How do I determine whether a given integer is between two other integers (e.g. greater than/equal to 10000
and less than/equal to 30000
)?
I'm using 2.3 IDLE and what I've attempted so far is not working:
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
This question is related to
python
You want the output to print the given statement if and only if the number falls between 10,000 and 30,000.
Code should be;
if number >= 10000 and number <= 30000:
print("you have to pay 5% taxes")
Your code snippet,
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
actually checks if number is larger than both 10000 and 30000.
Assuming you want to check that the number is in the range 10000 - 30000, you could use the Python interval comparison:
if 10000 <= number <= 30000:
print ("you have to pay 5% taxes")
This Python feature is further described in the Python documentation.
>>> 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
The condition should be,
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.
There are two ways to compare three integers and check whether b is between a and c:
if a < b < c:
pass
and
if a < b and b < c:
pass
The first one looks like more readable, but the second one runs faster.
Let's compare using dis.dis:
>>> 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:
~$ 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.
Your operator is incorrect. Should be if number >= 10000 and number <= 30000:
. Additionally, Python has a shorthand for this sort of thing, if 10000 <= number <= 30000:
.
Define the range between the numbers:
r = range(1,10)
Then use it:
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 <=
# 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
if number in xrange(10000, 30001): # ok you have to remember 30000 + 1 here :)
In Python3, you need to use range
instead of xrange
.
edit: People seem to be more concerned with microbench marks and how cool chaining operations. My answer is about defensive (less attack surface for bugs) programming.
As a result of a claim in the comments, I've added the micro benchmark here for Python3.5.2
$ 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
$ 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
. Mathematically speaking, if we want to determine if c
is between a
and b
, inclusively, one can use this formula:
(c - a) * (b - c) >= 0
or in Python:
> print((c - a) * (b - c) >= 0)
True
Source: Stackoverflow.com