CODEWITHSUNDEEP
pythonset
mpen
mpen

Reputation: 282885

How to maintain a set of sets?

I've seen the related question, but I can't use a frozenset for my inner-sets. I want everything to be mutable.

As I understand it, Python gives everything an ID, so why can't it use that as its hash? I need to maintain a list of references to all the inner-sets, even as they change.

Edit: Ok, I understand why, but in this case it would be preferable, as I only care about reference equality, not value equality.

How can I do this?

You're going to ask "why", so I'll give you some code:

def remove_captured(self):
    all_chains = set()
    chains = Grid(self.rows, self.cols)
    for m, n, stone in self.enumerate():
        if stone == self[m - 1, n]:
            chains[m, n] = chains[m - 1, n]
            if stone == self[m, n - 1]:
                all_chains.discard(chains[m, n - 1])
                chains[m, n].update(chains[m, n - 1])
                for s in chains[m, n - 1]:
                    chains[s] = chains[m, n]
        elif stone == self[m, n - 1]:
            chains[m, n] = chains[m, n - 1]
        else:
            chains[m, n] = set()
            all_chains.add(chains[m, n])
        chains[m, n].add((m,n))
    chains._print()
    print all_chains

I've essentially got a game board, and I want to divide the pieces on the board into groups (or "chains"). The above code works fine until I added all_chains -- it creates all the sets, but then I have no way of accessing each the sets its created without iterating over the whole board again.

So how do I maintain a list of all the sets it's created? Keep in mind that I need to remove sets too (which is why I want to use another set for the outter set).

Wrapping the reference in weakref.ref() didn't work either:

all_chains.add(weakref.ref(chains[m, n])) # TypeError: unhashable type: 'set'

Upvotes: 3

Views: 231

Answers (1)

mpen
mpen

Reputation: 282885

Decided to use a dictionary of sets instead.

def remove_captured(self):
    cdict = {}
    cid = 0
    chains = Grid(self.rows, self.cols)
    for m, n, stone in self.enumerate():
        if stone == self[m - 1, n]:
            chains[m, n] = chains[m - 1, n]
            if stone == self[m, n - 1] and chains[m, n] != chains[m, n - 1]:
                del_id = chains[m, n - 1]
                cdict[chains[m, n]].update(cdict[del_id])
                for c in cdict[del_id]:
                    chains[c] = chains[m, n]
                del cdict[del_id]
        elif stone == self[m, n - 1]:
            chains[m, n] = chains[m, n - 1]
        else:
            cdict[cid] = set()
            chains[m, n] = cid
            cid += 1
        cdict[chains[m, n]].add((m, n))
    chains._print()
    pprint(cdict)

It's not quite as pretty because I always have to look up the set in the dictionary before I can use it, but it seems to work.

Input:

0  .  W  W  W  W
1  W  B  B  B  .
2  .  .  .  W  .
3  .  B  B  W  W
4  .  .  B  W  .
   0  1  2  3  4

Output:

0 0 1 1 1 1
1 2 3 3 3 4
2 5 5 5 6 4
3 5 7 7 6 6
4 5 5 7 6 8
  0 1 2 3 4
{0: set([(0, 0)]),
 1: set([(0, 1), (0, 2), (0, 3), (0, 4)]),
 2: set([(1, 0)]),
 3: set([(1, 1), (1, 2), (1, 3)]),
 4: set([(1, 4), (2, 4)]),
 5: set([(2, 0), (2, 1), (2, 2), (3, 0), (4, 0), (4, 1)]),
 6: set([(2, 3), (3, 3), (3, 4), (4, 3)]),
 7: set([(3, 1), (3, 2), (4, 2)]),
 8: set([(4, 4)])}

Upvotes: 1

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