你能告诉我一种优化这段代码的方法吗？由于大型数据集，需要几十分钟才能完成...
df['sinistre'] = 0
for index_sin, row_sin in sinistre1.iterrows():
date_surv = row_sin['DATESURV']
quit_sin = df.loc[df['id_police'] == row_sin['id_police']]
for index, row in quit_sin.iterrows():
if row['DATEEFFE'] < date_surv < row['DATE_FIN']:
df['sinistre'][index] = 1
以下是DataFrames sinistre1和df的示例数据集：
>>> sinistre1
id_police id_sinistre DATESURV
0 p123 s123 30/05/2017
1 p123 s124 30/11/2017
2 p123 s125 29/02/2018
3 b123 s126 28/02/2018
4 b123 s127 30/05/2018
>>> df
id_police DATEEFFE DATE_FIN prime prime2
0 p123 24/01/2017 24/02/2017 0 0
1 p123 24/11/2017 24/12/2017 0 30
2 p123 25/02/2018 25/03/2018 10 10
3 b123 24/02/2018 24/03/2018 20 20
4 b123 24/03/2018 24/04/2018 30 0
这是预期的输出（这个想法是当sinistre1的DATESURV在DATEEFFE和DATE_FIN区间内，然后我标记sinistre）：
id_police DATEEFFE DATE_FIN prime prime2 sinistre
0 p123 24/01/2017 24/02/2017 0 0 0
1 p123 24/11/2017 24/12/2017 0 30 1
2 p123 25/02/2018 25/03/2018 10 10 1
3 b123 24/02/2018 24/03/2018 20 20 1
4 b123 24/03/2018 24/04/2018 30 0 0
如果我无法避免循环，请显示更好的循环方式...提前谢谢！
使用：
to_datetime用于转换所有日期时间列sinistre1以获得最后检查成员资格DataFrame.merge和left连接数据Series.between和inclusive=True过滤并获得sinistre1系列中的s列sinistre1用Series.isin检查最终s，转换为整数到True/False到1/0映射：df['DATEEFFE'] = pd.to_datetime(df['DATEEFFE'])
df['DATE_FIN'] = pd.to_datetime(df['DATE_FIN'])
sinistre1['DATESURV'] = pd.to_datetime(sinistre1['DATESURV'])
df['sinistre1'] = np.arange(len(df))
df1 = df.merge(sinistre1, on='id_police', how='left')
mask = df1['DATESURV'].between(df1['DATEEFFE'], df1['DATE_FIN'], inclusive=False)
s = df1.loc[mask, 'sinistre1']
print (s)
4 1
8 2
9 3
Name: sinistre1, dtype: int32
df['sinistre1'] = df['sinistre1'].isin(s).astype(int)
#alternative
#df['sinistre1'] = np.where(df['sinistre1'].isin(s), 1, 0)
print (df)
id_police DATEEFFE DATE_FIN prime prime2 sinistre1
0 p123 2017-01-24 2017-02-24 0 0 0
1 p123 2017-11-24 2017-12-24 0 30 1
2 p123 2018-02-25 2018-03-25 10 10 1
3 b123 2018-02-24 2018-03-24 20 20 1
4 b123 2018-03-24 2018-04-24 30 0 0
编辑：
df['DATEEFFE'] = pd.to_datetime(df['DATEEFFE'])
df['DATE_FIN'] = pd.to_datetime(df['DATE_FIN'])
sinistre1['DATESURV'] = pd.to_datetime(sinistre1['DATESURV'])
df1 = df.merge(sinistre1, on='id_police', how='left')
#filter all rows
df2 = df1[df1['DATESURV'].between(df1['DATEEFFE'], df1['DATE_FIN'], inclusive=False)]
print (df2)
id_police DATEEFFE DATE_FIN prime prime2 id_sinistre DATESURV
4 p123 2017-11-24 2017-12-24 0 30 s124 2017-11-30
8 p123 2018-02-25 2018-03-25 10 10 s125 2018-02-28
9 b123 2018-02-24 2018-03-24 20 20 s126 2018-02-28
#left join merge with inditator True
df = df.merge(df2.drop('DATESURV', axis=1), how='left', indicator=True)
#sinistre1 column - first rename and then compare both for 1, esle 0
df = df.rename(columns={'_merge':'sinistre1'})
df['sinistre1'] = df['sinistre1'].eq('both').astype(int)
print (df)
id_police DATEEFFE DATE_FIN prime prime2 id_sinistre sinistre1
0 p123 2017-01-24 2017-02-24 0 0 NaN 0
1 p123 2017-11-24 2017-12-24 0 30 s124 1
2 p123 2018-02-25 2018-03-25 10 10 s125 1
3 b123 2018-02-24 2018-03-24 20 20 s126 1
4 b123 2018-03-24 2018-04-24 30 0 NaN 0
正如我在评论中所说。接受的答案和合并现在没有意义，因为我认为OP想要比较两个数据帧中的每一行，因此需要数据帧id_sinistre中的关键df。或者想要像以下一样使用combine_first：
df_merge = df.merge(sinistre1, on='id_police', how='left')
df_merge['DATESURV'] = pd.to_datetime(df_merge['DATESURV'])
df_merge['sinistre'] = np.where(df_merge['DATESURV'].between(df_merge['DATEEFFE'], df_merge['DATE_FIN']), 1, 0)
df_merge = df_merge.drop(['DATESURV', 'id_sinistre'], axis=1)
print(df_merge)
DATEEFFE DATE_FIN id_police prime prime2 sinistre
0 2017-01-24 2017-02-24 p123 0 0 0
1 2017-11-24 2017-12-24 p123 0 30 1
2 2018-02-25 2018-03-25 p123 10 10 1
3 2018-02-24 2018-03-24 b123 20 20 1
4 2018-03-24 2018-04-24 b123 30 0 0
即（未经测试）：
t_table = pd.merge(sinistre1, df, how='left', on='id_police')
t_table['sinistre'] = [1 if row['DATEEFFE'] < ds< row['DATE_FIN'] else
0 for row,_ in t_table.iterrows()]