sql 查找相似字段

如何利用 SQL 查找相似字段

在进行数据库查询时，经常会遇到需要查找相似字段的情况。而利用 SQL 进行相似字段的查找是一种常见且有效的方法。本文将介绍如何利用 SQL 查询来搜索相似字段，以帮助您更好地处理数据库中复杂的数据。

使用 SQL LIKE 运算符进行相似字段查询

在 SQL 中，可以使用 LIKE 运算符来进行模糊查询，从而找到与指定模式匹配的数据。比如，如果您需要查找包含特定字符序列的字段，可以使用以下语法：

SELECT * FROM table_name WHERE column_name LIKE 'pattern';

其中，table_name 是您要查询的表名，column_name 是要进行相似字段匹配的字段名称，pattern 是要匹配的模式。

利用通配符进行更精确的相似字段匹配

通配符是在进行 SQL 查询时非常有用的工具，可以帮助您更精确地查找相似字段。常用的通配符包括：

• % 代表零个、一个或多个字符
• _ 代表单个字符

例如，如果您要查找以 'abc' 开头的字段，可以使用以下查询语句：

SELECT * FROM table_name WHERE column_name LIKE 'abc%';

结合通配符实现更复杂的相似字段查找

利用通配符结合在一起使用，可以实现更加复杂的相似字段查找。比如，如果您需要查找第二个字符是 'a'，第四个字符是 'b' 的字段，可以这样查询：

SELECT * FROM table_name WHERE column_name LIKE '_a__b%';

在上面的查询中，_ 匹配任意单个字符，因此可以实现所需的相似字段匹配条件。

使用 SQL 中的其他函数扩展相似字段查询功能

除了 LIKE 运算符外，SQL 还提供了其他函数和操作符，可以帮助您更灵活地进行相似字段查询。比如，您可以使用 CONCAT 函数将多个字段拼接在一起后再进行匹配：

SELECT * FROM table_name WHERE CONCAT(column1, column2) LIKE 'pattern';

通过灵活运用 SQL 中的这些函数和操作符，您可以更好地处理相似字段查询的需求，提高数据库查询的效率。

总结

通过本文的介绍，相信您对如何利用 SQL 查找相似字段有了更深入的了解。在实际应用中，灵活运用 LIKE 运算符、通配符以及其他函数，可以帮助您更准确、高效地进行相似字段的查询操作。希望以上内容能够对您在处理数据库查询时有所帮助！

 news_headline1 = "Elon Musk's Boring Co to build high-speed airport link in Chicago"
 news_headline2 = "Elon Musk's Boring Company to build high-speed Chicago airport link"
 news_headline3 = "Elon Musk’s Boring Company approved to build high-speed transit between downtown Chicago and O’Hare Airport"
 news_headline4 = "Both apple and orange are fruit"
 
 news_headlines = [news_headline1, news_headline2, news_headline3, news_headline4]
 print('Master Sentence: %s' % news_headlines[0])
 for i, news_headline in enumerate(news_headlines):
     score = sklearn.metrics.pairwise.euclidean_distances([transformed_results[i]], [transformed_results[0]])[0][0]
     print('-----')
     print('Score: %.2f, Comparing Sentence: %s' % (score, news_headline))

 Master Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 -----
 Score: 0.00, Comparing Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 -----
 Score: 1.73, Comparing Sentence: Elon Musk's Boring Company to build high-speed Chicago airport link
 -----
 Score: 4.36, Comparing Sentence: Elon Musk’s Boring Company approved to build high-speed transit between downtown Chicago and O’Hare Airport
 -----
 Score: 4.24, Comparing Sentence: Both apple and orange are fruit

 print('Master Sentence: %s' % news_headlines[0])
 for i, news_headline in enumerate(news_headlines):
     score = sklearn.metrics.pairwise.cosine_similarity([transformed_results[i]], [transformed_results[0]])[0][0]
     print('-----')
     print('Score: %.2f, Comparing Sentence: %s' % (score, news_headline))




 Master Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 -----
 Score: 1.00, Comparing Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 -----
 Score: 0.87, Comparing Sentence: Elon Musk's Boring Company to build high-speed Chicago airport link
 -----
 Score: 0.44, Comparing Sentence: Elon Musk’s Boring Company approved to build high-speed transit between downtown Chicago and O’Hare Airport
 -----
 Score: 0.00, Comparing Sentence: Both apple and orange are fruit

 """
     Finding the posistion (from lookup table) of word instead of using 1 or 0
     to prevent misleading of the meaning of "common" word
 """
 
 def calculate_position(values):
     x = []
     for pos, matrix in enumerate(values):
         if matrix > 0:
             x.append(pos)
     return x
 
 """
     Since scikit-learn can only compare same number of dimension of input. 
     Add padding to the shortest sentence.
 """
 def padding(sentence1, sentence2):
     x1 = sentence1.copy()
     x2 = sentence2.copy()
  
     diff = len(x1) - len(x2)
  
     if diff > 0:
         for i in range(0, diff):
             x2.append(-1)
     elif diff < 0:
         for i in range(0, abs(diff)):
             x1.append(-1)
  
     return x1, x2    
 
 y_actual = calculate_position(transformed_results[0])
 
 print('Master Sentence: %s' % news_headlines[0])
 for i, news_headline in enumerate(news_headlines):
     y_compare = calculate_position(transformed_results[i], )
     x1, x2 = padding(y_actual, y_compare)
     print(x1,x2)
     score = sklearn.metrics.jaccard_score(x1, x2,average='macro')
     print('-----')
     print(score, news_headline)
     print('Score: %.2f, Comparing Sentence: %s' % (score, news_headline))
Master Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 [0, 2, 4, 5, 7, 9, 11, 17, 20, 21, 22, 25] [0, 2, 4, 5, 7, 9, 11, 17, 20, 21, 22, 25]
 -----
 1.0 Elon Musk's Boring Co to build high-speed airport link in Chicago
 Score: 1.00, Comparing Sentence: Elon Musk's Boring Co to build high-speed airport link in Chicago
 [0, 2, 4, 5, 7, 9, 11, 17, 20, 21, 22, 25] [0, 2, 4, 6, 7, 9, 11, 17, 20, 22, 25, -1]
 -----
 0.5714285714285714 Elon Musk's Boring Company to build high-speed Chicago airport link
 Score: 0.57, Comparing Sentence: Elon Musk's Boring Company to build high-speed Chicago airport link
 [0, 2, 4, 5, 7, 9, 11, 17, 20, 21, 22, 25, -1, -1, -1, -1, -1, -1] [1, 2, 4, 6, 7, 8, 9, 10, 12, 14, 16, 17, 18, 20, 24, 25, 26, 27]
 -----
 0.125 Elon Musk’s Boring Company approved to build high-speed transit between downtown Chicago and O’Hare Airport
 Score: 0.12, Comparing Sentence: Elon Musk’s Boring Company approved to build high-speed transit between downtown Chicago and O’Hare Airport
 [0, 2, 4, 5, 7, 9, 11, 17, 20, 21, 22, 25] [3, 12, 13, 15, 19, 23, -1, -1, -1, -1, -1, -1]
 -----
 0.0 Both apple and orange are fruit
 Score: 0.00, Comparing Sentence: Both apple and orange are fruit