill this array until you have 265 values within it.
var lookup = ['A', 'B', 'C', 'D'];
var smallerValue = lookup[0x00];
This way you can have 265 possible values at a single position, AND you have used your byte to the fullest.
Please note that no real compression is happening here. We're rather utilising data types to be used more efficiently for your current use case.

If we make the assumption than the grid contains much more 0's than 1's, you may want to try this simple compression scheme:
convert the binary string to an hexadecimal string
convert '00' sub-strings to 'z' symbol
convert 'zz' sub-strings to 'Z' symbol
we could go further, but let's stop here for the demo
Below is an example with a 16x16 grid:
var bin =
'0000000000000000' +
'0000001000000000' +
'0000011100000000' +
'0000001000000000' +
'0000000000000000' +
'0000000000111000' +
'0000100000111000' +
'0000000000111000' +
'0000000000000000' +
'0000000000000000' +
'0000000010000000' +
'0000000101000000' +
'0000000010000000' +
'0000000000000000' +
'0000100000000000' +
'0000000000000000';
var packed = bin
.match(/(.{4})/g)
.map(function(x) {
return parseInt(x, 2).toString(16);
})
.join('')
.replace(/00/g, 'z')
.replace(/zz/g, 'Z');
This will produce the string "Z02z07z02ZZ380838z38ZZz8z14z08Zz8Zz".
The unpacking process is doing the exact opposite:
var bin = packed
.replace(/Z/g, 'zz')
.replace(/z/g, '00')
.split('')
.map(function(x) {
return ('000' + parseInt(x, 16).toString(2)).substr(-4, 4);
})
.join('');
Note that this code will only work correctly if the length of the input string is a multiple of 4. If it's not the case, you'll have to pad the input and crop the output.
EDIT : 2nd method
If the input is completely random -- with roughly as many 0's as 1's and no specific repeating patterns -- the best you can do is probably to convert the binary string to a BASE64 string. It will be significantly shorter (this time with a fixed compression ratio of about 17%) and can still be copied/pasted by the user.
Packing:
var bin =
'1110101110100011' +
'0000101111100001' +
'1010010101011010' +
'0000110111011111' +
'1111111001010101' +
'0111000011100001' +
'1011010100110001' +
'0111111110010100' +
'0111110110100101' +
'0000111101100111' +
'1100001111011100' +
'0101011100001111' +
'0110011011001101' +
'1000110010001001' +
'1010100010000011' +
'0011110000000000';
var packed =
btoa(
bin
.match(/(.{8})/g)
.map(function(x) {
return String.fromCharCode(parseInt(x, 2));
})
.join('')
);
Will produce the string "66ML4aVaDd/+VXDhtTF/lH2lD2fD3FcPZs2MiaiDPAA=".
Unpacking:
var bin =
atob(packed)
.split('')
.map(function(x) {
return ('0000000' + x.charCodeAt(0).toString(2)).substr(-8, 8);
})
.join('');
Or if you want to go a step further, you can consider using something like base91 instead, for a reduced encoding overhead.

LZ-string
Using LZ-string I was able to compress the "code" quite a bit.
By simply compressing it to base64 like this:
var compressed = LZString.compressToBase64(string)
Decompressing is also just as simple as this:
var decompressed = LZString.decompressFromBase64(compressed)
However the length of this compressed string is still pretty long given that you have about as many 0s as 1s (not given in the example)
example
But the compression does work.

ANSWER
For any of you who are wondering how exactly I ended up doing it, here's how:
First I made sure every string passed in would be padded with leading 0s untill it was devidable by 8. (saving the amount of 0s used to pad, since they're needed while decompressing)
I used Corstian's answer and functions to compress my string (interpreted as binary) into a hexadecimal string. Although i had to make one slight alteration.
Not every binary substring with a lenght of 8 will return exactly 2 hex characters. so for those cases i ended up just adding a 0 in front of the substring. The hex substring will have the same value but it's length will now be 2.
Next up i used a functionality from Arnaulds answer. Taking every double character and replacing it with a single character (one not used in the hexadecimal alphabet to avoid conflict). I did this twice for every hexadecimal character.
For example:
the hex string 11 will become h and hh will become H
01101111 will become 0h0H
Since most grids are gonna have more dead cells then alive ones, I made sure the 0s would be able to compress even further, using Arnaulds method again but going a step further.
00 -> g | gg -> G | GG -> w | ww -> W | WW -> x | xx -> X | XX-> y | yy -> Y | YY -> z | zz -> Z
This resulted in Z representing 4096 (binary) 0s
The last step of the compression was adding the amount of leading 0s in front of the compressed string, so we can shave those off at the end of decompressing.
This is how the returned string looks in the end.
amount of leading 0s-compressed string so a 64*64 empty grid, will result in 0-Z
Decompressing is practically doing everything the other way around.
Firstly splitting the number that represents how many leading 0s we've used as padding from the compressed string.
Then using Arnaulds functionality, turning the further "compressed" characters back into hexadecimal code.
Taking this hex string and turning it back into binary code. Making sure, as Corstian pointed out, that every binary substring will have a length of 8. (ifnot we pad the substrings with leading 0s untill the do, exactly, have a length of 8)
And then the last step is to shave off the leading 0s we've used as padding to make the begin string devidable by 8.
The functions
Function I use to compress:
/**
* Compresses the a binary string into a compressed string.
* Returns the compressed string.
*/
Codes.compress = function(bin) {
bin = bin.toString(); // To make sure the binary is a string;
var returnValue = ''; // Empty string to add our data to later on.
// If the lenght of the binary string is not devidable by 8 the compression
// won't work correctly. So we add leading 0s to the string and store the amount
// of leading 0s in a variable.
// Determining the amount of 'padding' needed.
var padding = ((Math.ceil(bin.length/8))*8)-bin.length;
// Adding the leading 0s to the binary string.
for (var i = 0; i < padding; i++) {
bin = '0'+bin;
}
for (var i = 0; i < parseInt(bin.length / 8); i++) {
// Determining the substring.
var substring = bin.substr(i*8, 8)
// Determining the hexValue of this binary substring.
var hexValue = parseInt(substring, 2).toString(16);
// Not all binary values produce two hex numbers. For example:
// '00000011' gives just a '3' while what we wand would be '03'. So we add a 0 in front.
if(hexValue.length == 1) hexValue = '0'+hexValue;
// Adding this hexValue to the end string which we will return.
returnValue += hexValue;
}
// Compressing the hex string even further.
// If there's any double hex chars in the string it will take those and compress those into 1 char.
// Then if we have multiple of those chars these are compressed into 1 char again.
// For example: the hex string "ff will result in a "v" and "ffff" will result in a "V".
// Also: "11" will result in a "h" and "1111" will result in a "H"
// For the 0s this process is repeated a few times.
// (string with 4096 0s) (this would represent a 64*64 EMPTY grid)
// will result in a "Z".
var returnValue = returnValue.replace(/00/g, 'g')
.replace(/gg/g, 'G')
// Since 0s are probably more likely to exist in our binary and hex, we go a step further compressing them like this:
.replace(/GG/g, 'w')
.replace(/ww/g, 'W')
.replace(/WW/g, 'x')
.replace(/xx/g, 'X')
.replace(/XX/g, 'y')
.replace(/yy/g, 'Y')
.replace(/YY/g, 'z')
.replace(/zz/g, 'Z')
//Rest of the chars...
.replace(/11/g, 'h')
.replace(/hh/g, 'H')
.replace(/22/g, 'i')
.replace(/ii/g, 'I')
.replace(/33/g, 'j')
.replace(/jj/g, 'J')
.replace(/44/g, 'k')
.replace(/kk/g, 'K')
.replace(/55/g, 'l')
.replace(/ll/g, 'L')
.replace(/66/g, 'm')
.replace(/mm/g, 'M')
.replace(/77/g, 'n')
.replace(/nn/g, 'N')
.replace(/88/g, 'o')
.replace(/oo/g, 'O')
.replace(/99/g, 'p')
.replace(/pp/g, 'P')
.replace(/aa/g, 'q')
.replace(/qq/g, 'Q')
.replace(/bb/g, 'r')
.replace(/rr/g, 'R')
.replace(/cc/g, 's')
.replace(/ss/g, 'S')
.replace(/dd/g, 't')
.replace(/tt/g, 'T')
.replace(/ee/g, 'u')
.replace(/uu/g, 'U')
.replace(/ff/g, 'v')
.replace(/vv/g, 'V');
// Adding the number of leading 0s that need to be ignored when decompressing to the string.
returnValue = padding+'-'+returnValue;
// Returning the compressed string.
return returnValue;
}
The function I use to decompress:
/**
* Decompresses the compressed string back into a binary string.
* Returns the decompressed string.
*/
Codes.decompress = function(compressed) {
var returnValue = ''; // Empty string to add our data to later on.
// Splitting the input on '-' to seperate the number of paddin 0s and the actual hex code.
var compressedArr = compressed.split('-');
var paddingAmount = compressedArr[0]; // Setting a variable equal to the amount of leading 0s used while compressing.
compressed = compressedArr[1]; // Setting the compressed variable to the actual hex code.
// Decompressing further compressed characters.
compressed = compressed// Decompressing the further compressed 0s. (even further then the rest of the chars.)
.replace(/Z/g, 'zz')
.replace(/z/g, 'YY')
.replace(/Y/g, 'yy')
.replace(/y/g, 'XX')
.replace(/X/g, 'xx')
.replace(/x/g, 'WW')
.replace(/W/g, 'ww')
.replace(/w/g, 'GG')
.replace(/G/g, 'gg')
.replace(/g/g, '00')
// Rest of chars...
.replace(/H/g, 'hh')
.replace(/h/g, '11')
.replace(/I/g, 'ii')
.replace(/i/g, '22')
.replace(/J/g, 'jj')
.replace(/j/g, '33')
.replace(/K/g, 'kk')
.replace(/k/g, '44')
.replace(/L/g, 'll')
.replace(/l/g, '55')
.replace(/M/g, 'mm')
.replace(/m/g, '66')
.replace(/N/g, 'nn')
.replace(/n/g, '77')
.replace(/O/g, 'oo')
.replace(/o/g, '88')
.replace(/P/g, 'pp')
.replace(/p/g, '99')
.replace(/Q/g, 'qq')
.replace(/q/g, 'aa')
.replace(/R/g, 'rr')
.replace(/r/g, 'bb')
.replace(/S/g, 'ss')
.replace(/s/g, 'cc')
.replace(/T/g, 'tt')
.replace(/t/g, 'dd')
.replace(/U/g, 'uu')
.replace(/u/g, 'ee')
.replace(/V/g, 'vv')
.replace(/v/g, 'ff');
for (var i = 0; i < parseInt(compressed.length / 2); i++) {
// Determining the substring.
var substring = compressed.substr(i*2, 2);
// Determining the binValue of this hex substring.
var binValue = parseInt(substring, 16).toString(2);
// If the length of the binary value is not equal to 8 we add leading 0s (js deletes the leading 0s)
// For instance the binary number 00011110 is equal to the hex number 1e,
// but simply running the code above will return 11110. So we have to add the leading 0s back.
if (binValue.length != 8) {
// Determining how many 0s to add:
var diffrence = 8 - binValue.length;
// Adding the 0s:
for (var j = 0; j < diffrence; j++) {
binValue = '0'+binValue;
}
}
// Adding the binValue to the end string which we will return.
returnValue += binValue
}
var decompressedArr = returnValue.split('');
returnValue = ''; // Emptying the return variable.
// Deleting the not needed leading 0s used as padding.
for (var i = paddingAmount; i < decompressedArr.length; i++) {
returnValue += decompressedArr[i];
}
// Returning the decompressed string.
return returnValue;
}
URL shortener
I still found the "compressed" strings a little long for sharing / pasting around. So i used a simple URL shortener (view here) to make this process a little easier for the user.
Now you might ask, then why did you need to compress this string anyway?
Here's why:
First of all, my project is hosted on github pages (gh-pages). The info page of gh-pages tells us that the url can't be any longer than 2000 chars. This would mean that the max grid size would be the square root of 2000 - length of the base url, which isn't that big. By using this "compression" we are able to share much larger grids.
Now the second reason why is that, it's a challange. I find dealing with problems like these fun and also helpfull since you learn a lot.
Live
You can view the live version of my project here. and/or find the github repository here.
Thankyou
I want to thank everyone who helped me with this problem. Especially Corstian and Arnauld, since i ended up using their answers to reach my final functions.
Sooooo.... thanks guys! apriciate it!

In the Game of Life there is a board of ones and zeros. I want to back up to previous generation - size 4800 - save each 16 cells as hexadecimal = 1/4 the size. http://innerbeing.epizy.com/cwebgl/gameoflife.html [g = Go] [b = Backup]
function drawGen(n) {
stop(); var i = clamp(n,0,brw*brh-1), hex = gensave[i].toString();
echo(":",i, n,nGEN); nGEN = i; var str = '';
for (var i = 0; i < parseInt(hex.length / 4); i++)
str = str + pad(parseInt(hex.substr(i*4,4), 16).toString(2),16,'0');
for (var j=0;j<Board.length;j++) Board[j] = intr(str.substr(j,1));
drawBoard();
}
function Bin2Hex(n) {
var i = n.indexOf("1"); /// leading Zeros = NAN
if (i == -1) return "0000";
i = right(n,i*-1);
return pad(parseInt(i,2).toString(16),4,'0');
}
function saveGen(n) {
var b = Board.join(''), str = ''; /// concat array to string 10101
for (var i = 0; i < parseInt(b.length / 16); i++)
str = str + Bin2Hex(b.substr(i*16,16));
gensave[n] = str;
}
function right(st,n) {
var s = st.toString();
if (!n) return s;
if (n < 0) return s.substr(n * -1,s.length + n);
return s.substr(s.length - n,n);
}
function pad(str, l, padwith) {
var s = str;
while (s.length < l) s = padwith + s;
return s;
}

Eval alternative

This code works as a calculator, but the scratch pad at codeacademy tells me that eval is evil. Is there another way to do the same thing without using eval?
var calculate = prompt("Enter problem");
alert(eval(calculate));

eval evaluates the string input as JavaScript and coincidentally JavaScript supports calculations and understands 1+1, which makes it suitable as a calculator.
If you don't want to use eval, which is good, you have to parse that string yourself and, finally, do the computation yourself (not exactly yourself though). Have a look at this math processor, which does what you want.
Basically what you do is:
Read the input string char by char (with this kind of problem it's still possible)
Building a tree of actions you want to do
At the end of the string, you evaluate the tree and do some calculations
For example you have "1+2/3", this could evaluate to the following data structure:
"+"
/ \
"1" "/"
/ \
"2" "3"
You could then traverse that structure from top to bottom and do the computations.
At first you've got the "+", which has a 1 on the left side and some expression on the right side,
so you have to evaluate that expression first. So you go to the "/" node, which has two numeric children. Knowing that, you can now compute 2/3 and replace the whole "/" node with the result of that. Now you can go up again and compute the result of the "+" node: 1 + 0.66. Now you replace that node with the result and all you've got left is the result of the expression.
Some pseudo code on how this might look in your code:
calculation(operator, leftValue, rightValue):
switch operator {
case '+': return leftValue + rightValue
case '-': return 42
}
action(node):
node.value = calculation(node.operator, action(node.left) action(node.right))
As you might have noticed, the tree is designed in such a way that it honors operator precedence. The / has a lower level than the +, which means it get's evaluated first.
However you do this in detail, that's basically the way to go.

You can use the expression parser that is included in the math.js library:
http://mathjs.org
Example usage:
mathjs.evaluate('1.2 / (2.3 + 0.7)'); // 0.4
mathjs.evaluate('5.08 cm in inch'); // 2 inch
mathjs.evaluate('sin(45 deg) ^ 2'); // 0.5
mathjs.evaluate('9 / 3 + 2i'); // 3 + 2i
mathjs.evaluate('det([-1, 2; 3, 1])'); // -7

You can use eval safely for a simple arithmetic calculator by filtering the input- if you only accept digits, decimal points and operators (+,-,*,/) you won't get in much trouble. If you want advanced Math functions, you are better off with the parser suggestions.
function calculate(){
"use strict";
var s= prompt('Enter problem');
if(/[^0-9()*+\/ .-]+/.test(s)) throw Error('bad input...');
try{
var ans= eval(s);
}
catch(er){
alert(er.message);
}
alert(ans);
}
calculate()

I write some functions when I had a problem like this. Maybe this can help:
data = [
{id:1,val1:"test",val2:"test2",val2:"test3"},
{id:2,val1:"test",val2:"test2",val2:"test3"},
{id:3,val1:"test",val2:"test2",val2:"test3"}
];
datakey = Object.keys(data[0]);
// here's a fix for e['datakey[f]'] >> e[x]
vix = function(e,f){
a = "string";
e[a] = datakey[f];
x = e.string;
end = e[x];
delete e.string;
return end;
};
// here's a fix to define that variable
vox = function(e,f,string){
a = "string";
e[a] = datakey[f];
x = e.string;
end = e[x] = string;
delete e.string;
};
row = 2 // 3th row ==> {id:3,val1:"test",val2:"test2",val2:"test3"}
column = 1 //datakey 2 ==> val1
vox(data[row],column,"new value");
alert(data[2].val1); //the value that we have changed

javascript parseFloat '500,000' returns 500 when I need 500000

How would it be a nice way of handling this?
I already thought on removing the comma and then parsing to float.
Do you know a better/cleaner way?
Thanks

parseFloat( theString.replace(/,/g,'') );

I don't know why no one has suggested this expression-
parseFloat( theString.replace(/[^\d\.]/g,'') );
Removes any non-numeric characters except for periods. You don't need custom functions/loops for this either, that's just overkill.

Nope. Remove the comma.

You can use the string replace method, but not in a one liner as a regexp allows.
while(str.indexOf(',')!=-1)str= str.replace(',','');
parseFloat(str);
Or to make a single expression without a regexp=
return parseFloat(str.split(',').join(''));
I'd use the regexp.

I don't have enough reputation to add a comment, but for anyone wondering on the performance for regex vs split/join, here's a quick fiddle: https://jsfiddle.net/uh3mmgru/
var test = "1,123,214.19";
var t0 = performance.now();
for (var i = 0; i < 1000000; i++)
{
var a = parseFloat(test.replace(/,/g,''));
}
var t1 = performance.now();
document.write('Regex took: ' + (t1 - t0) + ' ms');
document.write('<br>')
var t0 = performance.now();
for (var i = 0; i < 1000000; i++)
{
var b = parseFloat(test.split(',').join(''));
}
var t1 = performance.now();
document.write('Split/join took: ' + (t1 - t0) + ' ms');
The results I get are (for 1 million loops each):
Regex: 263.335 ms
Split/join: 1035.875 ms
So I think its safe to say that regex is the way to go in this scenario

Building on the idea from #kennebec, if you want to make sure that the commas are correct, and you don't want to replace commas, you could try something like this:
function myParse(num) {
var n2 = num.split(",")
out = 0
for(var i = 0; i < n2.length; i++) {
out *= 1000;
out += parseFloat(n2[i])
}
return out
}
alert(myParse("1,432,85"));
// Returns 1432085, as the comma is misplaced.
It may not be as fast, but you wanted alternatives :)

What about a simple function to solve most of the common problems?
function getValue(obj) {
Value = parseFloat( $(obj).val().replace(/,/g,'') ).toFixed(2);
return +Value;
}
The above function gets values from fields (using jQuery) assuming the entered values are numeric (I rather validate fields while user is entering data, so I know for sure field content is numeric).
In case of floating point values, if well formatted in the field, the function will return a float point value correctly.
This function is far from complete, but it quickly fix the "," (comma) issue for values entered as 1,234.56 or 1,234,567. It will return valid number as far the content is numeric.
The + (plus) sign in front of the variable Value in the return command is a "dirty trick" used in JavaScript to assure the variable content returned will be numeric.
it is easy to modify the function to other purposes, such as (for instance), convert strings to numeric values taking care of the "," (comma) issue:
function parseValue(str) {
Value = parseFloat( str.replace(/,/g,'') ).toFixed(2);
return +Value;
}
Both operations can even be combined in one function. I.e.:
function parseNumber(item,isField=false) {
Value = (isField) ? parseFloat( $(item).val().replace(/,/g,'') ).toFixed(2) : parseFloat( item.replace(/,/g,'') ).toFixed(2)
return +Value;
}
In such case, if function is called result = parseNumber('12,092.98'); it will parse the value as it is a String. But if called as result = parseNumber('#MyField', true); it will try to obtain the value from '#MyField'.
As I said before, such functions are far from complete, and can be expanded in many ways. One idea is to check the first character of the given parameter (string) and decide based on the string format where to obtain the value to be parsed (if 1st character is = '#' then it is an ID from a DOM object, otherwise, if it begins with a number, it must be a string to be parsed).
Try it... Happy coding.