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From: superpollo on 19 May 2010 15:58
.... 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?

bye
From: Jerry Hill on 19 May 2010 16:14
On Wed, May 19, 2010 at 3:58 PM, superpollo <utente(a)esempio.net> wrote:
> ... how many positive integers less than n have digits that sum up to m:
....
> any suggestion for pythonizin' it?

This is how I would do it:

def prttn(m, n):
"""How many positive integers less than n have digits that sum up to m"""
total = 0
for testval in range(n):
sumofdigits = sum(int(char) for char in str(testval))
if sumofdigits == m:
total += 1
return total

I added a docstring to the function, saying what it does, and what the
arguments are supposed to represent. I also moved the
convert-to-string-and-sum-the-digits logic into a single generator
expression that's passed to the builtin sum function. Oh, and I tried
to use slightly more expressive variable names.

--
Jerry
From: René 'Necoro' Neumann on 19 May 2010 16:24
Am 19.05.2010 21:58, schrieb superpollo:
> ... 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?
>
> bye

An idea would be:

>>> def prttn(m, n):
... return sum(1 for x in range(n) if sum(map(int, str(x))) == m)

A small oneliner :)

(I first had "return len([x for x in ...])" but the above avoids
creating an intermediate list)

- René

From: superpollo on 19 May 2010 16:25
Jerry Hill ha scritto:
> On Wed, May 19, 2010 at 3:58 PM, superpollo <utente(a)esempio.net> wrote:
>> ... how many positive integers less than n have digits that sum up to m:
> ...
>> any suggestion for pythonizin' it?
>
> This is how I would do it:
>
> def prttn(m, n):
> """How many positive integers less than n have digits that sum up to m"""
> total = 0
> for testval in range(n):
> sumofdigits = sum(int(char) for char in str(testval))

this line gives me this:

TypeError: 'int' object is not callable

is it some new feature in >2.5 ?

> if sumofdigits == m:
> total += 1
> return total
>
> I added a docstring to the function, saying what it does, and what the
> arguments are supposed to represent. I also moved the
> convert-to-string-and-sum-the-digits logic into a single generator
> expression that's passed to the builtin sum function. Oh, and I tried
> to use slightly more expressive variable names.
>
From: superpollo on 19 May 2010 16:30
René 'Necoro' Neumann ha scritto:
> Am 19.05.2010 21:58, schrieb superpollo:
>> ... 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?
>>
>> bye
>
> An idea would be:
>
>>>> def prttn(m, n):
> ... return sum(1 for x in range(n) if sum(map(int, str(x))) == m)

TypeError: 'int' object is not callable

on 2.5.4

>
> A small oneliner :)
>
> (I first had "return len([x for x in ...])" but the above avoids
> creating an intermediate list)
>
> - René
>
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