为什么比较两个属性类型`float`和`int`与相同的值在Python中得到`False`？

2023年5月25日 138次阅读

参见英文答案 >
Is floating point math broken?                                    28个

我们来考虑下面的代码

码：

#!/usr/bin/env python

class Foo():
    def __init__(self, b):
        self.a = 0.0
        self.b = b
    def count_a(self):
        self.a += 0.1

foo = Foo(1)
for i in range(0, 15):
    foo.count_a()
    print "a =", foo.a, "b =", foo.b, '"a == b" ->', foo.a == foo.b

输出：

a = 0.2 b = 1 "a == b" -> False
a = 0.4 b = 1 "a == b" -> False
a = 0.6 b = 1 "a == b" -> False
a = 0.8 b = 1 "a == b" -> False
a = 1.0 b = 1 "a == b" -> True
a = 1.2 b = 1 "a == b" -> False
a = 1.4 b = 1 "a == b" -> False
a = 1.6 b = 1 "a == b" -> False
a = 1.8 b = 1 "a == b" -> False
a = 2.0 b = 1 "a == b" -> False
a = 2.2 b = 1 "a == b" -> False
a = 2.4 b = 1 "a == b" -> False
a = 2.6 b = 1 "a == b" -> False
a = 2.8 b = 1 "a == b" -> False
a = 3.0 b = 1 "a == b" -> False

但是,如果我将第11行的代码更改为foo = Foo(2),则输出将变为：

a = 0.2 b = 2 "a == b" -> False
a = 0.4 b = 2 "a == b" -> False
a = 0.6 b = 2 "a == b" -> False
a = 0.8 b = 2 "a == b" -> False
a = 1.0 b = 2 "a == b" -> False
a = 1.2 b = 2 "a == b" -> False
a = 1.4 b = 2 "a == b" -> False
a = 1.6 b = 2 "a == b" -> False
a = 1.8 b = 2 "a == b" -> False
a = 2.0 b = 2 "a == b" -> False *
a = 2.2 b = 2 "a == b" -> False
a = 2.4 b = 2 "a == b" -> False
a = 2.6 b = 2 "a == b" -> False
a = 2.8 b = 2 "a == b" -> False
a = 3.0 b = 2 "a == b" -> False

你会看到输出a = 2.0 b = 2“a == b” – >虚假是完全奇怪的.我想我可能会误解Python中的一些OOP概念.请向我解释为什么会发生意外输出以及如何解决这个问题.

最佳答案 这与对象方向无关 – 它与计算机在内部表示浮点数的方式以及舍入错误有关.

http://floating-point-gui.de/basic/

这里的Python特性是浮点数的默认字符串表示形式,它将在小于小写位置的位置进行舍入,而不是用于漂亮打印的内部表示.

虽然对于需要正确比较的人来说,尊重浮点数的规模,Python已经引入了一个很好的机制PEP 485,它将math.isclose函数添加到标准库中.

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