I recently came up with this code while answering another StackOverflow question. Basically, on blur, this code will properly comma separate by thousands and leave the decimal at two digits (like how USD is written [7,745.56]).
I was wondering if there is more concise way of using regex to , separate and cut off excessive decimal places. I recently updated this post with my most recent attempt. Is there a better way of doing this with regex?
Input -> Target Output
7456 -> 7,456
45345 -> 45,345
25.23523534 -> 25.23
3333.239 -> 3,333.23
234.99 -> 234.99
2300.99 -> 2,300.99
23123123123.22 -> 23,123,123,123.22
Current Regex
var result;
var str = []
reg = new RegExp(/(\d*(\d{2}\.)|\d{1,3})/, "gi");
reversed = "9515321312.2323432".split("").reverse().join("")
while (result = reg.exec(reversed)) {
str.push(result[2] ? result[2] : result[0])
}
console.log(str.join(",").split("").reverse().join("").replace(",.","."))
As an alternative to the Regex, you could use the following approach
Number(num.toFixed(2)).toLocaleString('en-US')
or
num.toLocaleString('en-US', {maximumFractionDigits: 2})
You would still have the toFixed(2), but it's quite clean. toFixed(2) though won't floor the number like you want. Same with {maximumFractionDigits: 2} as the second parameter to toLocaleString as well.
var nums = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22]
for (var num of nums)
console.log(num, '->', Number(num.toFixed(2)).toLocaleString('en-US') )
Flooring the number like you showed is a bit tricky. Doing something like (num * 100 | 0) / 100 does not work. The calculation loses precision (e.g. .99 will become .98 in certain situations). (also |0 wouldn't work with larger numbers but even Math.floor() has the precision problem).
The solution would be to treat the numbers like strings.
function format(num) {
var num = num.toLocaleString('en-US')
var end = num.indexOf('.') < 0 ? num.length : num.indexOf('.') + 3
return num.substring(0, end)
}
var nums = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22]
for (var num of nums) console.log(num, '->', format(num))
function format(num) {
var num = num.toLocaleString('en-US')
var end = num.indexOf('.') < 0 ? num.length : num.indexOf('.') + 3
return num.substring(0, end)
}
(when changing to another format than 'en-US' pay attention to the . in numbers as some languages use a , as fractal separator)
For Compatibility, according to CanIUse toLocaleString('en-US') is
supported in effectively all browsers (since IE6+, Firefox 2+, Chrome
1+ etc)
If you really insist on doing this purely in regex (and truncate instead of round the fractional digits), the only solution I can think of is to use a replacement function as the second argument to .replace():
('' + num).replace(
/(\d)(?=(?:\d{3})+(?:\.|$))|(\.\d\d?)\d*$/g,
function(m, s1, s2){
return s2 || (s1 + ',');
}
);
This makes all your test cases pass:
function format(num){
return ('' + num).replace(
/(\d)(?=(?:\d{3})+(?:\.|$))|(\.\d\d?)\d*$/g,
function(m, s1, s2){
return s2 || (s1 + ',');
}
);
}
test(7456, "7,456");
test(45345, "45,345");
test(25.23523534, "25.23"); //truncated, not rounded
test(3333.239, "3,333.23"); //truncated, not rounded
test(234.99, "234.99");
test(2300.99, "2,300.99");
test(23123123123.22, "23,123,123,123.22");
function test(num, expected){
var actual = format(num);
console.log(num + ' -> ' + expected + ' => ' + actual + ': ' +
(actual === expected ? 'passed' : 'failed')
);
}
I added another layer where regex that drops the unwanted decimals below hundredths on top of your regex comma adding logic;
val.replace(/(\.\d{2})\d*/, "$1").replace(/(\d)(?=(\d{3})+\b)/g, "$1,")
doIt("7456");
doIt("45345");
doIt("25.23523534");
doIt("3333.239");
doIt("234.99");
doIt("2300.99");
doIt("23123123123.22");
doIt("5812090285.2817481974897");
function doIt(val) {
console.log(val + " -> " + val.replace(/(\.\d{2})\d*/, "$1").replace(/(\d)(?=(\d{3})+\b)/g, "$1,"));
}
If multiple calls of regex replace is OK, this answer should satisfy you, since it is only has regex replace logic and nothing else.
Try:
var n = 5812090285.2817481974897;
n = n.toFixed(2).replace(/(\d)(?=(\d{3})+\.)/g, '$1,');
console.log(n);
Outputs:
5,812,090,285.28
Note: .toFixed(2) returns a string. So in order to simplify this further you must add a way to turn n into a string before executing your regex. For example:
n.toString.replace(/(\d)(?=(\d{3})+\.)/g, '$1,'); //ofc with the additional regex
Although you would think it wouldn't matter in javascript, it apparently does in this situation. So I dont know how much 'less' messy it would be to not use.
Here is a way to do it without a regular expression:
value.toLocaleString("en-US", { maximumFractionDigits: 2 })
function formatValue() {
var source = document.getElementById("source");
var output = document.getElementById("output");
var value = parseFloat(source.value);
output.innerText = value.toLocaleString("en-US", { maximumFractionDigits: 2 });
}
<input id="source" type="text" />
<button onclick="formatValue()">Format</button>
<div id="output"></div>
RegEx to rescue again!
My solution has two parts :
.toFixed : Used to limit the decimal limit
/(\d)(?=(\d\d\d)+(?!\d))/g : It makes use of back reference with three digits at a time
Here's everything put together :
// .toFixed((/\./g.test(num)) ? 2 : 0) it tests if the input number has any decimal places, if so limits it to 2 digits and if not, get's rid of it altogether by setting it to 0
num.toFixed((/\./g.test(num)) ? 2 : 0).replace(/(\d)(?=(\d\d\d)+(?!\d))/g, "$1,"))
You can see it in action here :
var input = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22]
input.forEach(function(num) {
$('div')
.append(
$('<p>').text(num + ' => ' +
num.toFixed( (/\./g.test(num))?2:0 ).replace(/(\d)(?=(\d\d\d)+(?!\d))/g, "$1,"))
);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div> </div>
NOTE: I've only used jQuery to append the results
You can do like this
(parseFloat(num).toFixed(2)).replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1,").replace(".00","")
Here just convert number to formatted number with rounded down to 2 decimal places and then remove the .00 if exist.
This can be one approach you can use.
var format = function (num) {
return (parseFloat(num).toFixed(2)).replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1,").replace(".00","")
}
$(function () {
$("#principalAmtOut").blur(function (e) {
$(this).val(format($(this).val()));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="principalAmtOut" type="text" />
You can use Intl.NumberFormat with style set to "decimal" and maximumFractionDigits set to 2 at options object passed at second parameter
const nums = [7456, 45345, 25.23523534, 3333.239, 234.99, 2300.99, 23123123123.22];
const formatOptions = {style:"decimal", maximumFractionDigits:2};
const formatter = new Intl.NumberFormat("en-US", formatOptions);
const formatNums = num => formatter.format(num);
let formattedNums = nums.map(formatNums);
console.log(formattedNums);
I found a solution based on #Pierre's answer without using of toFixed:
function format(n) {
n = +n;
var d = Math.round(n * 100) % 100;
return (Math.floor(n) + '').replace(/(\d)(?=(\d{3})+$)/g, '$1,') + (d > 9 ? '.' + d : d > 0 ? '.0' + d : '');
}
console.log(format(7456));
console.log(format(7456.0));
console.log(format(7456.1));
console.log(format(7456.01));
console.log(format(7456.001));
console.log(format(45345));
console.log(format(25.23523534));
console.log(format(3333.239));
console.log(format(234.99));
console.log(format(2300.99));
console.log(format(23123123123.22));
console.log(format('23123123123.22'));
Related
I am using nerdamer.solve to solve roots but the roots are long and not truncated. I wanted to get truncated values upto 4 decimal places. How can I achieve this?
I am using the following code to solve and display output in html:
var r = nerdamer.solve(`1 - ${a} * x^(${p}) + ${b}`, 'x');
document.getElementById("polesAns").innerHTML= r.toString();
The folllowing is output:
[(138655807/135201312)*i+49385501/48155102,(-138655807/135201312)*i+49385501/48155102,(58886197/57419096)*i-49385501/48155102,(-58886197/57419096)*i-49385501/48155102,-560373381/386371730,172668482/119053157,(145619303/100403024)*i-5753750945848186/10000000000000000000000000000000,(-560373381/386371730)*i-5753750945848186/10000000000000000000000000000000]
There is no division performed also.
I tried the solution posted here:
How to use .toFixed() (or any alternative) on Nerdamer.solve solutions?
But how can I do this with my code? I tried the following:
var value = `1 - ${a} * x^(${p}) + ${b}`;
var toFixed = function(value, n) {
var img = Number(nerdamer.imagpart(value).text()).toFixed(n);
var real = Number(nerdamer.realpart(value).text()).toFixed(n);
// Format the number assuming i denotes imaginary in your case
var formatted = '';
if(real !== '0.0000') {
formatted += real;
}
if(img !== '0.0000') {
// Put the plus sign betweent the real and imaginary
if(img.charAt(0) !== '-' && formatted) {
formatted += '+';
}
// Assuming you're using i and not j for instance
formatted += img+'i';
}
return formatted;
};
sol_raw = this.nerdamer.solve(value,'s');
xs = this.nerdamer(sol_raw.toString()).each(function(solution) {
roundedSolutions.push(toFixed(solution, 4));
});
this.setState({
solution: roundedSolution.join(''),
equation:value})
document.getElementById("polesAns").value = solution.toString();
I don't understand the this.setState() part , should i declare sol_raw and xs as var?
Also the substitution of variable is used in the my above root equation from advice here javascript Solving equation with subsitution of variable value
thank you
Hi guys I have bytes say 007458415820874584158208042423283712.I want to convert this into GB, so tried to divide it by 1048576 i am getting a result of 7.112899609446129e+27. I want only the two numbers after the decimal point, so I have used .toFixed like below. It doesn't work, I am getting the same response as if I have not used the toFixed function. I just want the result to be just 7.1. help me out on this.
console.log((007458415820874584158208042423283712/1048576).toFixed(2));
You can use this prototype function for your solution.
Number.prototype.toFixedSpecial = function(n) {
var str = this.toFixed(n);
if (str.indexOf('e+') === -1)
return str;
// if number is in scientific notation, pick (b)ase and (p)ower
str = str.replace('.', '').split('e+').reduce(function(p, b) {
return p + Array(b - p.length + 2).join(0);
});
if (n > 0)
str += '.' + Array(n + 1).join(0);
return str;
};
var val = (007458415820874584158208042423283712/1048576);
console.log(val);
console.log(val.toFixedSpecial(2)) //"7112899609446129000000000000.00"
console.log( 1e21.toFixedSpecial(2) ); // "1000000000000000000000.00"
console.log( 2.1e24.toFixedSpecial(0) ); // "2100000000000000000000000"
console.log( 1234567..toFixedSpecial(1) ); // "1234567.0"
console.log( 1234567.89.toFixedSpecial(3) ); // "1234567.890"
Your problem is that this is scientific notation and toFixed() supports 20 decimal places. Your number is 7.112899609446129e+27 which technically (most likely) has decimal places but they are not visible due to scientific notation.
The solution would be to use toExponential() like so:
parseFloat((7458415820874584158208042423283712/1048576.0).toExponential(2))
Output:
7.11e+27
A more correct way is shown here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toExponential
But this gives "7.11e+27" (a string)
If you just want 7.11 then you can use slice(0,3) as follows:
var result_str = (7458415820874584158208042423283712/1048576).toExponential(2);
console.log(parseFloat(result_str.slice(0,3)));
Result: 7.1
i found the following code at http://rosettacode.org for the Vigenère cipher and i would like to undestand it better.
Could someone explain me what the single lines of code in function ordA(a) and in function(a) do?
function ordA(a) {
return a.charCodeAt(0) - 65;
}
// vigenere
function vigenere2(text, key, decode) {
var i = 0, b;
key = key.toUpperCase().replace(/[^A-Z]/g, '');
return text.toUpperCase().replace(/[^A-Z]/g, '').replace(/[A-Z]/g, function(a) {
b = key[i++ % key.length];
return String.fromCharCode(((ordA(a) + (decode ? 26 - ordA(b) : ordA(b))) % 26 + 65));
});
}
I'm not sure if that is supposed to be example code, but it mainly shows how not to program. Smart decisions are being made, but obviously the problem decomposition, variable naming and documentation leave a lot to be desired. Repeated code, convoluted lines, unexplained code fragments, the list goes on. Decode is a boolean, but the opposite of encryption is decryption not decoding. This code was made to not understand what is going on; what it does on the Rosetta site is mind-boggling in that respect.
returns an index in the English alphabet or ABC, assuming uppercase characters, 0 to 25 instead of 1 to 26 (because you can do modular calculations with zero indexing, not with one based indexing)
return a.charCodeAt(0) - 65;
function definition that takes a plaintext or ciphertext, a key which may be smaller than the plaintext and a Boolean to indicate if we're encoding or decoding
function vigenere2(text, key, decode)
index in plaintext and variable b, which will hold a character of the key for the index
var i = 0, b;
converts the key to uppercase and removed all characters not in the uppercase alphabet as well
key = key.toUpperCase().replace(/[^A-Z]/g, '');
this line is too long obviously; it converts the text to uppercase and removes the non-alphabet characters again
then it replaces the characters in the string using the function defined in the second argument of replace
return text.toUpperCase().replace(/[^A-Z]/g, '').replace(/[A-Z]/g, function(a) {
take the next character of the key in round robin fashion, using the modulus operator, update the index afterwards
b = key[i++ % key.length];
too much going on here, very bad program decomposition; in order of execution:
(decode ? 26 - ordA(b) : ordA(b)): calculate a number in the range to update the index of the plaintext character; use the opposite value for decryption (wrongly called "decoding" here)
(ordA(a) + (decode ? 26 - ordA(b) : ordA(b))) % 26 perform the addition with the calculated number, reduce to 0 to 25 (i.e. when reaching Z continue with A and vice versa)
((ordA(a) + (decode ? 26 - ordA(b) : ordA(b))) % 26 + 65) add 65 so the index is converted back into the ASCII index of uppercase characters, using two completely spurious parentheses
finally, returns a string from one character code result, otherwise + will be addition instead of concatenation
return String.fromCharCode(((ordA(a) + (decode ? 26 - ordA(b) : ordA(b))) % 26 + 65));
well, it needed to end
});
}
Let's show another way of programming this, using well named variables, functions for reused code and regular expressions that badly need a name to explain what they do.
var ALPHABET_SIZE = 'Z'.charCodeAt(0) - 'A'.charCodeAt(0) + 1;
var encrypted = vigenere(false, "B", "Zaphod Breeblebox");
document.body.append('<div>' + encrypted + '</div>');
var decrypted = vigenere(true, "B", encrypted);
document.body.append('<div>' + decrypted + '</div>');
function vigenere(decrypt, key, text) {
key = toJustUppercase(key);
text = toJustUppercase(text);
var textOffset = 0;
// iterate over all characters, performing the function on each of them
return text.replace(/[A-Z]/g, function(textChar) {
var keyChar = key[textOffset++ % key.length];
var cryptedChar = substituteCharacter(decrypt, keyChar, textChar);
return cryptedChar;
});
}
function substituteCharacter(decrypt, keyChar, textChar) {
var keyIndex = charToABCIndex(keyChar);
if (decrypt) {
// create the opposite of the encryption key index
keyIndex = ALPHABET_SIZE - keyIndex;
}
var textIndex = charToABCIndex(textChar);
// the actual Vigenere substitution, the rest is just indexing and conversion
var substitutedIndex = (textIndex + keyIndex) % ALPHABET_SIZE;
var substitutedChar = abcIndexToChar(substitutedIndex);
return substitutedChar;
}
function toJustUppercase(text) {
return text.toUpperCase().replace(/[^A-Z]/g, '')
}
function charToABCIndex(charValue) {
return charValue.charCodeAt(0) - 'A'.charCodeAt(0);
}
function abcIndexToChar(index) {
return String.fromCharCode(index + 'A'.charCodeAt(0));
}
Too many functions you say? Not really, I've not implemented ord and chr, or vigenereEncrypt and viginereDecrypt to make it even easier to read.
I'm facing an issue with Math.floor function of javascript for the below scenario:
1) from the value betwwen 8192 and 10484,
if I type 8192.8 -> The Math.floor converts it into 8192.79
if I type 8192.88 -> The Math.floor converts it into 8192.87
if I type 8192.3 -> The Math.floor converts it into 8192.29
Strange part is that except from the range given above the function works fine.
HTML:
<div data-bind="text: popIncrease"></div>
<input type="text" data-bind="value: userInput, valueUpdate: 'afterkeydown'" />
Javascript:
var ViewModel = function () {
var _self = this;
_self.userInput = ko.observable();
_self.popIncrease = ko.computed(function () {
return parseFloat((Math.floor(_self.userInput() * 100) / 100)).toFixed(2);
});
};
ko.applyBindings(new ViewModel());
jsfiddle:https://jsfiddle.net/91z5bdy4/1/
When I changed 100 with 1000 it solved the error but I do not understand why this happened on the first place?
You can just switch to this:
return parseFloat(_self.userInput()).toFixed(2);
Working version of your jsFiddle: https://jsfiddle.net/jfriend00/5rLL04Lk/
Or, if you want to work around some of the idiosyncrasies of .toFixed(), you can use this:
return (Math.round(_self.userInput() * 100) / 100).toFixed(2);
Working jsFiddle: https://jsfiddle.net/jfriend00/xx2aj2L0/
This solution passes all three of your test cases.
It's not Math.floor() that causes the problem, it is the inexactness of the floating point arithmetic. When you multiply 8192.8 by 100, you get 819279.9999999999.
Perhaps you should just manipulate it as a string:
function floorString(str) {
var pos = str.indexOf('.');
return (pos >= 0) ? ((str + '00').slice(0, pos + 3)) : (str + '.00');
}
jsfiddle
The order of your floor/parse seems out of order to me.
Try:
return Math.floor(parseFloat(_self.userInput())).toFixed(2);
Though be aware that 1.999999999999999999999999999999 gives 2.00 using the above; this is because floating point numbers aren't able to represent all values precisely.
another one without using a Math function (2 lines without formatting)
function floorString(str) {
var matches = str.match(/([\d]+(\.[\d]{0,2}))/);
return matches === null || matches[2].length === 1 ?
(str + ".00").replace("..", ".") :
matches[2].length < 3 ?
matches[0] + "00".substr(3 - matches[2].length) :
matches[0];
}
I have a string that looks like "(3) New stuff" where 3 can be any number.
I would like to add or subtract to this number.
I figured out the following way:
var thenumber = string.match((/\d+/));
thenumber++;
string = string.replace(/\(\d+\)/ ,'('+ thenumber +')');
Is there a more elegant way to do it?
Another way:
string = string.replace(/\((\d+)\)/ , function($0, $1) { return "(" + (parseInt($1, 10) + 1) + ")"; });
I believe Gumbo was on the right track
"(42) plus (1)".replace(/\((\d+)\)/g, function(a,n){ return "("+ (+n+1) +")"; });
Short of extending the String object, it looks good to me.
String.prototype.incrementNumber = function () {
var thenumber = string.match((/\d+/));
thenumber++;
return this.replace(/\(\d+\)/ ,'('+ thenumber +')');
}
Usage would then be:
alert("(2) New Stuff".incrementNumber());
I believe your method is the best elegant you can have for following reasons:
since the input is not a "clean" number, you do need to involve some sort of string parser. Using regular expressions is the very code-efficient method to do it
by looking at the code, it's clear what it does
short of wrapping this into a function, I don't think there's much more to be done
As galets says, I don't think your solution is a bad one but here is a function that will add a specified value to a number in a specified position in a string.
var str = "fluff (3) stringy 9 and 14 other things";
function stringIncrement( str, inc, start ) {
start = start || 0;
var count = 0;
return str.replace( /(\d+)/g, function() {
if( count++ == start ) {
return(
arguments[0]
.substr( RegExp.lastIndex )
.replace( /\d+/, parseInt(arguments[1])+inc )
);
} else {
return arguments[0];
}
})
}
// fluff (6) stringy 9 and 14 other things :: 3 is added to the first number
alert( stringIncrement(str, 3, 0) );
// fluff (3) stringy 6 and 14 other things :: -3 is added to the second number
alert( stringIncrement(str, -3, 1) );
// fluff (3) stringy 9 and 24 other things :: 10 is added to the third number
alert( stringIncrement(str, 10, 2) );