将值 "Good" 和 "Bad" 转换为布尔值以计算百分比（即，从所有记录中，80% 是 "Good"）- 使用 SQL

Question

假设我有一个示例数据集 p1，其中不同的用户对 4 种不同类型的食物进行了评分，如下所示：

Food_ID Rating
1       Good 
1       Good
2       Good
2       Bad
1       Bad
3       Bad
3       Good
4       Bad
1       Bad
4       Good
4       Good
4       Good
2       Bad
3       Bad

我如何编写代码，根据好与坏的评分告诉我 Food_ID 的接受度？

SELECT 
Food_ID, 
sum(Rating)/count(Rating) AS Avg_Rating

FROM p1
GROUP BY Food_ID

显然这段代码不起作用，因为评级是字符列...有没有快速的方法将其转换为布尔值？那么也许它可能？或者有更简单的方法吗？

谢谢！

编辑更新：

感谢下面的回答，我取得了进步。以下代码：

sqldf("
      SELECT food_id, 
sum(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) AS good_reviews, 
count(*) AS all_reviews, 
sum(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) / count(*) AS acceptance

FROM p1

GROUP BY food_id

  ")

其中returns输出如下：

  food_id good_reviews all_reviews acceptance
     1            2           4          0
     2            1           3          0
     3            0           3          0
     4            3           4          0

Answer 1

SELECT 
  food_id, 
  SUM(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) AS good_reviews, 
  COUNT(*) AS all_reviews, 
  CAST(Sum(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) AS Float) 
    / CAST(COUNT(*) AS Float) * 100 AS acceptance
FROM p1
GROUP BY food_id

Answer 2

SUM(CASE WHEN Rating='Good' THEN 1 ELSE 0 END AS BIT)/COUNT(*) AS numericalRating

Answer 3

使用 IF

select 
   food_id, 
     sum(IF rating='good' then 1 else 0 end)/count(*) as avg_rating
 from p1
 group by food_id

Answer 4

最终正确答案 -- 问题源于两个整数相除并产生非整数结果。谢谢@stickybit

最终工作代码：

sqldf("
  SELECT food_id, 
sum(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) AS good_reviews, 
count(*) AS all_reviews, 
CAST(sum(CASE WHEN rating = 'Good' THEN 1 ELSE 0 END) AS Float) / CAST(count(*) AS Float)*100 AS acceptance

FROM p1
GROUP BY food_id

  ")

将值 "Good" 和 "Bad" 转换为布尔值以计算百分比（即，从所有记录中，80% 是 "Good"）- 使用 SQL

Convert values "Good" and "Bad" to boolean in order to calculate percentage (i.e, from all records, 80% were "Good") - Using SQL

sql

select

group-by

proc-sql