如果您想使用sql解决方案，那么可以将csv数据放入一个规范化的表中，并按照id和word展开数据。
一旦你有了它，它就变成了一个自动加入表格的问题。把单词按字母顺序连在一起。
SQLFIDLE测试
不过，不确定这种方法在一个有1000k条记录的表上的速度有多快。
但这是一个有趣的难题。
样本数据：

DROP TABLE IF EXISTS test_words;
CREATE TABLE IF NOT EXISTS test_words (
  id int unsigned NOT NULL PRIMARY KEY,
  words varchar(60) NOT NULL
);
INSERT INTO test_words (id, words) VALUES
(1,'A C Z F G'),
(2,'G Q R C'),
(3,'Z G Q'),
(4,'C F'),
(5,'P O'),
(6,'O X Y J'),
(7,'A P X');

带数字的理货台：

DROP TABLE IF EXISTS tmp_numbers;
CREATE TEMPORARY TABLE IF NOT EXISTS tmp_numbers (
  n int unsigned NOT NULL PRIMARY KEY
);
INSERT INTO tmp_numbers (n) VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10);

展开文字：

DROP TABLE IF EXISTS test_words_unfolded;
CREATE TABLE test_words_unfolded (
  word varchar(10) NOT NULL,
  id int unsigned NOT NULL,
  PRIMARY KEY (word, id)
);
INSERT INTO test_words_unfolded (word, id)
SELECT DISTINCT SUBSTRING_INDEX(SUBSTRING_INDEX(t.words,' ', nr.n),' ',-1) as word, t.id
FROM test_words AS t
JOIN tmp_numbers AS nr
  ON CHAR_LENGTH(t.words) - CHAR_LENGTH(REPLACE(t.words,' ','')) >= nr.n - 1
 AND SUBSTRING_INDEX(SUBSTRING_INDEX(t.words,' ', nr.n),' ',-1) != '';

结果表：

DROP TABLE IF EXISTS test_result;
CREATE TABLE IF NOT EXISTS test_result (
  id int unsigned NOT NULL PRIMARY KEY,
  words varchar(60) NOT NULL
);

INSERT INTO test_result (id, words) 
SELECT q.id, GROUP_CONCAT(DISTINCT t3.word ORDER BY t3.word ASC SEPARATOR ' ') as words
FROM 
(
    SELECT t1.id, t2.id as id2
    FROM test_words_unfolded t1
    JOIN test_words_unfolded t2 ON t1.word = t2.word
    GROUP BY t1.id, t2.id
    HAVING COUNT(*) > 1 OR t1.id = t2.id
) q
LEFT JOIN test_words_unfolded t3 ON t3.id = q.id2
GROUP BY q.id
ORDER BY q.id;

SELECT *
FROM test_result
ORDER BY id;

结果：

id  words
--  -----
1   A C F G Q R Z
2   A C F G Q R Z
3   A C F G Q R Z
4   A C F G Z
5   O P
6   J O X Y
7   A P X

额外的
为了标记已添加的单词，填充结果表的查询变得有点复杂。

SELECT 
 q2.id, 
 GROUP_CONCAT(DISTINCT CASE WHEN q2.ori = 1 THEN q2.word ELSE CONCAT('[',q2.word,']') END ORDER BY q2.word ASC SEPARATOR ' ') as words
FROM
(
  SELECT
   q1.id, t3.word,
   MAX(CASE WHEN q1.id = t3.id THEN 1 ELSE 0 END) as ori
  FROM 
  (
      SELECT
       t1.id, t2.id as id2
      FROM test_words_unfolded t1
      JOIN test_words_unfolded t2 ON t1.word = t2.word
      GROUP BY t1.id, t2.id
      HAVING COUNT(*) > 1 OR t1.id = t2.id
  ) q1
  LEFT JOIN test_words_unfolded t3 ON t3.id = q1.id2
  GROUP BY q1.id, t3.word
) q2
GROUP BY q2.id
ORDER BY q2.id;

结果：

id  words
--  -----
1   A C F G [Q] [R] Z
2   [A] C [F] G Q R [Z]
3   [A] [C] [F] G Q [R] Z
4   [A] C F [G] [Z]
5   O P
6   J O X Y
7   A P X

这里还有其他的实验