En Thu, 14 Jun 2007 14:23:14 -0300, Evan Klitzke <[EMAIL PROTECTED]> escribió:
> On 6/14/07, HMS Surprise <[EMAIL PROTECTED]> wrote: >> >> Just wondered if there was some python idiom for moving a few items >> from one list to another. I often need to delete 2 or 3 items from one >> list and put them in another. Delete doesn't seem to have a return >> value. I don't care which items I get so now I just use a couple of >> pops or a for loop for more than two. > > I'm not sure if this is what you're asking, but if the elements in the > list are contiguous you can just use list slicing/addition, like this: > > a = [1, 2, 3, 4, 5] > b = [6, 7, 8, 9, 10] > > b = a[2:] + b > a = a[:2] > > Now the contents of a and b respectively are a = [1, 2] and b = [3, 4, > 5, 6, 7, 8, 9, 10]. When lists become large this is less convenient because it has to build three intermediate lists. At least on my box, pop+append is 5 orders of magnitude faster (but as shown on another posts, you should test on your box to see what happens): c:\temp>call python -m timeit -s "a=range(1000000);b=range(1000)" "b.append(a.po p())" 1000000 loops, best of 3: 1.35 usec per loop c:\temp>call python -m timeit -s "a=range(1000000);b=range(1000)" "b+=a[-1:];a=a [:-1]" 10 loops, best of 3: 107 msec per loop c:\temp>call python -m timeit -s "a=range(1000000);b=range(1000)" "b.append(a[-1 ]);a=a[:-1]" 10 loops, best of 3: 107 msec per loop c:\temp>call python -m timeit -s "a=range(1000000);b=range(1000)" "b.append(a[0] );a=a[1:]" 10 loops, best of 3: 110 msec per loop -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
