连续类指标计算方法，比如连续登录天数、连续领取天数等等指标~
下面以连续登录天数为例，说下解决方法；
输入
每日login表
| dt | user |
|---|---|
| 20170601 | A |
| 20170602 | A |
| 20170603 | A |
| 20170601 | B |
| 20170701 | B |
| 20170702 | B |
| 20170801 | C |
| 20170802 | C |
输出
| dt | user | lianxu_num |
|---|---|---|
| 20170601 | A | 1 |
| 20170602 | A | 2 |
| 20170603 | A | 3 |
| 20170601 | B | 1 |
| 20170701 | B | 1 |
| 20170702 | B | 2 |
| 20170801 | C | 1 |
| 20170802 | C | 2 |
解决方法
解决方法1： 写段累计的代码，以用户分组，以时间排序，如果本次时间和上次时间差值为1，则连续值+1
解决方法2：
| dt | user | row_number() over(partition by user order by dt) as rn | dt-rn+1 | 以user和（dt-rn+1）进行分组，以dt进行排序，获得row_number取值，即得到连续值 |
|---|---|---|---|---|
| 20170601 | A | 1 | 20170601 | 1 |
| 20170602 | A | 2 | 20170601 | 2 |
| 20170603 | A | 3 | 20170601 | 3 |
| 20170601 | B | 1 | 20170601 | 1 |
| 20170701 | B | 2 | 20170630 | 1 |
| 20170702 | B | 3 | 20170630 | 2 |
| 20170801 | C | 1 | 20170801 | 1 |
| 20170802 | C | 2 | 20170801 | 2 |