function that counts...

[Matteo Landi]

What about avoiding the string conversion and use mod/div operations
in order to create a list of digits for a number?

Now that you have the sequences it's easy to count which sums up to m.

On Mon, May 24, 2010 at 4:14 PM, Raymond Hettinger <python at rcn.com> wrote:
> On May 19, 12:58 pm, superpollo <ute... at esempio.net> wrote:
>> ... how many positive integers less than n have digits that sum up to m:
>>
>> In [197]: def prttn(m, n):
>>      tot = 0
>>      for i in range(n):
>>          s = str(i)
>>          sum = 0
>>          for j in range(len(s)):
>>              sum += int(s[j])
>>          if sum == m:
>>              tot += 1
>>      return tot
>>     .....:
>>
>> In [207]: prttn(25, 10000)
>> Out[207]: 348
>>
>> any suggestion for pythonizin' it?
>
> Your code is readable and does the job just fine.
> Not sure it is an improvement to reroll it into a one-liner:
>
> def prttn(m, n):
>    return sum(m == sum(map(int, str(x))) for x in range(n))
>>>> prttn(25, 10000)
> 348
>
> Some people find the functional style easier to verify because of
> fewer auxilliary variables and each step is testable independently.
> The m==sum() part is not very transparent because it relies on
> True==1, so it's only readable if it becomes a common idiom (which it
> could when you're answering questions of the form "how many numbers
> have property x").
>
> If speed is important, the global lookups can be localized:
>
> def prttn(m, n, map=itertools.imap, int=int, str=str, range=range):
>    return sum(m == sum(map(int, str(x))) for x in range(n))
>
> Raymond
>
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>



-- 
Matteo Landi
http://www.matteolandi.net/