jim stockford wrote: > > On Oct 12, 2007, at 11:48 AM, Kent Johnson wrote: > >> If all you want to do is copy the list, then >> lstB = lstA[:] >> is fine, or you can use >> lstB = list(lstA) > > why choose one over the other? is there a performance > or other difference?
The timeit module helps figure out if there is a performance difference. Slicing seems to have a slight edge over a wide range if list sizes: src $ python -m timeit -s 'l1=range(1000)' 'l2=list(l1)' 100000 loops, best of 3: 5.12 usec per loop src $ python -m timeit -s 'l1=range(1000)' 'l2=l1[:]' 100000 loops, best of 3: 4.27 usec per loop src $ python -m timeit -s 'l1=range(10)' 'l2=list(l1)' 1000000 loops, best of 3: 0.487 usec per loop src $ python -m timeit -s 'l1=range(10)' 'l2=l1[:]' 1000000 loops, best of 3: 0.258 usec per loop src $ python -m timeit -s 'l1=range(10000)' 'l2=list(l1)' 10000 loops, best of 3: 126 usec per loop src $ python -m timeit -s 'l1=range(10000)' 'l2=l1[:]' 10000 loops, best of 3: 108 usec per loop Other than that I think it is personal preference, whichever you prefer. Kent _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
