Determine whether integer is between two other integers
PythonPython Problem Overview
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)?
What I've attempted so far is not working:
if number >= 10000 and number >= 30000:
print ("you have to pay 5% taxes")
Python Solutions
Solution 1 - Python
if 10000 <= number <= 30000:
pass
For details, see the docs.
Solution 2 - Python
>>> 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
Solution 3 - Python
Your operator is incorrect. It should be if number >= 10000 and number <= 30000:. Additionally, Python has a shorthand for this sort of thing, if 10000 <= number <= 30000:.
Solution 4 - Python
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.
Solution 5 - Python
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.
Solution 6 - Python
if number >= 10000 and number <= 30000:
print ("you have to pay 5% taxes")
Solution 7 - Python
Define the range between the numbers:
r = range(1,10)
Then use it:
if num in r:
print("All right!")
Solution 8 - Python
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
Solution 9 - Python
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
Solution 10 - Python
Below are few possible ways, ordered from best to worse performance (i.e first one will perform best)
if 10000 <= b and b <=30000:
print ("you have to pay 5% taxes")
if 10000 <= number <= 30000:
print ("you have to pay 5% taxes")
if number in range(10000,30001):
print ("you have to pay 5% taxes")
Solution 11 - Python
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
Solution 12 - 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")
Solution 13 - Python
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!
Solution 14 - Python
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).
Solution 15 - Python
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.