Error despite Global keyword being used to access variable inside function

Question

I have created few global variables as shown in my code below. However when I try to use them inside the individual functions, I still get the same error.

Please find my code below

def create_df():
  global sheet_name, sheet_df 
  for s in sheets:
     sheet_name = s
     sheet_df = pd.read_excel(xls, sheet_name=s)
     sheet_df = sheet_df[sheet_df.columns.difference(sheet_df.filter(like='Derived').columns,sort=False)]
     print("Sheet " + str(s) + " is created as a dataframe successfully")
     transform_stage_1_df()

def transform_stage_1_df():
    global sheet_df 
    sheet_df = pd.melt(sheet_df, id_vars='subject_ID', var_name='headers', value_name='dates')
    sheet_df['header_extracted'] = [x.split("Date")[0] for x in sheet_df['headers']]
    sheet_df['day'] = sheet_df['header_extracted'].str.extract('(\d+)', expand=True).astype(int)
    sheet_df = sheet_df[sheet_df.groupby(['subject_ID','header_extracted'])['dates'].transform('count').ne(0)].copy()
    sheet_df = sheet_df.sort_values(by=['subject_ID','day'])
    sheet_df.drop(['header_extracted', 'day'], axis=1, inplace=True)
print("Stage 1 transformation is complete")


if __name__ == '__main__':
    print("Execution Started")
    print("File read successfully")
    sheets = xls.sheet_names
    sheet_name = sheet_df = Non_Cholesterol = None
    dataFramesDict = dict()
    create_df()
    add_units()
    Non_Cholesterol.to_csv('Output.csv')

Based on SO posts, I have already updated the global keyword but still I get the UnboundLocalError: local variable 'sheet_df' referenced before assignment

UnboundLocalError                         Traceback (most recent call last)
 <ipython-input-210-dc2f8412235d> in <module>
      7     sheet_df = None
       8     dataFramesDict = dict()
  ----> 9     create_df()
 10     Non_Cholesterol.to_csv('C:\\Users' + 
 str('Non_cholesterol.csv'),index=None)
 11     print("Export successful")

 <ipython-input-205-c93604f0da4f> in create_df()
  5         sheet_df =sheet_df[sheet_df.columns.difference(sheet_df.filter(like='Derived').columns,sort=False)]
  6         print("Sheet " + str(s) + " is created as a dataframe 
  successfully")
     ----> 7         transform_stage_1_df()

 <ipython-input-206-b59c70018a9b> in transform_stage_1_df()
  1 def transform_stage_1_df():
  ----> 2     sheet_df = pd.melt(sheet_df, id_vars='subject_ID', 
 var_name='headers', value_name='dates')
  3     sheet_df['header_extracted'] = [x.split("Date")[0] for x in 
 sheet_df['headers']]
  4     sheet_df['day'] = sheet_df['header_extracted'].str.extract('(\d+)', 
 expand=True).astype(int)
  5     sheet_df = 
 sheet_df[sheet_df.groupby(['subject_ID','header_extracted']) 
 ['dates'].transform('count').ne(0)].copy()

  UnboundLocalError: local variable 'sheet_df' referenced before assignment

you put the global in the wrong place, you need to do the global in the functions — Nullman
– Nullman, Commented Aug 1, 2019 at 10:54
You mean whereever I use the variable, I have to use global everywhere? I mean once in whichever function it is being used? — The Great
– The Great, Commented Aug 1, 2019 at 10:55
once at the top of each function declare the variables you want to survive as global. i want to point out that this is a pretty bad programming practice and unless you have some extremely specific need, don't do this — Nullman
– Nullman, Commented Aug 1, 2019 at 10:56
little bit of both. generally try to avoid using globals because that will very quickly make code unmaintainable. have your functions return what ever it is they changed, or pass mutables for them to change — Nullman
– Nullman, Commented Aug 1, 2019 at 11:12

Berkay Tekin Öz · Accepted Answer · 2019-08-01 11:13:13Z

1

You need to initialize the variables at the body of your script. When you say global variable_name it means that you will be accessing the variable outside of the function instead of a local variable named variable_name.

# Initialize the variables first
sheet_name = None
sheet_df = None

def create_df():
  global sheet_name, sheet_df 
  for s in sheets:
     sheet_name = s
     sheet_df = pd.read_excel(xls, sheet_name=s)
     sheet_df = sheet_df[sheet_df.columns.difference(sheet_df.filter(like='Derived').columns,sort=False)]
     print("Sheet " + str(s) + " is created as a dataframe successfully")
     transform_stage_1_df()

def transform_stage_1_df():
    global sheet_df 
    sheet_df = pd.melt(sheet_df, id_vars='subject_ID', var_name='headers', value_name='dates')
    sheet_df['header_extracted'] = [x.split("Date")[0] for x in sheet_df['headers']]
    sheet_df['day'] = sheet_df['header_extracted'].str.extract('(\d+)', expand=True).astype(int)
    sheet_df = sheet_df[sheet_df.groupby(['subject_ID','header_extracted'])['dates'].transform('count').ne(0)].copy()
    sheet_df = sheet_df.sort_values(by=['subject_ID','day'])
    sheet_df.drop(['header_extracted', 'day'], axis=1, inplace=True)
    print("Stage 1 transformation is complete")


if __name__ == '__main__':
   print("Execution Started")
   xls = pd.ExcelFile('C:\\Users\\All.xlsx')
   print("File read successfully")
   sheets = xls.sheet_names
   dataFramesDict = dict()
   create_df()

answered Aug 1, 2019 at 11:13

Berkay Tekin Öz

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3 Comments

Berkay Tekin Öz Over a year ago

Can you comment the full error and which line it throws? I can not access the image sadly.

The Great Over a year ago

Updated now. please check

Berkay Tekin Öz Over a year ago

It is supposed to work, if its stated like that. I thought you might be calling a function that contains code for main. Mistake on my part.

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