> > The original list 'a', isn't changed in any of these cases right? And > modifying b, c or d would not change 'a' either - or am I not understanding > this correctly? > > ## 1 ## > > creates a new list and copies all elements from a to b > > Yes.
> ## 2 ## > > take an already existing list (empty) and copies all elements from a to > it (no data is lost by c since c was empty to start out with) > > No. The 'already existing list' is simply completely discarded. You're creating it on one line, then destroying it. A brand new list is being created and returned and bound to the same name as the already existing list. > ## 3 ## > > d is a list, and all of its contents (if it had any) would be > filled with that of a .. but changes to a still are not reflected in d > .. this is confusing... > Not exactly, if I understand you correctly. D is a list, but its contents (if it had any) would be completely replaced. Any previous content would be gone, and it would now contain a copy of a. > Semi-aside, if I wanted to make local copy of a list sent to me as a > parameter, which of these is the most appropriate to use (I don't want > changes to change the original list sent). > my_copy = original[:] --S
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