python list by value not by reference [duplicate]

Answer 1

b = list(a)

See http://henry.precheur.org/python/copy_list.

Answer 2

To create a copy of a list do this:

b = a[:]

Answer 3

Also, you can do:

b = list(a)

This will work for any sequence, even those that don't support indexers and slices...

Answer 4

I found that we can use extend() to implement the function of copy()

a=['help', 'copyright', 'credits', 'license']
b = []
b.extend(a)
b.append("XYZ") 

Answer 5

As mentioned by phihag in his answer,

b = a[:]

will work for your case since slicing a list creates a new memory id of the list (meaning you are no longer referencing the same object in your memory and the changes you make to one will not be reflected in the other.)

However, there is a slight problem. If your list is multidimensional, as in lists within lists, simply slicing will not solve this problem. Changes made in the higher dimensions, i.e. the lists within the original list, will be shared between the two.

Do not fret, there is a solution. The module copy has a nifty copying technique that takes care of this issue.

from copy import deepcopy

b = deepcopy(a)

will copy a list with a new memory id no matter how many levels of lists it contains!