pandas列包含对象列表，根据键名拆分此列，并将值存储为逗号分隔值

Question

我有一个包含列的数据帧：

A
[{"A": 28, "B": "abc"},{"A": 29, "B": "def"},{"A": 30, "B": "hij"}]
[{"A": 31, "B": "hij"},{"A": 32, "B": "abc"}]
[{"A": 28, "B": "abc"}]
[{"A": 28, "B": "abc"},{"A": 29, "B": "def"},{"A": 30, "B": "hij"}]
[{"A": 28, "B": "abc"},{"A": 29, "B": "klm"},{"A": 30, "B": "nop"}]
[{"A": 28, "B": "abc"},{"A": 29, "B": "xyz"}]

输出应为：

^{pr2}$

如何根据键名将对象列表拆分为列，并将它们存储为上面所示的逗号分隔值。在

Answer 1

我假设A是一个dict列表

A = [
    [{"A": 28, "B": "abc"},{"A": 29, "B": "def"},{"A": 30, "B": "hij"}],
    [{"A": 31, "B": "hij"},{"A": 32, "B": "abc"}],
    [{"A": 28, "B": "abc"}],
    [{"A": 28, "B": "abc"},{"A": 29, "B": "def"},{"A": 30, "B": "hij"}],
    [{"A": 28, "B": "abc"},{"A": 29, "B": "klm"},{"A": 30, "B": "nop"}],
    [{"A": 28, "B": "abc"},{"A": 29, "B": "xyz"}]
]

我要做的第一件事就是用理解来编一本新词典。然后','.join在groupby内

^{pr2}$

Answer 2

我想我要试试这个。首先，永远不要在可以避免的地方使用eval。更好的解决方案是使用ast：

import ast
df.A = df.A.apply(ast.literal_eval)

接下来，展开列：

^{pr2}$

现在，使用i的间隔执行groupby。在

idx = pd.cut(df.index, bins=np.append([0], i), include_lowest=True, right=False)
df = df.groupby(idx, as_index=False).agg(','.join)

df

          A            B
0  28,29,30  abc,def,hij
1     31,32      hij,abc
2        28          abc
3  28,29,30  abc,def,hij
4  28,29,30  abc,klm,nop
5     28,29      abc,xyz

得到了巴拉斯的一点帮助。在

---

替代IntervalIndex（proposed by Wen）的另一个很酷的方法是使用np.put：

i = df.A.str.len().cumsum()  
df = pd.DataFrame.from_dict(np.concatenate(df.A).tolist())
df.A = df.A.astype(str)

v = pd.Series(0, index=df.index)
np.put(v, i-1, [1] * len(i))

df = df.groupby(v[::-1].cumsum()).agg(','.join)[::-1].reset_index(drop=True)

df

          A            B
0  28,29,30  abc,def,hij
1     31,32      hij,abc
2        28          abc
3  28,29,30  abc,def,hij
4  28,29,30  abc,klm,nop
5     28,29      abc,xyz

---

性能

df = pd.concat([df] * 1000, ignore_index=True)

%%timeit 
df.A.apply(pd.Series).stack().\
     apply(pd.Series).groupby(level=0).\
        agg(lambda x :','.join(x.astype(str)))

1 loop, best of 3: 8.76 s per loop

%%timeit 
A = df.A.values.tolist()
B = {
    (i, j, k): v
    for j, row in enumerate(A)
    for i, d in enumerate(row)
    for k, v in d.items()
}    
pd.Series(B).astype(str).groupby(level=[1, 2]).apply(','.join).unstack()

1 loop, best of 3: 2.08 s per loop

%%timeit
i = df.A.str.len().cumsum() 
df2 = pd.DataFrame.from_dict(np.concatenate(df.A).tolist())
df2.A = df2.A.astype(str)
idx = pd.cut(df2.index, bins=np.append([0], i), include_lowest=True, right=False)
df2.groupby(idx, as_index=False).agg(','.join)

1 loop, best of 3: 810 ms per loop

%%timeit
i = df.A.str.len().cumsum() 
df2 = pd.DataFrame.from_dict(np.concatenate(df.A).tolist())
df2.A = df2.A.astype(str)
v = pd.Series(0, index=df2.index)
np.put(v, i-1, [1] * len(i))
df2.groupby(v[::-1].cumsum()).agg(','.join)[::-1].reset_index(drop=True)

1 loop, best of 3: 548 ms per loop

Answer 3

通过使用stack，然后使用groupby

df.A.apply(pd.Series).stack().\
     apply(pd.Series).groupby(level=0).\
        agg(lambda x :','.join(x.astype(str)))
Out[457]: 
          A            B
0  28,29,30  abc,def,hij
1     31,32      hij,abc
2        28          abc
3  28,29,30  abc,def,hij
4  28,29,30  abc,klm,nop

数据输入：

^{pr2}$

对于您的附加问题，请阅读csv

import ast
df=pd.read_csv(r'your.csv',dtype={'A':object})

df['A'] = df['A'].apply(ast.literal_eval)