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题解 | #10月的新户客单价和获客成本#

select round(avg(total_amount),1), round(avg(b-total_amount),1) from (select order_id, total_amount from( select order_id, uid, event_time, total_amo...

2023-03-20

0 252

题解 | #零食类商品中复购率top3高的商品#

with tb2 as( select product_id, uid from (select *, max(date(event_time))over() dt from tb_order_overall where status=1) tb1 left join tb_order_detail...

2023-03-18

0 303

题解 | #某店铺的各商品毛利率及店铺整体毛利率#

with tb2 as ( select product_id, in_price, price, cnt from( (select * from tb_order_overall where status=1 and date(event_time)>='2021-10-01') as ...

2023-03-17

0 245

题解 | #使用字典计数#

a = list(input()) b = [a.count(i) for i in a] print(dict(zip(a,b)))

2023-03-16

0 194

题解 | #喜欢的颜色#

a = {'Allen': ['red', 'blue', 'yellow'],'Tom': ['green', 'white', 'blue'],'Andy': ['black', 'pink']} for i in sorted(a): print("%s's favorite colo...

2023-03-16

0 201

题解 | #增加元组的长度#

a = tuple(list(range(1,6))) print(a,len(a),sep='\n') b = tuple(list(range(6,11))) c = a+b print(c,len(c),sep='\n')

2023-03-15

0 269

题解 | #跳过列表的某个元素#

a=[i for i in range(1,16) if i%13 != 0] for i in a: print(i, end=' ')

2023-03-14

0 184

题解 | #连续签到领金币#

思路： ①如何判断连签：用签到日期-日期排序=连签日期（连签日期一致表明连续签到），对连签日期计数即得到连签天数。 ②连签天数以7天一个周期，每个周期里第3天、第7天得3分、7分，其余得1分，用%7进行取余判断。 with tb1 as( select uid, date(in_time) dt,...

2023-03-14

0 488

题解 | #每天的日活数及新用户占比#

with tb1 as( select distinct uid, date(in_time) dt, min(date(in_time))over(partition by uid) a from tb_user_log union select distinct uid,...

2023-03-13

0 180

题解 | #被5整除的数字#

a = list(range(1,51)) for i in a: if i%5 ==0: print(i) else: continue

2023-03-12

0 185