Re: problem with subset one image by another

negra wrote:
....
> max_x = min(file1_xmap[1], file_xmap[1])
> min_y = max(file1_ymap[0], file_ymap[0])
> max_y = min(file1_ymap[1], file_ymap[1])

As Jean has already pointed out, you'd have to organize things 
differently to use MAX() and MIN() here. However, it's far simpler to 
avoid MAX() and MIN() altogether.

max_x = file1_xmap[1] < file_xmap[1]
min_y = file1_ymap[0] > file_ymap[0]
max_y = file1_ymap[1] < file_ymap[1]

It all looks pretty confusing if you're used to a language (there are 
quite a few) where '<' and '>' are comparison operators, but in IDL the 
corresponding comparison operators are called 'lt' and 'gt'. The '<' and 
'>' operators return the minimum and maximum, respectively, of their 
operands.