Frank Millman wrote: > I have found two ways of doing it that seem to work. > > 1 - row = table[23][:] > > 2 - row = [] > row[:] = table[23] > > Are these effectively identical, or is there a subtle distinction which > I should be aware of. > > I did some timing tests, and 2 is quite a bit faster if 'row' > pre-exists and I just measure the second statement.
quite a bit ? maybe if you're using very short rows, and all rows have the same length, but hardly in the general case: python -mtimeit -s "data=[range(100)]*100; row = []" "row[:] = data[23]" 100000 loops, best of 3: 5.35 usec per loop python -mtimeit -s "data=[range(100)]*100" "row = data[23][:]" 100000 loops, best of 3: 4.81 usec per loop (for constant-length rows, the "row[:]=" form saves one memory allocation, since the target list can be reused as is. for longer rows, other things seem to dominate) </F> -- http://mail.python.org/mailman/listinfo/python-list
