文本字段上单词频率计数的 clickhouse 方法答案

【问题标题】：clickhouse approach for word frequency count on textual field文本字段上单词频率计数的 clickhouse 方法
【发布时间】：2020-03-09 09:39:16
【问题描述】：

我有一个 Clickhouse 表，其中一个字段包含文字说明（约 300 个字）。

例如评论：

Rev_id    Place_id    Stars    Category    Text
1         12           3        Food       Nice food but a bad dirty place.
2         31           4        Sport      Not bad, they have everything.
3         55           1        Bar        Poor place,bad audience.

我想做一些字数分析，例如一般字频计数（每个单词出现了多少次）或每个类别的前 K 个单词。

在示例中：

word    count
bad     3
place   2

... 有没有办法只在 ClickHouse 中完成而不涉及编程语言？

【问题讨论】：

标签： word-count clickhouse

【解决方案1】：

SELECT
    arrayJoin(splitByChar(' ', replaceRegexpAll(x, '[.,]', ' '))) AS w,
    count()
FROM
(
    SELECT 'Nice food but a bad dirty place.' AS x
    UNION ALL
    SELECT 'Not bad, they have everything.'
    UNION ALL
    SELECT 'Poor place,bad audience.'
)
GROUP BY w
ORDER BY count() DESC

┌─w──────────┬─count()─┐
│            │       4 │
│ bad        │       3 │
│ place      │       2 │
│ have       │       1 │
│ Poor       │       1 │
│ food       │       1 │
│ Not        │       1 │
│ they       │       1 │
│ audience   │       1 │
│ Nice       │       1 │
│ but        │       1 │
│ dirty      │       1 │
│ a          │       1 │
│ everything │       1 │
└────────────┴─────────┘


SELECT CATEGORY, ....
GROUP BY CATEGORY, w

【讨论】：

【解决方案2】：

如果它适用于您的情况，我会考虑使用alphaTokens 作为更有效的方法。

SELECT
    category,
    arrayJoin(arrayFilter(x -> NOT has(['a', 'the', 'but' /*.. exclude stopwords */], x), alphaTokens(text))) token,
    count() count
FROM
(
    /* test data */
    SELECT data.1 AS rev_id, data.2 AS place_id, data.3 AS stars, data.4 AS category, data.5 AS text
    FROM
    (
        SELECT arrayJoin([
          (1, 12, 3, 'Food', 'Nice      food but a bad dirty place.'), 
          (4, 12, 3, 'Food', ' the the the the good food   ..'), 
          (2, 31, 4, 'Sport', 'Not bad,,, they have everything.'), 
          (3, 55, 1, 'Bar', 'Poor place,bad audience..')]) AS data
    )
)
GROUP BY category, token
ORDER BY count DESC
LIMIT 5;
/*
┌─category─┬─token────┬─count─┐
│ Food     │ food     │     2 │
│ Food     │ bad      │     1 │
│ Bar      │ audience │     1 │
│ Food     │ Nice     │     1 │
│ Bar      │ Poor     │     1 │
└──────────┴──────────┴───────┘
*/

topK使用示例：

SELECT
    category,
    arrayReduce('topK(3)', 
                arrayFilter(x -> (NOT has(['a', 'the', 'but' /*.. exclude stopwords */], x)), groupArrayArray(alphaTokens(text)))) AS result
FROM
(
    /* test data */
    SELECT data.1 AS rev_id, data.2 AS place_id, data.3 AS stars, data.4 AS category, data.5 AS text
    FROM
    (
        SELECT arrayJoin([
          (1, 12, 3, 'Food', 'Nice      food but a bad dirty place.'), 
          (4, 12, 3, 'Food', ' the the the the good food   ..'), 
          (2, 31, 4, 'Sport', 'Not bad,,, they have everything.'), 
          (3, 55, 1, 'Bar', 'Poor place,bad audience..')]) AS data
    )
)
GROUP BY category;
/* result
┌─category─┬─result─────────────────┐
│ Bar      │ ['Poor','place','bad'] │
│ Food     │ ['food','Nice','bad']  │
│ Sport    │ ['Not','bad','they']   │
└──────────┴────────────────────────┘
*/

ps：在处理之前可能对lower所有字符串/令牌有意义

【讨论】：