如何基于共享公共密钥的另一个数据帧的值来“更新”df？python

Question

如何基于共享公共密钥的另一个数据帧的值来“更新”a12？ 在下面的示例中，公共键是列a

a12=

  a  b  c
  0  1  1
  1  na na

try10=

  a  b  c
  1  1  1

当我使用合并，我得到这样的东西。你知道吗

你知道吗pd.合并（a12，try10）=

  a  b  c  b_y  c_y
  0  1  1   na  na
  1  na na  1   1

我决定手动操作，但我想一定有比下面更具Python风格的方法。我很感激你的帮助。你知道吗

for i, val in a12.iterrows():
    for x, xval in try10.iterrows():
        if xval['Firm1'] == val['Firm']: 
            try10.ix[x]['AMranking'] =  val['AMranking']
            try10.ix[x]['numlawyers'] = val['numlawyers']
            try10.ix[x]['grossprofits'] = val['grossprofits']  

del try12['firm']

Answer 1

您可以在merge之前将dropna应用于a12：

In [53]:

a12 = pd.DataFrame({'a':[0, 1], 'b':[1, np.nan], 'c':[1, np.nan]})
try10= pd.DataFrame({'a':[1], 'b':[1], 'c':[1]})
In [54]:

print pd.merge(a12, try10, how='outer', left_on='a', right_on='a')
   a  b_x  c_x  b_y  c_y
0  0    1    1  NaN  NaN
1  1  NaN  NaN    1    1
In [55]:

print pd.merge(a12.dropna(0), try10, how='outer')
   a  b  c
0  0  1  1
1  1  1  1

如果a是索引：

In [57]:

print pd.merge(a12.dropna(0).reset_index(), try10.reset_index(), how='outer').set_index('a')
   b  c
a      
0  1  1
1  1  1