有点老了，但我是被指引到这里的。有办法把它按多个不同的列分组吗?

"column1", "column2", "column3"
"foo", "val1", 3
"foo", "val2", 0
"foo", "val2", 3
"bar", "other", 99

:

"column1", "column2", "column3"
"foo", "val1", [ 3 ]
"foo", "val2", [ 0, 3 ]
"bar", "other", [ 99 ]

只是把之前的答案加起来，在我的情况下，我想要列表和其他函数，如min和max。这样做的方法是:

df = pd.DataFrame({
    'a':['A','A','B','B','B','C'], 
    'b':[1,2,5,5,4,6]
})

df=df.groupby('a').agg({
    'b':['min', 'max',lambda x: list(x)]
})

#then flattening and renaming if necessary
df.columns = df.columns.to_flat_index()
df.rename(columns={('b', 'min'): 'b_min', ('b', 'max'): 'b_max', ('b', '<lambda_0>'): 'b_list'},inplace=True)

如果性能是重要的，下降到numpy级别:

import numpy as np

df = pd.DataFrame({'a': np.random.randint(0, 60, 600), 'b': [1, 2, 5, 5, 4, 6]*100})

def f(df):
         keys, values = df.sort_values('a').values.T
         ukeys, index = np.unique(keys, True)
         arrays = np.split(values, index[1:])
         df2 = pd.DataFrame({'a':ukeys, 'b':[list(a) for a in arrays]})
         return df2

测试:

In [301]: %timeit f(df)
1000 loops, best of 3: 1.64 ms per loop

In [302]: %timeit df.groupby('a')['b'].apply(list)
100 loops, best of 3: 5.26 ms per loop

我们用df。带有列表和系列构造函数的groupby

pd.Series({x : y.b.tolist() for x , y in df.groupby('a')})
Out[664]: 
A       [1, 2]
B    [5, 5, 4]
C          [6]
dtype: object

有点老了，但我是被指引到这里的。有办法把它按多个不同的列分组吗?

"column1", "column2", "column3"
"foo", "val1", 3
"foo", "val2", 0
"foo", "val2", 3
"bar", "other", 99

:

"column1", "column2", "column3"
"foo", "val1", [ 3 ]
"foo", "val2", [ 0, 3 ]
"bar", "other", [ 99 ]

你可以使用groupby对感兴趣的列进行分组，然后将list应用到每个组:

In [1]: df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6]})
        df

Out[1]: 
   a  b
0  A  1
1  A  2
2  B  5
3  B  5
4  B  4
5  C  6

In [2]: df.groupby('a')['b'].apply(list)
Out[2]: 
a
A       [1, 2]
B    [5, 5, 4]
C          [6]
Name: b, dtype: object

In [3]: df1 = df.groupby('a')['b'].apply(list).reset_index(name='new')
        df1
Out[3]: 
   a        new
0  A     [1, 2]
1  B  [5, 5, 4]
2  C        [6]