d0 = d0.groupby(['key2'])['data2'].agg(lambda x:max(x)).reset_index()
以key2分组，取data2的最小值

d1['min_date_received'] = d1.interval.apply(lambda s:min([int(d) for d in s.split(':')]))
interval的内容为x:x:x,此处对s拆分，并取出最小值

t2['receive_number'] = t2.date_received.apply(lambda s:len(s.split(':')))
date_received的内容为x:x:x,此处对s拆分，并取出个数

dataset3['day_of_week'] = dataset3.date_received.astype('str').apply(lambda x:date(int(x[0:4]),int(x[4:6]),int(x[6:8])).weekday()+1)
取出日期对应的星期数，周一，周二，周三。。。

dataset3['day_of_month'] = dataset3.date_received.astype('str').apply(lambda x:int(x[6:8]))
取出日期是每月的多少天

dataset2['days_distance'] = dataset2.date_received.astype('str').apply(lambda x:(date(int(x[0:4]),int(x[4:6]),int(x[6:8]))-date(2016,5,14)).days)
计算日期相对于某一天的间隔天数

dataset3['is_weekend'] = dataset3.day_of_week.apply(lambda x:1 if x in (6,7) else 0)
是否周末。注意此处lambda的写法。1 if xxx else 0