On 2007-02-02, ardief <[EMAIL PROTECTED]> wrote:
> Hi everyone
> Here is my problem:
> I have a list that looks like this -
> [['a', '13'], ['a', '3'], ['b', '6'], ['c', '12'], ['c', '15'], ['c',
> '4'], ['d', '2'], ['e', '11'], ['e', '5'], ['e', '16'], ['e', '7']]
>
> and I would like to end up with something like this, i.e. with
> the only one list per letter:
>
> [['a', ['13' '3']], ['b', '6'], ['c', ['12', '15', '4']], ['d', '2'],
> ['e', ['11', '5', '16', '7']]]
>
> I have the feeling it's trivial, and I've scoured the group
> archives - sets might be a possibility, but I'm not sure how to
> operate on a list of lists with sets.
This is a job for... duhn-duhn-DAAAAH! Captain CHAOS!
Er... I mean itertools.groupby.
I took the liberty of not assuming the alist was sorted already.
This could be a one-liner if you wanted to be evil.
def bundle_alist(seq):
""" Bundle together some alist tails.
>>> seq = [['a', '13'], ['a', '3'], ['b', '6'], ['c', '12'], ['c', '15'],
['c', '4'], ['d', '2'], ['e', '11'], ['e', '5'], ['e', '16'], ['e', '7']]
>>> bundle_alist(seq)
[['a', ['13', '3']], ['b', ['6']], ['c', ['12', '15', '4']], ['d', ['2']],
['e', ['11', '5', '16', '7']]]
"""
from itertools import groupby
def key_func(t):
return t[0]
groups = groupby(sorted(seq, key=key_func), key_func)
seq = []
for item in groups:
seq.append([item[0], [a[1] for a in item[1]]])
return seq
--
Neil Cerutti
Music gets more chromatic and heavy towards the end of the century. One of the
popular questions of the day was, "Why?" --Music Lit Essay
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