dcm*_*m88 5 python merge dataframe pandas
我有2个数据帧,两个都有一个可能有重复的键列,但数据帧大多数都有相同的重复键.我想在这个键上合并这些数据帧,但是当这两个数据帧具有相同的重复时,这些重复项将分别合并.此外,如果一个数据帧的密钥重复多于另一个,我希望它的值可以填充为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
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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
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所以基本上,我想将重复的K2键视为K2_1,K2_2,......然后在数据帧上进行how ='outer'合并.我有什么想法可以做到这一点?
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)
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更快的答案; 不可扩展
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)
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老答案
df1['cc'] = df1.groupby('key').cumcount()
df2['cc'] = df2.groupby('key').cumcount()
df1.merge(df2, how='outer').drop('cc', 1)
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