包含對象列表的pandas列，根據鍵名拆分此列，並將值存儲為逗號分隔的值

Question

我有一個包含列的數據框：

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"}]

輸出應為：

A              B
28,29,30       abc,def,hij
31,32          hij,abc
28             abc
28,29,30       abc,def,hij
28,29,30       abc,klm,nop
28,29          abc,xyz

我如何根據鍵名將對象列表分為幾列，並將它們存儲為逗號分隔的值，如上所示。

Answer 1

通過使用stack然后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

數據輸入：

df=pd.DataFrame({'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"}]]})

對於您的其他問題，請從csv中閱讀

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

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

Answer 2

我以為A是字典列表

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"}]
]

我要做的第一件事是使用理解能力來創建新詞典。 然后','.join groupby

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()

          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

Answer 3

以為我會為此開槍。 首先， 切勿在可以避免使用eval地方使用它。 更好的解決方案是使用ast ：

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

接下來，將您的列展平：

i = df.A.str.len().cumsum()   # we'll need this later
df = pd.DataFrame.from_dict(np.concatenate(df.A).tolist())
df.A = df.A.astype(str)

df

     A    B
0   28  abc
1   29  def
2   30  hij
3   31  hij
4   32  abc
5   28  abc
6   28  abc
7   29  def
8   30  hij
9   28  abc
10  29  klm
11  30  nop
12  28  abc
13  29  xyz

現在，使用由i間隔執行groupby 。

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

在這里有巴拉特的幫助。

---

IntervalIndex （ 由Wen提出 ）的一個很酷的替代方案是使用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