使用 isinstance 比较 boolean 和 int

2022-01-14 00:00:00 python comparison boolean int isinstance

问题描述

有人能给我解释一下为什么 isinstance() 在以下情况下返回 True 吗?在编写代码时,我期望 False.

Can someone give me an explanation why isinstance() returns True in the following case? I expected False, when writing the code.

print isinstance(True, (float, int))
True

我的猜测是它的 Python 的内部子类化,零和一 - 无论是浮点数还是整数 - 都在用作布尔值时进行评估,但不知道确切的原因.

My guess would be that its Python's internal subclassing, as zero and one - whether float or int - both evaluate when used as boolean, but don't know the exact reason.

解决这种情况的最pythonic方法是什么?我可以使用 type() 但在大多数情况下,这被认为不那么 Pythonic.

What would be the most pythonic way to solve such a situation? I could use type() but in most cases this is considered less pythonic.


解决方案

由于历史原因,boolint 的子类,所以 Trueint 的一个实例.(最初,Python 没有 bool 类型,返回真值的东西返回 1 或 0.当他们添加 bool 时,True 和 False 必须尽可能地替换 1 和 0 以实现向后兼容性,因此是子类化.)

For historic reasons, bool is a subclass of int, so True is an instance of int. (Originally, Python had no bool type, and things that returned truth values returned 1 or 0. When they added bool, True and False had to be drop-in replacements for 1 and 0 as much as possible for backward compatibility, hence the subclassing.)

解决"这个问题的正确方法取决于您认为问题是什么.

The correct way to "solve" this depends on exactly what you consider the problem to be.

  • 如果你想让 True 不再是 int,那太糟糕了.这不会发生.
  • 如果您想检测布尔值并以不同于其他整数的方式处理它们,您可以这样做:

  • If you want True to stop being an int, well, too bad. That's not going to happen.
  • If you want to detect booleans and handle them differently from other ints, you can do that:

if isinstance(whatever, bool):
    # special handling
elif isinstance(whatever, (float, int)):
    # other handling

  • 如果你想检测特定类正好是floatint的对象,拒绝子类,你可以这样做:

  • If you want to detect objects whose specific class is exactly float or int, rejecting subclasses, you can do that:

    if type(whatever) in (float, int):
        # Do stuff.
    

  • 如果您想检测所有浮点数和整数,您已经在这样做了.
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