给熊猫数据框标题导出为ex​​cel

Question

我有6个数据框（名为a到f），我想导入到ms.excel，为了使这更容易阅读我想给每个数据框标签，我尝试工作with cell notation，但我得到的错误mesage，这里是我的代码：

writer = pd.ExcelWriter('Summary Data Request 8 weeks.xlsx') 
a.to_excel(writer, sheet_name='Sheet1', startrow=1) 
b.to_excel(writer, sheet_name='Sheet1', startcol=7) 
c.to_excel(writer, sheet_name='Sheet1', startrow=11) 
d.to_excel(writer, sheet_name='Sheet1', startrow=11, startcol=6) 
e.to_excel(writer, sheet_name='Sheet1', startrow=21) 
f.to_excel(writer, sheet_name='Sheet1', startrow=21, startcol=6) 
writer.write('A1', 'Retail Kelontong') 
writer.write('G1', 'Pulsa') 
writer.write('A11', 'Minimarket') 
writer.write('G11', 'Retail dengan barcode') 
writer.write('A21', 'Retail dengan barcode 2') 
writer.write('G21', 'Food and Beverage') 
writer.save() 

这里的错误消息：

AttributeError       Traceback (most recent call last) 
<ipython-input-30-2354b95670c9> in <module>() 
     6 e.to_excel(writer, sheet_name='Sheet1', startrow=21) 
     7 f.to_excel(writer, sheet_name='Sheet1', startrow=21, startcol=6) 
----> 8 writer.write('A1', 'Retail Kelontong') 
     9 writer.write('G1', 'Pulsa') 
    10 writer.write('A11', 'Minimarket') 


AttributeError: '_XlsxWriter' object has no attribute 'write' 

我仍然不知道我做错了什么，我试图改变作家Sheet1和错误消息是

name 'Sheet1' is not defined 

我没有想法什么尝试

Answer 1

的根据文档，你必须使用纸张的write方法，而不是作家：

writer = pd.ExcelWriter('Summary Data Request 8 weeks.xlsx') 
a.to_excel(writer, sheet_name='Sheet1', startrow=1) 
b.to_excel(writer, sheet_name='Sheet1', startcol=7) 
c.to_excel(writer, sheet_name='Sheet1', startrow=11) 
d.to_excel(writer, sheet_name='Sheet1', startrow=11, startcol=6) 
e.to_excel(writer, sheet_name='Sheet1', startrow=21) 
f.to_excel(writer, sheet_name='Sheet1', startrow=21, startcol=6) 
# Get sheet1 
sheet1 = writer.sheets['Sheet1'] 
sheet1.write('A1', 'Retail Kelontong') 
sheet1.write('G1', 'Pulsa') 
sheet1.write('A11', 'Minimarket') 
sheet1.write('G11', 'Retail dengan barcode') 
sheet1.write('A21', 'Retail dengan barcode 2') 
sheet1.write('G21', 'Food and Beverage') 
writer.save()