简单 通过率:74.78% 时间限制:1秒 空间限制:256M
现有某乎问答创作者回答情况表answer_tb如下(其中answer_date表示创作日期、author_id指创作者编号、issue_id表示问题id、char_len表示回答字数):
answer_date | author_id | issue_id | char_len |
---|---|---|---|
2021-11-01 | 101 | E001 | 150 |
2021-11-01 | 101 | E002 | 200 |
2021-11-01 | 102 | C003 | 50 |
2021-11-01 | 103 | P001 | 35 |
2021-11-01 | 104 | C003 | 120 |
2021-11-01 | 105 | P001 | 125 |
2021-11-01 | 102 | P002 | 105 |
2021-11-02 | 101 | P001 | 201 |
2021-11-02 | 110 | C002 | 200 |
2021-11-02 | 110 | C001 | 225 |
2021-11-02 | 110 | C002 | 220 |
2021-11-03 | 101 | C002 | 180 |
2021-11-04 | 109 | E003 | 130 |
2021-11-04 | 109 | E001 | 123 |
2021-11-05 | 108 | C001 | 160 |
2021-11-05 | 108 | C002 | 120 |
2021-11-05 | 110 | P001 | 180 |
2021-11-05 | 106 | P002 | 45 |
2021-11-05 | 107 | E003 | 56 |
请你统计11月份日人均回答量(回答问题数量/答题人数),按回答日期排序,结果保留两位小数,以上例子的输出结果如下:
answer_date | per_num |
---|---|
2021-11-01 | 1.40 |
2021-11-02 | 2.00 |
2021-11-03 | 1.00 |
2021-11-04 | 2.00 |
2021-11-05 | 1.25 |
select answer_date, round(count(issue_id) / count(distinct author_id), 2) as per_num
from answer_tb
where month(answer_date)=11
group by answer_date
order by answer_date
本题要点在于count作者的时候 distinct
的运用。
然后按日期group by就可以了。