问题描述
我无法将整数 1
添加到现有集合中.在交互式外壳中,这就是我正在做的事情:
I cannot add the integer number 1
to an existing set. In an interactive shell, this is what I am doing:
>>> st = {'a', True, 'Vanilla'}
>>> st
{'a', True, 'Vanilla'}
>>> st.add(1)
>>> st
{'a', True, 'Vanilla'} # Here's the problem; there's no 1, but anything else works
>>> st.add(2)
>>> st
{'a', True, 'Vanilla', 2}
这个问题是两个月前发布的,但我认为它被误解了.我正在使用 Python 3.2.3.
This question was posted two months ago, but I believe it was misunderstood. I am using Python 3.2.3.
推荐答案
>>> 1 == True
True
我相信你的问题是 1
和 True
是相同的值,所以 1 是已经在集合中".
I believe your problem is that 1
and True
are the same value, so 1 is "already in the set".
>>> st
{'a', True, 'Vanilla'}
>>> 1 in st
True
在数学运算中,True
本身被视为 1
:
In mathematical operations True
is itself treated as 1
:
>>> 5 + True
6
>>> True * 2
2
>>> 3. / (True + True)
1.5
虽然 True 是 bool 而 1 是 int:
Though True is a bool and 1 is an int:
>>> type(True)
<class 'bool'>
>>> type(1)
<class 'int'>
因为 1 in st
返回 True,我认为您应该没有任何问题.这是一个非常奇怪的结果.如果您有兴趣进一步阅读,@Lattyware 指向 PEP 285深入解释了这个问题.
Because 1 in st
returns True, I think you shouldn't have any problems with it. It is a very strange result though. If you're interested in further reading, @Lattyware points to PEP 285 which explains this issue in depth.
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