The Python Oracle

Is there a ceiling equivalent of // operator in Python?

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Chapters
00:00 Is There A Ceiling Equivalent Of // Operator In Python?
00:21 Answer 1 Score 446
02:09 Accepted Answer Score 71
02:21 Answer 3 Score 70
03:41 Answer 4 Score 33
03:56 Answer 5 Score 24
04:20 Thank you

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Full question
https://stackoverflow.com/questions/1482...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #python3x

#avk47



ANSWER 1

Score 454


No, but you can use upside-down floor division:¹

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).

Here's a demonstration:

>>> from __future__ import division     # for Python 2.x compatibility
>>> import math
>>> def ceildiv(a, b):
...     return -(a // -b)
...
>>> b = 3
>>> for a in range(-7, 8):
...     q1 = math.ceil(a / b)   # a/b is float division
...     q2 = ceildiv(a, b)
...     print("%2d/%d %2d %2d" % (a, b, q1, q2))
...
-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

Why this instead of math.ceil?

math.ceil(a / b) can quietly produce incorrect results, because it introduces floating-point error. For example:

>>> from __future__ import division     # Python 2.x compat
>>> import math
>>> def ceildiv(a, b):
...     return -(a // -b)
...
>>> x = 2**64
>>> y = 2**48
>>> ceildiv(x, y)
65536
>>> ceildiv(x + 1, y)
65537                       # Correct
>>> math.ceil(x / y)
65536
>>> math.ceil((x + 1) / y)
65536                       # Incorrect!

In general, it's considered good practice to avoid floating-point arithmetic altogether unless you specifically need it. Floating-point math has several tricky edge cases, which tends to introduce bugs if you're not paying close attention. It can also be computationally expensive on small/low-power devices that do not have a hardware FPU.


¹In a previous version of this answer, ceildiv was implemented as return -(-a // b) but it was changed to return -(a // -b) after commenters reported that the latter performs slightly better in benchmarks. That makes sense, because the dividend (a) is typically larger than the divisor (b). Since Python uses arbitrary-precision arithmetic to perform these calculations, computing the unary negation -a would almost always involve equal-or-more work than computing -b.




ACCEPTED ANSWER

Score 71


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




ANSWER 3

Score 33


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




ANSWER 4

Score 24


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