[英]How to check if a number is in a interval
Suppose I got: 假设我得到了:
first_var = 1
second_var = 5
interval = 2
I want an interval from second_var like second_var ± interval
(from 3 to 7). 我想要second_var的间隔,如
second_var ± interval
(从3到7)。 I wank to check if first_var is in that interval. 我想知道first_var是否在那段时间内。 So in this specific case I want
False
If first_var = 4
, I want True
所以在这种特殊情况下我想要
False
如果first_var = 4
,我想要True
I can do this: 我可以做这个:
if (first_var > second_var-interval) and (first_var < second_var+interval):
#True
Is there a more pythonic way to do this? 是否有更多的pythonic方式来做到这一点?
You can use math-like sequence as Python supports that 您可以使用类似数学的序列,因为Python支持它
if (second_var-interval < first_var < second_var+interval):
# True
Note that comments in python begin with a #
请注意,python中的注释以
#
开头
I use a class with __contains__
to represent the interval: 我使用带
__contains__
的类来表示间隔:
class Interval(object):
def __init__(self, middle, deviation):
self.lower = middle - abs(deviation)
self.upper = middle + abs(deviation)
def __contains__(self, item):
return self.lower <= item <= self.upper
Then I define a function interval
to simplify the syntax: 然后我定义一个函数
interval
来简化语法:
def interval(middle, deviation):
return Interval(middle, deviation)
Then we can call it as follows: 然后我们可以这样称呼它:
>>> 8 in interval(middle=6, deviation=2)
True
>>> 8 in interval(middle=6, deviation=1)
False
With Python 2 this solution is more efficient than using range
or xrange
as they don't implement __contains__
and they have to search for a matching value. 使用Python 2,这个解决方案比使用
range
或xrange
更有效,因为它们没有实现__contains__
并且它们必须搜索匹配的值。
Python 3 is smarter and range
is a generating object which is efficient like xrange
, but also implements __contains__
so it doesn't have to search for a valid value. Python 3更智能,
range
是一个生成对象,它像xrange
一样高效,但也实现__contains__
因此它不必搜索有效值。 xrange
doesn't exist in Python 3. Python 3中不存在
xrange
。
This solution also works with floats. 此解决方案也适用于浮动。
Also, note, if you use range
you need to be careful of off-by-1 errors. 另外,请注意,如果使用
range
,则需要注意1分之一的错误。 Better to encapsulate it, if you're going to be doing it more than once or twice. 最好将其封装起来,如果你要做的不止一次或两次。
You can use lambdas: 你可以使用lambdas:
lmd = lambda fv, sv, inval: print('True') if \
sv - inval < fv < sv + inval else print('False')
and use it like: 并使用它像:
lmd(first_var, second_var, interval)
but it's a little bit long! 但它有点长!
if (first_var in range(second_var-interval, second_var+interval+1)):
#Do something
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.