Pandas: Looking up the list of sheets in an excel file

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 (sheet_name) as None. like this:

 df = pandas.read_excel("/yourPath/FileName.xlsx", sheet_name=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

The easiest way to retrieve the sheet-names from an excel (xls., xlsx) is:

tabs = pd.ExcelFile("path").sheet_names 
print(tabs)

Then to read and store the data of a particular sheet (say, sheet names are "Sheet1", "Sheet2", etc.), say "Sheet2" for example:

data = pd.read_excel("path", "Sheet2") 
print(data)

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")]