Question

0

192

Visualizações

How to check same value is matching to how many Ids using MySQL or R

I have below-mentioned table in MySQL/Dataframe (MySQL Version - 5.7.18):

Table_1:

ID      Date                  uid
I-1     2020-01-01 10:12:15   K-1
I-2     2020-01-02 10:12:15   K-1
I-3     2020-02-01 10:12:15   K-2
I-4     2020-02-02 10:12:15   K-3
I-5     2020-02-04 10:12:15   K-4
I-6     2019-11-01 10:12:15   K-4
I-7     2019-11-01 10:12:15   K-3
I-8     2018-12-13 10:12:15   K-5
I-9     2019-05-17 10:12:15   K-4
I-19    2020-03-11 10:12:15   K-7

Table_2:

 ID        city           code
I-1        New York       123
I-2        Washington     122
I-3        Tokyo          123
I-4        London         144
I-5        Dubai          101
I-6        Dubai          101
I-7        London         144
I-8        Tokyo          143
I-9        Dubai          101
I-19       Dubai          150

Using the above-mentioned table, I want to fetch records between 1st Jan 2020 to 29th Feb 2002 and compare those ID in entire database to check whether both city and code together match with other ID and categorize it further to check how many have the same uid and how many have different.

Where,

Match - combination of city and code match with other ID in database
Same_uid - classification of Match ids to identify how many ID have similar uid
different_uid - classification of Match ids to identify how many ID doesn't have similar uid
uid_count - count of similar uid of that particular ID in entire database

Required Output

ID      Date                  city         code   uid   Match   Same_uid   different_uid  uid_count
I-1     2020-01-01 10:12:15   New York     123    K-1    No      0          0              2
I-2     2020-01-02 10:12:15   Washington   122    K-1    No      0          0              2
I-3     2020-02-01 10:12:15   Tokyo        123    K-2    No      0          0              1   
I-4     2020-02-02 10:12:15   London       144    K-3    Yes     1          0              2
I-5     2020-02-04 10:12:15   Dubai        101    K-4    Yes     2          0              3

over 4 years ago · Santiago Trujillo

1 Respostas

Responde à pergunta

0

Responde à pergunta

Encontrar trabalhos remotos

Accepted Answer · 2022-05-11T21:44:12.381Z

It seems (after all corrections) that you need

SELECT t1.ID, 
       t1.`Date`, 
       t1. city, 
       t1.code, 
       t1.uid, 
       CASE WHEN SUM((t1.city = t2.city) * (t1.code = t2.code)) - 1
            THEN 'Yes'
            ELSE 'No' END `Match`,
       SUM((t1.city = t2.city) * (t1.code = t2.code) * (t1.uid = t2.uid)) - 1 same_uid,
       SUM((t1.city = t2.city) * (t1.code = t2.code) * (t1.uid != t2.uid)) different_uid,
       SUM(t1.uid = t2.uid) uid_count
FROM cities t1
CROSS JOIN cities t2
WHERE t1.`Date` >= '2020-01-01' AND t1.`Date` < '2020-03-01'
GROUP BY t1.ID, t1.`Date`, t1. city, t1.code, t1.uid
ORDER BY t1.ID

fiddle

PS. Multilpyings in SUM()s may be replaced with AND.

what if I have these information in two different table

SELECT t11.ID, 
       t11.`Date`, 
       t21.city, 
       t21.code, 
       t11.uid, 
       CASE WHEN SUM((t21.city = t22.city) * (t21.code = t22.code)) - 1
            THEN 'Yes'
            ELSE 'No' END `Match`,
       SUM((t21.city = t22.city) * (t21.code = t22.code) * (t11.uid = t12.uid)) - 1 same_uid,
       SUM((t21.city = t22.city) * (t21.code = t22.code) * (t11.uid != t12.uid)) different_uid,
       SUM(t11.uid = t12.uid) uid_count
FROM /* cities t1 */
     (Table_1 t11 NATURAL JOIN Table_2 t21)
CROSS JOIN /* cities t2 */
           (Table_1 t12 NATURAL JOIN Table_2 t22)
WHERE t11.`Date` >= '2020-01-01' AND t11.`Date` < '2020-03-01'
GROUP BY t11.ID, t11.`Date`, t21.city, t21.code, t11.uid
ORDER BY t11.ID

fiddle

0

192

How to check same value is matching to how many Ids using MySQL or R

1 Respostas

0

Encontrar trabalhos remotos

Andres GPT