如何根据条件从数据框中删除行

Question

I have the following dataframe with ("ID", "Month" and "status"). Status is regarding "Churn"= 1 and 'Not Churn" = 2. I want to delete all rows for ID's who are already churned except the first appearance. For example:

Dataframe

    ID      Month   Status
    2310    201708  2
    2310    201709  2
    2310    201710  1
    2310    201711  1
    2310    201712  1
    2310    201801  1
    2311    201704  2
    2311    201705  2
    2311    201706  2
    2311    201707  2
    2311    201708  2
    2311    201709  2
    2311    201710  1
    2311    201711  1
    2311    201712  1
    2312    201708  2
    2312    201709  2
    2312    201710  2
    2312    201711  1
    2312    201712  1
    2312    201801  1

After deleting I should have the following dataframe

    ID      Month   Status
    2310    201708  2
    2310    201709  2
    2310    201710  1

    2311    201704  2
    2311    201705  2
    2311    201706  2
    2311    201707  2
    2311    201708  2
    2311    201709  2
    2311    201710  1

    2312    201708  2
    2312    201709  2
    2312    201710  2
    2312    201711  1

I tried the following- first to find min date for each customer ID and status=1

    df1=df[df.Status==1].groupby('ID')['Month'].min()

then I have to delete all rows for each ID with status 1 greater than min value for MOnth.

Answer 1

If you're familiar with DataFrame.idxmin to get the indices of the elements of the most recent month, you could try:

# find minimum months
min_df = df.groupby(['ID','Status'])['Month'].idxmin().reset_index(drop=True)
# find indices of status 2 rows
df2 = df[df['Status'].eq(2)].index.to_series()
# append indices together
idx_df = min_df.append(df2).drop_duplicates()
# filter indices
df_new = df.iloc[idx_df].sort_index()

---

print(df_new)                                                                        
      ID   Month  Status
0   2310  201708       2
1   2310  201709       2
2   2310  201710       1
6   2311  201704       2
7   2311  201705       2
8   2311  201706       2
9   2311  201707       2
10  2311  201708       2
11  2311  201709       2
12  2311  201710       1
15  2312  201708       2
16  2312  201709       2
17  2312  201710       2
18  2312  201711       1

Update

Or, you may think about using GroupBy.apply :

df1 = df.groupby(['ID','Status']).apply(lambda x: (x['Status'].eq(2)) | (x['Month'].eq(x['Month'].min())))
df1 = df1.reset_index(level=['ID','Status'], drop=True)
df_new = df.loc[df1]

---

print(df_new)                                                                                                                                              
      ID   Month  Status
0   2310  201708       2
1   2310  201709       2
2   2310  201710       1
6   2311  201704       2
7   2311  201705       2
8   2311  201706       2
9   2311  201707       2
10  2311  201708       2
11  2311  201709       2
12  2311  201710       1
15  2312  201708       2
16  2312  201709       2
17  2312  201710       2
18  2312  201711       1

Update 2

However, if you're simply wanting to remove all of the status 1 rows that come after the row with the earliest month, then you could simply sort_values and transform :

df = df.sort_values(by=['ID','Month']).reset_index(drop=True) 
df = df[df.groupby('ID')['Status'].transform(lambda x: ~(x.duplicated() & (x == 1)))]

---

print(df)                                                              
      ID   Month  Status
0   2310  201708       2
1   2310  201709       2
2   2310  201710       1
6   2311  201704       2
7   2311  201705       2
8   2311  201706       2
9   2311  201707       2
10  2311  201708       2
11  2311  201709       2
12  2311  201710       1
15  2312  201708       2
16  2312  201709       2
17  2312  201710       2
18  2312  201711       1

Answer 2

IIUC, you can use groupby with transform with boolean logic and then boolean indexing:

df[df.groupby('ID')['Status'].transform(lambda x: ~(x.duplicated() & (x == 1)))]

Output:

      ID   Month  Status
0   2310  201708       2
1   2310  201709       2
2   2310  201710       1
6   2311  201704       2
7   2311  201705       2
8   2311  201706       2
9   2311  201707       2
10  2311  201708       2
11  2311  201709       2
12  2311  201710       1
15  2312  201708       2
16  2312  201709       2
17  2312  201710       2
18  2312  201711       1