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トップMySQLに関する質問Mysql：カラム毎の割合を抽出したい

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回答編集履歴

5

修正

2018/04/10 07:04

投稿

sazi
sazi

スコア25430

answer CHANGED
@@ -1,10 +1,10 @@
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1
  もう少し簡略化できそうな気もしますが、取り敢えずこんな感じでしょうか。
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  ```SQL
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  select `name`
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- , (cnt/q1y) * 100 as q1y_per
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+ , (q1y/cnt) * 100 as q1y_per
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- , (cnt/q2y) * 100 as q2y_per
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+ , (q2y/cnt) * 100 as q2y_per
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- , (cnt/q3y) * 100 as q3y_per
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+ , (q3y/cnt) * 100 as q3y_per
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- , (cnt/q4y) * 100 as q4y_per
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+ , (q4y/cnt) * 100 as q4y_per
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  from (
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  select `name`
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  , sum(case when q1='y' then 1 else 0 end) as q1y
@@ -21,10 +21,10 @@
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  boolean型を数値として扱って簡略化すると
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  ```SQL
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  select `name`
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- , (cnt/q1y) * 100 as q1y_per
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+ , (q1y/cnt) * 100 as q1y_per
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- , (cnt/q2y) * 100 as q2y_per
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+ , (q2y/cnt) * 100 as q2y_per
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- , (cnt/q3y) * 100 as q3y_per
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+ , (q3y/cnt) * 100 as q3y_per
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- , (cnt/q4y) * 100 as q4y_per
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+ , (q4y/cnt) * 100 as q4y_per
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  from (
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  select `name`
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  , sum((q1='y')) as q1y
@@ -39,10 +39,10 @@
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  いっそまとめて
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  ```SQL
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  select `name`
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- , count(*) / sum((q1='y')) * 100 as q1y_per
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+ , sum((q1='y')) / count(*) * 100 as q1y_per
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- , count(*) / sum((q2='y')) * 100 as q2y_per
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+ , sum((q2='y')) / count(*) * 100 as q2y_per
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- , count(*) / sum((q3='y')) * 100 as q3y_per
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+ , sum((q3='y')) / count(*) * 100 as q3y_per
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- , count(*) / sum((q4='y')) * 100 as q4y_per
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+ , sum((q4='y')) / count(*) * 100 as q4y_per
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  from ckeck
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  group by `name`
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  ```

4

修正

2018/04/10 07:03

投稿

sazi
sazi

スコア25430

answer CHANGED
@@ -1,6 +1,6 @@
1
1
  もう少し簡略化できそうな気もしますが、取り敢えずこんな感じでしょうか。
2
2
  ```SQL
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- slect `name`
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+ select `name`
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  , (cnt/q1y) * 100 as q1y_per
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  , (cnt/q2y) * 100 as q2y_per
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  , (cnt/q3y) * 100 as q3y_per
@@ -20,7 +20,7 @@
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  ---
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  boolean型を数値として扱って簡略化すると
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  ```SQL
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- slect `name`
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+ select `name`
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  , (cnt/q1y) * 100 as q1y_per
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  , (cnt/q2y) * 100 as q2y_per
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  , (cnt/q3y) * 100 as q3y_per
@@ -38,7 +38,7 @@
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  ```
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  いっそまとめて
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  ```SQL
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- slect `name`
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+ select `name`
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  , count(*) / sum((q1='y')) * 100 as q1y_per
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  , count(*) / sum((q2='y')) * 100 as q2y_per
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  , count(*) / sum((q3='y')) * 100 as q3y_per

3

推敲

2018/04/10 06:44

投稿

sazi
sazi

スコア25430

answer CHANGED
@@ -39,10 +39,10 @@
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  いっそまとめて
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  ```SQL
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  slect `name`
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- , cout(*)/ sum((q1='y')) * 100 as q1y_per
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+ , count(*) / sum((q1='y')) * 100 as q1y_per
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- , cout(*)/ sum((q2='y')) * 100 as q2y_per
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+ , count(*) / sum((q2='y')) * 100 as q2y_per
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- , cout(*)/ sum((q3='y')) * 100 as q3y_per
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+ , count(*) / sum((q3='y')) * 100 as q3y_per
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- , cout(*)/ sum((q4='y')) * 100 as q4y_per
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+ , count(*) / sum((q4='y')) * 100 as q4y_per
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  from ckeck
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  group by `name`
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  ```

2

修正

2018/04/10 06:42

投稿

sazi
sazi

スコア25430

answer CHANGED
@@ -19,7 +19,6 @@
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  追記
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  ---
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  boolean型を数値として扱って簡略化すると
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- ```
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  ```SQL
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  slect `name`
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  , (cnt/q1y) * 100 as q1y_per

1

追記

2018/04/10 06:41

投稿

sazi
sazi

スコア25430

answer CHANGED
@@ -15,4 +15,35 @@
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  from ckeck
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  group by `name`
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  ) clc
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+ ```
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+ 追記
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+ ---
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+ boolean型を数値として扱って簡略化すると
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+ ```
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+ ```SQL
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+ slect `name`
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+ , (cnt/q1y) * 100 as q1y_per
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+ , (cnt/q2y) * 100 as q2y_per
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+ , (cnt/q3y) * 100 as q3y_per
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+ , (cnt/q4y) * 100 as q4y_per
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+ from (
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+ select `name`
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+ , sum((q1='y')) as q1y
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+ , sum((q2='y')) as q2y
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+ , sum((q3='y')) as q3y
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+ , sum((q4='y')) as q4y
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+ , count(*) as cnt
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+ from ckeck
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+ group by `name`
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+ ) clc
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+ ```
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+ いっそまとめて
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+ ```SQL
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+ slect `name`
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+ , cout(*)/ sum((q1='y')) * 100 as q1y_per
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+ , cout(*)/ sum((q2='y')) * 100 as q2y_per
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+ , cout(*)/ sum((q3='y')) * 100 as q3y_per
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+ , cout(*)/ sum((q4='y')) * 100 as q4y_per
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+ from ckeck
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+ group by `name`
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  ```