Python中是否有一个//运算符的上限？

I found out about the // operator in Python which in Python 3 does division with floor.

Is there an operator which divides with ceil instead? (I know about the / operator which in Python 3 does floating point division.)

There is no operator which divides with ceil. You need to import math and use math.ceil

You can just do upside-down floor division:

def ceildiv(a, b):
    return -(-a // b)

This works because Python’s division operator does floor division (unlike in C, where integer division truncates the fractional part).

This also works with Python’s big integers, because there’s no (lossy) floating-point conversion.

Here’s a demonstration:

>>> from __future__ import division   # a/b is float division
>>> from math import ceil
>>> b = 3
>>> for a in range(-7, 8):
...     print(["%d/%d" % (a, b), int(ceil(a / b)), -(-a // b)])
... 
['-7/3', -2, -2]
['-6/3', -2, -2]
['-5/3', -1, -1]
['-4/3', -1, -1]
['-3/3', -1, -1]
['-2/3', 0, 0]
['-1/3', 0, 0]
['0/3', 0, 0]
['1/3', 1, 1]
['2/3', 1, 1]
['3/3', 1, 1]
['4/3', 2, 2]
['5/3', 2, 2]
['6/3', 2, 2]
['7/3', 3, 3]

You could do (x + (d-1)) // d when dividing x by d, i.e. (x + 4) // 5.

Solution 1: Convert floor to ceiling with negation

def ceiling_division(n, d):
    return -(n // -d)

Reminiscent of the Penn & Teller levitation trick, this “turns the world upside down (with negation), uses plain floor division (where the ceiling and floor have been swapped), and then turns the world right-side up (with negation again)”

Solution 2: Let divmod() do the work

def ceiling_division(n, d):
    q, r = divmod(n, d)
    return q + bool(r)

The divmod() function gives (a // b, a % b) for integers (this may be less reliable with floats due to round-off error). The step with bool(r) adds one to the quotient whenever there is a non-zero remainder.

Solution 3: Adjust the numerator before the division

def ceiling_division(n, d):
    return (n + d - 1) // d

Translate the numerator upwards so that floor division rounds down to the intended ceiling. Note, this only works for integers.

Solution 4: Convert to floats to use math.ceil()

def ceiling_division(n, d):
    return math.ceil(n / d)

The math.ceil() code is easy to understand, but it converts from ints to floats and back. This isn’t very fast and it may have rounding issues. Also, it relies on Python 3 semantics where “true division” produces a float and where the ceil() function returns an integer.

You can always just do it inline as well

((foo - 1) // bar) + 1

In python3, this is just shy of an order of magnitude faster than forcing the float division and calling ceil(), provided you care about the speed. Which you shouldn’t, unless you’ve proven through usage that you need to.

>>> timeit.timeit("((5 - 1) // 4) + 1", number = 100000000)
1.7249219375662506
>>> timeit.timeit("ceil(5/4)", setup="from math import ceil", number = 100000000)
12.096064013894647

Note that math.ceil is limited to 53 bits of precision. If you are working with large integers, you may not get exact results.

The gmpy2 libary provides a c_div function which uses ceiling rounding.

Disclaimer: I maintain gmpy2.

Simple solution: a // b + 1