Why is .pop() behaving like this? (Python)

Question

I don't understand why the original list "strg" is being modified.
I applied .pop() to its copy "g", yet "strg" is also changed (I wanted to keep "strg" unchanged ).

strg=["a","a","b","b"]
g=strg
g.pop(1) #applied to "g" only
print(strg)
print(g)

The output is ["a","b","b"] and ["a","b","b"]

Answer 1

Alvaro Vallejos ,

you can find some more explanations in the file about creating *copies* from a list by using an assignment, using function copy() or function deepcopy ().

https://code.sololearn.com/cQPb1y2A5quw/?ref=app

Answer 2

Try this

strg=["a","a","b","b"]
g=strg.copy()
g.pop(1) #applied to "g" only
print(strg)
print(g)

#Explanation
The copy() method returns a copy of the specified list.

Answer 3

Alvaro Vallejos This is a strange, but intended, behaviour of mutable objects in Python.
When you assign a mutable object - like a list or a dictionary - to multiple variables, they don't get copies of the object. They get references to the same object. So, if you have:
a = <Some list>
b = a
Both a and b have the same object. So, changing either variable actually changes the object referenced by both.

For fun:
a = [1, 2]
b = a # Same object
c = a.copy() # Copy to another object
a.append(3) # What changes?
print(b)
print(c)

Answer 4

Thank you!
I will look up this method.

Answer 5

Thank you everybody, very helpful

Answer 6

The issue in more detail explained here (code for reading, run for seeing the examples play out).

https://code.sololearn.com/c89ejW97QsTN/?ref=app