variable gets redefined inside a python function without explicitly changing it

Question

   def f():
        list_0 = [1,2,3,4]
        p = [0,1,2,3,4,5,6,7,8]
        for i in list_0:
            x = p
            x[i] ="L"
            print(x)
            print(p)
    f()

Wanted the result to be: 0,"L",2,3,4,5,6,7,8 , 0,1,"L",3,4...... and so on , but got , 0,"L",2,3,4,5,6,7,8 , 0,"L","L",3,4,5,6,7,8.. so one, the "L" was getting saved . How do i change the code such that the variable p does not change and remains the same

`x = p` make x and p refer to the *same objects*. The *variable* isn't re-defined, you merely are mutating the same object, referenced by two different variables. — juanpa.arrivillaga, Apr 28 '20 at 08:38
The problem simply happens because `x` gets assigned to the same `p`. So, when `x` changes, `p` gets changed as well. To fix that, you need to create a copy from `p` like so: `x = p.copy()` — Anwarvic, Apr 28 '20 at 08:42
read the following: https://nedbatchelder.com/text/names.html — juanpa.arrivillaga, Apr 28 '20 at 08:42

Dhaval Taunk · Answer 1 · 2020-04-28T08:42:18.943

0

Write your code in this way:-

def f():
    list_0 = [1,2,3,4]
    p = [0,1,2,3,4,5,6,7,8]
    for i in list_0:
        x = p.copy()
        x[i] ="L"
        print(x)
        print(p)
f()

edited Apr 28 '20 at 08:42

answered Apr 28 '20 at 08:39

Dhaval Taunk

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1

there is no need to `import copy`, `list` objects (and most built-in containers) have a `.copy()` method – juanpa.arrivillaga Apr 28 '20 at 08:40
1

You don't even need `copy` module as `list` has a method called `copy()` already You need to use this `x = p.copy()` – Anwarvic Apr 28 '20 at 08:40

variable gets redefined inside a python function without explicitly changing it

1 Answers1