Python ().is_integer() 不工作
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().is_integer() not working
提问by aaaa
Whats wrong with this code:
这段代码有什么问题：
n = 10
((n/3)).is_integer()
I do not understand why I cannot set n = any number and check if it is an integer or not.
我不明白为什么我不能设置 n = 任何数字并检查它是否是整数。
Thanks for your help!
谢谢你的帮助！
python 2.7.4
蟒蛇 2.7.4
error:
错误：
Traceback (most recent call last):
File "/home/userh/Arbeitsfl?che/übung.py", line 2, in <module>
print ((n/3)).is_integer()
AttributeError: 'int' object has no attribute 'is_integer'
回答by Ohlin
...
...
When I wrote this answer there was no information about language.
当我写这个答案时，没有关于语言的信息。
But in python2 you can use the following to check if it's an integer or not
但是在python2中，您可以使用以下内容来检查它是否是整数
isinstance( <var>, ( int, long ) )
回答by Morwenn
You are using Python 2.7. Unless you use from __future__ import division, dividing two integers will return you and integer. is_integerexists only in float, hence your error.
您正在使用 Python 2.7。除非您使用from __future__ import division，否则将两个整数相除将返回您和整数。is_integer仅存在于 中float，因此您的错误。
回答by Sebastian ?rleryd
The reason you get this error is because you divide the integer 10 by 3 using integer division, getting the integral number 3 in the form of an intinstance as a result. You then try to call the method is_integer()on that result but that method is in the floatclass and not in the intclass, just as the error message says.
出现此错误的原因是因为您使用整数除法将整数 10 除以 3，int结果以实例的形式得到整数 3 。然后，您尝试is_integer()对该结果调用该方法，但该方法在float类中而不是在int类中，正如错误消息所述。
A quick fix would be to change your code and divide by 3.0instead of 3which would result in floating point division and give you a floatinstance on which you can call the is_integer()method like you are trying to. Do this:
一个快速的解决方法是更改您的代码并除以除法，3.0而不是3将导致浮点除法，并为您提供一个float实例，您可以在该实例上is_integer()像尝试一样调用该方法。做这个：
n = 10
((n/3.0)).is_integer()
回答by andrew cooke
the other answers say this but aren't very clear (imho).
其他答案是这样说的，但不是很清楚（恕我直言）。
in python 2, the /sign means "integer division" when the arguments are integers. that gives you just the integer part of the result:
在 python 2 中，当参数为 integers 时，/符号表示“整数除法” 。只给你结果的整数部分：
>>> 10/3
3
which means that in (10/3).is_integer()you are calling is_integer()on 3, which is an integer. and that doesn't work:
这意味着(10/3).is_integer()您正在调用is_integer()上3，这是一个整数。这不起作用：
>>> (3.0).is_integer()
True
>>> (3).is_integer()
AttributeError: 'int' object has no attribute 'is_integer'
what you probably want is to change one of the numbers to a float:
您可能想要的是将其中一个数字更改为浮点数：
>>> (10/3.0).is_integer()
False
this is fixed in python 3, by the way (which is the future, and a nicer language in many small ways).
顺便说一句，这是在python 3中修复的（这是未来，并且在许多小方面是一种更好的语言）。
回答by Ravikiran Reddy Kotapati
You can use isdigit, this is a good function provided by Python itself
可以使用isdigit，这是Python本身提供的一个很好的函数
You can refer documentation here https://docs.python.org/2/library/stdtypes.html#str.isdigit
您可以在此处参考文档https://docs.python.org/2/library/stdtypes.html#str.isdigit
if token.isdigit():
return int(token)
