PYTHON why extend() change my original variable?

Question

I understand that it is something simple but can't understand it

why is my variable 'a' modified?

a=["A","E"]
for x in range(0,2):
    b=a
    b.extend(["I","O","U"])   
    print(b)   
print(a)

Output:

b(first loop)=['A', 'E', 'I', 'O', 'U']
b(second loop)=['A', 'E', 'I', 'O', 'U', 'I', 'O', 'U']
a=['A', 'E', 'I', 'O', 'U', 'I', 'O', 'U']

My variable 'a' would have to be ["A","E"] at the end of the loop and each repetition 'b' would have to be ["A","E","I,"O","U"]

@Naveen While true, this is not actually at play here. This is more about assignment by reference rather than assignment by instantiation. — Adam Smith, Feb 05 '21 at 04:18

score 1 · Accepted Answer · answered Feb 05 '21 at 04:06

1

Because both variables a and b reference the same object.

a = ["A","E"]
b = a

print(id(a))
print(id(b))
print(a is b)

output:

140387638915552
140387638915552
True

answered Feb 05 '21 at 04:06

sarartur

what would be the solution in this case? – Fernando Feb 05 '21 at 04:13
The solution would be to copy the object with:```b = a.copy()```. If the object is nested with other objects then ```from copy import deepcopy()``` and ```b = deepcopy(a)``` – sarartur Feb 05 '21 at 04:14
1

for lists it's idiomatic to simply slice without bounds. `b = a[:]` (but this will not deep copy) – Adam Smith Feb 05 '21 at 04:17
You can also do ```b = [*a]``` – sarartur Feb 05 '21 at 04:18
@sarartur also, `deepcopy` is a bit heavy-handed, I'll often just write `[sub.copy() for sub in nested_list]` – juanpa.arrivillaga Feb 05 '21 at 04:24
@sarartur true, though only in Python 3.5+ – Adam Smith Feb 05 '21 at 05:55

1 Answers1