字典中 pandas DataFrame 的列映射错误

Question

I am trying to map 2 DataFrames

L1

ship  city   code  

 NaN   aa      12    
 NaN   bb      23    
 NaN   cc      13    
 NaN   dd      43  

B1

 ship  city    
 
 21     dd     
 32     bb      
 43     aa      
 654    cc      
 34     bb     
 54     aa

 

I want to map column code from L1 to B1 . I tried mapping using a dictionary, so I would get a result like this

Expected_Result =

ship  city   code  
 
 21    dd      43    
 32    bb      23    
 43    aa      12    
 654   cc      13    
 34    bb      23    
 54    aa      12



 code_dict = dict(zip(L1['city'],L1['code'])) 
 B1['code'] = L1['city'].map(code_dict)
 print(B1)

The result that I got is not what I expected.

Please help me fix this issue.

 ship  city    code  
 
 21    dd       12    
 32    bb       23    
 43    aa       13    
 654   cc       43    
 34    bb       NaN   
 54    aa       NaN 

Answer 1

I think you're looking for:

from pandas import DataFrame


df1 = DataFrame([
    {'city': 'aa', 'code': 12},
    {'city': 'bb', 'code': 23},
    {'city': 'cc', 'code': 13},
    {'city': 'dd', 'code': 43}
])

df2 = DataFrame([
    {'ship': '21', 'city': 'dd'},
    {'ship': '32', 'city': 'bb'},
    {'ship': '43', 'city': 'aa'},
    {'ship': '654', 'city': 'cc'},
    {'ship': '34', 'city': 'bb'},
    {'ship': '54', 'city': 'aa'}
])

expected_result = df2.merge(df1, on='city')
print(expected_result)

Result:

  ship city  code
0   21   dd    43
1   32   bb    23
2   34   bb    23
3   43   aa    12
4   54   aa    12
5  654   cc    13

Answer 2

Two options.

data setup:

ship=[np.nan]*4
city=['aa','bb','cc','ddd']
code=[12,23,13,43]

ship2=[21,32,43,654,34,54]
city2=['dd','bb','aa','cc','bb','aa']

l1=pd.DataFrame({"ship":ship,"city":city,"code":code})
b1=pd.DataFrame({"ship":ship2,"city":city2})

1st through merge on city:

l1.merge(b1,on=['city'],how='inner')[['ship_y','city','code']].rename({"ship_y":"ship"},axis=1)

2nd use map/dictionary:

ship_city_d = dict(zip(l1.city, l1.code)) # build the dictionary
b1['code']=b1['city'].apply(lambda x: str(ship_city_d.get(x,"Not Found")) )

note: b1 will have some entries if key not found, you can drop those record as next step if not needed.