计算MySQL中每个项目的平均行数

Question

I had trouble finding a clear, generic way of phrasing this question, so apologies if it's a repeat. Here's the situation:

I have a table recording collaborative tagging data, with each row storing an annotation (ie a particular user tagging a particular item with a particular tag at a particular time). Here's a sample for clarity:

+---------+---------+--------+------------+
| user_id | item_id | tag_id | tag_month  |
+---------+---------+--------+------------+
| 1040740 |    2653 |   1344 | 2005-07-01 |
| 1040740 |    3602 |   1344 | 2005-07-01 |
| 1040740 |   17746 |    217 | 2005-07-01 |
| 1040740 |   21426 |   1344 | 2005-07-01 |
| 1040740 |   22224 |    180 | 2005-07-01 |
+---------+---------+--------+------------+

...and so on. What I'm trying to calculate is, on a month-by-month basis, the mean number of annotations per item, across all items. In other words, for each month, what is the average number of rows per unique item for that month? My dataset spans 94 total months, so the output of the query I'm wanting should be 94 rows, each with average number of annotations per item for that month. Note that the "user_id" column is completely irrelevant to this.

Answer 1

I think you need just to do the corresponding COUNT's:

SELECT 
  COUNT(DISTINCT item_id),
  YEAR(tag_month),
  MONTH(tag_month)
FROM
  t
GROUP BY
  YEAR(tag_month),
  MONTH(tag_month)

not sure if you want to get item_id , but, if you need, then:

SELECT 
  COUNT(1),
  item_id,
  YEAR(tag_month),
  MONTH(tag_month)
FROM
  t
GROUP BY
  item_id,
  YEAR(tag_month),
  MONTH(tag_month)

Answer 2

Slight deviation to Alma Do Mundo's answer to give the average number of tags per item per month.

SELECT 
  COUNT(*) / COUNT(DISTINCT item_id) as tag_average,
  YEAR(tag_month),
  MONTH(tag_month)
FROM
  t
GROUP BY
  YEAR(tag_month),
  MONTH(tag_month)