You can just flatten the list using itertools and collate the results using
defaultdict:
import itertools
from collections import defaultdict
sequence =
((0,'a','b'),(1,'c','d'),(2,'e','f'),(3,'g','h'),(1,'i','j'),(2,'k','l'),(4,'m','n'),(2,'o','p'),(4,'q','r'),(5,'s','t'))
# flatten the list to (0,'a','b',1,'c','d'...
chain = itertools.chain.from_iterable(sequence)
# use defaultdict to set up the final list and deal with initial empty
dictionary key
final_list = defaultdict(list)
for i in chain:
if isinstance(i, int):
number = i
else:
final_list[number].append(i)
print (final_list.items())
HTH,
Nick
On Sunday, September 26, 2010 2:05:13 PM UTC+10, Xah Lee wrote:
> here's a interesting toy list processing problem.
>
> I have a list of lists, where each sublist is labelled by
> a number. I need to collect together the contents of all sublists
> sharing
> the same label. So if I have the list
>
> ((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n) (2 o p) (4 q
> r) (5 s t))
>
> where the first element of each sublist is the label, I need to
> produce:
>
> output:
> ((a b) (c d i j) (e f k l o p) (g h) (m n q r) (s t))
>
> a Mathematica solution is here:
> http://xahlee.org/UnixResource_dir/writ/notations_mma.html
>
> R5RS Scheme lisp solution:
> http://xahlee.org/UnixResource_dir/writ/Sourav_Mukherjee_sourav.work_gmail.scm
> by Sourav Mukherjee
>
> also, a Common Lisp solution can be found here:
> http://groups.google.com/group/comp.lang.lisp/browse_frm/thread/5d1ded8824bc750b?
>
> anyone care to give a solution in Python, Perl, javascript, or other
> lang? am guessing the scheme solution can be much improved... perhaps
> using some lib but that seems to show scheme is pretty weak if the lib
> is non-standard.
>
> Xah ∑ xahlee.org ☄
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