Remove duplicate based on condition - sql

Question

My table value:
COLUMN1 COLUMN2 COLUMN3
WF1 Email 1640
WF1 Email 1641
WF1 Email N/A
WF3 Email N/A
Expected Result:
COLUMN1 COLUMN2 COLUMN3
WF1 Email 1640
WF3 Email N/A
I need to retrieve all records which column2 = 'Email' and if column1 contains duplicate value, i have to choose the record which column3 <> 'N/A'.
I read tutorial about partition by but still not sure how to get the result.
Any help is appreciated.
CREATE TABLE TABLE1
(
COLUMN1 varchar2(20),
COLUMN2 varchar2(20),
COLUMN3 varchar2(20)
);
INSERT INTO TABLE1
(COLUMN1, COLUMN2, COLUMN3)
VALUES
('WF1', 'Email', '1640');
INSERT INTO TABLE1
(COLUMN1, COLUMN2, COLUMN3)
VALUES
('WF1', 'Email', '1641');
INSERT INTO TABLE1
(COLUMN1, COLUMN2, COLUMN3)
VALUES
('WF1', 'Email', 'N/A');
INSERT INTO TABLE1
(COLUMN1, COLUMN2, COLUMN3)
VALUES
('WF3', 'Email', 'N/A');

Answer 1

Try something like this:
SELECT column1, column2, column3
from(
SELECT column1, column2, column3,
row_number() over (partition BY column1, column2 ORDER BY CASE WHEN column3 = 'N/A' THEN 999999999 ELSE to_number(column3) END ) rn
FROM table1)
WHERE rn = 1
Here is a sqlfiddle

Answer 2

select column1,column2,max(case when column3 = 'N/A' then '0' else column3 end) column3
from table1
group by column1,column2;

Answer 3

If column1 can be not unique in the result:
select column1, column3 from table1 t1
where column3 != 'N/A' or
column3 = 'N/A' and not exists
(select * from table1 t2
where t2.column1 = t1.column1 and t2.column3 != 'N/A')
If column1 must be unique in the result:
select column1, column3 from table1 t1
where column3 != 'N/A' and not exists
(select * from table1 t2
where t2.column1 = t1.column1 and t2.column3 != 'N/A' and
to_number(t2.column3, '9999') > to_number(t1.column3, '9999')
) union
select column1, column3 from table1 t1
where column3 = 'N/A' and not exists
(select * from table1 t2 where t2.column1 = t1.column1 and t2.column3 != 'N/A')
Note that we use union instead of union all.