On 1/2/2011 8:31 AM Stefan Sonnenberg-Carstens said...
A last one: l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b'] dict((x[1],x[0]) for x in ((l.pop(),l.pop()) for x in xrange(len(l)/2)))
This also works: l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b'] pop=l.pop dict([(pop(),pop()) for i in l]) -- http://mail.python.org/mailman/listinfo/python-list
