Group items of a list with a step size python?

Question

Given an input list

 l = [1 2 3 4 5 6 7 8 9 10]

and group size grp and step step

grp = 3; step = 2

I would like return a list. Note the repetition at the end

1 2 3
3 4 5
5 6 7
7 8 9
9 10 1

or if

grp= 4; step = 2

The output should be

1 2 3 4
3 4 5 6
5 6 7 8
7 8 9 10

This is the code I came up with it does not do the cyclic thing. But would like to know if there is a smaller or a simpler solution

def grouplist(l,grp,step):
    oplist = list()
    for x in range(0,len(l)):
        if (x+grp<len(l)):
        oplist.append(str(l[x:x+grp]))
    return oplist

Answer 1

You can take advantage of the step function in xrange or range depending on what version of python you are using. Then to wrap back around just mod by the length of the list like so

import sys

def grouplist(l,grp,step):
    newlist=[]
    d = len(l)
    for i in xrange(0,len(l),step):
        for j in xrange(grp):
            newlist.append(l[(i+j)%d])
            sys.stdout.write(str(l[(i+j)%d]) + ' ')
        print

l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print grouplist(l,3,2)
1 2 3 
3 4 5 
5 6 7 
7 8 9 
9 10 1 
[1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 1]

print grouplist(l,4,2)
1 2 3 4 
3 4 5 6 
5 6 7 8 
7 8 9 10 
9 10 1 2
[1, 2, 3, 4, 3, 4, 5, 6, 5, 6, 7, 8, 7, 8, 9, 10, 9, 10, 1, 2]