合并熊猫数据帧与关键副本

Question

我有两个数据帧，都有一个键列，可能有重复项，但数据帧大部分都有相同的重复键。我希望在该键上合并这些数据帧，但这样做的方式是，当两者具有相同的副本时，这些副本将分别合并。此外，如果一个数据帧的键的副本比另一个多，我希望它的值填充为NaN。例如：

df1 = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K2', 'K2', 'K3'],
                    'A':   ['A0', 'A1', 'A2', 'A3', 'A4', 'A5']}, 
                   columns=['key', 'A'])
df2 = pd.DataFrame({'B':   ['B0', 'B1', 'B2', 'B3', 'B4', 'B5', 'B6'],
                    'key': ['K0', 'K1', 'K2', 'K2', 'K3', 'K3', 'K4']}, 
                   columns=['key', 'B'])

  key   A
0  K0  A0
1  K1  A1
2  K2  A2
3  K2  A3
4  K2  A4
5  K3  A5

  key   B
0  K0  B0
1  K1  B1
2  K2  B2
3  K2  B3
4  K3  B4
5  K3  B5
6  K4  B6

我正在尝试获得以下输出

   key    A   B
0   K0   A0  B0
1   K1   A1  B1
2   K2   A2  B2
3   K2   A3  B3
6   K2   A4  NaN
8   K3   A5  B4
9   K3  NaN  B5
10  K4  NaN  B6

所以基本上，我想把复制的K2键当作K2_1，K2_2。。。然后在数据帧上进行how='outer'合并。你知道我怎样才能做到这一点吗？

Answer 1

再快一点

%%cython
# using cython in jupyter notebook
# in another cell run `%load_ext Cython`
from collections import defaultdict
import numpy as np

def cg(x):
    cnt = defaultdict(lambda: 0)

    for j in x.tolist():
        cnt[j] += 1
        yield cnt[j]


def fastcount(x):
    return [i for i in cg(x)]

df1['cc'] = fastcount(df1.key.values)
df2['cc'] = fastcount(df2.key.values)

df1.merge(df2, how='outer').drop('cc', 1)

更快的回答；不可伸缩

def fastcount(x):
    unq, inv = np.unique(x, return_inverse=1)
    m = np.arange(len(unq))[:, None] == inv
    return (m.cumsum(1) * m).sum(0)

df1['cc'] = fastcount(df1.key.values)
df2['cc'] = fastcount(df2.key.values)

df1.merge(df2, how='outer').drop('cc', 1)

老答案

df1['cc'] = df1.groupby('key').cumcount()
df2['cc'] = df2.groupby('key').cumcount()

df1.merge(df2, how='outer').drop('cc', 1)