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 <pyt...@rcn.com> wrote: > On May 19, 12:58 pm, superpollo <ute...@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/ -- http://mail.python.org/mailman/listinfo/python-list
