为什么Python的节目`0.2 + 0.1`为'0.30000000000000004`？ [英] Why does Python show `0.2 + 0.1` as `0.30000000000000004`?

问题描述

我写了下面的code用于生成与花车范围：

I have written the following code for generating a range with floats:

def drange(start, stop, step):
    result = []
    value = start
    while value <= stop:
        result.append(value)
        value += step
    return result

当调用本声明此功能：

print drange(0.1,1.0,0.1)

我希望获得此：

[0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

不过，我得到以下，而不是：

But I obtain the following, instead:

[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7, 0.7999999999999999, 0.8999999999999999, 0.9999999999999999]

这是为什么？我该如何解决？

Why is this?, and how can I fix it?

谢谢！

推荐答案

这是如何浮点数的工作。不能重新present实数的无限数目在一个有限数目的位，所以有一些截短。你应该看一看 什么每个程序员应该知道关于浮点运算 ：

That's how floating-point numbers work. You can't represent an infinite number of real numbers in a finite number of bits, so there is some truncation. You should take a look at What Every Programmer Should Know About Floating-Point Arithmetic:

为什么不是我的号码，像0.1 + 0.2加起来一个漂亮的圆0.3，而是我得到一个怪异的结果像0.30000000000000004？

由于内部，电脑使用的格式（二进制浮点）不能准确地重新present一个像0.1，0.2或0.3的。

Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.

当code编译或间preTED，你的0.1已经四舍五入到该格式的最接近的数字，甚至计算发生之前导致小的舍入误差。

When the code is compiled or interpreted, your "0.1" is already rounded to the nearest number in that format, which results in a small rounding error even before the calculation happens.

使用圆（周数，k）小数点后四舍五入给定的浮点值 K 数字（所以你的情况下，使用轮（数字，1）为一个数字）。

Use round(number, k) to round a given floating-point value to k digits after the decimal (so in your case, use round(number, 1) for one digit).

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