|
|
 |
 |
 |
 |
Python Programming Language
|
 |
 |
 |
 |
|
|
|
|
Integer division
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
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))
Ie ceil(3/2) = floor((3+(2-1))/2) = 2. Correct.
But ceil(4/2) = floor((4+(2-1))/2) = 2 also. Correct.
Hamish
On Jun 7, 2:15 pm, Hamish Moffatt <ham@cloud.net.au> wrote:
> 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
>>> 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
jm.sur @no.spam.gmail.com <jm.sur @gmail.com> wrote: > 3/2 => 1 # Usual behavior > some_func(3, 2) => 2 # Wanted
return -(-a/b)
And people complain about Python's behaviour regarding division of
negative integers.
--
\S -- s@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 become se bera eadward ofdun hlddre heafdes bce bump bump bump
jm.sur @no.spam.gmail.com wrote: > On Jun 7, 2:15 pm, Hamish Moffatt <ham @cloud.net.au> wrote: >> 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)
def div_ceil(a, b):
return ((a+(b-1))/b)
Hamish
Hamish Moffatt <ham@cloud.net.au> wrote:
>jm.sur @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)
$ 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 -- s@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 become se bera eadward ofdun hlddre heafdes bce bump bump bump
"Sion Arrowsmith" <s @chiark.greenend.org.uk> wrote in message
| jm.sur @no.spam.gmail.com <jm.sur @gmail.com> wrote:
| > 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
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
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.
On Jun 7, 8:30 pm, Some Other Guy <bga@microsoft.com> wrote:
> 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.
at how few Pythonistas are using it six years later.
"Some Other Guy" <bga@microsoft.com> wrote in message
news:bb9ae$4668b133$cef88832$11811@TEKSAVVY.COM...
| 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?
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
In <1181275686.205607.53@q69g2000hsb.googlegroups.com>, Dan Bishop
wrote:
> On Jun 7, 8:30 pm, Some Other Guy <bga @microsoft.com> wrote: >> 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.
if used with at least one `float` as operands:
In [1]: 10.0 // 2
Out[1]: 5.0
Ciao,
Marc 'BlackJack' Rintsch
|
|
|
|
|
|
Add to del.icio.us |
Digg this |
Stumble it |
Powered by Megasolutions Inc