Is it possible to get an Excel document's row count without loading the entire document into memory?

Adding on to what Hubro said, apparently get_highest_row() has been deprecated. Using the max_row and max_column properties returns the row and column count. For example:

    wb = load_workbook(path, use_iterators=True)
    sheet = wb.worksheets[0]

    row_count = sheet.max_row
    column_count = sheet.max_column

The solution suggested in this answer has been deprecated, and might no longer work.

Taking a look at the source code of OpenPyXL (IterableWorksheet) I've figured out how to get the column and row count from an iterator worksheet:

wb = load_workbook(path, use_iterators=True)
sheet = wb.worksheets[0]

row_count = sheet.get_highest_row() - 1
column_count = letter_to_index(sheet.get_highest_column()) + 1

IterableWorksheet.get_highest_column returns a string with the column letter that you can see in Excel, e.g. "A", "B", "C" etc. Therefore I've also written a function to translate the column letter to a zero based index:

def letter_to_index(letter):
    """Converts a column letter, e.g. "A", "B", "AA", "BC" etc. to a zero based
    column index.

    A becomes 0, B becomes 1, Z becomes 25, AA becomes 26 etc.

    Args:
        letter (str): The column index letter.
    Returns:
        The column index as an integer.
    """
    letter = letter.upper()
    result = 0

    for index, char in enumerate(reversed(letter)):
        # Get the ASCII number of the letter and subtract 64 so that A
        # corresponds to 1.
        num = ord(char) - 64

        # Multiply the number with 26 to the power of `index` to get the correct
        # value of the letter based on it's index in the string.
        final_num = (26 ** index) * num

        result += final_num

    # Subtract 1 from the result to make it zero-based before returning.
    return result - 1

I still haven't figured out how to get the column sizes though, so I've decided to use a fixed-width font and automatically scaled columns in my application.

Python 3

import openpyxl as xl

wb = xl.load_workbook("Sample.xlsx", enumerate)

#the 2 lines under do the same. 
sheet = wb.get_sheet_by_name('sheet') 
sheet = wb.worksheets[0]

row_count = sheet.max_row
column_count = sheet.max_column

#this works fore me.

This might be extremely convoluted and I might be missing the obvious, but without OpenPyXL filling in the column_dimensions in Iterable Worksheets (see my comment above), the only way I can see of finding the column size without loading everything is to parse the xml directly:

from xml.etree.ElementTree import iterparse
from openpyxl import load_workbook
wb=load_workbook("/path/to/workbook.xlsx", use_iterators=True)
ws=wb.worksheets[0]
xml = ws._xml_source
xml.seek(0)

for _,x in iterparse(xml):

    name= x.tag.split("}")[-1]
    if name=="col":
        print "Column %(max)s: Width: %(width)s"%x.attrib # width = x.attrib["width"]

    if name=="cols":
        print "break before reading the rest of the file"
        break

Options using pandas.

1. Gets all sheetnames with count of rows and columns.

import pandas as pd
xl = pd.ExcelFile('file.xlsx')
sheetnames = xl.sheet_names
for sheet in sheetnames:
    df = xl.parse(sheet)
    dimensions = df.shape
    print('sheetname', ' --> ', dimensions)

2. Single sheet count of rows and columns.

import pandas as pd
xl = pd.ExcelFile('file.xlsx')
sheetnames = xl.sheet_names
df = xl.parse(sheetnames[0])   # [0] get first tab/sheet.
dimensions = df.shape
print(f'sheetname: "{sheetnames[0]}" - -> {dimensions}')

output sheetname "Sheet1" --> (row count, column count)