将pandas dataframe单元格中的字典解析为新的行单元格(新列)

Question

我有一个Pandas Dataframe,其中包含一列包含key:value对字典的单元格,如下所示:

{"name":"Test Thorton","company":"Test Group","address":"10850 Test #325\r\n","city":"Test City","state_province":"CA","postal_code":"95670","country":"USA","email_address":"test@testtest.com","phone_number":"999-888-3333","equipment_description":"I'm a big red truck\r\n\r\nRSN# 0000","response_desired":"week","response_method":"email"}

我正在尝试解析字典,因此生成的Dataframe包含每个键的新列,并使用每列的结果值填充该行,如下所示:

//Before

1  2  3  4  5
a  b  c  d  {6:y, 7:v}

//After

1  2  3  4  5           6  7
a  b  c  d  {6:y, 7:v}  y  v

建议非常感谢.

Answer 1

考虑 df

df = pd.DataFrame([
        ['a', 'b', 'c', 'd', dict(F='y', G='v')],
        ['a', 'b', 'c', 'd', dict(F='y', G='v')],
    ], columns=list('ABCDE'))

df

   A  B  C  D                     E
0  a  b  c  d  {'F': 'y', 'G': 'v'}
1  a  b  c  d  {'F': 'y', 'G': 'v'}

选项1
使用pd.Series.apply,分配新列

df.E.apply(pd.Series)

   F  G
0  y  v
1  y  v

像这样分配它

df[['F', 'G']] = df.E.apply(pd.Series)
df.drop('E', axis=1)

   A  B  C  D  F  G
0  a  b  c  d  y  v
1  a  b  c  d  y  v

选项2
使用该pd.DataFrame.assign方法管理整个事物

df.drop('E', 1).assign(**pd.DataFrame(df.E.values.tolist()))

   A  B  C  D  F  G
0  a  b  c  d  y  v
1  a  b  c  d  y  v

Answer 2

我认为你可以使用concat：

df = pd.DataFrame({1:['a','h'],2:['b','h'], 5:[{6:'y', 7:'v'},{6:'u', 7:'t'}] })

print (df)
   1  2                 5
0  a  b  {6: 'y', 7: 'v'}
1  h  h  {6: 'u', 7: 't'}

print (df.loc[:,5].values.tolist())
[{6: 'y', 7: 'v'}, {6: 'u', 7: 't'}]

df1 = pd.DataFrame(df.loc[:,5].values.tolist())
print (df1)
   6  7
0  y  v
1  u  t

print (pd.concat([df, df1], axis=1))
   1  2                 5  6  7
0  a  b  {6: 'y', 7: 'v'}  y  v
1  h  h  {6: 'u', 7: 't'}  u  t

时间( len(df)=2k):

In [2]: %timeit (pd.concat([df, pd.DataFrame(df.loc[:,5].values.tolist())], axis=1))
100 loops, best of 3: 2.99 ms per loop

In [3]: %timeit (pir(df))
1 loop, best of 3: 625 ms per loop

df = pd.concat([df]*1000).reset_index(drop=True)

print (pd.concat([df, pd.DataFrame(df.loc[:,5].values.tolist())], axis=1))


def pir(df):
    df[['F', 'G']] = df[5].apply(pd.Series)
    df.drop(5, axis=1)
    return df

print (pir(df))