How to avoid unintentional changing a global variable when using it as a local variable (inside a function)?

Question

Problem:

I am using a function that takes a (global) variable as an input, then performs operations on that variable (locally), and then returns that same variable. I do only want the variable to change locally, but instead my function is changing the global variable as well.

Code to reproduce:

data = {'A' : [1,2,3],
        'B' : [4,5,6],
        'C' : [7,8,9]}
df = pd.DataFrame(data)

def func(df):
    df['D'] = df['A'] * df['B'] / df['C']
    return df

func(df) # running function, without assigning it to original variable

print(df)

Returns:

Running the code shows that the original dataframe has been changed and a column was added.

   A  B  C   D
0  1  4  7  12
1  2  5  8  15
2  3  6  9  18

Expected behaviour:

My intention is to run the function without adding the column to the global variable, only add it locally within the function.

    A   B   C
0   1   4   7
1   2   5   8
2   3   6   9

Set-up:

Python 3.7
Pandas 0.25.3
Windows 10

df is a local variable, but it refers to the same object as the global variable data. All function arguments behave this way. The general rule is: don't use a mutating method on a function argument unless you intend to modify the object. — chepner
– chepner, Commented Mar 11, 2020 at 19:34
or do ret = df.copy() and work with the ret dataframe in your function. — Quang Hoang
– Quang Hoang, Commented Mar 11, 2020 at 19:38
You need to make your own local copy of df if you intend to add a new column temporarily, or (if other threads using the data frame aren't a concern) you can remove the new column with del df['D'] before returning. — chepner
– chepner, Commented Mar 11, 2020 at 19:40
Don't focus on the variables - variables are just ways to refer to objects, and objects are what really matter in a Python program. — user2357112
– user2357112, Commented Mar 11, 2020 at 19:45

Błotosmętek · Accepted Answer · 2020-03-11 19:37:55Z

2

You can make a local copy:

def func(d):
    df = d.copy()
    df['D'] = df['A'] * df['B'] / df['C']
    return df

answered Mar 11, 2020 at 19:37

Błotosmętek

13k23 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

MarKre Over a year ago

Thanks, this works! I'm still wondering, why doesn't my code behave as I expected?

Błotosmętek Over a year ago

Because passing parameters (and in fact all assigments) in Python is done by reference. When you do x = y, you're not copying the value of y - instead, you're making x refer to the same object as y does. Therefore df in your function refers to the same object the actual parameter (which you somewhat unfortunately also named df) refers to,

MarKre Over a year ago

Okay, so to make sure I understand correctly, moving variables into a different namespace does not automatically create a copy of the variable into that new namespace. Thus changing the variable in one namespace (local) will also change the variable in the other namespace (global). Is that right?

Collectives™ on Stack Overflow

How to avoid unintentional changing a global variable when using it as a local variable (inside a function)?

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related