如何根据规则提取独特的行

Question

我有一个像这样的数据框架

id_1，date_1，id_2，date_2

我需要数据框，其中行（date_1 15天）＆lt;date_2如果此规则匹配，我只需要首先出现

仅使用布尔面具无法解决问题

所以我认为我可能需要使用某种for index, row in df.iterrows():并创建新的DataFrame

Answer 1

import pandas as pd
from datetime import timedelta
df = pd.DataFrame(data=dd)
df[['date_1', 'date_2']] = df[['date_1', 'date_2']].apply(pd.to_datetime)
df['date_1_15'] = df['date_1'] + timedelta(4)
def apply_mask(row):
    if row['date_1_15'] < row['date_2']:
        row['mask'] = True
    else:
        row['mask'] = False
    return row
df = df.apply(lambda row: apply_mask(row), axis=1)
dx = df.loc[df['mask'] == True]
dx = dx.groupby(['date_1']).first()
dx['mask_first'] = True
dx = dx.reset_index()
dx = dx[['date_1', 'date_2', 'mask_first']]
df = pd.merge(df, dx, on=['date_1', 'date_2'], how='outer')

Answer 2

import pandas as pd
from datetime import timedelta
df = pd.DataFrame(data={'id_1':[1,2,3,4], 
                        'date1': ['2018-01-10', '2018-02-05', '2018-02-20', '2018-02-21'],
                        'date2': ['2018-01-11', '2018-02-15', '2018-02-27', '2018-02-22']})

df[['date1', 'date2']] = df[['date1', 'date2']].apply(pd.to_datetime)

df['date1_15'] = df['date1'] + timedelta(15)
df = df.loc[df['date1_15'] < df['date2']].head(1)