On May 7, 12:30 pm, Paul Melis <[EMAIL PROTECTED]> wrote:
> [EMAIL PROTECTED] wrote:
> > Here is my code for a letter frequency counter. It seems bloated to
> > me and any suggestions of what would be a better way (keep in my mind
> > I'm a beginner) would be greatly appreciated..
>
> Not bad for a beginner I think :)
>
>
>
>
>
> > def valsort(x):
> > res = []
> > for key, value in x.items():
> > res.append((value, key))
> > return res
>
> > def mostfreq(strng):
> > dic = {}
> > for letter in strng:
> > if letter not in dic:
> > dic.setdefault(letter, 1)
> > else:
> > dic[letter] += 1
> > newd = dic.items()
> > getvals = valsort(newd)
> > getvals.sort()
> > length = len(getvals)
> > return getvals[length - 3 : length]
>
> > thanks much!!
>
> Slightly shorter:
>
> def mostfreq(strng):
> dic = {}
> for letter in strng:
> if letter not in dic:
> dic[letter] = 0
> dic[letter] += 1
> # Swap letter, count here as we want to sort on count first
> getvals = [(pair[1],pair[0]) for pair in dic.iteritems()]
> getvals.sort()
> return getvals[-3:]
>
> I'm not sure if you wanted the function mostfreq to return the 3 most
> frequent letters of the first 3 letters? It seems to do the latter. The
> code above uses the former, i.e. letters with highest frequency.
>
> Paul- Hide quoted text -
>
> - Show quoted text -
I think I'd try to get a deque on disk. Constant indexing. Store
disk addresses in b-trees. How long does 'less than' take? Is a
sector small, and what's inside?
--
http://mail.python.org/mailman/listinfo/python-list
