python - 如何分组 ndarray？

Question

我有 DataFrame(只是一个例子)

D = pd.DataFrame({i: {"name": str(i),
                      "vector": np.arange(i + i % 4, i + i % 4 + 10),
                      "sq": i ** 2,
                      "gp": i % 2} for i in range(10)}).T

  gp name  sq                                    vector
0  0    0   0            [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
1  1    1   1          [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
2  0    2   4        [4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
3  1    3   9      [6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
4  0    4  16        [4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
5  1    5  25      [6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
6  0    6  36    [8, 9, 10, 11, 12, 13, 14, 15, 16, 17]
7  1    7  49  [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
8  0    8  64    [8, 9, 10, 11, 12, 13, 14, 15, 16, 17]
9  1    9  81  [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

我想按列向量分组，然后按 gp 列分组。我该怎么做？

from dfply import *
D >>\
    groupby(X.vector, X.gp) >>\
    summarize(b=X.sq.sum())

结果

TypeError: unhashable type: 'numpy.ndarray'

Answer 1

我认为您需要先在 pandas 中将列 vector 转换为元组:

print(D['sq'].groupby([D['vector'].apply(tuple), D['gp']]).sum().reset_index())
                                     vector  gp   sq
0            (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)   0    0
1          (2, 3, 4, 5, 6, 7, 8, 9, 10, 11)   1    1
2        (4, 5, 6, 7, 8, 9, 10, 11, 12, 13)   0   20
3      (6, 7, 8, 9, 10, 11, 12, 13, 14, 15)   1   34
4    (8, 9, 10, 11, 12, 13, 14, 15, 16, 17)   0  100
5  (10, 11, 12, 13, 14, 15, 16, 17, 18, 19)   1  130

另一种解决方案是先转换列:

D['vector'] = D['vector'].apply(tuple)
print(D.groupby(['vector','gp'])['sq'].sum().reset_index())
                                     vector  gp   sq
0            (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)   0    0
1          (2, 3, 4, 5, 6, 7, 8, 9, 10, 11)   1    1
2        (4, 5, 6, 7, 8, 9, 10, 11, 12, 13)   0   20
3      (6, 7, 8, 9, 10, 11, 12, 13, 14, 15)   1   34
4    (8, 9, 10, 11, 12, 13, 14, 15, 16, 17)   0  100
5  (10, 11, 12, 13, 14, 15, 16, 17, 18, 19)   1  130

Anf 如果需要，最后转换为 array 返回:

D['vector'] = D['vector'].apply(tuple)
df = D.groupby(['vector','gp'])['sq'].sum().reset_index()
df['vector'] = df['vector'].apply(np.array)
print (df)
                                     vector  gp   sq
0            [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]   0    0
1          [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]   1    1
2        [4, 5, 6, 7, 8, 9, 10, 11, 12, 13]   0   20
3      [6, 7, 8, 9, 10, 11, 12, 13, 14, 15]   1   34
4    [8, 9, 10, 11, 12, 13, 14, 15, 16, 17]   0  100
5  [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]   1  130

print (type(df['vector'].iat[0]))
<class 'numpy.ndarray'>

我尝试使用您的代码并且对我有用:

from dfply import *

D['vector'] = D['vector'].apply(tuple)
a = D >> groupby(X.vector, X.gp) >> summarize(b=X.sq.sum())
a['vector'] = a['vector'].apply(np.array)
print (a)
   gp                                    vector    b
0   0            [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]    0
1   1          [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]    1
2   0        [4, 5, 6, 7, 8, 9, 10, 11, 12, 13]   20
3   1      [6, 7, 8, 9, 10, 11, 12, 13, 14, 15]   34
4   0    [8, 9, 10, 11, 12, 13, 14, 15, 16, 17]  100
5   1  [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]  130