I have a df.
import pandas as pd
df = pd.DataFrame({'id_c':[1] * 4 + [2] * 3 + [3] * 4,
'Run':[7,8,5,4,3,2,1,2,3,4,5],
'Date_diff':[4,12,0,0,2,2,10,1,1,3,3]})
id_c Run Date_diff
1 7 4
1 8 12
1 5 0
1 4 0
2 3 2
2 2 2
2 1 10
3 2 1
3 3 1
3 4 3
3 5 3
For each unique value of id_c, if Date_diff equals to 0, 1, 2 for two consecutive rows, I want to keep the row with the maximum value in Run.
I tried:
df.groupby(['id_c' , 'Date_diff'])['Run'].idxmax()]
But it also selects maximum values for values of Date_diff different than 0, 1, 2.
The desired output would be:
id_c Run Date_diff
1 7 4
1 8 12
1 5 0
2 3 2
2 1 10
3 3 1
3 4 3
3 5 3
Thanks!
IIUC, compute a custom group and get the max index per group, then slice:
# get values not in 0/1/2
mask = ~df['Date_diff'].isin([0,1,2])
# group the consecutive 0/1/2 and get id of max Run
idx = df.groupby(['id_c', (mask|mask.shift()).cumsum()])['Run'].idxmax().values
# slice output with max ids
out = df.loc[idx]
output:
id_c Run Date_diff
0 1 7 4
1 1 8 12
2 1 5 0
4 2 3 2
6 2 1 10
8 3 3 1
9 3 4 3
10 3 5 3
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