Simple List division problem
thebjorn
BjornSteinarFjeldPettersen at gmail.com
Sun Jan 13 13:05:31 CET 2008
On Jan 12, 8:33 pm, Fredrik Lundh <fred... at pythonware.com> wrote:
> marcstuart wrote:
> > How do I divide a list into a set group of sublist's- if the list is
> > not evenly dividable ? consider this example:
>
> > x = [1,2,3,4,5,6,7,8,9,10]
> > y = 3 # number of lists I want to break x into
> > z = y/x
>
> > what I would like to get is 3 sublists
>
> > print z[0] = [1,2,3]
> > print z[2] = [4,5,6]
> > print z[3] = [7,8,9,10]
>
> > obviously not even, one list will have 4 elements, the other 2 will
> > have 3.,
>
> here's one way to do it:
>
> # chop it up
> n = len(x) / y
> z = [x[i:i+n] for i in xrange(0, len(x), n)]
>
> # if the last piece is too short, add it to one before it
> if len(z[-1]) < n and len(z) > 1:
> z[-2].extend(z.pop(-1))
>
> </F>
Eh...
def chop(lst, length):
n = len(lst) / length
z = [lst[i:i+n] for i in xrange(0, len(lst), n)]
if len(z[-1]) < n and len(z) > 1:
z[-2].extend(z.pop(-1))
return z
gives
>>> chop(range(1,9), 3)
[[1, 2], [3, 4], [5, 6], [7, 8]]
>>> chop(range(1,8), 3)
[[1, 2], [3, 4], [5, 6, 7]]
>>> chop(range(1,6), 3)
[[1], [2], [3], [4], [5]]
>>> chop([1], 3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "beforemeth.py", line 9, in chop
if len(z[-1]) < n and len(z) > 1:
ValueError: xrange() arg 3 must not be zero
Perhaps something like this?
def chop(lst, length):
from itertools import islice
it = iter(lst)
z = [list(islice(it, length)) for i in xrange(1 + len(lst) //
length)]
if len(z) > 1:
z[-2].extend(z.pop()) # the last item will be empty or contain
"overflow" elements.
return z
-- bjorn
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