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

With the load_workbook readonly option, what was earlier seen as a execution seen visibly waiting for many seconds happened with milliseconds. The solution could however be still improved. import pandas as pd from openpyxl import load_workbook class ExcelFile: def __init__(self, **kwargs): ........ ..... self._SheetNames = list(load_workbook(self._name,read_only=True,keep_links=False).sheetnames) The Excelfile.parse takes the same time as reading the complete xls in order of 10s of sec. This result was obtained with windows 10 operating system with below package versions C:\>python -V Python 3.9.1 C:\>pip list Package Version --------------- ------- et-xmlfile 1.0.1 numpy 1.20.2 openpyxl 3.0.7 pandas 1.2.3 pip 21.0.1 python-dateutil 2.8.1 pytz 2021.1 pyxlsb 1.0.8 setuptools 49.2.1 six 1.15.0 xlrd 2.0.1

我尝试过xlrd、pandas、openpyxl和其他类似的库，所有这些库似乎都需要指数级的时间，因为它们读取整个文件时，文件大小会增加。上面提到的其他解决方案，他们使用“on_demand”不适合我。如果您只想最初获得表名，下面的函数适用于xlsx文件。

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

由于所有xlsx基本上都是压缩文件，我们提取底层xml数据并直接从工作簿中读取表名，这与库函数相比只需几分之一秒的时间。

基准测试:(在一个6mb的xlsx文件上，有4张纸) 熊猫，xlrd: 12秒 Openpyxl: 24秒 建议方法:0.4秒

因为我的要求只是读取表名，读取整个时间的不必要开销让我很困扰，所以我选择了这种方法。

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)

from openpyxl import load_workbook

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

对于我正在使用的5MB Excel文件，没有read_only标志的load_workbook花了8.24秒。对于read_only标志，只需要39.6 ms。如果您仍然希望使用Excel库而不使用xml解决方案，那么这比解析整个文件的方法要快得多。