0 
company  | email | phone | website | address
Amar CO LLC | amar@gmail.com | 123 | NULL | India
Amar CO | amar@gmail.com | NULL | NULL | IND
Stacks CO | stack@gmail.com | 910 | stacks.com | United Kingdom
Stacks CO LLC | stack@gmail.com | NULL | NULL | UK

我想删除公司名称，CO LLC而是希望保留Amar CO所有列，Amar CO LLC因为它具有minimum NULL值或最大列数据。

简而言之：对记录进行重复数据删除，删除以 LLC 结尾或匹配的公司名称（不区分大小写），但保留两个具有最大信息列的记录中的值。

预期产出

Amar CO | amar@gmail.com | 123 | NULL | India
Stacks CO | stack@gmail.com | 910 | stacks.com | United Kingdom
4

2 回答 2

1

你需要group by和replace如下：

select replace(company,' LLC','') as company, max(email) as email, max(phone) as phone,
       max(website) as website, max(address) as address
  from your_table t
group by replace(company,' LLC','')

我可以看到您需要两行的所有数据，但应优先考虑 LLC 记录（India, IND --> India），然后您可以按如下方式使用它：

select t.company, 
       coalesce(tt.email,t.emial) as email, 
       coalesce(tt.phone,t.phone) as phone
       coalesce(tt.website,t.website) as website,
       coalesce(tt.address,t.address) as address
  from your_table t join your_table tt 
    on concat(t.company,' LLC') = tt.company

如果您想更新数据然后删除记录本身，我建议您使用以下delete和update.

delete from your_table where t.company = 'Amar CO';

update your_table t
set t.comapny = replace(company,' LLC','') -- or use 'Amar CO'
where t.company = 'Amar CO LLC';

- 更新

您想优先考虑具有最小空值的记录，然后您可以使用以下查询：

select t.company,
       case when tt_nulls > t_nulls then ttemail else temail end as email,
       case when tt_nulls > t_nulls then ttphone else tphone end as phone,
       case when tt_nulls > t_nulls then ttwebsite else twebsite end as website,
       case when tt_nulls > t_nulls then taddress else taddress end as address
from    
(select t.company, 
        count(case when t.email IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when t.phone IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when t.website IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when t.address IS NULL THEN 1 end) over (partition by t.company)  
        as t_nulls,
        count(case when tt.email IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when tt.phone IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when tt.website IS NULL THEN 1 end) over (partition by t.company) 
        + count(case when tt.address IS NULL THEN 1 end) over (partition by t.company)  
        as tt_nulls
        t.email as temail, 
        t.phone as tphone,
        t.website as twebsite,
        t.address as taddress,
        tt.email as ttemail, 
        tt.phone as ttphone,
        tt.website as ttwebsite,
        tt.address as ttaddress
   from your_table t join your_table tt 
     on concat(t.company,' LLC') = tt.company) t
于 2020-12-30T05:29:10.557 回答
1

优先考虑具有最小空值的记录...

以下是 BigQuery 标准 SQL（查询#1）

#standardSQL
select 
  array_agg(t 
    order by array_length(regexp_extract_all(to_json_string(t), ':null')) 
    limit 1
  )[offset(0)].* 
  replace(regexp_replace(company, r'(?i)CO LLC', 'CO') as company) 
from `project.dataset.table` t
group by company

如果应用于您问题的样本数据 - 输出是

在此处输入图像描述

如果您想填写所有记录中的所有字段 - 您可以在下面使用（查询＃2）

select regexp_replace(company, r'(?i)CO LLC', 'CO') as company,
  max(email) email,
  max(phone) phone,
  max(website) website,
  max(address) address
from `project.dataset.table`
group by company

最后 - 如果您仍然想优先考虑具有最小空值的记录，但其余的空值替换为其他行的值 - 使用下面（查询＃3）

select company, 
  ifnull(email, max_email) email,
  ifnull(phone, max_phone) phone,
  ifnull(website, max_website) website,
  ifnull(address, max_address) address
from (
  select array_agg(t 
      order by array_length(regexp_extract_all(to_json_string(t), ':null')) 
      limit 1
    )[offset(0)].* 
    replace(regexp_replace(company, r'(?i)CO LLC', 'CO') as company),
    max(email) max_email, 
    max(phone) max_phone,
    max(website) max_website,
    max(address) max_address
  from `project.dataset.table` t
  group by company 
)

您可以通过将它们应用于以下虚拟数据来测试/检查此选项与上一个选项之间的区别

with `project.dataset.table` as (
  select 'Amar CO LLC' company, 'amar@gmail.com' email, 123 phone, NULL website, 'India' address union all
  select 'Amar CO', NULL, 222, 'amar.com', NULL union all
  select 'Stacks CO LLC', 'stack@gmail.com', NULL, NULL, 'UK' union all
  select 'Stacks CO', 'stack@gmil.com', 910, 'stacks.com', 'United Kingdom'
)

最后一个查询（query#3）给出

在此处输入图像描述

而先前的（查询＃2）只会在所有行中给出最大值

在此处输入图像描述

于 2020-12-30T06:49:27.893 回答