Python List & for-each access (Find/Replace in built-in list)

Question

I originally thought Python was a pure pass-by-reference language.

Coming from C/C++ I can\'t help but think about memory management, and it\'s hard to put it out of my

Answer 1

Python is not Java, nor C/C++ -- you need to stop thinking that way to really utilize the power of Python.

Python does not have pass-by-value, nor pass-by-reference, but instead uses pass-by-name (or pass-by-object) -- in other words, nearly everything is bound to a name that you can then use (the two obvious exceptions being tuple- and list-indexing).

When you do spam = "green", you have bound the name spam to the string object "green"; if you then do eggs = spam you have not copied anything, you have not made reference pointers; you have simply bound another name, eggs, to the same object ("green" in this case). If you then bind spam to something else (spam = 3.14159) eggs will still be bound to "green".

When a for-loop executes, it takes the name you give it, and binds it in turn to each object in the iterable while running the loop; when you call a function, it takes the names in the function header and binds them to the arguments passed; reassigning a name is actually rebinding a name (it can take a while to absorb this -- it did for me, anyway).

With for-loops utilizing lists, there are two basic ways to assign back to the list:

for i, item in enumerate(some_list):
    some_list[i] = process(item)

or

new_list = []
for item in some_list:
    new_list.append(process(item))
some_list[:] = new_list

Notice the [:] on that last some_list -- it is causing a mutation of some_list's elements (setting the entire thing to new_list's elements) instead of rebinding the name some_list to new_list. Is this important? It depends! If you have other names besides some_list bound to the same list object, and you want them to see the updates, then you need to use the slicing method; if you don't, or if you do not want them to see the updates, then rebind -- some_list = new_list.