Ceil and floor equivalent in Python 3 without Math module?

Question

I need to ceil and floor 3/2 result (1.5) without using import math.

math.floor(3/2) => 3//2 math.ceil(3/2) => ?

OK, here is the problem: to sum all numbers 15 + 45 + 15 + 45 + 15 ... with N items.

sum = (n//2) * 5 + int(n/2) * 15

Solution 1

>>> 3/2
1.5
>>> 3//2 # floor
1
>>> -(-3//2) # ceil
2

Solution 2

Try

def ceil(n):
    return int(-1 * n // 1 * -1)

def floor(n):
    return int(n // 1)

I used int() to make the values integer. As ceiling and floor are a type of rounding, I thought integer is the appropriate type to return.

The integer division //, goes to the next whole number to the left on the number line. Therefore by using -1, I switch the direction around to get the ceiling, then use another * -1 to return to the original sign. The math is done from left to right.

Solution 3

Try:

def ceil(n):
    res = int(n)
    return res if res == n or n < 0 else res+1

def floor(n):
    res = int(n)
    return res if res == n or n >= 0 else res-1

Comments

[zooks]

wrong result for 4/2

[Ffisegydd]

Why not just use the math library?

[zooks]

for education purposes

[zooks]

Believe me, I tried. BTW I used PHP before, now I'm learning Python :)

[Mark Dickinson]

Use the ceiling division operator, --0--! This converts floor division to ceiling division. For example, --0-- 3//2 gives the ceiling of 3/2. Try it if you don't believe me! (Okay, so you could spell it without the leading --, but it looks better with it.)

[Mark Dickinson]

@Ffisegydd: There's at least one good reason to avoid the math library here, which is that by going via floating-point you can lose precision (and get the wrong answer as a result).

[Mark Dickinson]

@zooks: More seriously, use the fact that ceiling(x) == -floor(-x) for any real number x.

[user202729]

Does this answer your question? Is there a ceiling equivalent of // operator in Python?

[Mark Dickinson]

This floor definition is broken: it won't work correctly for negative integers.

[soshial]

funny that >>>math.ceil(4.0000000000000001/2) is 2, but >>>math.ceil(4.000000000000001/2) is 3

[FiddleStix]

But not surprising, when you consider that 4.0000000000000001 == 4.0 is True but 4.000000000000001 == 4.0 is False