问题
I am trying to truncate decimal numbers to decimal places. Something like this:
5.467 -> 5.46
985.943 -> 985.94
toFixed(2) does just about the right thing but it rounds off the value. I don\'t need the value rounded off. Hope this is possible in javascript.
回答1:
upd:
So, after all it turned out, rounding bugs will always haunt you, no matter how hard you try to compensate them. Hence the problem should be attacked by representing numbers exactly in decimal notation.
Number.prototype.toFixedDown = function(digits) {
var re = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)"),
m = this.toString().match(re);
return m ? parseFloat(m[1]) : this.valueOf();
};
[ 5.467.toFixedDown(2),
985.943.toFixedDown(2),
17.56.toFixedDown(2),
(0).toFixedDown(1),
1.11.toFixedDown(1) + 22];
// [5.46, 985.94, 17.56, 0, 23.1]
Old error-prone solution based on compilation of others':
Number.prototype.toFixedDown = function(digits) {
var n = this - Math.pow(10, -digits)/2;
n += n / Math.pow(2, 53); // added 1360765523: 17.56.toFixedDown(2) === "17.56"
return n.toFixed(digits);
}
回答2:
Dogbert's answer is good, but if your code might have to deal with negative numbers, Math.floor by itself may give unexpected results.
E.g. Math.floor(4.3) = 4, but Math.floor(-4.3) = -5
Use a helper function like this one instead to get consistent results:
truncateDecimals = function (number) {
return Math[number < 0 ? 'ceil' : 'floor'](number);
};
// Applied to Dogbert's answer:
var a = 5.467;
var truncated = truncateDecimals(a * 100) / 100; // = 5.46
Here's a more convenient version of this function:
truncateDecimals = function (number, digits) {
var multiplier = Math.pow(10, digits),
adjustedNum = number * multiplier,
truncatedNum = Math[adjustedNum < 0 ? 'ceil' : 'floor'](adjustedNum);
return truncatedNum / multiplier;
};
// Usage:
var a = 5.467;
var truncated = truncateDecimals(a, 2); // = 5.46
// Negative digits:
var b = 4235.24;
var truncated = truncateDecimals(b, -2); // = 4200
If that isn't desired behaviour, insert a call to Math.abs on the first line:
var multiplier = Math.pow(10, Math.abs(digits)),
EDIT: shendz correctly points out that using this solution with a = 17.56 will incorrectly produce 17.55. For more about why this happens, read What Every Computer Scientist Should Know About Floating-Point Arithmetic. Unfortunately, writing a solution that eliminates all sources of floating-point error is pretty tricky with javascript. In another language you'd use integers or maybe a Decimal type, but with javascript...
This solution should be 100% accurate, but it will also be slower:
function truncateDecimals (num, digits) {
var numS = num.toString(),
decPos = numS.indexOf('.'),
substrLength = decPos == -1 ? numS.length : 1 + decPos + digits,
trimmedResult = numS.substr(0, substrLength),
finalResult = isNaN(trimmedResult) ? 0 : trimmedResult;
return parseFloat(finalResult);
}
For those who need speed but also want to avoid floating-point errors, try something like BigDecimal.js. You can find other javascript BigDecimal libraries in this SO question: "Is there a good Javascript BigDecimal library?" and here's a good blog post about math libraries for Javascript
回答3:
var a = 5.467;
var truncated = Math.floor(a * 100) / 100; // = 5.46
回答4:
You can fix the rounding by subtracting 0.5 for toFixed, e.g.
(f - 0.005).toFixed(2)
回答5:
Consider taking advantage of the double tilde: ~~.
Take in the number. Multiply by significant digits after the decimal so that you can truncate to zero places with ~~. Divide that multiplier back out. Profit.
function truncator(numToTruncate, intDecimalPlaces) {
var numPower = Math.pow(10, intDecimalPlaces); // "numPowerConverter" might be better
return ~~(numToTruncate * numPower)/numPower;
}
I'm trying to resist wrapping the ~~ call in parens; order of operations should make that work correctly, I believe.
alert(truncator(5.1231231, 1)); // is 5.1
alert(truncator(-5.73, 1)); // is -5.7
alert(truncator(-5.73, 0)); // is -5
JSFiddle link.
EDIT: Looking back over, I've unintentionally also handled cases to round off left of the decimal as well.
alert(truncator(4343.123, -2)); // gives 4300.
The logic's a little wacky looking for that usage, and may benefit from a quick refactor. But it still works. Better lucky than good.
回答6:
I thought I'd throw in an answer using | since it is simple and works well.
truncate = function(number, places) {
var shift = Math.pow(10, places);
return ((number * shift) | 0) / shift;
};
回答7:
Nice one-line solution:
function truncate (num, places) {
return Math.trunc(num * Math.pow(10, places)) / Math.pow(10, places);
}
Then call it with:
truncate(3.5636232, 2); // returns 3.56
truncate(5.4332312, 3); // returns 5.433
truncate(25.463214, 4); // returns 25.4632
回答8:
@Dogbert's answer can be improved with Math.trunc, which truncates instead of rounding.
There is a difference between rounding and truncating. Truncating is clearly the behaviour this question is seeking. If I call truncate(-3.14) and receive -4 back, I would definitely call that undesirable. – @NickKnowlson
var a = 5.467;
var truncated = Math.trunc(a * 100) / 100; // = 5.46
var a = -5.467;
var truncated = Math.trunc(a * 100) / 100; // = -5.46
回答9:
Truncate using bitwise operators:
~~0.5 === 0
~~(-0.5) === 0
~~14.32794823 === 14
~~(-439.93) === -439
回答10:
I found a problem: considering the next situation: 2.1 or 1.2 or -6.4
What if you want always 3 decimals or two or wharever, so, you have to complete the leading zeros to the right
// 3 decimals numbers
0.5 => 0.500
// 6 decimals
0.1 => 0.10000
// 4 decimales
-2.1 => -2.1000
// truncate to 3 decimals
3.11568 => 3.115
This is the fixed function of Nick Knowlson
function truncateDecimals (num, digits)
{
var numS = num.toString();
var decPos = numS.indexOf('.');
var substrLength = decPos == -1 ? numS.length : 1 + decPos + digits;
var trimmedResult = numS.substr(0, substrLength);
var finalResult = isNaN(trimmedResult) ? 0 : trimmedResult;
// adds leading zeros to the right
if (decPos != -1){
var s = trimmedResult+"";
decPos = s.indexOf('.');
var decLength = s.length - decPos;
while (decLength <= digits){
s = s + "0";
decPos = s.indexOf('.');
decLength = s.length - decPos;
substrLength = decPos == -1 ? s.length : 1 + decPos + digits;
};
finalResult = s;
}
return finalResult;
};
https://jsfiddle.net/huttn155/7/
回答11:
function toFixed(number, digits) {
var reg_ex = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)")
var array = number.toString().match(reg_ex);
return array ? parseFloat(array[1]) : number.valueOf()
}
var test = 10.123456789
var __fixed = toFixed(test, 6)
console.log(__fixed)
// => 10.123456
回答12:
I think this function could be a simple solution:
function trunc(decimal,n=2){
let x = decimal + ''; // string
return x.lastIndexOf('.')>=0?parseFloat(x.substr(0,x.lastIndexOf('.')+(n+1))):decimal; // You can use indexOf() instead of lastIndexOf()
}
console.log(trunc(-241.31234,2));
console.log(trunc(241.312,5));
console.log(trunc(-241.233));
console.log(trunc(241));回答13:
The one that is mark as the solution is the better solution I been found until today, but has a serious problem with 0 (for example, 0.toFixedDown(2) gives -0.01). So I suggest to use this:
Number.prototype.toFixedDown = function(digits) {
if(this == 0) {
return 0;
}
var n = this - Math.pow(10, -digits)/2;
n += n / Math.pow(2, 53); // added 1360765523: 17.56.toFixedDown(2) === "17.56"
return n.toFixed(digits);
}
回答14:
Here is simple but working function to truncate number upto 2 decimal places.
function truncateNumber(num) {
var num1 = "";
var num2 = "";
var num1 = num.split('.')[0];
num2 = num.split('.')[1];
var decimalNum = num2.substring(0, 2);
var strNum = num1 +"."+ decimalNum;
var finalNum = parseFloat(strNum);
return finalNum;
}
回答15:
Number.prototype.trim = function(decimals) {
var s = this.toString();
var d = s.split(".");
d[1] = d[1].substring(0, decimals);
return parseFloat(d.join("."));
}
console.log((5.676).trim(2)); //logs 5.67
回答16:
The resulting type remains a number...
/* Return the truncation of n wrt base */
var trunc = function(n, base) {
n = (n / base) | 0;
return base * n;
};
var t = trunc(5.467, 0.01);
回答17:
The answer by @kirilloid seems to be the correct answer, however, the main code needs to be updated. His solution doesn't take care of negative numbers (which someone did mention in the comment section but has not been updated in the main code).
Updating that to a complete final tested solution:
Number.prototype.toFixedDown = function(digits) {
var re = new RegExp("([-]*\\d+\\.\\d{" + digits + "})(\\d)"),
m = this.toString().match(re);
return m ? parseFloat(m[1]) : this.valueOf();
};
Sample Usage:
var x = 3.1415629;
Logger.log(x.toFixedDown(2)); //or use whatever you use to log
Fiddle: JS Number Round down
PS: Not enough repo to comment on that solution.
回答18:
Lodash has a few Math utility methods that can round, floor, and ceil a number to a given decimal precision. This leaves off trailing zeroes.
They take an interesting approach, using the exponent of a number. Apparently this avoids rounding issues.
(Note: func is Math.round or ceil or floor in the code below)
// Shift with exponential notation to avoid floating-point issues.
var pair = (toString(number) + 'e').split('e'),
value = func(pair[0] + 'e' + (+pair[1] + precision));
pair = (toString(value) + 'e').split('e');
return +(pair[0] + 'e' + (+pair[1] - precision));
Link to the source code
回答19:
I wrote an answer using a shorter method. Here is what I came up with
function truncate(value, precision) {
var step = Math.pow(10, precision || 0);
var temp = Math.trunc(step * value);
return temp / step;
}
The method can be used like so
truncate(132456.25456789, 5)); // Output: 132456.25456
truncate(132456.25456789, 3)); // Output: 132456.254
truncate(132456.25456789, 1)); // Output: 132456.2
truncate(132456.25456789)); // Output: 132456
Or, if you want a shorter syntax, here you go
function truncate(v, p) {
var s = Math.pow(10, p || 0);
return Math.trunc(s * v) / s;
}
回答20:
Here my take on the subject:
convert.truncate = function(value, decimals) {
decimals = (decimals === undefined ? 0 : decimals);
return parseFloat((value-(0.5/Math.pow(10, decimals))).toFixed(decimals),10);
};
It's just a slightly more elaborate version of
(f - 0.005).toFixed(2)
回答21:
just to point out a simple solution that worked for me
convert it to string and then regex it...
var number = 123.45678;
var number_s = '' + number;
var number_truncated_s = number_s.match(/\d*\.\d{4}/)[0]
var number_truncated = parseFloat(number_truncated_s)
It can be abbreviated to
var number_truncated = parseFloat(('' + 123.4568908).match(/\d*\.\d{4}/)[0])
回答22:
Here is what I use:
var t = 1;
for (var i = 0; i < decimalPrecision; i++)
t = t * 10;
var f = parseFloat(value);
return (Math.floor(f * t)) / t;
回答23:
Here is an ES6 code which does what you want
const truncateTo = (unRouned, nrOfDecimals = 2) => {
const parts = String(unRouned).split(".");
if (parts.length !== 2) {
// without any decimal part
return unRouned;
}
const newDecimals = parts[1].slice(0, nrOfDecimals),
newString = `${parts[0]}.${newDecimals}`;
return Number(newString);
};
// your examples
console.log(truncateTo(5.467)); // ---> 5.46
console.log(truncateTo(985.943)); // ---> 985.94
// other examples
console.log(truncateTo(5)); // ---> 5
console.log(truncateTo(-5)); // ---> -5
console.log(truncateTo(-985.943)); // ---> -985.94回答24:
Number.prototype.truncate = function(places) {
var shift = Math.pow(10, places);
return Math.trunc(this * shift) / shift;
};
来源：https://stackoverflow.com/questions/4912788/truncate-not-round-off-decimal-numbers-in-javascript