Andreas Tawn wrote:
>> > > This is purely sport question. I don't really intend to use the
>> > > answer in my code, but I am wondering, if such a feat could be done.
>> > >
>> > > I have a following problem: I have a list based upon which I would
>> > > like to construct a different one. I could simply use list
>> > > comprehensions, but there is an additional trick: for some elements
>> > > on this list, I would like to return two objects. For example I have
>> > > a list of 0s and 1s and for 0 I would like to add 1 'a' and for 1 I
>> > > would like to add 2 'b', like this:
>> > >
>> > > [1, 0, 0, 1] -> ['b', 'b', 'a', 'a', 'b', 'b']
>> > >
>> > > The easy way is to return a tuple ('b', 'b') for 1s and then flatten
>> > > them. But this doesn't seem very right - I'd prefer to create a nice
>> > > iterable right away. Is it possible to achieve this? Curiosly, the
>> > > other way round is pretty simple to achieve, because you can filter
>> > > objects using if in list comprehension.
>> > >
>> > If you'll allow me a prior "import itertools",
>> >
>> > >>> [i for e in [1,0,0,1] for i in itertools.repeat('ab'[e], e+1)]
>> >
>> > does the job in 62 characters.
>>
>> list("".join([("a","b"*2)[x] for x in [1,0,0,1]])
>>
>> 50 characters. Do I win £5?
>
> list("".join([("a","bb")[x] for x in [1,0,0,1]])
>
> Or 49 :o)
>>> len("""sum(([["a"],["b","b"]][i]for i in [1,0,0,1]),[])""")
48
>>> sum(([["a"],["b","b"]][i]for i in [1,0,0,1]),[])
['b', 'b', 'a', 'a', 'b', 'b']
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