运用pandas将字典的列表转化为独立的数据列

在Stack Overflow看到的一个帖子

table0.csv数据集如下：

name	status	number	message
matt	active	12345	[job:  , money: none, wife: none]
james	active	23456	[group: band, wife: yes, money: 10000]
adam	inactive	34567	[job: none, money: none, wife:  , kids: one, group: jail]

但现在我想处理数据得到如下的table:

方法一：

首先通过replace（\s+代表一个及以上空格），将list of dict 转化为set of dict 然后使用ast

import ast
df.message = df.message.replace([':\s+,','\[', '\]', ':\s+', ',\s+'], ['":"none","', '{"', '"}', '":"', '","'], regex=True)
df.message = df.message.apply(ast.literal_eval)
df1 = pd.DataFrame(df.pop('message').values.tolist(), index=df.index)

print (df1)

   kids  money group   job  money  wife
0   NaN   none   NaN  none    NaN  none
1   NaN    NaN  band   NaN  10000   yes
2   one    NaN  jail  none   none  none

问题来了因为‘money’在第二行的message中是第三个dict，不同于其他两行在第二个dict,

因此会产生两列‘money’。这时候需要我们手动修改，不展开了。

所以按正常的操作得到如下：

df=pd.concat([df,df1],axis=1)

print(df)

    name    status  number  kids  money group   job  money  wife
0   matt    active   12345   NaN   none   NaN  none    NaN  none
1  james    active   23456   NaN    NaN  band   NaN  10000   yes
2   adam  inactive   34567   one    NaN  jail  none   none  none

方法二：

使用yaml包

import yaml
df.message = df.message.replace(['\[','\]'],['{','}'], regex=True).apply(yaml.load)
df1 = pd.DataFrame(df.pop('message').values.tolist(), index=df.index)
print (df1)

  group   job kids  money  wife
0   NaN  None  NaN   none  none
1  band   NaN  NaN  10000  True
2  jail  none  one   none  None

df = pd.concat([df, df1], axis=1)

print (df)

    name    status  number group   job kids  money  wife
0   matt    active   12345   NaN  None  NaN   none  none
1  james    active   23456  band   NaN  NaN  10000  True
2   adam  inactive   34567  jail  none  one   none  Non

源地址：

https://stackoverflow.com/questions/43032182/pandas-list-of-dictionary-to-separate-columns