+ 1
List
a=[3,5,7,9] b=a a[1]=8 print(b[1]) Output: 8 Why when we changes the value in a-list it changes in b list also and vice versa?
3 Respostas
+ 7
What you did in your code is a so called shallow copy. Both lists a and b have the same object id, so they are referencing the same object.
To get 2 independent objects you can use copy() when you create b in this case.
b=a.copy()
An other way to make an independent copy is to use a slice:
b = a[:]
If your source list is a compound list (means that it contains e.g. lists), you have to use deep copy. So the code has to be:
from copy import deepcopy
a = [1,2,3,[11,9]]
b = deepcopy(a)
In this case b will be independent from a, and also [11,9] will be independent from sub list in a.
+ 3
Lothar is right
Because a and b have the same object id ,
so if you change the value of a ,it is going to change the value of b ,because it is a shallow copy.
So,there are many ways to create an independent copy like this
a = [1,2,3]
b = list(a) # this is going to create an independent copy
or by using copy() method like this
a = [1,2,3]
b = a.copy()
You can check that by using id()
Id shows us that a and b have different addresses
id gives us the address of the object and everything is just an object in python and lists are also objects.
And if you have a compound list that is lists of lists or in other words a list that contains lists
Then we can also use list comprehensions for deep copies like this
a =[[1,2],[3,4]
b = [[x for x in lis] for lis in a]
And if you have any doubt about all this ,
feel free to message me 😃
+ 1
@Lothar, Thanks a lot!