如何展平熊猫数据框多列中的字典列表

Posted 2023-02-23

【中文标题】如何展平熊猫数据框多列中的字典列表

【英文标题】：How to flatten list of dictionaries in multiple columns of pandas dataframe

【发布时间】：2021-08-19 18:06:57

【问题描述】：

我有一个数据框，每条记录都存储一个字典列表，如下所示：

row prodect_id recommend_info
  0 XQ002      ["recommend_key":"XXX567","recommend_point":50,
                "recommend_key":"XXX236","recommend_point":20,
                "recommend_key":"XXX090","recommend_point":35]
  1 XQ003      ["recommend_key":"XXX089","recommend_point":30,
                "recommend_key":"XXX567","recommend_point":20]

我想展平字典列表，使其看起来像这样

row prodect_id recommend_info_recommend_key recommend_info_recommend_point
  0 XQ002      XXX567                       50
  1 XQ002      XXX236                       20
  2 XQ002      XXX090                       35
  3 XQ003      XXX089                       30
  4 XQ003      XXX567                       20

我知道如何仅将一个字典列表转换为数据框。 像这样：

d = ["recommend_key":"XXX089","recommend_point":30,
     "recommend_key":"XXX567","recommend_point":20]

df = pd.DataFrame(d)

row recommend_key recommend_point
  0 XXX089        30
  1 XXX567        20

但是当有一列存储字典列表，或者有多列存储字典列表时，我不知道如何对数据框执行此操作

row  col_a  col_b                  col_c
  0  B001   ["a":"b","a":"c"]  ["y":11,"a":"c"]
  1  D009   ["c":"o","g":"c"]  ["y":11,"a":"c","l":"c"]   
  2  G068   ["c":"b","a":"c"]  ["a":56,"d":"c"]
  3  C004   ["d":"a","b":"c"]  ["c":22,"a":"c","b":"c"]
  4  F011   ["h":"u","d":"c"]  ["h":27,"d":"c"]

【问题讨论】：

如果您喜欢我的回答并认为它有用，请将其标记为已接受

没问题！感谢您的帮助

【参考方案1】：

试试：

pd.concat([df.explode('recommend_info').drop(['recommend_info'], axis=1),
           df.explode('recommend_info')['recommend_info'].apply(pd.Series)],
          axis=1)

你可以对每一列一遍又一遍地做同样的事情

这是一个例子：

>>> df = pd.DataFrame('a': [[3: 4, 5: 6, 3:8, 5: 1],
...                          [3:2, 5:4, 3: 8, 5: 10]],
...                    'b': ['X', "Y"])
>>> df
                               a  b
0   [3: 4, 5: 6, 3: 8, 5: 1]  X
1  [3: 2, 5: 4, 3: 8, 5: 10]  Y
>>> df = pd.concat([df.explode('a').drop(['a'], axis=1),
...                 df.explode('a')['a'].apply(pd.Series)],
...                axis=1)
>>> df
   b  3   5
0  X  4   6
0  X  8   1
1  Y  2   4
1  Y  8  10

【讨论】：

感谢您的回答里卡多！它也适用于非结构化的字典列表吗？