浮点比较-为什么在Python中比较具有相同值的两个属性类型“float”和“int”会得到“False”

Question

让我们考虑下面的代码

代码：

#!/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11行上的代码更改为foo = Foo(2)，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

您将看到输出a = 2.0 b = 2 "a == b" -> False完全是怪异的。我想我可能误解了Python中OOP的一些概念。请向我解释为什么会出现这种意外输出，以及如何解决这个问题。

Answer 1

这与对象定向无关，而是与计算机内部表示浮点数的方式以及舍入误差有关。http://floating-point-gui.de/basic/

Python的特殊性是浮点数字的默认字符串表示，与内部表示相比，浮点数字将以更少的小数位数进行四舍五入，以进行漂亮的打印。

尽管对于需要正确比较的人来说，考虑到浮点数的小数位数，Python已经引入了一个很好的机制PEP485，它将math.isclose函数添加到了标准库中。

Answer 2

除了jsbueno的正确解释外，请记住Python通常允许将"基本类型"转换为自己。

即str（"a"）=="a"

因此，如果除此之外还需要一个变通方法，只需将int/foat混合转换为所有浮点并测试它们即可。

a = 2.0 
b = 2 
print "a == b", float(a) == float(b)

输出：

a == b True