a sequence question

[Nick Coghlan]

Duncan Booth wrote:
> Try this:
> 
> l = [1, 2, 3, 4]
> for a, b in zip(*[iter(l)]*2):
>    print a, b
> 
> zip(*[iter(seq)]*N) will group by N (but if there are any odd items at the 
> end it will ignore them).
> 
> map(None, *[iter(seq)]*N) will group by N padding the last item with None 
> if it needs to.

For anyone else who was as bemused as I was that Duncan's and F. Petitjean's 
suggestions actually *work*, this was what I had to do to figure out *why* they 
work:

Py> l = [1, 2, 3, 4]
Py> itr = iter(l)
Py> zip(itr) # Put all items from iterator in position 1
[(1,), (2,), (3,), (4,)]
Py> itr = iter(l)
Py> zip(itr, itr) # Put every second item in position 2
[(1, 2), (3, 4)]

Using zip(*[iter(l)]*N) or zip(*(iter(l),)*N) simply extends the above to the 
general case.

I'd definitely recommend hiding this trick inside a function. Perhaps something 
like (using Michael's function name):

from itertools import izip, repeat, chain

def partition(seq, part_len):
   return izip(*((iter(seq),) * part_len))

def padded_partition(seq, part_len, pad_val=None):
   itr = iter(seq)
   if (len(seq) % part_len != 0):
     padding = repeat(pad_val, part_len)
     itr = chain(itr, padding)
   return izip(*((itr,) * part_len))

Py> list(partition(range(10), 2))
[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]
Py> list(partition(range(10), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8)]
Py> list(padded_partition(range(10), 2))
[(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]
Py> list(padded_partition(range(10), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]
Py> list(padded_partition(range(10), 3, False))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, False, False)]
Py> zip(*padded_partition(range(10), 3))
[(0, 3, 6, 9), (1, 4, 7, None), (2, 5, 8, None)]

Not sure how useful that last example is, but I thought it was cute :)

Cheers,
Nick.

-- 
Nick Coghlan   |   ncoghlan at email.com   |   Brisbane, Australia
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