在 Python 中检查数字是否不在范围内
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/36507957/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Checking if a number is not in range in Python
提问by What's in a Google Search
Ok, so I have Python code at the moment which does something like this:
好的，所以我现在有 Python 代码，它执行以下操作：
if plug in range(1, 5):
print "The number spider has disappeared down the plughole"
But what I actually want to do is check if the number is notin range. I've googled and had a look at the Python docs but I can't find anything. Any ideas?
但我真正想做的是检查数字是否不在范围内。我用谷歌搜索并查看了 Python 文档，但我找不到任何东西。有任何想法吗？
Additional data: When running this code:
附加数据：运行此代码时：
if not plug in range(1, 5):
print "The number spider has disappeared down the plughole"
I get the following error:
我收到以下错误：
Traceback (most recent call last):
File "python", line 33, in <module>
IndexError: list assignment index out of range
I also tried:
我也试过：
if plug not in range(1,5):
print "The number spider has disappeared down the plughole"
Which returned the same error.
返回相同的错误。
回答by Markus Meskanen
If your range has a stepof one, it's performance-wise much faster to use:
如果您的范围step为 1，则使用它的性能要快得多：
if not 1 <= plug < 5:
Than it would be to use the notmethod suggested by others:
比使用not其他人建议的方法：
if plug not in range(1, 5)
Proof:
证明：
>>> import timeit
>>> timeit.timeit('1 <= plug < 5', setup='plug=3') # plug in range
0.053391717400628654
>>> timeit.timeit('1 <= plug < 5', setup='plug=12') # plug not in range
0.05137874743129345
>>> timeit.timeit('plug not in r', setup='plug=3; r=range(1, 5)') # plug in range
0.11037584743321105
>>> timeit.timeit('plug not in r', setup='plug=12; r=range(1, 5)') # plug not in range
0.05579263413291358
And this is not even taking into account the time spent on creating the range.
这甚至没有考虑到创建range.
回答by Ebrahim Byagowi
This seems work as well:
这似乎也有效：
if not 2 < 3 < 4:
print('3 is not between 2 and 4') # which it is, and you will not see this
if not 2 < 10 < 4:
print('10 is not between 2 and 4')
Exact answer to the original question would be if not 1 <= plug < 5:I guess
if not 1 <= plug < 5:我猜原始问题的确切答案是
回答by Kunal Joshi
Use:
用：
if plug not in range(1,5):
print "The number spider has disappeared down the plughole"
It will print given line whenever variable plug is out of range 1 to 5
每当变量插头超出范围 1 到 5 时，它将打印给定的行
回答by Yana Sosnovskaya
if (int(5.5) not in range(int(3.0), int(6.9))):
print('False')
else:
print('True')
value should type casting in integer, otherwise not in rangegives strange result.
value 应该在整数中进行类型转换，否则not in range会产生奇怪的结果。
回答by jusx
if not plug in range(1,5):
#bla
