Specify filename in python script to open Excel workbook

Question

this is a really dumb question, but Im trying to learn python and I got stuck on example with reading Excel files using xlrd. I found this script online but I cant figure out where am I supposed to fill my filename to get it open.

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   

def get_excel_sheet_object(fname, idx=0):
if not isfile(fname):
print ('File doesn't exist: ', fname)
# Open the workbook and 1st sheet
xl_workbook = xlrd.open_workbook(fname)
xl_sheet = xl_workbook.sheet_by_index(0)
print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)
return xl_sheetsdf

Answer 1

You can enter the filename :

• When you call the function

  get_excel_sheet_object("myfile.xlsx")

         OR
  

  fname = "myfile.xlsx"

  get_excel_sheet_object(fname)

• Raw in your program :

  def get_excel_sheet_object(idx=0):
  
        fname = "myfile.xlsx"
  
        if not isfile(fname):
           print ("File doesn't exist: ", fname)
        # Open the workbook and 1st sheet
        xl_workbook = xlrd.open_workbook(fname)
        xl_sheet = xl_workbook.sheet_by_index(0)
        print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)
  
        return xl_sheet      
  

Answer 2

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   


def get_excel_sheet_object(fname, idx=0):
    if not isfile(fname):
        print ('File doesn't exist: ', fname)
            # Open the workbook and 1st sheet
    xl_workbook = xlrd.open_workbook(fname)
    xl_sheet = xl_workbook.sheet_by_index(0)
    print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

    return xl_sheet

xl_sheet_obj = get_excel_sheet_object('FILE_NAME_HERE')

Do anything you want with xl_sheet_obj after that this object is the excel sheet object.