What is the difference between int() and floor() in Python 3?

Question

In Python 2, floor() returned a float value. Although not obvious to me, I found a few explanations clarifying why it may be useful to have floor() return float (for cases like float('inf') and float('nan')).

However, in Python 3, floor() returns integer (and returns overflow error for the special cases mentioned before).

So what is the difference, if any, between int() and floor() now?

Answer 1

floor() rounds down. int() truncates. The difference is clear when you use negative numbers:

>>> import math
>>> math.floor(-3.5)
-4
>>> int(-3.5)
-3

Rounding down on negative numbers means that they move away from 0, truncating moves them closer to 0.

Putting it differently, the floor() is always going to be lower or equal to the original. int() is going to be closer to zero or equal.

Answer 2

I tested the time complexity of both methods, they are the same.

from time import time
import math
import random

r = 10000000
def floorTimeFunction():
    for i in range(r):
        math.floor(random.randint(-100,100))

def intTimeFunction():
    for i in range(r):
        int(random.randint(-100,100))

t0 = time()
floorTimeFunction()
t1 = time()
intTimeFunction()
t2 = time()

print('function floor takes %f' %(t1-t0))
print('function int     takes %f' %(t2-t1))

Output is:

# function floor takes 11.841985
# function int   takes 11.841325