Pandas Advanced Groupby And Filter By Date
Create the output dataframe from input, how to filter for rows when target == 1 for the first time for each id, or in order words removing consecutive occurrence for each ids where
Solution 1:
You could keep only the rows in the groupby where the cumsum of target is <= 1, then group again and make sure that a zero after a one is dropped using .ne
import pandas as pd
df = pd.DataFrame({'ID': ['a1', 'a1', 'a1', 'a1', 'a1', 'a2', 'a2', 'a2', 'a2'],
 'date': ['2019-11-01',
  '2019-12-01',
  '2020-01-01',
  '2020-02-01',
  '2020-03-01',
  '2019-11-01',
  '2019-12-01',
  '2020-03-01',
  '2020-04-01'],
 'target': [0, 0, 1, 1, 0, 0, 1, 0, 1]})
df = df.loc[df.groupby('ID')['target'].cumsum()<=1]
df = df.loc[df.groupby('ID')['target'].shift(1).ne(1)]
Output
IDdatetarget0a12019-11-01  01a12019-12-01  02a12020-01-01  15a22019-11-01  06a22019-12-01  1Solution 2:
from io import stringIO
data = StringIO("""
uid,  date,         target
a1,   2019-11-01,   0
a1,   2019-12-01,   0
a1,   2020-01-01,   1
a1,  2020-02-01,   1
a1,   2020-03-01,   0
a2,   2019-11-01,   0
a2,   2019-12-01,   1
a2,   2020-03-01,   0
a2,   2020-04-01,   1
"""
)
df = pd.read_csv(data).rename(columns=lambda x: x.strip())
deffilter_in_group(df: pd.DataFrame):
  ind = np.argmax(df.target)
  return df.loc[:, ['date', 'target']].iloc[:ind+1]
df_filtered = (
df
.groupby('uid')
.apply(lambda x: filter_in_group(x))
.reset_index()
.drop('level_1', axis=1)
)
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