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'
请您参考如下方法:
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
