Dealing with float precision in Javascript [duplic

问题:

This question already has an answer here:

• How to deal with floating point number precision in JavaScript? 33 answers

I have a large amount of numeric values y in javascript. I want to group them by rounding them down to the nearest multiple of x and convert the result to a string.

How do I get around the annoying floating point precision?

For example:

0.2 + 0.4 = 0.6000000000000001

Two things I have tried:

>>> y = 1.23456789 
>>> x = 0.2 
>>> parseInt(Math.round(Math.floor(y/x))) * x; 
1.2000000000000002

and:

>>> y = 1.23456789 
>>> x = 0.2 
>>> y - (y % x)
1.2000000000000002

回答1:

From this post: How to deal with floating point number precision in JavaScript?

You have a few options:

• Use a special datatype for decimals, like decimal.js
• Format your result to some fixed number of significant digits, like this: (Math.floor(y/x) * x).toFixed(2)
• Convert all your numbers to integers

回答2:

> var x = 0.1
> var y = 0.2
> var cf = 10
> x * y
0.020000000000000004
> (x * cf) * (y * cf) / (cf * cf)
0.02

Quick solution:

var _cf = (function() {
  function _shift(x) {
    var parts = x.toString().split(\'.\');
    return (parts.length < 2) ? 1 : Math.pow(10, parts[1].length);
  }
  return function() { 
    return Array.prototype.reduce.call(arguments, function (prev, next) { return prev === undefined || next === undefined ? undefined : Math.max(prev, _shift (next)); }, -Infinity);
  };
})();

Math.a = function () {
  var f = _cf.apply(null, arguments); if(f === undefined) return undefined;
  function cb(x, y, i, o) { return x + f * y; }
  return Array.prototype.reduce.call(arguments, cb, 0) / f;
};

Math.s = function (l,r) { var f = _cf(l,r); return (l * f - r * f) / f; };

Math.m = function () {
  var f = _cf.apply(null, arguments);
  function cb(x, y, i, o) { return (x*f) * (y*f) / (f * f); }
  return Array.prototype.reduce.call(arguments, cb, 1);
};

Math.d = function (l,r) { var f = _cf(l,r); return (l * f) / (r * f); };

> Math.m(0.1, 0.2)
0.02

You can check the full explanation here.

回答3:

You could do something like this:

> +(Math.floor(y/x)*x).toFixed(15);
1.2

回答4:

Tackling this task, I\'d first find the number of decimal places in x, then round y accordingly. I\'d use:

y.toFixed(x.toString().split(\".\")[1].length);

It should convert x to a string, split it over the decimal point, find the length of the right part, and then y.toFixed(length) should round y based on that length.

回答5:

Check out this link.... It helped me a lot

http://www.w3schools.com/jsref/jsref_toprecision.asp

the toPrecision(no_of_digits_required) function returns a string so don\'t forge to use the parseFloat() function to convert to decimal point of req precision