Question

我正在尝试计算可变时间线上每小时，一周中每天的平均通话次数。到目前为止，我已经能够解析来电并为他们分配一周中的一天。这是我要处理的表的摘要：

+------+------+------+------+------+
| day  | H1   | H2   | H3   | H4   |
+------+------+------+------+------+
| Tue  |    1 |    1 |    3 |    5 |
| Wed  |    3 |    2 |    3 |    4 |
| Thu  |    0 |    3 |    0 |    5 |
| Fri  |    3 |    2 |    4 |   11 |
| Sat  |    0 |    0 |    0 |    1 |
| Sun  |    0 |    0 |    0 |    2 |
| Mon  |    1 |    2 |    2 |    8 |
| Tue  |    1 |    1 |    2 |    6 |
| Wed  |    1 |    1 |    4 |    3 |
| Thu  |    0 |    1 |    2 |    3 |
| Fri  |    0 |    3 |    3 |    6 |
| Sat  |    0 |    0 |    0 |    4 |
| Sun  |    0 |    0 |    0 |    0 |
| Mon  |    1 |    2 |    2 |    8 |
| Tue  |    1 |    4 |    2 |    7 |
| Wed  |    1 |    3 |    2 |    6 |
| Thu  |    1 |    1 |    2 |    6 |

我很难找到一种方法，可以将每天的每小时一栏汇总起来，然后对它们进行平均。

最终结果应该类似于

Day  H1   H2   H3   H4
Tue  1    2    2.33 6
Wed  1.66 2    3    4.33
Thu  0.33 1.66 1.33 4.66

每天加起来的每一列，然后除以3，即当天的实例数。请记住，每天实例的数量会发生变化，例如星期一和星期五仅在表上显示两次。

我一直在努力寻找一种干净的方法，但是我也很难弄清楚如何在给定的时间范围内每天进行计数。

请谢谢！ DW

Answer 1

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(id SERIAL PRIMARY KEY
,day  CHAR(3) NOT NULL
,H INT NOT NULL
,val INT NOT NULL
);

INSERT INTO my_table (day,h,val) VALUES
('Tue' , 1 ,  1),
('Wed' , 1 ,  3),
('Thu' , 1 ,  0),
('Fri' , 1 ,  3),
('Sat' , 1 ,  0),
('Sun' , 1 ,  0),
('Mon' , 1 ,  1),
('Tue' , 1 ,  1),
('Wed' , 1 ,  1),
('Thu' , 1 ,  0),
('Fri' , 1 ,  0),
('Sat' , 1 ,  0),
('Sun' , 1 ,  0),
('Mon' , 1 ,  1),
('Tue' , 1 ,  1),
('Wed' , 1 ,  1),
('Thu' , 1 ,  1),
('Tue' , 2 ,  1),
('Wed' , 2 ,  2),
('Thu' , 2 ,  3),
('Fri' , 2 ,  2),
('Sat' , 2 ,  0),
('Sun' , 2 ,  0),
('Mon' , 2 ,  2),
('Tue' , 2 ,  1),
('Wed' , 2 ,  1),
('Thu' , 2 ,  1),
('Fri' , 2 ,  3),
('Sat' , 2 ,  0),
('Sun' , 2 ,  0),
('Mon' , 2 ,  2),
('Tue' , 2 ,  4),
('Wed' , 2 ,  3),
('Thu' , 2 ,  1),
('Tue' , 3 ,  3),
('Wed' , 3 ,  3),
('Thu' , 3 ,  0),
('Fri' , 3 ,  4),
('Sat' , 3 ,  0),
('Sun' , 3 ,  0),
('Mon' , 3 ,  2),
('Tue' , 3 ,  2),
('Wed' , 3 ,  4),
('Thu' , 3 ,  2),
('Fri' , 3 ,  3),
('Sat' , 3 ,  0),
('Sun' , 3 ,  0),
('Mon' , 3 ,  2),
('Tue' , 3 ,  2),
('Wed' , 3 ,  2),
('Thu' , 3 ,  2),
('Tue' , 4 ,  5),
('Wed' , 4 ,  4),
('Thu' , 4 ,  5),
('Fri' , 4 , 11),
('Sat' , 4 ,  1),
('Sun' , 4 ,  2),
('Mon' , 4 ,  8),
('Tue' , 4 ,  6),
('Wed' , 4 ,  3),
('Thu' , 4 ,  3),
('Fri' , 4 ,  6),
('Sat' , 4 ,  4),
('Sun' , 4 ,  0),
('Mon' , 4 ,  8),
('Tue' , 4 ,  7),
('Wed' , 4 ,  6),
('Thu' , 4 ,  6);

SELECT day,h,AVG(val)x FROM my_table GROUP BY day,h;
+-----+---+--------+
| day | h | x      |
+-----+---+--------+
| Fri | 1 | 1.5000 |
| Fri | 2 | 2.5000 |
| Fri | 3 | 3.5000 |
| Fri | 4 | 8.5000 |
| Mon | 1 | 1.0000 |
| Mon | 2 | 2.0000 |
| Mon | 3 | 2.0000 |
| Mon | 4 | 8.0000 |
| Sat | 1 | 0.0000 |
| Sat | 2 | 0.0000 |
| Sat | 3 | 0.0000 |
| Sat | 4 | 2.5000 |
| Sun | 1 | 0.0000 |
| Sun | 2 | 0.0000 |
| Sun | 3 | 0.0000 |
| Sun | 4 | 1.0000 |
| Thu | 1 | 0.3333 |
| Thu | 2 | 1.6667 |
| Thu | 3 | 1.3333 |
| Thu | 4 | 4.6667 |
| Tue | 1 | 1.0000 |
| Tue | 2 | 2.0000 |
| Tue | 3 | 2.3333 |
| Tue | 4 | 6.0000 |
| Wed | 1 | 1.6667 |
| Wed | 2 | 2.0000 |
| Wed | 3 | 3.0000 |
| Wed | 4 | 4.3333 |
+-----+---+--------+

在设定时间内每天平均每小时的小时数

1 个答案: