Taking the floor of a float

问题

I have found two ways of taking floors in Python:

3.1415 // 1

and

import math
math.floor(3.1415)

The problem with the first approach is that it return a float (namely 3.0). The second approach feels clumsy and too long.

Are there alternative solutions for taking floors in Python?

回答1:

As long as your numbers are positive, you can simply convert to an int to round down to the next integer:

>>> int(3.1415)
3

For negative integers, this will round up, though.

回答2:

You can call int() on the float to cast to the lower int (not obviously the floor but more elegant)

int(3.745)  #3

Alternatively call int on the floor result.

from math import floor

f1 = 3.1415
f2 = 3.7415

print floor(f1)       # 3.0
print int(floor(f1))  # 3
print int(f1)         # 3
print int(f2)         # 3 (some people may expect 4 here)
print int(floor(f2))  # 3

http://docs.python.org/library/functions.html#int

回答3:

The second approach is the way to go, but there's a way to shorten it.

from math import floor
floor(3.1415)

回答4:

Cast it to int if you don't want a float

int(3.1415 // 1)

回答5:

Beware that taking the floor and casting to an int are not the same thing with negative numbers. If you really want the floor as an integer, you should cast to an int after calling math.floor().

>>> int(-0.5)
0
>>> math.floor(-0.5)
-1.0
>>> int(math.floor(-0.5))
-1

回答6:

from math import floor


def ff(num, step=0):
    if not step:
        return floor(num)
    if step < 0:
        mplr = 10 ** (step * -1)
        return floor(num / mplr) * mplr
    ncnt = step
    if 1 > step > 0:
        ndec, ncnt = .0101, 1
        while ndec > step:
            ndec *= .1
            ncnt += 1
    mplr = 10 ** ncnt
    return round(floor(num * mplr) / mplr, ncnt)

You can use positive/negative numbers and float points .1, .01, .001...

来源：https://stackoverflow.com/questions/9404967/taking-the-floor-of-a-float