On Jan 12, 8:33 pm, Fredrik Lundh <[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list