熊猫：在Excel文件中查找工作表列表

The new version of Pandas uses the following interface to load Excel files:

read_excel('path_to_file.xls', 'Sheet1', index_col=None, na_values=['NA'])

but what if I don’t know the sheets that are available?

For example, I am working with excel files that the following sheets

Data 1, Data 2 …, Data N, foo, bar

but I don’t know N a priori.

Is there any way to get the list of sheets from an excel document in Pandas?

You can still use the ExcelFile class (and the sheet_names attribute):

xl = pd.ExcelFile('foo.xls')

xl.sheet_names  # see all sheet names

xl.parse(sheet_name)  # read a specific sheet to DataFrame

see docs for parse for more options…

You should explicitly specify the second parameter (sheetname) as None. like this:

 df = pandas.read_excel("/yourPath/FileName.xlsx", None);

“df” are all sheets as a dictionary of DataFrames, you can verify it by run this:

df.keys()

result like this:

[u'201610', u'201601', u'201701', u'201702', u'201703', u'201704', u'201705', u'201706', u'201612', u'fund', u'201603', u'201602', u'201605', u'201607', u'201606', u'201608', u'201512', u'201611', u'201604']

please refer pandas doc for more details: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.read_excel.html

This is the fastest way I have found, inspired by @divingTobi’s answer. All The answers based on xlrd, openpyxl or pandas are slow for me, as they all load the whole file first.

from zipfile import ZipFile
from bs4 import BeautifulSoup  # you also need to install "lxml" for the XML parser

with ZipFile(file) as zipped_file:
    summary = zipped_file.open(r'xl/workbook.xml').read()
soup = BeautifulSoup(summary, "xml")
sheets = [sheet.get("name") for sheet in soup.find_all("sheet")]

Building on @dhwanil_shah ‘s answer, you do not need to extract the whole file. With zf.open it is possible to read from a zipped file directly.

import xml.etree.ElementTree as ET
import zipfile

def xlsxSheets(f):
    zf = zipfile.ZipFile(f)

    f = zf.open(r'xl/workbook.xml')

    l = f.readline()
    l = f.readline()
    root = ET.fromstring(l)
    sheets=[]
    for c in root.findall('{http://schemas.openxmlformats.org/spreadsheetml/2006/main}sheets/*'):
        sheets.append(c.attrib['name'])
    return sheets

The two consecutive readlines are ugly, but the content is only in the second line of the text. No need to parse the whole file.

This solution seems to be much faster than the read_excel version, and most likely also faster than the full extract version.

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. If you just want to get the sheet names initially, 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'],
                'name': sheet['@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

Since my requirement was just reading the sheet names, the unnecessary overhead of reading the entire time was bugging me so I took this route instead.

from openpyxl import load_workbook

sheets = load_workbook(excel_file, read_only=True).sheetnames

For a 5MB Excel file I’m working with, load_workbook without the read_only flag took 8.24s. With the read_only flag it only took 39.6 ms. If you still want to use an Excel library and not drop to an xml solution, that’s much faster than the methods that parse the whole file.