## The Question :

*457 people think this question is useful*

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")

*The Question Comments :*

## The Answer 1

*1193 people think this answer is useful*

if 10000 <= number <= 30000:
pass

For details, see the docs.

## The Answer 2

*88 people think this answer is useful*

>>> 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 Answer 3

*54 people think this answer is useful*

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:`

.

## The Answer 4

*30 people think this answer is useful*

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.

## The Answer 5

*11 people think this answer is useful*

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

## The Answer 6

*11 people think this answer is useful*

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.

## The Answer 7

*9 people think this answer is useful*

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

## The Answer 8

*9 people think this answer is useful*

Define the range between the numbers:

r = range(1,10)

Then use it:

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

## The Answer 9

*1 people think this answer is useful*

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

## The Answer 10

*0 people think this answer is useful*

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 Answer 11

*-3 people think this answer is useful*

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.