Question

我有以下类型的数据集：

user_id   country1  city1      country2  city2
1         usa       new york   france    paris 
2         usa       dallas     japan     tokyo 
3         india     mumbai     italy     rome 
4         france    paris      usa       new york 
5         brazil    sao paulo  russia    moscow

我想将country1，city1，country2和city2的组合分组到哪个顺序（country1或country2 ）无所谓。通常，我会尝试：

SELECT   country1 
       , city1
       , country2
       , city2 
       , COUNT(*) 
FROM dataset
GROUP BY country1 
       , city1
       , country2
       , city2

但是，此代码段将user_id=1和user_id=4的行视为两个独立的情况，我希望它们被视为等效。

任何知道如何解决此问题的人？

提前致谢！

Answer 1

通常，您使用least()和greatest()来解决此类问题，但您有两列，而不是一列。那么，让我们通过比较城市来做到这一点。我猜测city比country更独特：

select (case when city1 < city2 then country1 else country2 end) as country1,
       (case when city1 < city2 then city1 else city2 end) as city1,
       (case when city1 < city2 then country2 else country1 end) as country2,
       (case when city1 < city2 then city2 else city1 end) as city2,
       count(*)
from dataset
group by (case when city1 < city2 then country1 else country2 end),
       (case when city1 < city2 then city1 else city2 end),
       (case when city1 < city2 then country2 else country1 end),
       (case when city1 < city2 then city2 else city1 end)

SQL / Vertica - 对多属性组合进行分组

1 个答案: