python – Determine Whether Integer Is Between Two Other Integers?

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 :
  • Check your boolean operators, of course a number will be greater than 10000 if it’s greater than 30000. Look at the little details and you will catch far more mistakes.
  • Comparisons can be chained docs.python.org/2/reference/expressions.html#comparisons
  • Pls change >= 30000 to <= 30000
  • The last edit made on this question is just putting “the solution” into the problem code. (makes the question somewhat invalid, defeats the purpose of this post I think.)

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.

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