Postgresql ORDER BY spaces

Don't know, may be something like this:

with cte as (
   select col1, regexp_split_to_array(col1, ' ') as d
   from Table1

)
select col1
from cte
order by
    d[1] ~ '^([0-9]+[.]?[0-9]*|[.][0-9]+)$' desc,
    case
        when d[1] ~ '^([0-9]+[.]?[0-9]*|[.][0-9]+)$' then
            d[1]::numeric
    end,
    d[2]

sql fiddle demo

this one splits string by spaces to array, convert first entry to numeric and sorts results by this number and remaining string

This should do it:

order by regexp_replace(the_column, '[^0-9\.]', '', 'g')::numeric DESC

It removes all character that are non-numeric (leaves only digits and the .) from the value and then converts it into a number. That number is then used to sort descending

The only thing is that 5.0 will be sorted after 5.1 with the above.

If you need to take values into account which don't have any digits, something like this:

order by 
      case 
         when regexp_replace(the_column, '[^0-9\.]', '', 'g') <> 
            then regexp_replace(the_column, '[^0-9\.]', '', 'g')::numeric
      end DESC NULLS LAST, 
      the_column DESC -- to sort the "NULL" values alphabetically

If you don't want to repeat the regex expression you could do something like this:

with clean_data as (
   select the_column, 
          regexp_replace(the_column, '[^0-9\.]', '', 'g') as clean_column, 
          ....
   from ...
)
select *
from clean_data
order by case 
           when clean_column <> '' then clean_column::numeric 
         end desc nulls last, 
         the_column desc;