David Hirschfield <[EMAIL PROTECTED]> writes: > So sequentialChunks([1,2,3,5,6,8,12]) returns: > [[1,2,3],[5,6],[8],[12]]
Ugly and not too efficient: find the break points and use them to make sub-lists. def sequentialChunks(l, stride=1): p = [0] + [i for i in xrange(1,len(l)) if l[i]-l[i-1] != stride] + [len(l)] return [l[p[i]:p[i+1]] for i in xrange(len(p)-1)] print sequentialChunks([1,2,3,5,6,8,12]) -- http://mail.python.org/mailman/listinfo/python-list
