# Determine whether integer is between two other integers

## Question:

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`)?

``````if number >= 10000 and number <= 30000:
print ("you have to pay 5% taxes")
``````

Use `if number >= 10000 and number <= 30000:`. Alternately, Python has a shorthand for this sort of thing, `if 10000 <= number <= 30000:`.

``````if 10000 <= number <= 30000:
pass
``````

For details, see the docs.

To check that the number is in the range 10000 – 30000, 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
``````

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, 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
``````

Define the range between the numbers:

``````r = range(1,10)
``````

Then use it:

``````if num in r:
print("All right!")
``````

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')
4 COMPARE_OP               0 (<)
6 JUMP_IF_FALSE_OR_POP    14
12 COMPARE_OP               0 (<)
>>   14 RETURN_VALUE
>>> dis.dis('a < b < c')
4 DUP_TOP
6 ROT_THREE
8 COMPARE_OP               0 (<)
10 JUMP_IF_FALSE_OR_POP    18
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.

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
``````

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")
``````

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.

Try this simple function; it checks if `A` is between `B` and `C` (`B` and `C` may not be in the right order):

``````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)`.

You used >=30000, so if number is 45000 it will go into the loop, but we need it to be more than 10000 but less than 30000. Changing it to <=30000 will do it!

While `10 <= number <= 20` works in Python, I find this notation using `range()` more readable:

``````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:

``````>>> list(range(10, 21))
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
``````

However prefer the `range()` approach only if it’s not running on some performance critical path. A single call is still fast enough for most requirements, but if run 10,000,000 times, we clearly notice nearly 3 times slower performance compared to `a <= x < b`:

``````> { time python3 -c "for i in range(10000000): x = 50 in range(1, 100)"; } 2>&1 | sed -n 's/^.*cpu (.*) total\$/1/p'
1.848

> { time python3 -c "for i in range(10000000): x = 1 <= 50 < 100"; } 2>&1 | sed -n 's/^.*cpu (.*) total\$/1/p'
0.630
``````

Below are few possible ways, ordered from best to worse performance (i.e first one will perform best)

``````     # Old school check
if 10000 >= b and b <=30000:
print ("you have to pay 5% taxes")
# Python range check
if 10000 <= number <= 30000:
print ("you have to pay 5% taxes")
# As suggested by others but only works for integers and is slow
if number in range(10000,30001):
print ("you have to pay 5% taxes")
``````

I’m adding a solution that nobody mentioned yet, using Interval class from sympy library:

``````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")
``````
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