On 03/20/2013 04:21 AM, Peter Otten wrote:
Phil wrote:
On 20/03/13 15:09, Mitya Sirenef wrote:
<cut>
By the way, you can further simplify it by doing:
def histogram2(s):
return {c: d.get(c,0)+1 for c in s}
That will work in python 3, in python 2 you need:
return dict((c: d.get(c,0)+1) for c in s)
Thanks again Mitya, although I'm not sure it's a simplification at my
present level.
Especially as Mitya's code doesn't work.
Ah, yes - I messed up here.. I agree the loop is the best option here,
vs. the example below. -m
d = {}
d.update((c, d.get(c, 0)+1) for c in "abba")
d
{'a': 2, 'b': 2}
but frankly, I don't see how that is better than the for loop.
So as your experience with Python grows you may continue to use a loop or
switch to the standard library's collections.Counter (Python3 only).
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