[英]Determine whether integer is between two other integers
How do I determine whether a given integer is between two other integers (eg greater than/equal to 10000
and less than/equal to 30000
)?如何确定给定整数是否介于其他两个整数之间(例如大于/等于
10000
和小于/等于30000
)?
What I've attempted so far is not working:到目前为止我所尝试的是行不通的:
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
>>> r = range(1, 4)
>>> 1 in r
True
>>> 2 in r
True
>>> 3 in r
True
>>> 4 in r
False
>>> 5 in r
False
>>> 0 in r
False
Your operator is incorrect.您的运营商不正确。 Should be
if number >= 10000 and number <= 30000:
.应该是
if number >= 10000 and number <= 30000:
。 Additionally, Python has a shorthand for this sort of thing, if 10000 <= number <= 30000:
.此外,Python 对这类事情有一个简写,
if 10000 <= number <= 30000:
。
Your code snippet,你的代码片段,
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
actually checks if number is larger than both 10000 and 30000.实际上检查数字是否大于 10000 和 30000。
Assuming you want to check that the number is in the range 10000 - 30000, you could use the Python interval comparison:假设您要检查数字是否在 10000 - 30000 范围内,您可以使用 Python 间隔比较:
if 10000 <= number <= 30000:
print ("you have to pay 5% taxes")
This Python feature is further described in the Python documentation . Python 文档 中进一步描述了此 Python 功能。
There are two ways to compare three integers and check whether b is between a and c :有两种方法可以比较三个整数并检查b是否在a和c之间:
if a < b < c:
pass
and和
if a < b and b < c:
pass
The first one looks like more readable, but the second one runs faster .第一个看起来更具可读性,但第二个运行得更快。
Let's compare using dis.dis :让我们使用dis.dis进行比较:
>>> dis.dis('a < b and b < c')
1 0 LOAD_NAME 0 (a)
2 LOAD_NAME 1 (b)
4 COMPARE_OP 0 (<)
6 JUMP_IF_FALSE_OR_POP 14
8 LOAD_NAME 1 (b)
10 LOAD_NAME 2 (c)
12 COMPARE_OP 0 (<)
>> 14 RETURN_VALUE
>>> dis.dis('a < b < c')
1 0 LOAD_NAME 0 (a)
2 LOAD_NAME 1 (b)
4 DUP_TOP
6 ROT_THREE
8 COMPARE_OP 0 (<)
10 JUMP_IF_FALSE_OR_POP 18
12 LOAD_NAME 2 (c)
14 COMPARE_OP 0 (<)
16 RETURN_VALUE
>> 18 ROT_TWO
20 POP_TOP
22 RETURN_VALUE
>>>
and using timeit :并使用timeit :
~$ python3 -m timeit "1 < 2 and 2 < 3"
10000000 loops, best of 3: 0.0366 usec per loop
~$ python3 -m timeit "1 < 2 < 3"
10000000 loops, best of 3: 0.0396 usec per loop
also, you may use range , as suggested before, however it is much more slower.此外,您可以使用range ,如前所述,但它要慢得多。
Define the range between the numbers:定义数字之间的范围:
r = range(1,10)
Then use it:然后使用它:
if num in r:
print("All right!")
if number >= 10000 and number <= 30000:
print ("you have to pay 5% taxes")
The trouble with comparisons is that they can be difficult to debug when you put a >=
where there should be a <=
比较的麻烦在于,当您将
>=
放在应该有<=
位置时,它们可能难以调试
# v---------- should be <
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
Python lets you just write what you mean in words Python 让你只用文字表达你的意思
if number in xrange(10000, 30001): # ok you have to remember 30000 + 1 here :)
In Python3, you need to use range
instead of xrange
.在 Python3 中,您需要使用
range
而不是xrange
。
edit: People seem to be more concerned with microbench marks and how cool chaining operations.编辑:人们似乎更关心微基准测试和链接操作有多酷。 My answer is about defensive (less attack surface for bugs) programming.
我的答案是关于防御性(漏洞的攻击面较少)编程。
As a result of a claim in the comments, I've added the micro benchmark here for Python3.5.2由于评论中的声明,我在此处为 Python3.5.2 添加了微基准测试
$ python3.5 -m timeit "5 in range(10000, 30000)"
1000000 loops, best of 3: 0.266 usec per loop
$ python3.5 -m timeit "10000 <= 5 < 30000"
10000000 loops, best of 3: 0.0327 usec per loop
If you are worried about performance, you could compute the range once如果您担心性能,您可以计算一次范围
$ python3.5 -m timeit -s "R=range(10000, 30000)" "5 in R"
10000000 loops, best of 3: 0.0551 usec per loop
Suppose there are 3 non-negative integers: a
, b
, and c
.假设有 3 个非负整数:
a
、 b
和c
。 Mathematically speaking, if we want to determine if c
is between a
and b
, inclusively, one can use this formula:从数学上讲,如果我们要确定
c
是否介于a
和b
之间,可以使用以下公式:
(c - a) * (b - c) >= 0
(c - a) * (b - c) >= 0
or in Python:或在 Python 中:
> print((c - a) * (b - c) >= 0)
True
I'm adding a solution that nobody mentioned yet, using Interval class from sympy library:我添加了一个尚未提及的解决方案,使用 sympy 库中的 Interval 类:
from sympy import Interval
lower_value, higher_value = 10000, 30000
number = 20000
# to decide whether your interval shhould be open or closed use left_open and right_open
interval = Interval(lower_value, higher_value, left_open=False, right_open=False)
if interval.contains(number):
print("you have to pay 5% taxes")
You want the output to print the given statement if and only if the number falls between 10,000 and 30,000.当且仅当数字介于 10,000 和 30,000 之间时,您希望输出打印给定的语句。
Code should be;代码应该是;
if number >= 10000 and number <= 30000:
print("you have to pay 5% taxes")
您使用了 >=30000,因此如果 number 为 45000,它将进入循环,但我们需要它大于 10000 但小于 30000。将其更改为 <=30000 即可!
While 10 <= number <= 20
works in Python, I find this notation using range()
more readable:虽然
10 <= number <= 20
在 Python 中有效,但我发现使用range()
这种表示法更具可读性:
if number in range(10, 21):
print("number is between 10 (inclusive) and 21 (exclusive)")
else:
print("outside of range!")
Keep in mind that the 2nd, upper bound parameter is not included in the range set as can be verified with:请记住,第二个上限参数不包含在范围集中,可以通过以下方式进行验证:
>>> list(range(10, 21))
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Below are few possible ways, ordered from best to worse performance (ie first one will perform best)以下是几种可能的方法,按性能从好到坏排序(即第一个性能最好)
if 10000 <= b and b <=30000:
print ("you have to pay 5% taxes")
if 10000 <= number <= 30000:
print ("you have to pay 5% taxes")
if number in range(10000,30001):
print ("you have to pay 5% taxes")
How do I determine whether a given integer is between two other integers (eg greater than/equal to 10000
and less than/equal to 30000
)?如何确定给定的整数是否介于两个其他整数之间(例如,大于/等于
10000
且小于/等于30000
)?
I'm using 2.3 IDLE and what I've attempted so far is not working:我正在使用 2.3 IDLE 并且到目前为止我尝试过的不起作用:
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
Try this simple function;试试这个简单的功能; it checks if
A
is between B
and C
( B
and C
may not be in the right order):它检查
A
是否在B
和C
之间( B
和C
的顺序可能不正确):
def isBetween(A, B, C):
Mi = min(B, C)
Ma = max(B, C)
return Mi <= A <= Ma
so isBetween(2, 10, -1)
is the same as isBetween(2, -1, 10)
.所以
isBetween(2, 10, -1)
与isBetween(2, -1, 10)
。
The condition should be,条件应该是,
if number == 10000 and number <= 30000:
print("5% tax payable")
reason for using number == 10000
is that if number's value is 50000 and if we use number >= 10000
the condition will pass, which is not what you want.使用
number == 10000
原因是如果 number 的值为 50000 并且如果我们使用number >= 10000
条件将通过,这不是您想要的。
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