四舍五入至显著位任意数目 [英] rounding to an arbitrary number of significant digits

问题描述

你怎么能轮的任意的数量（不只是整数> 0）到N显著的数字？

How can you round any number (not just integers > 0) to N significant digits?

例如，如果我要舍入到3显著的数字，我在寻找一个公式，可以采取：

For example, if I want to round to 3 significant digits, I'm looking for a formula that could take:

1239451，返回1,240,000

1,239,451 and return 1,240,000

12.1257并​​返回12.1

12.1257 and return 12.1

0.0681，并回到0.0681

.0681 and return .0681

5，并返回5

自然算法不应该是硬codeD只能处理3 N，尽管这将是一个开始。

Naturally the algorithm should not be hard-coded to only handle N of 3, although that would be a start.

推荐答案

下面是同样的code在Java中没有12.100000000000001错误其他答案有

Here's the same code in Java without the 12.100000000000001 bug other answers have

我也删除重复code，改变了动力来一个整数类型，以prevent浮动的问题时， N - ð完成，并取得了很长的中间更加清晰

I also removed repeated code, changed power to a type integer to prevent floating issues when n - d is done, and made the long intermediate more clear

的缺陷已通过用小数目乘以造成大量。相反，我把同样大小的两个数字。

The bug was caused by multiplying a large number with a small number. Instead I divide two numbers of similar size.

修改
修正了更多的错误。加入检查0，因为这将导致为NaN。所取得的实际功能与负数（原code不处理负数，因为日志的负数是复数）

EDIT
Fixed more bugs. Added check for 0 as it would result in NaN. Made the function actually work with negative numbers (The original code doesn't handle negative numbers because a log of a negative number is a complex number)

public static double roundToSignificantFigures(double num, int n) {
    if(num == 0) {
        return 0;
    }

    final double d = Math.ceil(Math.log10(num < 0 ? -num: num));
    final int power = n - (int) d;

    final double magnitude = Math.pow(10, power);
    final long shifted = Math.round(num*magnitude);
    return shifted/magnitude;
}

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