[newbie] copying identical list for a function argument

[Jean Dupont]

Op maandag 3 februari 2014 23:19:39 UTC+1 schreef Steven D'Aprano:
> On Mon, 03 Feb 2014 13:36:24 -0800, Jean Dupont wrote:
> > I have a list like this:
> > [1,2,3]
> > 
> > The argument of my function should be a repeated version e.g.
> > [1,2,3],[1,2,3],[1,2,3],[1,2,3] (could be a different number of times
> > repeated also)
> > 
> > what is the prefered method to realize this in Python?
>
> I don't really understand your question. It could mean any of various 
> things, so I'm going to try to guess what you mean. If my guesses are 
> wrong, please ask again, giving more detail, and possibly an example of 
> what you want to do and the result you expect.
> I think you mean that you have some function that needs to take (say) 
> five arguments, and you want to avoid writing:
> result = function([1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 23], [1, 2, 3])
> because it's too easy to make a mistake (as I did, deliberately, above -- 
> can you see it?).
> If my guess is correct, try this:
> mylist = [1, 2, 3]
> result = function(mylist, mylist, mylist, mylist, mylist)
>
> That's perfectly reasonable for two or three arguments, but not so much 
> for five. Instead, here's a trick: first we make five identical 
> references to the same list:
> [mylist]*5  # same as [mylist, mylist, mylist, mylist, mylist]
> then expand them as arguments to the function:
> mylist = [1, 2, 3]
> list_of_lists = [mylist]*5
> result = function(*list_of_lists)
> (The * operator means multiplication when used between two arguments, and 
> inside a function call a leading * also does argument expansion.)
>
> But wait... there's something slightly weird here. Even though there are 
> five distinct references to mylist, they're all the same list! Change 
> one, change all. This may be what you want, or it may be a problem. Hard 
> to tell from your question.
> Think about references as being a little bit like names. A *single* 
> person could be known as "son", "Dad", "Mr Obama", "Barack", "Mr 
> President", "POTUS", and more. In this case, we have a single list, [1, 
> 2, 3], which is known by six references: the name "mylist", and five 
> additional references list_of_lists index 0, list_of_lists index 1, and 
> so on up to list_of_lists index 4.
> We can prove that they all refer to the same list by running a bit of 
> code in the interactive interpreter:
>
> py> mylist = [1, 2, 3]
> py> list_of_lists = [mylist]*5
> py> list_of_lists
> [[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
> py> mylist.append(99)
> py> list_of_lists
> [[1, 2, 3, 99], [1, 2, 3, 99], [1, 2, 3, 99], [1, 2, 3, 99], [1, 2, 3, 
> 99]]
>
> So rather than having five references to the same, identical, list, you 
> might want five *copies*. You can copy a list using slicing:
> mylist = [1, 2, 3]
> copy = mylist[:]
> Instead of using list multiplication to repeat five identical lists, we 
> make five copies using a list comprehension:
> list_of_lists = [mylist[:] for i in range(5)]
> then expand it in the function call as before:
> result = function(*list_of_lists)
>
> Hope this helps,
Yes it does, thanks a lot to you and all the others who responded, "the
missing link" which until now I wasn't aware of but which was essential
for the solution was the "*" in
result = function(*list_of_lists)

kind regards,
jean