Using regexp_replace to remove letters from numbers

Let's leave the ranges (like 1200-1300) to the side, since - even regardless of any kind of programming - it is not clear what you would want to "extract" from that. And, you may also have problems with things like '5 ft 10 in' or similar, if they are possible in your data. (And it is not clear what the whole thing means if all altitudes aren't using the same unit of measurement anyway - some are in meters, some in feet, the info disappears when you just keep the number).

To remove all the non-digits from a string and to keep the digits, you do NOT need regular expressions, which may be quite slow (an order of magnitude slower!) than standard string functions.

One way to remove all non-digit characters uses the TRANSLATE function. Like so:

translate(input_string, '0123456789' || input_string, '0123456789')

The function "translates" (replaces) 0 with 0, 1 with 1, etc., and any character in the input string that hasn't already appeared earlier in the second argument (which in this case means "non-digit") to nothing (null, zip, disappears, is removed).

Example (note the use of TO_NUMBER to also convert to actual numbers):

with
  data (input_string) as (
    select '1500'   from dual union all
    select '2100 m' from dual union all
    select '535 ft' from dual
  )
select input_string,
       to_number(translate(input_string, '0123456789' || input_string, 
                                         '0123456789')) as extracted_number
from   data;

INPUT_STRING EXTRACTED_NUMBER
------------ ----------------
1500                     1500
2100 m                   2100
535 ft                    535