Change one list item in place

[Jean-Michel Pichavant]

Gnarlodious wrote:
> This works for me:
>
> def sendList():
>     return ["item0", "item1"]
>
> def query():
>     l=sendList()
>     return ["Formatting only {0} into a string".format(l[0]), l[1]]
>
> query()
>
>
> However, is there a way to bypass the
>
> l=sendList()
>
> and change one list item in-place? Possibly a list comprehension
> operating on a numbered item?
>
> -- Gnarlie
>   
what about

def query():
    return ["Formating only {0} into a string".format(sendList()[0])] + 
sendList()[1:]


JM