每月玩此游戏的玩家百分比 Redshift

Question

此查询从活动表中获取每月所有活跃玩家的数量：

SELECT
  date_trunc('month', createdat) as month,
  count(distinct playerid) as play_all
FROM
    activity
group by month
order by 1

这个查询得到每个月玩“蜜蜂”游戏的玩家数量：

SELECT
  date_trunc('month', createdat) as month,
  count(distinct playerid) as play_bee
FROM
    activity
where gamename = 'bee'    
group by month
order by 1

如何获取每个月玩“蜜蜂”游戏的玩家百分比？

Answer 1

这可能有效：

SELECT
  DATE_TRUNC('month', createdat) AS month,
  COUNT(DISTINCT playerid) AS play_all,
  COUNT(DISTINCT CASE WHEN gamename = 'bee' THEN playerid END) AS play_bee,
  100. * COUNT(DISTINCT CASE WHEN gamename = 'bee' THEN playerid END)
       / COUNT(DISTINCT playerid) AS percent_play_bee,
FROM
    activity
group by month
order by 1

它使用了这样一个事实：如果gamename = 'bee，CASE WHEN gamename = 'bee' THEN playerid END 将返回 playerid，但如果不是，则返回 NONE。 （我不确定 NONE 响应是否会包含在 COUNT(DISTINCT..) 中，因此它可能会关闭 1。）

基本上，CASE 是针对每一行单独评估的。然后，将这些值设为 DISTINCT，然后对它们进行计数。