How to obtain sheet names from XLS files without loading the whole file?

I have tried xlrd, pandas, openpyxl and other such libraries and all of them seem to take exponential time as the file size increase as it reads the entire file. The other solutions mentioned above where they used 'on_demand' did not work for me. The following function works for xlsx files.

def get_sheet_details(file_path):
    sheets = []
    file_name = os.path.splitext(os.path.split(file_path)[-1])[0]
    # Make a temporary directory with the file name
    directory_to_extract_to = os.path.join(settings.MEDIA_ROOT, file_name)
    os.mkdir(directory_to_extract_to)

    # Extract the xlsx file as it is just a zip file
    zip_ref = zipfile.ZipFile(file_path, 'r')
    zip_ref.extractall(directory_to_extract_to)
    zip_ref.close()

    # Open the workbook.xml which is very light and only has meta data, get sheets from it
    path_to_workbook = os.path.join(directory_to_extract_to, 'xl', 'workbook.xml')
    with open(path_to_workbook, 'r') as f:
        xml = f.read()
        dictionary = xmltodict.parse(xml)
        for sheet in dictionary['workbook']['sheets']['sheet']:
            sheet_details = {
                'id': sheet['sheetId'], # can be @sheetId for some versions
                'name': sheet['name'] # can be @name
            }
            sheets.append(sheet_details)

    # Delete the extracted files directory
    shutil.rmtree(directory_to_extract_to)
    return sheets

Since all xlsx are basically zipped files, we extract the underlying xml data and read sheet names from the workbook directly which takes a fraction of a second as compared to the library functions.

Benchmarking: (On a 6mb xlsx file with 4 sheets)
Pandas, xlrd: 12 seconds
openpyxl: 24 seconds
Proposed method: 0.4 seconds

you can use the xlrd library and open the workbook with the "on_demand=True" flag, so that the sheets won't be loaded automaticaly.

Than you can retrieve the sheet names in a similar way to pandas:

import xlrd
xls = xlrd.open_workbook(r'<path_to_your_excel_file>', on_demand=True)
print xls.sheet_names() # <- remeber: xlrd sheet_names is a function, not a property