What you want is openpyxl.utils.coordinate_from_string() and openpyxl.utils.column_index_from_string() from openpyxl.utils.cell import coordinate_from_string, column_index_from_string xy = coordinate_from_string(‘A4’) # returns (‘A’,4) col = column_index_from_string(xy[0]) # returns 1 row = xy[1]
Why can’t you just iterate over column ‘C’ (version 2.4.7): for cell in ws[‘C’]: print cell.value
Like Charlie Clark already suggest you can set data_only on True when you load your workbook: from openpyxl import load_workbook wb = load_workbook(“file.xlsx”, data_only=True) sh = wb[“Sheet_name”] print(sh[“x10”].value)
I believe the issue is that you’re trying to assign a fill object to a style. ws[‘A1’].fill = redFill should work fine.
Use the sheetnames property: sheetnames Returns the list of the names of worksheets in this workbook. Names are returned in the worksheets order. Type: list of strings print (wb.sheetnames) You can also get worksheet objects from wb.worksheets: ws = wb.worksheets[0]
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
You should use wb[sheetname] from openpyxl import load_workbook wb2 = load_workbook(‘test.xlsx’) ws4 = wb2[“New Title”] PS: You should check if your sheet in sheet names wb.sheetnames print(wb2.sheetnames) [‘Sheet2’, ‘New Title’, ‘Sheet1’]
If you don’t use conda, just use : pip install openpyxl If you use conda, I’d recommend : conda install -c anaconda openpyxl instead of simply conda install openpyxl Because there are issues right now with conda updating (see GitHub Issue #8842) ; this is being fixed and it should work again after the next … Read more
wb = openpyxl.load_workbook(filename, data_only=True) The data_only flag helps.
You should be using pandas own ExcelWriter class: from pandas import ExcelWriter # from pandas.io.parsers import ExcelWriter Then the save_xls function works as expected: def save_xls(list_dfs, xls_path): with ExcelWriter(xls_path) as writer: for n, df in enumerate(list_dfs): df.to_excel(writer,’sheet%s’ % n)