Change filename of Excel-files based on the content of a cell

Question

I have hundreds of Excel-sheets looking something like this:

enter image description here

_{I can't use the actual sheet, and not the exact document numbers, since they are confidential.}

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The name of the files are something like

2301 Item 1.xlsx
2014 Item 2.xlsx
5321 Item 3.xlsx
3212 Item 200.xlsx

I want to change the names of these files, based on the document name that's given in cell C5 in each Excel-sheet. Note: the cells C5 - E5 are merged.

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I've written the following Python 3 script:

import os
import xlrd
def rename_excel_files():
 path = r"C:\Users\Stewie\Documents\excel_folder\\"
 for filename in os.listdir(path):
 excel_file = xlrd.open_workbook(path + filename)
 work_sheet = excel_file.sheet_by_index(0)
 new_filename = work_sheet.cell_value(4,2)
 os.rename(path + '\\' + filename, path + new_filename + '.xlsx')
rename_excel_files()

It works perfectly fine... But I guess it's not perfect..?

Any tips?

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Note, this finishes almost instantly, so I don't really want to improve the execution time if that results in code that's harder to read, write and understand.

Also note: It's always cell C5. I don't need the cell-number to be dynamic. This is a fixed template that's used on literally thousands of files. I will only use it on the files on that template.

Answer 1

The good practice is to finish your code with

if __name__ == '__main__':
 sys.exit(main())

which has the advantage of returning an error code to the system (for example for batch processing of related programs).

It means 3 things for changing your code:

1. import sys

2. moving your command rename_excel_files(r"C:\Users\Stewie\Documents\excel_folder") to the new created main() function.

3. adding an return statement to the main() function with value 0 in the case of the success (and optionally something else for the failure):

So the result will be something as

import os
import sys
import xlrd
 ...
 ...
 ...
def main():
 rename_excel_files(r"C:\Users\Stewie\Documents\excel_folder")
 return 0
if __name__ == '__main__':
 sys.exit(main())

(See "3.3.1 Use sys.exit in your script to return proper error codes" in the book Writing Idiomatic Python by Jeff Knupp, 2013 - or directly from our BDFL - in All Things Pythonic - Python main() functions by Guido van van Rossum May 15, 2003)

• Grajdeanu Alex: Could you link to the source where it's specified that what you suggested is a good practice? IMO it kinda' looks verbose and unnecessary and I've never read anywhere about it :)
• MarianD: @MrGrj thanks for your comment - I added some links at the end of my answer.

Answer 2

Instead of hardcoding the path to the files to be renamed, I would use a parameter. This will let you expand the script to allow to pass that path on the command-line or via a GUI or whatever later.

I would also use os.path.join instead of manually concatenating the path, the intent feels cleaner.

Lastly, an if __name__ == '__main__': guard would allow you to import that file without executing any logic:

import os
import xlrd
def rename_excel_files(path):
 """TODO: add a docstring here"""
 for filename in os.listdir(path):
 filepath = os.path.join(path, filename)
 excel_file = xlrd.open_workbook(filepath)
 work_sheet = excel_file.sheet_by_index(0)
 new_filename = work_sheet.cell_value(4,2)
 os.rename(filepath, os.path.join(path, new_filename + '.xlsx'))
if __name__ == '__main__':
 rename_excel_files(r'C:\Users\Stewie\Documents\excel_folder')

Answer 3

• os.rename could raise OSError if the destination directory already exists on Windows. So, perhaps you want to handle that. Or you could use os.replace to get cross-platform destination file overriding.
• You should group standard library and third party imports separately.
• If you're on Python 3.4+ you could also use pathlib library.

import os
from pathlib import Path
import xlrd
def rename_excel_files(path):
 path = Path(path)
 for filename in os.listdir(path):
 file_path = path.joinpath(filename)
 excel_file = xlrd.open_workbook(file_path)
 work_sheet = excel_file.sheet_by_index(0)
 dest_path = path.joinpath(work_sheet.cell_value(4, 2) + '.xlsx')
 os.replace(file_path, dest_path)
if __name__ == '__main__':
 rename_excel_files(r"C:\Users\Stewie\Documents\excel_folder")

• Cristian Ciupitu: I think os.replace(file_path, dest_path) could be replaced with file_path.replace(dest_path).