我正在处理一个处理大量Excel 2007文件的应用程序,而我正在使用OpenPyXL来完成它.OpenPyXL有两种不同的读取Excel文件的方法 - 一种是"普通"方法,其中整个文档一次加载到内存中,另一种方法是使用迭代器逐行读取.
问题是,当我使用迭代器方法时,我没有得到任何文档元数据,如列宽和行/列数,我真的需要这些数据.我假设这些数据存储在靠近顶部的Excel文档中,因此不必将整个10MB文件加载到内存中以访问它.
那么,有没有办法获得行/列数和列宽,而无需先将整个文档加载到内存中?
dra*_*m90 92
除了Hubro所说的,显然get_highest_row()
已经被弃用了.使用max_row
和max_column
属性返回行和列计数.例如:
wb = load_workbook(path, use_iterators=True)
sheet = wb.worksheets[0]
row_count = sheet.max_row
column_count = sheet.max_column
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Hub*_*bro 16
看一下OpenPyXL(IterableWorksheet)的源代码,我已经找到了如何从迭代器工作表中获取列数和行数:
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
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IterableWorksheet.get_highest_column
返回一个字符串,其中包含您可以在Excel中看到的列字母,例如"A","B","C"等.因此,我还编写了一个函数来将列字母转换为基于零的索引:
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
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我仍然没有弄清楚如何获得列大小,所以我决定使用固定宽度的字体并在我的应用程序中自动缩放列.
小智 8
蟒蛇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.
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