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])