试试这个：

# Convert 'Date' column to datetime format (to be sure)
df['Date'] = pd.to_datetime(df['Date'])

# Define cutoff date
cutoff_date = datetime(2023, 5, 1)

# Create a mask to filter IDs flagged before the cutoff date
mask = (df['Flagged'] == 1) & (df['Date'] >= cutoff_date)

# Get the list of IDs to be removed
ids_to_remove = df.loc[mask, 'ID'].unique()

# Filter out the IDs from the DataFrame
filtered_df = df[~df['ID'].isin(ids_to_remove)]

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首先找到在指定日期之前已标记的任何ID，然后只找到既已标记且不具有其中一个坏ID的值：

import pandas as pd
import datetime 

# create dataframe:
df = pd.DataFrame(
    {'ID' : [101, 101, 102, 102, 102, 102, 103, 103, 104, 104, 104, 104],
     'Date' : ['6/4/2023' ,'7/23/2023','4/28/2023','5/2/2023' ,'6/30/2023','7/11/2023','6/23/2023','7/12/2023','4/17/2023','5/12/2023','6/17/2023','7/22/2023'],
     'Flagged' : [0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1]})

# Convert dates
df['Date'] = pd.to_datetime(df.Date, format = '%m/%d/%Y')

# Find bad IDs (ones that had a flag before specific date)
bad_ids = df[(df.Flagged.eq(1)) & (df.Date < datetime.datetime(2023, 5, 1))].ID.unique()

# Find all values that are flagged and do not have a bad ID
df[(df.Flagged.eq(1)) & ~(df.ID.isin(bad_ids))]

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输出量：

ID  Date        Flagged
6   103 2023-06-23  1
7   103 2023-07-12  1
9   104 2023-05-12  1
10  104 2023-06-17  1
11  104 2023-07-22  1

型

使用布尔索引：

#convert to datetime if needed
df["Date"] = pd.to_datetime(df["Date"],format="%m/%d/%Y")

#get flagged IDs to ignore
flagged = df[df["Flagged"].eq(1)&df["Date"].lt(pd.Timestamp.today()-pd.DateOffset(90))]

>>> df[~df["ID"].isin(flagged["ID"])&df["Flagged"].eq(1)]

     ID       Date  Flagged
6   103 2023-06-23        1
7   103 2023-07-12        1
9   104 2023-05-12        1
10  104 2023-06-17        1
11  104 2023-07-22        1

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