#It will work for Both '.xls' and '.xlsx' by using pandas

import pandas as pd
excel_Sheet_names = (pd.ExcelFile(excelFilePath)).sheet_names

#for '.xlsx' use only  openpyxl

from openpyxl import load_workbook
excel_Sheet_names = (load_workbook(excelFilePath, read_only=True)).sheet_names
                                      

这是我发现的最快的方法，灵感来自@divingTobi的答案。所有基于xlrd、openpyxl或pandas的答案对我来说都很慢，因为它们都先加载整个文件。

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")]

你仍然可以使用ExcelFile类(和sheet_names属性):

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

xl.sheet_names  # see all sheet names

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

更多选项参见文档解析…

import pandas as pd

path = "\\DB\\Expense\\reconcile\\"

file_name = "202209-v01.xlsx"

df = pd.read_excel(path + file_name, None)
print(df)

sheet_names = list(df.keys())

# print last sheet name
print(sheet_names[len(sheet_names)-1])

last_month = df.get(sheet_names[len(sheet_names)-1])
print(last_month)

如果你读excel文件

dfs = pd.ExcelFile('file')

然后使用

dfs.sheet_names
dfs.parse('sheetname')

另一种变体

df = pd.read_excel('file', sheet_name='sheetname')

您应该显式地将第二个参数(sheetname)指定为None。是这样的:

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

"df"都是一个数据帧字典，你可以通过运行这个来验证:

df.keys()

结果是这样的:

[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']

详情请参考熊猫文档:https://pandas.pydata.org/pandas-docs/stable/generated/pandas.read_excel.html