Dataframe 来自字典列表的字典？

Question

我有一个字典列表的字典。 将其转换为DataFrame中的 DataFrame 的最有效方法是什么？

data = {
     "0a2":[{"a":1,"b":1},{"a":1,"b":1,"c":1},{"a":1,"b":1}],
     "279":[{"a":1,"b":1,"c":1},{"a":1,"b":1,"d":1}],
     "ae2":[{"a":1,"b":1},{"a":1,"d":1},{"a":1,"b":1},{"a":1,"d":1}], 
     #...
}
import pandas as pd
pd.DataFrame(data, columns=["a","b","c","d"])

我试过的：

一种解决方案是通过复制“id”键来像这样对数据进行非规范化：

bad_data = [
      {"a":1,"b":1,"id":"0a2"},{"a":1,"b":1,"c":1,"id":"0a2"},{"a":1,"b":1,"id":"0a2"},
      {"a":1,"b":1,"c":1,"id":"279"},{"a":1,"b":1,"d":1,"id":"279"},
      {"a":1,"b":1,"id":"ae2"},{"a":1,"d":1,"id":"ae2"},{"a":1,"b":1,"id":"ae2"},{"a":1,"d":1,"id":"ae2"}
]
pd.DataFrame(bad_data, columns=["a","b","c","d","id"])

但是我的数据非常大，所以我更喜欢其他一些分层索引解决方案。

Answer 1

IIUC，你可以做（推荐）

new_df = pd.concat((pd.DataFrame(d) for d in data.values()), keys=data.keys())

Output：

       a    b    c    d
0a2 0  1  1.0  NaN  NaN
    1  1  1.0  1.0  NaN
    2  1  1.0  NaN  NaN
279 0  1  1.0  1.0  NaN
    1  1  1.0  NaN  1.0
ae2 0  1  1.0  NaN  NaN
    1  1  NaN  NaN  1.0
    2  1  1.0  NaN  NaN
    3  1  NaN  NaN  1.0

或者

pd.concat(pd.DataFrame(v).assign(ID=k) for k,v in data.items())

Output：

   a    b    c   ID    d
0  1  1.0  NaN  0a2  NaN
1  1  1.0  1.0  0a2  NaN
2  1  1.0  NaN  0a2  NaN
0  1  1.0  1.0  279  NaN
1  1  1.0  NaN  279  1.0
0  1  1.0  NaN  ae2  NaN
1  1  NaN  NaN  ae2  1.0
2  1  1.0  NaN  ae2  NaN
3  1  NaN  NaN  ae2  1.0