I have this data frame : basically each row being a transaction carried out by one customer on a day. there are multiple transactions by same customer on same day and on different dates. I want to get a column for a customers number of previous visits.

id  date   purchase 

id1 date1  $10    

id1 date1  $50    

id1 date2  $30     

id2 date1  $10     

id2 date1  $10     

id3 date3  $10     

after adding visits column:

id  date   purchase  visit

id1 date1  $10         0 

id1 date1  $50         0

id1 date2  $30         1

id2 date1  $10         0

id2 date2  $10         1

id2 date3  $10         2 

I do this in pandas using factorize :

df.visits = 1 
df.visits = df.groupby('id')['date'].transform(lambda x: pd.factorize(x)[0]) 

I want to do it through SQL, what would the query be like ?

1 Answers

0
Erfan On Best Solutions

You need DENSE_RANK() with PARTITION BY:

Creation of example dataset:

IF OBJECT_ID('Source', 'U') IS NOT NULL 
  DROP TABLE Source; 

CREATE TABLE Source
(
  id varchar(30),
  Date varchar(30),
  purchase varchar(30)
)

INSERT INTO Source
VALUES
('id1', 'date1', '$10'),   
('id1', 'date1', '$50'),   
('id1', 'date2', '$30'),    
('id2', 'date1', '$10'),   
('id2', 'date2', '$10'),  
('id2', 'date3', '$10')

SELECT *, 
  DENSE_RANK() OVER (PARTITION BY id ORDER BY date) - 1 AS visit
FROM Source

Output

Output