MySQL中的回归分析

Question

在我的项目中介绍我正在保存FacebookPages及其类似数量,以及每个国家/地区的类似数量.我有一个FacebookPages表,一个用于语言,一个用于facebook页面和语言之间的关联(并计算喜欢的)和一个表将这些数据保存为历史记录.我想要做的是在特定时间段内获得最强烈增加的页面.

要使用的数据

我正在从创建查询中剥离不相关的信息.

包含所有Facebook页面的表

CREATE TABLE `pages` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `facebook_id` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
  `facebook_name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
  `facebook_likes` int(11) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

示例数据:

INSERT INTO `facebook_pages` (`id`, `facebook_id`, `facebook_name`, `facebook_likes`)
VALUES
    (1, '552825254796051', 'Mesut Özil', 28593755),
    (2, '134904013188254', 'Borussia Dortmund', 13213354),
    (3, '310111039010406', 'Marco Reus', 12799627);

---

包含所有语言的表

CREATE TABLE `languages` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `language` varchar(5) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

示例数据

INSERT INTO `languages` (`id`, `language`)
VALUES
    (1, 'ID'),
    (2, 'TR'),
    (3, 'BR');

---

包含相关性的表

CREATE TABLE `language_page_likes` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `language_id` int(10) unsigned NOT NULL,
  `facebook_page_id` int(10) unsigned NOT NULL,
  `likes` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`),
  // Foreign key stuff
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

示例数据

INSERT INTO `language_page_likes` (`id`, `language_id`, `facebook_page_id`)
VALUES
    (1, 1, 1),
    (2, 2, 1),
    (3, 3, 1),
    (47, 3, 2),
    (51, 1, 2),
    (53, 2, 2),
    (92, 3, 3),
    (95, 2, 3),
    (97, 1, 3);

---

包含历史记录的表

CREATE TABLE `language_page_likes_history` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `language_page_likes_id` int(10) unsigned NOT NULL,
  `likes` int(11) NOT NULL,
  `created_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`),
  // Foreign key stuff
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

示例数据

INSERT INTO `language_page_likes_history` (`id`, `language_page_likes_id`, `likes`, `created_at`)
VALUES
    (1, 1, 3272484, '2015-09-11 08:40:23'),
    (132014, 1, 3272827, '2015-09-14 08:31:00'),
    (2, 2, 1581361, '2015-09-11 08:40:23'),
    (132015, 2, 1580392, '2015-09-14 08:31:00'),
    (3, 3, 1467090, '2015-09-11 08:40:23'),
    (132016, 3, 1467329, '2015-09-14 08:31:00'),
    (47, 47, 828736, '2015-09-11 08:40:23'),
    (132060, 47, 828971, '2015-09-14 08:31:00'),
    (51, 51, 602747, '2015-09-11 08:40:23'),
    (132064, 51, 603071, '2015-09-14 08:31:00'),
    (53, 53, 545484, '2015-09-11 08:40:23'),
    (132066, 53, 545092, '2015-09-14 08:31:00'),
    (92, 92, 916570, '2015-09-11 08:40:24'),
    (132105, 92, 917032, '2015-09-14 08:31:01'),
    (95, 95, 537382, '2015-09-11 08:40:24'),
    (132108, 95, 537395, '2015-09-14 08:31:01'),
    (97, 97, 419175, '2015-09-11 08:40:24'),
    (132110, 97, 419484, '2015-09-14 08:31:01');

如您所见,我获得了9月14日和11日的数据.现在我希望得到这个网站,增加最多的喜欢.在我使用名为last_like_count的列完成之前,但问题是,我在日期范围内不能动态.使用"正常"回归函数,我可以对每个日期范围都是动态的.

解决方案发现
我已经设法做的是构建所有存在的关系

SELECT p.id, p.facebook_name, plh.likes, l.language FROM facebook_pages p
INNER JOIN language_page_likes pl ON pl.facebook_page_id = p.id
INNER JOIN language_page_likes_history plh ON plh.language_page_likes_id = pl.id
INNER JOIN languages l ON l.id = pl.language_id
WHERE pl.language_id = 5 OR pl.language_id = 46 OR pl.language_id = 68

通过该查询,我可以获得特定语言系统历史记录中的所有相关数据.但是,我如何在该部分建立回归分析？

我已经在这里找到了这个链接

使用SQL查询识别趋势

但我的数学和MySQL技能不够高,无法将SQL转换为MySQL.有帮助吗？

Answer 1

这可能是,你在寻找什么:

SELECT SUM((X-AVG_X)*(Y-AVG_Y)) / SUM((X-AVG_X)*(X-AVG_X)) AS Slope,
       PageId, LanguageId
FROM
(
SELECT Q0.Y, 
       Q0.X, 
       Q1.AVG_Y,
       Q1.AVG_X,
       Q1.PageId,
       Q1.LanguageId
FROM   (SELECT T0.likes AS Y,
               UNIX_TIMESTAMP(T0.created_at) AS X,
               T1.facebook_page_id AS PageId,
               T1.language_id AS LanguageId
        FROM   language_page_likes_history T0 INNER JOIN
               language_page_likes T1 ON 
               (T0.language_page_likes_id = T1.id)
        WHERE  T0.created_at > '2015-09-11 00:00:00' AND
               T0.created_at < '2015-09-15 00:00:00') Q0 INNER JOIN
       (SELECT AVG(T2.likes) AS AVG_Y,
               AVG(UNIX_TIMESTAMP(T2.created_at)) AS AVG_X,
               T3.facebook_page_id AS PageId,
               T3.language_id AS LanguageId
        FROM   language_page_likes_history T2 INNER JOIN
               language_page_likes T3 ON 
               (T2.language_page_likes_id = T3.id)
        WHERE  T2.created_at > '2015-09-11 00:00:00' AND
               T2.created_at < '2015-09-15 00:00:00'
        GROUP BY T3.facebook_page_id, T3.language_id) Q1
        ON (Q0.PageId = Q1.PageId) AND (Q0.LanguageId = Q1.LanguageId)
) Q2
GROUP BY PageId, LanguageId
ORDER BY Slope DESC

它返回每页和语言的线性回归的斜率.列Slope表示每秒的喜欢次数.在您的样本数据中,两种情况下的喜欢量减少了.我不知道为什么.输出应该如下所示.测试了SQL语句,我手动检查了两行计算以获得正确的输出.

|           Slope | PageId | LanguageId |
|-----------------|--------|------------|
|  0.001786287345 |      3 |          3 |
|  0.001326183029 |      1 |          1 |
|  0.001252720995 |      2 |          1 |
|  0.001194724653 |      3 |          1 |
|  0.000924075055 |      1 |          3 |
|  0.000908609364 |      2 |          3 |
|  0.000050263497 |      3 |          2 |
| -0.001515637747 |      2 |          2 |
| -0.003746563717 |      1 |          2 |

如果表中没有数据,则可能存在问题.因此,可能需要添加ISNULL检查.

当您只想知道绝对值时,它更简单.您可以采取以下声明:

SELECT PageId, LanguageId,
       (likes_last_in_period - likes_before_period) AS Likes
FROM
(SELECT T1.facebook_page_id AS PageId,
       T1.language_id AS LanguageId,
       (SELECT likes 
        FROM   language_page_likes_history
        WHERE  created_at < '2015-09-12 00:00:00' AND
               language_page_likes_id = T1.id
        ORDER BY created_at DESC LIMIT 1) likes_before_period,
       (SELECT likes 
        FROM   language_page_likes_history
        WHERE  created_at >= '2015-09-12 00:00:00' AND
               language_page_likes_id = T1.id
        ORDER BY created_at ASC LIMIT 1) likes_first_in_period,
       (SELECT likes 
        FROM   language_page_likes_history
        WHERE  created_at <= '2015-09-15 00:00:00' AND
               language_page_likes_id = T1.id
        ORDER BY created_at DESC LIMIT 1) likes_last_in_period,
       (SELECT likes 
        FROM   language_page_likes_history
        WHERE  created_at > '2015-09-15 00:00:00' AND
               language_page_likes_id = T1.id
        ORDER BY created_at ASC LIMIT 1) likes_after_period

        FROM   language_page_likes T1) Q0
ORDER BY Likes DESC

它有4个子查询.您只需要选择两个.我已经选择使用期间之前的喜欢数量和期间的最后一些喜欢来计算差异.结果如下:

| PageId | LanguageId | Likes |
|--------|------------|-------|
|      3 |          3 |   462 |
|      1 |          1 |   343 |
|      2 |          1 |   324 |
|      3 |          1 |   309 |
|      1 |          3 |   239 |
|      2 |          3 |   235 |
|      3 |          2 |    13 |
|      2 |          2 |  -392 |
|      1 |          2 |  -969 |