[EMAIL PROTECTED] writes:
> def valsort(x):
> res = []
> for key, value in x.items():
> res.append((value, key))
> return res
Note: all code below is untested and may have errors ;-)
I think the above is misnamed because it doesn't actually sort.
Anyway, you could write it as a list comprehension:
def valsort(d):
return [(value, key) for (key, value) in d]
> def mostfreq(strng):
> dic = {}
> for letter in strng:
> if letter not in dic:
> dic.setdefault(letter, 1)
> else:
> dic[letter] += 1
I would write that with the defaultdict module:
from collections import defaultdict
def mostfreq(strng):
dic = defaultdict(int)
for letter in strng:
dic[letter] += 1
Alternatively with regular dicts, you could say:
def mostfreq(strng):
dic = {}
for letter in strng:
dic[letter] = dic.get(letter, 0) + 1
> newd = dic.items()
> getvals = valsort(newd)
> getvals.sort()
> length = len(getvals)
> return getvals[length - 3 : length]
Someone else suggested the heapq module, which is a good approach
though it might be considered a little bit high-tech. If you
want to use sorting (conceptually simpler), you could use the
sorted function instead of the in-place sorting function:
# return the second element of a 2-tuple. Note how we
# use tuple unpacking: this is really a function of one argument
# (the tuple) but we specify the arg as (a,b) so the tuple
# is automatically unpacked on entry to the function.
# this is a limited form of the "pattern matching" found in
# languages like ML.
def snd((a,b)): return b
return sorted(dic.iteritems, key=snd, reverse=True)[-3:]
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