Ric Deez a écrit :
> Hi there,
>
> I have a list:
> L1 = [1,1,1,2,2,3]
>
> How can I easily turn this into a list of tuples where the first element
> is the list element and the second is the number of times it occurs in
> the list (I think that this is referred to as a histogram):
>
> i.e.:
>
> L2 = [(1,3),(2,2),(3,1)]
>
> I was doing something like:
>
> myDict = {}
> for i in L1:
> myDict.setdefault(i,[]).append(i)
>
> then doing this:
>
> L2 = []
> for k, v in myDict.iteritems():
> L2.append((k, len(v)))
>
> This works but I sort of feel like there ought to be an easier way,
If you don't care about order (but your solution isn't garanteed to
preserve order either...):
L2 = dict([(item, L1.count(item)) for item in L1]).items()
But this may be inefficient is the list is large, so...
def hist(seq):
d = {}
for item in seq:
if not item in d:
d[item] = seq.count(item)
return d.items()
> I also tried this trick, where locals()['_[1]'] refers to the list
Not sure to understand how that one works... But anyway, please avoid
this kind of horror unless your engaged in WORN context with a
perl-monger !-).
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