python - 如何在一行数据帧中解压列表内的多个字典对象？

Question

我有一个数据框，其中每行和每行的单个列表中包含以下字典，该列表的大小不同，它们的大小不同，如下所示:

ID    unnest_column

1     [{'abc': 11, 'def': 1},{'abc': 15, 'def': 1},
     {'abc': 16, 'def': 1},
     {'abc': 17, 'def': 1},
     {'abc': 18, 'def': 1, 'ghi': 'abc'},
     {'abc': 23, 'def': 'xxx', 'def': 1},
     {'abc': 23, 'def': 'xxx', 'def': 2},
     {'abc': 23, 'def': 'xxx', 'def': 4}]


2   [{'abc': 11, 'def': 1}]

如何解压列表中的字典并创建键值列？

可能有新的 df，不确定它到底是什么样子，只需要将键插入列:

id    abc    def     ghi

1       2             3         abc

Answer 1

IIUC，来自

df = pd.DataFrame()
df['x'] = [[{'QuestionId': 11, 'ResponseId': 1},{'QuestionId': 15, 'ResponseId': 1},
 {'QuestionId': 16, 'ResponseId': 1},
 {'QuestionId': 17, 'ResponseId': 1},
 {'QuestionId': 18, 'ResponseId': 1, 'Value': 'abc'},
 {'QuestionId': 23, 'DataLabel': 'xxx', 'ResponseId': 1},
 {'QuestionId': 23, 'DataLabel': 'xxx', 'ResponseId': 2},
 {'QuestionId': 23, 'DataLabel': 'xxx', 'ResponseId': 4}],
[{'QuestionId': 11, 'ResponseId': 1}]]

您可以sum列表来聚合它们，并使用DataFrame构造函数

new_df = pd.DataFrame(df.x.values.sum())


    DataLabel   QuestionId  ResponseId  Value
0   NaN         11          1           NaN
1   NaN         15          1           NaN
2   NaN         16          1           NaN
3   NaN         17          1           NaN
4   NaN         18          1           abc
5   xxx         23          1           NaN
6   xxx         23          2           NaN
7   xxx         23          4           NaN
8   NaN         11          1           NaN

如果您想维护原始索引，可以构建一个 inds 列表并将其作为参数传递给构造函数:

inds = [index for _ in ([i] * len(v) for i,v in df.x.iteritems()) for index in _]
pd.DataFrame(df.x.values.sum(), index=inds)

    DataLabel   QuestionId  ResponseId  Value
0   NaN         11          1           NaN
0   NaN         15          1           NaN
0   NaN         16          1           NaN
0   NaN         17          1           NaN
0   NaN         18          1           abc
0   xxx         23          1           NaN
0   xxx         23          2           NaN
0   xxx         23          4           NaN
1   NaN         11          1           NaN