Python: writing over a new list with less variables

Question

I'm having trouble thinking through this problem.

I have two lists as shown below.

my_orig_list = [(None, None), (None, None), (None, None), (None, None),(None, None), (None, None)]
new_values = [('a', 'a'), ('b', 'b'), ('c', 'c'), ('d', 'd'), ('e', 'e')]

How can I get it to have (None, None) as the final output as shown in the following desired output?

my_new_list = [('a', 'a'), ('b', 'b'), ('c', 'c'), ('d', 'd'), ('e', 'e'), (None, None)]

I should also state that the length of new_values will change from time to time.

I thought about determining the length of the list and then writing a function that would swap those values, but I cant seem to wrap my head around this relatively simple process.

If this has already been asked, I apologize. Maybe I don't know the terminology well enough to know what this would be called.

Answer 1

just slicing makes it easy ..

my_orig_list[:len(new_values)] = new_values[:]

keep in mind that new_values[:] copies the list ie make another copy of it, whereas just doing.

my_orig_list[:len(new_values)] = new_values

will give a reference to it, both will work in this case as long as you don't manipulate the new_values , how ever i would recommend new_values[:]

Answer 2

You can iterate through your new values, and use the index of the each array to update your original values like this:

for i in range(len(new_values)):
    my_orig_values[i] = new_values[i]

This will set the first value of my_orig_values to the first value of new_values, and so on, until the end of new_values

Answer 3

可能在下面：

new_values.extend(my_orig_list[:-len(new_values)])