行号/排名“特定分区”
桌子是这个
这是错误的row_number，我想计算它以便按真实分组日期几周（从星期一到周日），即 1对于前4行，2行从5到8，为9到13的行，等等。也许这可以通过等级解决，但是请注意，所需的列并不总是取决于Year_week的更改。结果表将是：
The table is this
The row_number in this is wrong, I want to to compute it in order to group the dates by real weeks (from monday to sunday), i.e.
1 for the first 4 rows, 2 for rows from 5 to 8, 3 for rows from 9 to 13 and so on. Maybe this can be solve with rank, but notice the desired column not always is determined by the change in year_week. The resulting table would be:
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只是为了结束答案...
搜索后，我发现此问题可以通过函数conditdital_change_event（）解决，这并非在所有SQL lenguages中实现，而是可以被解释的。检查 ？有关更多信息。
Just for close the answer...
After some searching I found this problem can ve solve with the function CONDITIONAL_CHANGE_EVENT(), this is not implemented in all SQL lenguages, but can be explicited. Check Is there any alternative to Vertica's conditional_true_event in RedShift? for more information.