Specific iterator in one line

[Andreas Tawn]

> > > 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)