On Thu, Dec 8, 2011 at 5:03 PM, Steven D'Aprano <st...@pearwood.info> wrote:
> Robert Berman wrote:
>
>> Hi,
>>
>> Assuming a list similar to this: l1=[['a',1],['b',2],['c',3]] and I want
>> to get the index of 'c'.
>>
>
> You will need to explain what you mean by "the index of 'c'".
>
> Do you mean 0, because 'c' is in position 0 of the sub-list ['c', 3]?
>
> Or do you mean 2, because 'c' is in the sub-list at position 2?
>
> What happens if there is a sub-list ['d', 'c']? Should that also count?
> What about sub-sub-lists, should they be checked too?
>
> Here is a version which checks each sub-list in turn, and returns the
> index of any 'c' it finds of the first such sub-list.
>
> def inner_find(list_of_lists):
> for sublist in list_of_lists:
> try:
> return sublist.index('c')
> except ValueError:
> pass # go to the next one
> # If not found at all:
> raise ValueError('not found')
>
>
> Here's a version which finds the index of the first sub-list that begins
> with 'c' as the zeroth element:
>
> def match_sublist(list_of_lists):
> for i, sublist in enumerate(list_of_lists):
> if sublist and sublist[0] == 'c':
> return i
> raise ValueError('not found')
>
>
>
>
> Other variations on these two techniques are left for you to experiment
> with.
>
>
>
> --
> Steven
>
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>
l1=[['a',1],['b',2],['c',3]]
>>> r = [s for s in l1 if s[0] == 'c']
>>> r
[['c', 3]]
>
This doesn't get you all the way, but maybe its a start. If you had more
than one sublist like ['c', 8] is would include that too
--
Joel Goldstick
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