Question

有这个：

                                  items, name
0   { [{'a': 2, 'b': 1}, {'a': 4, 'b': 3}], this }
1   { [{'a': 2, 'b': 1}, {'a': 4, 'b': 3}], that }

但是希望将字典对象列表分解成（展平？）到这样的实际行中：

    a, b, name
0   { 2, 1, this}
1   { 4, 3, this}
0   { 2, 1, that}
1   { 4, 3, that}

尝试使用melt但没有运气，有什么想法吗？建议？

Answer 1

pd.concat([pd.DataFrame(df1.iloc[0]) for x,df1 in df.groupby('name').group],keys=df.name)\
     .reset_index().drop('level_1',1)
Out[63]: 
   name  a  b
0  this  2  1
1  this  4  3
2  that  2  1
3  that  4  3

数据输入

df = pd.DataFrame({ "group":[[{'a': 2, 'b': 1}, {'a': 4, 'b': 3}],[{'a': 2, 'b': 1}, {'a': 4, 'b': 3}]],
                   "name": ['this', 'that']})

Answer 2

使用concat的另一种方式可能更干净：

In [11]: pd.concat(df.group.apply(pd.DataFrame).tolist(), keys=df["name"])
Out[11]:
        a  b
name
this 0  2  1
     1  4  3
that 0  2  1
     1  4  3

In [12]: pd.concat(df.group.apply(pd.DataFrame).tolist(), 
                        keys=df["name"]).reset_index(level="name")
Out[12]:
   name  a  b
0  this  2  1
1  this  4  3
0  that  2  1
1  that  4  3

Answer 3

ab = pd.DataFrame.from_dict(np.concatenate(df['items']).tolist())
lens = df['items'].str.len()
rest = df.drop('items', 1).iloc[df.index.repeat(lens)].reset_index(drop=True)
ab.join(rest)

   a  b  name
0  2  1  this
1  4  3  this
2  2  1  that
3  4  3  that

Answer 4

tmp_list = list()
for index, row in a.iterrows():
    for list_item in row['items']:
        tmp_list.append(dict(list_item.items()+[('name', row['name'])]))
pd.DataFrame(tmp_list)

   a  b  name
0  2  1  this
1  4  3  this
2  2  1  that
3  4  3  that

Pandas将字典列表分成行

4 个答案: