问题描述
我有一个 User 表,该表具有 id , first_name ,姓氏,街道地址,城市,状态,邮政编码,公司, user_identifier , created_at , update_at 。
I have a User table, that has id, first_name, last_name, street_address, city, state, zip-code, firm, user_identifier, created_at, update_at.
此表有很多重复项,例如同一用户已作为新用户多次输入,因此示例
This table has a lot of duplication like the same users have been entered multiple times as a new user, so example
id first_name last_name street_address user_identifier --------------------------------------------------------- 11 Mary Doe 123 Main Ave M2111111 --------------------------------------------------------- 21 Mary Doe 123 Main Ave M2344455 --------------------------------------------------------- 13 Mary Esq Doe 123 Main Ave M1233444
我想知道是否有一种方法可以对此表进行模糊匹配
I would like to know if there is a way of doing fuzzy matching on this table.
基本上,我想找到所有具有相同名称,相同地址但可能会有细微差别的用户,也许地址相同但具有不同
Basically I would like to find all the users that have the same name, same address but can be with a slight difference, maybe the address is the same but has different apartment number, or has a middle name and the other duplicates don't.
我正在考虑创建一个新列,该列将名字,姓氏,街道地址串联在一起,然后对该列进行模糊匹配。
I was thinking to create a new column that has concatenated first_name, last_name, street_address and do a fuzzy match on that column.
我尝试了将first_name和last_name串联为 full_name $ c $的levenshtein距离c>
,但似乎没有赶上具有中间名的名称
I tried levenshtein distance on concatenated first_name and last_name as full_namebut doesn't seem to catch up the name that has a middle name
select * from users where levenshtein('Mary Doe', full_name) <=1;
我正在使用Databricks和PostgreSQL。
I am using a Databricks and PostgreSQL.
谢谢!
推荐答案
在postgres中,您可以使用 fuzzystrmatch 软件包。它提供了 levenshtein 函数,该函数返回两个文本之间的距离,然后可以使用以下示例性谓词执行模糊匹配:
In postgres you can use fuzzystrmatch package. It provies a levenshtein function, that returns distance between two texts, you can then perform fuzzy matching with the following exemplary predicate:
where levenshtein(street_address, '123 Main Avex') <= 1
这将匹配所有记录,因为'123 Main Ave'和'123 Main Avex'之间的距离是1(插入1个)。
This will match all records, because the distance between '123 Main Ave' and '123 Main Avex' is 1 (1 insertion).
Of当然,这里的值 1 只是一个例子,将非常严格地执行匹配(相差一个字符)。您应该使用更大的数字,或者使用@IVO GELOV表示的数字-使用相对距离(距离除以长度)。
Of course, value 1 here is just an example and will perform matching quite strictly (difference by only one character). You should either use larger number or, what @IVO GELOV sugests - use relative distance (distance divided by the length).
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