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

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")] 

#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 

If you:

• care about performance
• don't need the data in the file at execution time.
• want to go with conventional libraries vs rolling your own solution

Below was benchmarked on a ~10Mb xlsx, xlsb file.

xlsx, xls

from openpyxl import load_workbook def get_sheetnames_xlsx(filepath): wb = load_workbook(filepath, read_only=True, keep_links=False) return wb.sheetnames 

Benchmarks: ~ 14x speed improvement

# get_sheetnames_xlsx vs pd.read_excel 225 ms ± 6.21 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 3.25 s ± 140 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 

xlsb

from pyxlsb import open_workbook def get_sheetnames_xlsb(filepath): with open_workbook(filepath) as wb: return wb.sheets 

Benchmarks: ~ 56x speed improvement

# get_sheetnames_xlsb vs pd.read_excel 96.4 ms ± 1.61 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 5.36 s ± 162 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 

Notes:

• This is a good resource - http://www.python-excel.org/
• xlrd is no longer maintained as of 2020

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.

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.

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.

if you read excel file

dfs = pd.ExcelFile('file') 

then use

dfs.sheet_names dfs.parse('sheetname') 

another variant

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

If using pd.read_excel() with sheet_name=None to read sheet names, setting nrows=0 can significantly boost speeds.

Benchmarking on an 8MB XLSX file with nine sheets:

%timeit pd.read_excel('foo.xlsx', sheet_name=None, nrows=0).keys() >>> 145 ms ± 50.4 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) %timeit pd.read_excel('foo.xlsx', sheet_name=None).keys() >>> 16.5 s ± 2.04 s per loop (mean ± std. dev. of 7 runs, 1 loop each) 

This takes advantage of the fact that (as you might expect) pandas stops reading a sheet once it reaches nrows (reference).

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) 

1. 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) 

2. 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