I’m just curious about which is the most efficient way of doing this kind of node enumiration:

for i in something():
    o=[var1,var2,var3,varN][i]
    o.new()
    o.do_something_based_on_number_of_loops()
    add_child(o)

or

for i in something():
    match i:
        0:
            o=var1
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        1:
            o=var2
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        2:
            o=var3
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        N-1:
            o=varN
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)

or

var items = [var1,var2,var3,varN]
for i in something():
    o=items[i]
    o.new()
    o.do_something_based_on_number_of_loops()
    add_child(o)

Or is there a more efficient way of doing it?

Edit: Sorry if that wasn’t clear. Is it better to constantly get something from an “unstored list”, store the list in a variable, or not use a list and use a match statement instead? Do they have any advantages/disadvantages that make them better in certain situations?

  • Trantarius
    7•1 year ago

    Use:

    items=[...]
for o in items:
    ...

    This is the most direct way of doing what you want. The first option might allocate a new array each iteration, which is unnecessary. The match statement is both a pain in the ass to write and less direct, which at best compiles to the same thing and at worst has you doing a bunch of totally unneeded comparisons.

    If this ‘i’ variable you used isn’t just an incrementing counter, use the last option. If it is though, it’s an extra counter you don’t actually need.

    The performance difference here would be so small I doubt you could even observe it. So, you really shouldn’t worry about this particular pattern. Compiler optimizations are more likely to trigger on simple, direct code, so writing it as directly as possible is probably the fastest option anyway.