我有以下DataFrame:
purchase_1 = pd.Series({'Name': 'Chris',
'Item Purchased': 'Dog Food',
'Cost': 22.50})
purchase_2 = pd.Series({'Name': 'Kevyn',
'Item Purchased': 'Kitty Litter',
'Cost': 2.50})
purchase_3 = pd.Series({'Name': 'Vinod',
'Item Purchased': 'Bird Seed',
'Cost': 5.00})
df = pd.DataFrame([purchase_1, purchase_2, purchase_3], index=['Store 1', 'Store 1', 'Store 2'])
df['Location'] = df.index
df = df.set_index(['Location', 'Name'])
df2 = df.copy()
print(df2)
Cost Item Purchased
Location Name
Store 1 Chris 22.5 Dog Food
Kevyn 2.5 Kitty Litter
Store 2 Vinod 5.0 Bird Seed
然后我有以下系列:
purchase_4 = pd.Series({'Name': 'Kevyn',
'Item Purchased': 'Kitty Food',
'Cost': 3.00,
'Location': 'Store 2'})
当我尝试将此系列添加到我的DF时,它可以工作但是有大量的NaN:
df2 = df2.append(purchase_4, ignore_index=True)
Cost Item Purchased Location Name
0 22.5 Dog Food NaN NaN
1 2.5 Kitty Litter NaN NaN
2 5.0 Bird Seed NaN NaN
3 3.0 Kitty Food Store 2 Kevyn
答案 0 :(得分:1)
您可以使用concat
:
df2 = pd.concat([df2, purchase_4.to_frame().T.set_index(df.index.names)])
print (df2)
Cost Item Purchased
Location Name
Store 1 Chris 22.5 Dog Food
Kevyn 2.5 Kitty Litter
Store 2 Vinod 5 Bird Seed
Kevyn 3 Kitty Food
或loc
,设置Multiindex使用()
:
df2.loc[(purchase_4['Location'], purchase_4['Name']),:] = purchase_4
print (df2)
Cost Item Purchased
Location Name
Store 1 Chris 22.5 Dog Food
Kevyn 2.5 Kitty Litter
Store 2 Vinod 5.0 Bird Seed
Kevyn 3.0 Kitty Food
答案 1 :(得分:1)
替代解决方案:
In [237]: df.append(purchase_4.to_frame().T.set_index(df.index.names))
Out[237]:
Cost Item Purchased
Location Name
Store 1 Chris 22.5 Dog Food
Kevyn 2.5 Kitty Litter
Store 2 Vinod 5 Bird Seed
Kevyn 3 Kitty Food
答案 2 :(得分:-2)
df = df.set_index([df.index,'Name']) df.index.names = ['位置','名称'] df = df.append(pd.Series(数据= {'成本':3.00,'购买的物品':'凯蒂食品'},名称=('商店2','凯文'))) df