如何拆分字典以分隔熊猫中的行

Question

>我有一个这样的数据帧：

| col1 | d
-------------------------------------------
0 | A    | {'student99': [[0.83, "nice"]]}
1 | B    | {'student99': [[0.84, "great"], [0.89, "cool"]]}
2 | C    | {'student98': [[0.85, "amazing"]], 'student97': [[0.9, "crazy"]]}

我正在尝试转换为数据帧，例如：

| col1 | student  | grade | comment
---------------------------------------------
0 | A    | student99| 0.83  | nice
1 | B    | student99| 0.84  | great
2 | B    | student99| 0.89  | cool 
3 | C    | student98| 0.85  | amazing 
4 | C    | student97| 0.9   | crazy

如您所见，我需要d列拆分为student、grade和comment列，我需要根据d列中的键数(如上面的 C 行(和每个键的列表数量(如上面的 B 行(将行拆分为一些行。

我该怎么做？

在评论之后，我注意到数据以下一个格式作为 JSON 到达(我将其转换为数据帧(：

{"A": {"d" : {'student99': [[0.83, "nice"]]}}, 
"B": {"d" : {'student99': [[0.84, "great"], [0.89, "cool"]]},
"C": {"d" : {'student98': [[0.85, "amazing"]], 'student97': [[0.9, "crazy"]]}
}

Answer 1

我们可以对pd.Series执行explode，然后重新创建数据帧join它回来

s=df.pop('d').apply(pd.Series).stack().explode()
df_add=pd.DataFrame({'student':s.index.get_level_values(1),
'grade':s.str[0].values,
'comment':s.str[1].values},
index=s.index.get_level_values(0))
df=df.join(df_add,how='right')
df
Out[169]: 
col1    student  grade  comment
0  A      student99   0.83     nice
1  B      student99   0.84    great
1  B      student99   0.89     cool
2  C      student98   0.85  amazing
2  C      student97   0.90    crazy

Answer 2

@YOBEN_S的解决方案很棒;这是对更快解决方案的尝试：

from itertools import product, chain
#chain.... is long... flatten is shorter
#and still gets the point accross
flatten = chain.from_iterable
#flatten the product of each key,value pair 
#in the dictionary
m = [(first, flatten(product([key], value) for key, value in last.items()))
for first, last in emp]
#flatten again
phase2 = flatten(product(first, last) for first, last in m)
#at this point we have 
#the column entry("A","B",...)
#and the flattened entries in the dict
#so, unpacking solves this
phase3 = [(first,second, *last) for first, (second,last) in phase2]
result = pd.DataFrame(phase3, columns = ["col1","student","grade","comment"])
result

col1    student grade   comment
0   A   student99   0.83    nice
1   B   student99   0.84    great
2   B   student99   0.89    cool
3   C   student98   0.85    amazing
4   C   student97   0.90    crazy