Sorting a list while maintaining associations [SOLVED]

Scripting Issues
1

I’ve run into a little dilemma.

In my game, I have two kinds of actors in play: active and passive. Active kids belong to clubs. Passive kids do not. Passive kids keep a list that represents their level of loyalty to all the clubs in play. If a passive kid is loyal enough to a club, he/she will follow members of that club. This level of loyalty could exist for multiple clubs at once.

To determine who the passive kid will follow I need to make a list of all the kids they might pick and sort that list from lowest loyalty to highest (or highest to lowest). There are conditions where a passive kid won’t be able to follow an active kid, so I need to be able to walk through the list and make second or third best picks if the best pick isn’t available.

The problem is that if I make a list containing loyalties and sort it, I don’t know what loyalty belongs to what club anymore.

I made this code to solve the problem:

# Passive kids will only follow an active kid who put an up arrow on them.
				
uAOwners = []
# creates a list to contain the owners of all up arrows on this passive kid.				
				
uAOwners.append(self.iconManager.upArrow[0].getPythonTag("owner"))
# Adds the first up arrow owner to the list.
								
for uA in self.iconManager.upArrow:
# walks through all of the up arrows on this passive kid.
				
	for O in uAOwners:
	# walks through the owners currently in the list.
					
		if(self.loyalties[uA.getPythonTag("owner").clubID] >= self.loyalties[O.clubID]):
		# Checks if the loyalty to the current up arrow's owner is higher than or equal to 
		# the loyalty to the current owner in the uAOwners list.
						
			uAOwners.insert(uAOwners.index(O), uA.getPythonTag("owner"))
			# If so, the owner of the current arrow is inserted into the list before
			# the owner that he was just compared against.
							
			break
			# We don't want to add the current up arrow owner to the list more than once, so
			# after adding him we break the loop.
							
		elif(uAOwners.index(O) == len(uAOwners) - 1):
		# Checks to see if we are at the end of the owner list.
						
			uAOwners.append(uA.getPythonTag("owner")
			# If the current up arrow owner's loyalty isn't higher than or equal to any other's
			# we just put him at the end of the list.

This code creates a sorted list to begin with, filled with the active kids that the passive kid could follow, sorted in descending order of loyalty. I’m confident this will work, but it isn’t a very elegant solution. I’d like to find a simpler way to solve this problem, if anyone has any ideas.

2

Instead of sorting the list of loyalties, consider creating a list of 2-tuples: (loyalty, clubID), and sort that list.

David

3

If I use tuples will only the first item in the tuple be considered in the sorting?

4

Technically, the second item is used too, but only if the first items are the same. So, for the most part, yes.

David

5

Alright, that should be a much nicer solution. Thanks David.