Rename the file names in the folder with the help of Python Script, using name map from Excel sheet

Question

There are many CSV files in a folder which I want it to be renamed. There is an excel sheet which contains name of files to be renamed to folder.

The files in folder are named as

TestData_30April.csv
TestData_20April.csv
TestData_18April.csv etc

while the excel sheet contains the name as

0.25-TestData_30April
0.98-TestData_20April
0.33-TestData_20April etc

Also first row in the excel sheet contains Header name while row 2 on wards contains the file name to be renamed.

My Aim is to rename TestData_30April.csv to 0.25-TestData_30April.csv similarly for all other files as well.

Here is the Code:

#Excel Sheet containing name of files to be renamed in that folder
path="C:\\Users\\Desktop\\Test_Data\\Test_Summary.csv"

#Folder Containg all orginal file names
dir = "C:\\Users\\Desktop\\Wear_Data"

wb = xlrd.open_workbook(path) 
sheet = wb.sheet_by_index(0)
sheet.cell_value(0, 0)

#In excel sheet column X or col_values(23) contains the file name to be renamed
print(sheet.col_values(23)) 

list_of_filename_in_folder = [] # name of the files in the folder
list_of_filename_in_excel = [] #name of the files in excel
path_to_folder = ''  # base path of folder
for name in list_of_filename_in_excel:
    excel_file_name = os.path.join(path_to_folder, name,'.csv')
    newname = name
    if '-' in name:
        newname = name.split('-')[1]
    dir_file_name = os.path.join(path_to_folder,newname,'.csv' )

    if os.path.exists(dir_file_name):
        print('changing file name {} to {}'.format(newname,name))
        os.rename(dir_file_name, excel_file_name)
    else:
        print('no file {} with name found in location'.format(newname+'.csv')

Here is the error:

XLRDError: Unsupported format, or corrupt file: Expected BOF record; 

Kindly Help in resolving this error.

Answer 1

Although you might be able to open csv files with Excel, .csv files are not the same as usual excel files (ending with .xlsx ). Python comes with a very convenient way of handling csv files: The csv module .

Assuming your data looks like in your example, you could do the following:

import csv
import os

path= 'C:\\Users\\Desktop\\Test_Data\\Test_Summary.csv'
dir = 'C:\\Users\\Desktop\\Wear_Data'
# open the .csv file with the csv module
with open(path, 'r') as f:
    csv_file = csv.reader(f)
    # read the new file name from every row 
    for row in csv_file:
        # assuming the new file path is stored in the first column (= row[0])
        new_file_name = row[0] + '.csv'
        # your old file should always have the same pattern according to your example
        old_file_name = new_file_name.split('-')[1] + '.csv'
        old_file = os.path.join(dir, old_file_name)
        new_file = os.path.join(dir, new_file_name)
        # rename the file
        os.rename(old_file, new_file)

I did not test this snippet, but I think it shows the basic principles of how this could work using the csv module.