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Integer division

P: n/a
Hello all,
I have two integers and I want to divide one by another, and want to
get an integer result which is the higher side whenever the result is
a fraction.
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted

Any easier solution other than int(math.ceil(float(3)/2))

-
Suresh

Jun 7 '07 #1
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9 Replies


P: n/a
jm*******@no.spam.gmail.com wrote:
Hello all,
I have two integers and I want to divide one by another, and want to
get an integer result which is the higher side whenever the result is
a fraction.
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted

Any easier solution other than int(math.ceil(float(3)/2))
The normal solution is to add (divisor-1) to the dividend before division.

Ie ceil(3/2) = floor((3+(2-1))/2) = 2. Correct.

But ceil(4/2) = floor((4+(2-1))/2) = 2 also. Correct.

Hamish
Jun 7 '07 #2

P: n/a
On Jun 7, 2:15 pm, Hamish Moffatt <ham...@cloud.net.auwrote:
jm.sur...@no.spam.gmail.com wrote:
Hello all,
I have two integers and I want to divide one by another, and want to
get an integer result which is the higher side whenever the result is
a fraction.
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted
Any easier solution other than int(math.ceil(float(3)/2))

The normal solution is to add (divisor-1) to the dividend before division.

Ie ceil(3/2) = floor((3+(2-1))/2) = 2. Correct.

But ceil(4/2) = floor((4+(2-1))/2) = 2 also. Correct.

Hamish
What about this?
>>def div_ceil(a, b):
.... if a%b:
.... return ((a/b)+1)
.... else:
.... return (a/b)
....
>>div_ceil(3,2)
2
>>div_ceil(-3,2)
-1
>>-3%2
1
>>-(3%2)
-1
>>div_ceil(-5,2)
-2
>>div_ceil(5,2)
3

Jun 7 '07 #3

P: n/a
jm*******@no.spam.gmail.com <jm*******@gmail.comwrote:
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted
def some_func(a, b):
return -(-a/b)

And people complain about Python's behaviour regarding division of
negative integers.

--
\S -- si***@chiark.greenend.org.uk -- http://www.chaos.org.uk/~sion/
"Frankly I have no feelings towards penguins one way or the other"
-- Arthur C. Clarke
her nu becomež se bera eadward ofdun hlęddre heafdes bęce bump bump bump
Jun 7 '07 #4

P: n/a
jm*******@no.spam.gmail.com wrote:
On Jun 7, 2:15 pm, Hamish Moffatt <ham...@cloud.net.auwrote:
>jm.sur...@no.spam.gmail.com wrote:
>>Hello all,
I have two integers and I want to divide one by another, and want to
get an integer result which is the higher side whenever the result is
a fraction.
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted
Any easier solution other than int(math.ceil(float(3)/2))
The normal solution is to add (divisor-1) to the dividend before division.

Ie ceil(3/2) = floor((3+(2-1))/2) = 2. Correct.
But ceil(4/2) = floor((4+(2-1))/2) = 2 also. Correct.

What about this?
>>>def div_ceil(a, b):
... if a%b:
... return ((a/b)+1)
... else:
... return (a/b)
Yes, although it's not as short or as fast (probably as my version):

def div_ceil(a, b):
return ((a+(b-1))/b)

Hamish
Jun 7 '07 #5

P: n/a
Hamish Moffatt <ha****@cloud.net.auwrote:
>jm*******@no.spam.gmail.com wrote:
>>>>def div_ceil(a, b):
... if a%b:
... return ((a/b)+1)
... else:
... return (a/b)

Yes, although it's not as short or as fast (probably as my version):

def div_ceil(a, b):
return ((a+(b-1))/b)
If that's what you care about:

$ python -mtimeit -s 'def divc1(a,b): return (a+b-1)/b' 'divc1(3,2)'
1000000 loops, best of 3: 0.443 usec per loop
$ python -mtimeit -s 'def divc2(a,b): return -(-a/b)' 'divc2(3,2)'
1000000 loops, best of 3: 0.331 usec per loop

Also, note:
>>divc2(sys.maxint, 2)
1073741824
>>divc1(sys.maxint, 2)
1073741824L

which is going to cause problems with sys.version_info < (2, 3) .
(Or do I mean (2, 2)? I don't have a 2.2 to hand.)

--
\S -- si***@chiark.greenend.org.uk -- http://www.chaos.org.uk/~sion/
"Frankly I have no feelings towards penguins one way or the other"
-- Arthur C. Clarke
her nu becomež se bera eadward ofdun hlęddre heafdes bęce bump bump bump
Jun 7 '07 #6

P: n/a

"Sion Arrowsmith" <si***@chiark.greenend.org.ukwrote in message
news:Y9*******@news.chiark.greenend.org.uk...
| jm*******@no.spam.gmail.com <jm*******@gmail.comwrote:
| 3/2 =1 # Usual behavior
| some_func(3, 2) =2 # Wanted
|
| def some_func(a, b):
| return -(-a/b)
|
| And people complain about Python's behaviour regarding division of
| negative integers.

Nice. This goes on my 'why didn't I think of that' list;-)

I was thinking of the rather pedestrian

d,r = divmod(a,b)
if r: d+=1

tjr

Jun 7 '07 #7

P: n/a
On Jun 7, 8:30 pm, Some Other Guy <bga...@microsoft.comwrote:
jm.sur...@no.spam.gmail.com wrote:
Hello all,
I have two integers and I want to divide one by another, and want to
get an integer result which is the higher side whenever the result is
a fraction.
3/2 =1 # Usual behavior
some_func(3, 2) =2 # Wanted

Are you trying to accomplish int((a/b) + 0.5), but cheaply?

If so, consider that that is the same as (a/b) + (b/2b),
which is (2a/2b) + (b/2b), which is (2a+b)/2b.

Since this just involves doubling you can avoid multiplying altogether
and just use this:

def rounddiv(a,b):
return int((a+a+b)/(b+b))

That's 3 integer adds and 1 integer divide. The int() cast is just
there is case somebody passes in floats for a or b; if you know you're
only going to have integer arguments you don't need it.
The // operator was added to the language for a reason. I'm surprised
at how few Pythonistas are using it six years later.

Jun 8 '07 #8

P: n/a

"Some Other Guy" <bg****@microsoft.comwrote in message
news:bb***************************@TEKSAVVY.COM...
| jm*******@no.spam.gmail.com wrote:
|
| Hello all,
| I have two integers and I want to divide one by another, and want to
| get an integer result which is the higher side whenever the result is
| a fraction.
| 3/2 =1 # Usual behavior
| some_func(3, 2) =2 # Wanted
|
| Are you trying to accomplish int((a/b) + 0.5), but cheaply?

No, the OP wanted 'ceiling' division rather than 'floor' division.
In other words, always round up instead of down.

| If so, consider that that is the same as (a/b) + (b/2b),
| which is (2a/2b) + (b/2b), which is (2a+b)/2b.

This is nice for rounded division.

tjr

Jun 8 '07 #9

P: n/a
In <11*********************@q69g2000hsb.googlegroups. com>, Dan Bishop
wrote:
On Jun 7, 8:30 pm, Some Other Guy <bga...@microsoft.comwrote:
>Since this just involves doubling you can avoid multiplying altogether
and just use this:

def rounddiv(a,b):
return int((a+a+b)/(b+b))

That's 3 integer adds and 1 integer divide. The int() cast is just
there is case somebody passes in floats for a or b; if you know you're
only going to have integer arguments you don't need it.

The // operator was added to the language for a reason. I'm surprised
at how few Pythonistas are using it six years later.
It would just document intent here because it still gives `float` results
if used with at least one `float` as operands:

In [1]: 10.0 // 2
Out[1]: 5.0

Ciao,
Marc 'BlackJack' Rintsch
Jun 8 '07 #10

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