two rows of delete with different values

Hello

I wouldn't lines with the same id and have values of E and F 10258932.

I would like the following data:
A        B  C      D     E N
-------- -- ------ ----- - -
10258927 1  103,35 0
10258929 3  284,85 89,52 E N
10258929 4  323,85 89,52 E N
10258930 5  478,80 91,53 E N
10258931 6  436,67 78,09 E N
But I have the following lines:
SQL> with my_table as
  2  (
  3  select '10258927' a, '1'  b, '103,35'  c, '0'      d,   NULL  e from dual union all    
  4  select '10258928' a, '2'  b, '0'       c, '15,19'  d,  'E'    e from dual union all
  5  select '10258929' a, '3'  b, '284,85'  c, '89,52'  d,  'E'    e from dual union all
  6  select '10258929' a, '4'  b, '323,85'  c, '89,52'  d,  'E'    e from dual union all
  7  select '10258930' a, '5'  b, '478,80'  c, '91,53'  d,  'E'    e from dual union all
  8  select '10258931' a, '6'  b, '436,67'  c, '78,09'  d,  'E'    e from dual union all
  9  select '10258932' a, '7'  b, '784,88'  c, '23,19'  d,  'E'    e from dual union all
 10  select '10258932' a, '8'  b, '100,88'  c, '11,11'  d,  'E'    e from dual union all
 11  select '10258932' a, '9'  b, '300,88'  c, '22,22'  d,  'F'    e from dual union all
 12  select '10258932' a, '10' b, '468,25'  c, '78,33'  d,  'F'    e from dual)
 13  select a, b, c, d, e, no_ds from
 14  (
 15    select a, b, c, d, e,
 16           case when c > 0 and e != 'F'
 17                then 'N'
 18           end no_ds
 19    from my_table)
 20  where no_ds = 'N' or e is null
 21  order by 1
 22  /

A        B  C      D     E N
-------- -- ------ ----- - -
10258927 1  103,35 0
10258929 3  284,85 89,52 E N
10258929 4  323,85 89,52 E N
10258930 5  478,80 91,53 E N
10258931 6  436,67 78,09 E N
10258932 7  784,88 23,19 E N
10258932 8  100,88 11,11 E N

7 Zeilen ausgewählt.

SQL> 
Does anyone have an idea?
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Oh, I read the title again once: 'remove double lines with different values.
If it means that he is not specifically E and F values, but only if there is more than one distinct value, then you can do the following:

with my_table as
(
select '10258927' a, '1'  b, '103,35'  c, '0'      d,   NULL  e from dual union all
select '10258928' a, '2'  b, '0'       c, '15,19'  d,  'E'    e from dual union all
select '10258929' a, '3'  b, '284,85'  c, '89,52'  d,  'E'    e from dual union all
select '10258929' a, '4'  b, '323,85'  c, '89,52'  d,  'E'    e from dual union all
select '10258930' a, '5'  b, '478,80'  c, '91,53'  d,  'E'    e from dual union all
select '10258931' a, '6'  b, '436,67'  c, '78,09'  d,  'E'    e from dual union all
select '10258932' a, '7'  b, '784,88'  c, '23,19'  d,  'E'    e from dual union all
select '10258932' a, '8'  b, '100,88'  c, '11,11'  d,  'E'    e from dual union all
select '10258932' a, '9'  b, '300,88'  c, '22,22'  d,  'F'    e from dual union all
select '10258932' a, '10' b, '468,25'  c, '78,33'  d,  'F'    e from dual
)
--
-- end-of-test-data
--
select a, b, c, d, e,
       num_distinct_values
from
(
  select a, b, c, d, e,
         count(distinct e) over (partition by a) num_distinct_values
  from my_table)
where num_distinct_values <= 1
order by 1
/ 

The county notes how the distinct non-null values in column e for each one.
Where clause then selects those with count = 0 (this is where e = null) and count = 1.
If the number is greater than 1, different values exist for this id.

Choose what fits your condition ;-)

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