[EMAIL PROTECTED] wrote: > Tim Williams: > >>You could also use a list comprehension for your case >> >>>>>alist = [1 ,2 ,3] >>>>>alist = [x for x in alist if x != 2] >>>>>alist >> >>[1, 3] > > > The list comprehension filtering is the simpler and often the best > solution. For memory-conscious people this is another possible > (un-pythonic) solution, usable in less dynamic languages too (not much > tested), expecially if the good elements are few (Psyco may help too). > I haven't tested its speed, but in Python it may be slower than the > simpler comprehension filtering solution: > > array = [1,2,3,1,5,1,6,0] > good = lambda x: x > 2 > slow = 0 > for fast, item in enumerate(array): > print item, > if good(item): > if slow != fast: > array[slow] = array[fast] > slow += 1 > print "\n", array > del array[slow:] > print array > > Output: > 1 2 3 1 5 1 6 0 > [3, 5, 6, 1, 5, 1, 6, 0] > [3, 5, 6] > This would be a way-premature optimisation until you had proven the need for it in the context of a specific program.
regards Steve -- Steve Holden +44 150 684 7255 +1 800 494 3119 Holden Web LLC/Ltd http://www.holdenweb.com Skype: holdenweb http://holdenweb.blogspot.com Recent Ramblings http://del.icio.us/steve.holden -- http://mail.python.org/mailman/listinfo/python-list
