Pandas：用另一個 DataFrame 替換 NaN

Question

我正在嘗試解決這個問題，所以請幫助我，我有這個數據集：

df1= pd.DataFrame(data={'col1': ['a','b','c','d'],
                              'col2': [1,2,np.nan,4]})
df2=pd.DataFrame(data={'col1': ['a','b','b','a','f','c','e','d','e','a'],
                       'col2':[1,3,2,3,6,4,1,2,5,2]})

df1

  col1  col2
0    a   1.0
1    b   2.0
2    c   NaN
3    d   4.0

df2

  col1  col2
0    a     1
1    b     3
2    b     2
3    a     3
4    f     6
5    c     4
6    e     1
7    d     2
8    e     5
9    a     2

我試過這個

df1[df1['col2'].isna()] = pd.merge(df1, df2, on=['col1'], how='left')

我期待這個

  col1  col2
0    a   1.0
1    b   2.0
2    c   4
3    d   4.0

但相反，我得到了這個

  col1  col2
0    a   1.0
1    b   2.0
2    a   NaN
3    d   4.0

然后我嘗試了這個

for x in zip(df1,df2):
    if x in df1['col2'] == x in df2['col2']:
        df1['col1'][df1['col1'].isna()] = df2['col1'].where(df1['col2'][x] == df2['col2'][x])

但得到了這個

  col1  col2
0    a   1.0
1    b   2.0
2    c   NaN
3    d   4.0

我也試過這個答案

但仍然沒有

Answer 1

使用Series.map將col1 col1 Series匹配DataFrame.drop_duplicates並僅用Series.fillna替換缺失值：

s = df2.drop_duplicates('col1').set_index('col1')['col2']
print (s)
col1
a    1
b    3
f    6
c    4
e    1
d    2
Name: col2, dtype: int64

print (df1['col1'].map(s))
0    1
1    3
2    4
3    2
Name: col1, dtype: int64

df1['col2'] = df1['col2'].fillna(df1['col1'].map(s))
print (df1)
  col1  col2
0    a   1.0
1    b   2.0
2    c   4.0
3    d   4.0