我正在尝试assign组中unique pandas的值df 3。对于下面的df，我有一个Column的值出现了很多次。我计算了这些值中有多少是当前发生的。也就是说，如果它们再次出现，则将它们视为 on 。
如果出现新值，它将增加打开的值的数量。如果没有再出现一个值，则会减少启用的值的数量。
这是我在下面的尝试。我可以把它分成三组。但是它不考虑unique的值。
import pandas as pd
import numpy as np
d = ({
'Place' : ['House 1','House 2','House 3','House 4','House 1','House 2','House 3','House 4'],#,'House 1','House 2']#,'House 4','House 5','House 6','House 7'],
'On' : [1,2,3,4,4,3,2,1],
})
df = pd.DataFrame(data=d)
df["Ind"] = np.ceil(df["On"]/3)
Output:
Place On P
0 House 1 1 1.0
1 House 2 2 1.0
2 House 3 3 1.0
3 House 4 4 2.0
4 House 1 4 2.0
5 House 2 3 1.0
6 House 3 2 1.0
7 House 4 1 1.0
预期输出：
Place On P
0 House 1 1 1.0
1 House 2 2 1.0
2 House 3 3 1.0
3 House 4 4 2.0
4 House 1 4 1.0
5 House 2 3 1.0
6 House 3 2 1.0
7 House 4 1 1.0
说明：
区别在于Index 4。应将Houses 1分配给1，就像它们最初分配给此integer一样。
On列的说明：
Index 0: House 1 is inserted and it appears again = 1
Index 1: House 2 is inserted and it appears again = 2
Index 2: House 3 is inserted and it appears again = 3
Index 3: House 4 is inserted and it appears again = 4
Index 4: House 1 doesn't appear again so it will decrease number of values on on the next row = 4
Index 5: House 2 doesn't appear again so it will decrease number of values on = 3
Index 6: House 3 doesn't appear again so it will decrease number of values on = 2
Index 7: House 4 doesn't appear again so it will decrease number of values on = 1