如何在 sphinx 中对多面体的 MVA 字段进行分组
How to group MVA field for faceted in sphinx
我有一个索引，其中一些数据重复，所有字段都相似，除了纬度、经度和 ID（字段 ID 不是真正的 ID，刚刚生成 row_number() OVER () AS id）。
示例：
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 6 | 919 | 15,243 | 0.825162 | 0.692828 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 8 | 918 | 8,154 | 0.825162 | 0.692828 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 10 | 920 | 17,283,288 | 0.958915 | 1.282215 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 12 | 924 | 12,208 | 0.973336 | 0.658237 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 14 | 923 | 21,365 | 0.973336 | 0.658237 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 16 | 922 | 20,359 | 0.973336 | 0.658237 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 18 | 921 | 19,346 | 0.973336 | 0.658237 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
现在我想按 vacancy_id
对数据进行分组
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy group by vacancy_id;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
| 21 | 961 | 4,105 | 0.959217 | 1.280721 |
| 23 | 960 | 8,155 | 0.959217 | 1.280721 |
| 25 | 959 | 12,208 | 0.959217 | 1.280721 |
| 27 | 928 | 1,60 | 0.963734 | 1.070297 |
| 29 | 927 | 32,513 | 0.963734 | 1.070297 |
| 31 | 929 | 6,140 | 0.786553 | 0.678649 |
| 33 | 932 | 1,40,46 | 0.824627 | 0.694182 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
结果太棒了！但是当我想使用 faceted
获取所有分组数据时，问题就开始了
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+-----------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 |
+------+------------+-----------------+----------+-----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
分面结果不正确。因为实际上 prof_area_ids=199 处的数据计数必须是 5 而不是 12。那么我如何为分面分组字段？
另外
我在这里 http://sphinxsearch.com/blog/2013/06/21/faceted-search-with-sphinx/ 但只是写了“如果你有一个 MVA 方面，你需要使用 GROUPBY() 函数，returns 进行分组的实际值。”没有考试。
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude,GROUPBY() as selected,COUNT(*) from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+----------+----------+
| id | vacancy_id | prof_area_ids | latitude | longitude | selected | count(*) |
+------+------------+-----------------+----------+-----------+----------+----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 | 917 | 1 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 | 1004 | 2 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 | 1072 | 3 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 | 1136 | 3 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 | 1097 | 3 |
+------+------------+-----------------+----------+-----------+----------+----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
分面结果也是错误的。
Faceted result is incorrect. Because in fact data's count where prof_area_ids=199 must be 5 and not 12. So how I can group field for faceted?
您似乎误解了 FACET 的工作原理。在我看来，您认为它以主查询的结果为基础，但实际上它只是进行了另一个分组。例如。这里：
mysql> select g, t from idx_mva where t = 11 group by g facet t;
+------+----------+
| g | t |
+------+----------+
| 1 | 11,12 |
| 2 | 11,13,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
对于 t=11 你可以看到，在你的情况下，它在第一个查询的结果中找到了 3 次，但在 FACET 的查询结果中它的计数是 6 次。这是因为它在索引中实际出现了6次：
mysql> select * from idx_mva where t = 11;
+------+------+----------+
| id | g | t |
+------+------+----------+
| 2 | 1 | 11,12 |
| 3 | 1 | 11,15 |
| 3 | 2 | 11,13,15 |
| 6 | 3 | 9,11 |
| 8 | 5 | 11,12,15 |
| 11 | 2 | 3,11,15 |
+------+------+----------+
6 rows in set, 1 warning (0.00 sec)
在第一种情况下它发生了 3 次，只是因为每个组只返回一次 t 的值。您可以使用 group_concat() 查看同一组中的更多值：
mysql> select g, group_concat(to_string(t)) from idx_mva where t = 11 group by g facet t;
+------+----------------------------+
| g | group_concat(to_string(t)) |
+------+----------------------------+
| 1 | 11,12,11,15 |
| 2 | 11,13,15,3,11,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------------------------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
如果你想了解更多关于分面的知识，这里有一个互动课程 - https://play.manticoresearch.com/faceting/
似乎，想要在 FACET 上有效地 COUNT(DISTINCT vacancy_id) 而不是默认的 COUNT(*)，但结果是
... FACET prof_area_ids,COUNT(DISTINCT vacancy_id) AS vacancies BY prof_area_ids
不起作用。 BY 之前的位只支持属性，不支持自定义函数。
...只需要写很长的路，用完整的查询...
select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy
where prof_area_ids=199 group by vacancy_id;
SELECT GROUPBY() AS prof_area_id, COUNT(DISTINCT vacancy_id) FROM jobVacancy
WHERE prof_area_ids=199 GROUP BY prof_area_id;
相同的结果，只是稍微更冗长。 ie 而不是使用 FACET shorthand，写它
完整地作为多个单独的查询。
我有一个索引，其中一些数据重复，所有字段都相似，除了纬度、经度和 ID（字段 ID 不是真正的 ID，刚刚生成 row_number() OVER () AS id）。
示例：
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 6 | 919 | 15,243 | 0.825162 | 0.692828 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 8 | 918 | 8,154 | 0.825162 | 0.692828 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 10 | 920 | 17,283,288 | 0.958915 | 1.282215 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 12 | 924 | 12,208 | 0.973336 | 0.658237 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 14 | 923 | 21,365 | 0.973336 | 0.658237 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 16 | 922 | 20,359 | 0.973336 | 0.658237 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 18 | 921 | 19,346 | 0.973336 | 0.658237 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
现在我想按 vacancy_id
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy group by vacancy_id;
+------+------------+---------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+---------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 2 | 916 | 17,283,288 | 0.973178 | 0.743566 |
| 3 | 915 | 17,288 | 0.973178 | 0.743566 |
| 4 | 914 | 30,482 | 0.973178 | 0.743566 |
| 5 | 919 | 15,243 | 0.825153 | 0.692837 |
| 7 | 918 | 8,154 | 0.825153 | 0.692837 |
| 9 | 920 | 17,283,288 | 0.958914 | 1.282161 |
| 11 | 924 | 12,208 | 0.97333 | 0.658246 |
| 13 | 923 | 21,365 | 0.97333 | 0.658246 |
| 15 | 922 | 20,359 | 0.97333 | 0.658246 |
| 17 | 921 | 19,346 | 0.97333 | 0.658246 |
| 19 | 926 | 12,17,208,292 | 0.88396 | 2.389868 |
| 20 | 925 | 12,208 | 0.88396 | 2.389868 |
| 21 | 961 | 4,105 | 0.959217 | 1.280721 |
| 23 | 960 | 8,155 | 0.959217 | 1.280721 |
| 25 | 959 | 12,208 | 0.959217 | 1.280721 |
| 27 | 928 | 1,60 | 0.963734 | 1.070297 |
| 29 | 927 | 32,513 | 0.963734 | 1.070297 |
| 31 | 929 | 6,140 | 0.786553 | 0.678649 |
| 33 | 932 | 1,40,46 | 0.824627 | 0.694182 |
+------+------------+---------------+----------+-----------+
20 rows in set (0.00 sec)
结果太棒了！但是当我想使用 faceted
获取所有分组数据时，问题就开始了mysql> select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+
| id | vacancy_id | prof_area_ids | latitude | longitude |
+------+------------+-----------------+----------+-----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 |
+------+------------+-----------------+----------+-----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
分面结果不正确。因为实际上 prof_area_ids=199 处的数据计数必须是 5 而不是 12。那么我如何为分面分组字段？
另外
我在这里 http://sphinxsearch.com/blog/2013/06/21/faceted-search-with-sphinx/ 但只是写了“如果你有一个 MVA 方面，你需要使用 GROUPBY() 函数，returns 进行分组的实际值。”没有考试。
mysql> select id,vacancy_id,prof_area_ids,latitude,longitude,GROUPBY() as selected,COUNT(*) from jobVacancy where prof_area_ids=199 group by vacancy_id facet prof_area_ids;
+------+------------+-----------------+----------+-----------+----------+----------+
| id | vacancy_id | prof_area_ids | latitude | longitude | selected | count(*) |
+------+------------+-----------------+----------+-----------+----------+----------+
| 1 | 917 | 11,199,202 | 0.973178 | 0.743566 | 917 | 1 |
| 191 | 1004 | 11,196,199 | 0.925335 | 2.768874 | 1004 | 2 |
| 313 | 1072 | 1,11,60,197,199 | 0.963968 | 1.070624 | 1072 | 3 |
| 318 | 1136 | 11,196,199 | 0.96071 | 1.448998 | 1136 | 3 |
| 374 | 1097 | 11,199 | 0.785255 | 0.678504 | 1097 | 3 |
+------+------------+-----------------+----------+-----------+----------+----------+
5 rows in set (0.00 sec)
+---------------+----------+
| prof_area_ids | count(*) |
+---------------+----------+
| 202 | 1 |
| 199 | 12 |
| 11 | 12 |
| 196 | 5 |
| 197 | 3 |
| 60 | 3 |
| 1 | 3 |
+---------------+----------+
7 rows in set (0.02 sec)
分面结果也是错误的。
Faceted result is incorrect. Because in fact data's count where prof_area_ids=199 must be 5 and not 12. So how I can group field for faceted?
您似乎误解了 FACET 的工作原理。在我看来，您认为它以主查询的结果为基础，但实际上它只是进行了另一个分组。例如。这里：
mysql> select g, t from idx_mva where t = 11 group by g facet t;
+------+----------+
| g | t |
+------+----------+
| 1 | 11,12 |
| 2 | 11,13,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
对于 t=11 你可以看到，在你的情况下，它在第一个查询的结果中找到了 3 次，但在 FACET 的查询结果中它的计数是 6 次。这是因为它在索引中实际出现了6次：
mysql> select * from idx_mva where t = 11;
+------+------+----------+
| id | g | t |
+------+------+----------+
| 2 | 1 | 11,12 |
| 3 | 1 | 11,15 |
| 3 | 2 | 11,13,15 |
| 6 | 3 | 9,11 |
| 8 | 5 | 11,12,15 |
| 11 | 2 | 3,11,15 |
+------+------+----------+
6 rows in set, 1 warning (0.00 sec)
在第一种情况下它发生了 3 次，只是因为每个组只返回一次 t 的值。您可以使用 group_concat() 查看同一组中的更多值：
mysql> select g, group_concat(to_string(t)) from idx_mva where t = 11 group by g facet t;
+------+----------------------------+
| g | group_concat(to_string(t)) |
+------+----------------------------+
| 1 | 11,12,11,15 |
| 2 | 11,13,15,3,11,15 |
| 3 | 9,11 |
| 5 | 11,12,15 |
+------+----------------------------+
4 rows in set (0.00 sec)
+------+----------+
| t | count(*) |
+------+----------+
| 12 | 2 |
| 11 | 6 |
| 15 | 4 |
| 13 | 1 |
| 9 | 1 |
| 3 | 1 |
+------+----------+
6 rows in set (0.00 sec)
如果你想了解更多关于分面的知识，这里有一个互动课程 - https://play.manticoresearch.com/faceting/
似乎，想要在 FACET 上有效地 COUNT(DISTINCT vacancy_id) 而不是默认的 COUNT(*)，但结果是
... FACET prof_area_ids,COUNT(DISTINCT vacancy_id) AS vacancies BY prof_area_ids
不起作用。 BY 之前的位只支持属性，不支持自定义函数。
...只需要写很长的路，用完整的查询...
select id,vacancy_id,prof_area_ids,latitude,longitude from jobVacancy
where prof_area_ids=199 group by vacancy_id;
SELECT GROUPBY() AS prof_area_id, COUNT(DISTINCT vacancy_id) FROM jobVacancy
WHERE prof_area_ids=199 GROUP BY prof_area_id;
相同的结果，只是稍微更冗长。 ie 而不是使用 FACET shorthand，写它 完整地作为多个单独的查询。