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)?

Answer 1

if 10000 <= number <= 30000:
    pass

For details, see the docs.

Answer 2

>>> 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

Answer 3

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

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

Define the range between the numbers:

r = range(1,10)

Then use it:

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

Answer 8

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

Answer 9

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")

Answer 10

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

Answer 11

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

Answer 12

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")

Answer 13

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")

Answer 14

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!

Answer 15

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).

Answer 16

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.