Dan Sommers wrote: > > counter = 0 > for an_element in the_list: > print an_element, > counter = counter + 1 > if counter == n: > print > counter = 0 >
Yes, often simple old-fashioned ways are the best. A little verbose, though. And I'd do some old-fashioned testing -- the above needs "if counter: print" after the loop has finished, to get the final '\n' for cases where there are not an even multiple of n elements. >>> def wrapn(alist, n): ... ctr = 0 ... for item in alist: ... print item, ... ctr = (ctr + 1) % n ... if not ctr: ... print ... if ctr: ... print ... >>> for k in range(8): ... print '--- %d ---' % k ... wrapn(range(k), 3) ... --- 0 --- --- 1 --- 0 --- 2 --- 0 1 --- 3 --- 0 1 2 --- 4 --- 0 1 2 3 --- 5 --- 0 1 2 3 4 --- 6 --- 0 1 2 3 4 5 --- 7 --- 0 1 2 3 4 5 6 >>> Cheers, John -- http://mail.python.org/mailman/listinfo/python-list
