处理 Javascript 中的浮点精度
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Dealing with float precision in Javascript
提问by Jeroen
I have a large amount of numeric values yin javascript. I want to group them by rounding them down to the nearest multiple of xand convert the result to a string.
我y在 javascript 中有大量的数值。我想通过将它们四舍五入到最接近的倍数来对它们进行分组x，并将结果转换为字符串。
How do I get around the annoying floating point precision?
如何解决烦人的浮点精度？
For example:
例如：
0.2 + 0.4 = 0.6000000000000001
Two things I have tried:
我尝试过的两件事：
>>> y = 1.23456789
>>> x = 0.2
>>> parseInt(Math.round(Math.floor(y/x))) * x;
1.2000000000000002
and:
和：
>>> y = 1.23456789
>>> x = 0.2
>>> y - (y % x)
1.2000000000000002
回答by Rusty Fausak
From this post: How to deal with floating point number precision in JavaScript?
来自这篇文章：如何处理 JavaScript 中的浮点数精度？
You have a few options:
您有几个选择：
- 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
- 使用特殊的小数数据类型，如decimal.js
- 将结果格式化为一些固定数量的有效数字，如下所示：
(Math.floor(y/x) * x).toFixed(2) - 将所有数字转换为整数
回答by philipvr
You could do something like this:
你可以这样做：
> +(Math.floor(y/x)*x).toFixed(15);
1.2
回答by Hamza Alayed
> 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:
快速解决方案：
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.
您可以在此处查看完整说明。
回答by prahaladp
Check out this link.. It helped me a lot.
看看这个链接..它帮了我很多。
http://www.w3schools.com/jsref/jsref_toprecision.asp
http://www.w3schools.com/jsref/jsref_toprecision.asp
The toPrecision(no_of_digits_required)function returns a stringso don't forget to use the parseFloat()function to convert to decimal point of required precision.
该toPrecision(no_of_digits_required)函数返回一个string所以不要忘记使用该parseFloat()函数转换为所需精度的小数点。
回答by whaatt
Tackling this task, I'd first find the number of decimal places in x, then round yaccordingly. I'd use:
处理这个任务时，我会先找到 中的小数位数x，然后y相应地四舍五入。我会用：
y.toFixed(x.toString().split(".")[1].length);
It should convert xto a string, split it over the decimal point, find the length of the right part, and then y.toFixed(length)should round ybased on that length.
它应该转换x为一个字符串，将其拆分为小数点，找到正确部分的长度，然后根据该长度进行y.toFixed(length)舍入y。
