pd.DataFrame 将连续满足条件的多行数据进行比较，返回较大值和较小值

import pandas as pd

## the time column doesn't matter in your problem
df = pd.DataFrame({
    'time':['2021-3-19','2021-3-20','2021-3-21','2021-3-22',
    '2021-3-23','2021-3-24','2021-3-25','2021-3-26','2021-3-27'],
    'value':[10,11,9,5,4,2,4,9,5],
    'status':['X']*3+['Y']+['X']+['Y']*2+['X']*2
})

df_new = pd.DataFrame(columns=df.columns)
## perform a groupby on consecutive values
for _, g in df.groupby([(df.status != df.status.shift()).cumsum()]):
    g = g.sort_values(by='value')
    ## keep the highest value for X
    if g.status.values[0] == 'X':
        g = g.drop_duplicates(subset=['status'], keep='last')

    ## keep the lowest value for Y
    elif g.status.values[0] == 'Y':
        g = g.drop_duplicates(subset=['status'], keep='first')

    else:
        pass
    df_new = pd.concat([df_new, g])
df_new = df_new.reset_index(drop=True)

def inp(x):
    if x.time in df_new.time.values:
        return x.value

               
df["value_1"]=df.apply(inp,axis=1)
#参照df_new.time,给相应df.value_1赋值.

df.loc[:0,'value_1']=df.head(1).value
#第一行必然: df.value_1= df.value

df["value_1"]=df.value_1.fillna(method='ffill')
#对空值NaN 参照上一行的value_1赋值.

print(df)
exit()