Referencing local variable within the same function

Question

I want to merge 2 DataFrames using a function. The function creates DataFrame df1 when called with variable 'x=1', and then another, df2, when called with 'x != 1', based on an if-statement within the function - code snippet below for further clarity.

Upon reaching the "df3 = pd.concat" line, I get the error "UnboundLocalError: local variable 'df1' referenced before assignment".

I would like to understand how to achieve the result of concatenating df1 and df2 into df3.

def Concat(url, x):
    if x == 1:
        df1 = pd.read_json(url)
    else:
        df2 = pd.read_json(url)
        df3 = pd.concat([df1, df2], ignore_index=True)


def main():
    Concat('*url*', 1)
    Concat('*url*', 2)

In the else block, you never define df1, so you get the UnboundLocalError. — leaf_soba
– leaf_soba, Commented Aug 23, 2021 at 1:35
both df1 and df2 are generated that's where you're wrong. You have code for generating both but it will never run together. — Jan Wilamowski
– Jan Wilamowski, Commented Aug 23, 2021 at 1:37

U13-Forward · Accepted Answer · 2021-08-23 10:48:45Z

2

You should tweak it a bit, to be:

def Concat(url, x):
    for i in x:
        if i == 1:
            df1 = pd.read_json(url)
        else:
            df2 = pd.read_json(url)
            df3 = pd.concat([df1, df2], ignore_index=True)


def main():
    Concat('*url*', [1, 2])

edited Aug 23, 2021 at 10:48

answered Aug 23, 2021 at 1:37

U13-Forward

71.8k15 gold badges100 silver badges125 bronze badges

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

6 Comments

Flybat Ramirez Over a year ago

Comments and answer make sense - I understand why I'm getting the error. Just to make sure I understand the Answer, does Concat run twice from main(), once with x=1 and once with x=2?

U13-Forward Over a year ago

@FlybatRamirez Nope, now it doesn't, it runs once

Flybat Ramirez Over a year ago

Ah, I see, you pass a list - it didn't occur to me to do that. Quite elegant.

Flybat Ramirez Over a year ago

I've tested your code, however I'm still getting the same error = "UnboundLocalError: local variable 'df1' referenced before assignment". It turns out df1 never gets created at all, so x ==1 is never True for some reason.

U13-Forward Over a year ago

@FlybatRamirez Well that is probably an issue with your other code you didn't show us, Not my code...

|

Collectives™ on Stack Overflow

Referencing local variable within the same function

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related