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alice-cool avatar alice-cool commented on July 18, 2024 2

I think maybe the irregular list A in the dataset.py the get_item() makes the mistake.

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alice-cool avatar alice-cool commented on July 18, 2024 2

in dataset.py

A = [[] for k in range(self.n_node)] 
for triple in data[i]["graph"]:
        A[triple[0]].append((triple[1], triple[2]))

it is confusing. Because the first line says the number of elements A is equal to self.n_node( the maximum number of nodes in one graph) But the last two lines say that the len of A will be modified by the id of node, such as triple[0], that is the maximum of id of node.

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alice-cool avatar alice-cool commented on July 18, 2024
    @staticmethod
    def find_max_syz(data, num):
        max_syz = 0
        for i in range(len(data)):
            listnum = [0 for k in range(num)]
            for j in range(len(data[i]['graph'])):
                listnum[data[i]['graph'][j][0]] = listnum[data[i]['graph'][j][0]]+1
            if max_syz < max(listnum):
                max_syz = max(listnum)

        return max_syz

    @staticmethod
    def find_max_node_id(data):
        max_num_id = 0
        for i in range(len(data)):
            for triple in data[i]["graph"]:
                if triple[0] > max_num_id:
                    max_num_id = triple[0]
                if triple[2] > max_num_id:
                    max_num_id = triple[2]
        return max_num_id

self.n_node_types = self.find_max_node_id(data)

            A = [[] for k in range(self.n_node_types)]

            for triple in data[i]["graph"]:
                A[triple[0]].append((triple[1], triple[2]))

            print("syz:",self.syz_num)
            #padding syz
            for i in range(len(A)):
                if A[i]==[]:
                    for k in range(self.syz_num):
                        A[i].append((0, 0))
                elif len(A[i]) < self.syz_num:
                    cc = self.syz_num - len(A[i])
                    for k in range(cc):
                        A[i].append((0, 0))

            A_list.append(A)
            data_idx.append(i)

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entslscheia avatar entslscheia commented on July 18, 2024

in dataset.py

A = [[] for k in range(self.n_node)] 
for triple in data[i]["graph"]:
        A[triple[0]].append((triple[1], triple[2]))

it is confusing. Because the first line says the number of elements A is equal to self.n_node( the maximum number of nodes in one graph) But the last two lines say that the len of A will be modified by the id of node, such as triple[0], that is the maximum of id of node.

Hi! Sorry about the trouble. I have not been maintaining this repo for like two years, so I don't actually remember this kind of debugging-level details at this point. I would suggest you trying some toy data under the data/ directory and see how it works. But for the code you pasted here, I don't quite think the claim "last two lines say that the len of A will be modified by the id of node" makes sense. The last two lines only update the content of A, without changing the size of A (i.e., len(A)).
Hope it helps!

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alice-cool avatar alice-cool commented on July 18, 2024

from ggnn_reasoning.

alice-cool avatar alice-cool commented on July 18, 2024

I run my modified code that throw the memory error. SO my way will produce a big sparse A list. I said the len(A)will be modified because if evey graph at most has 3 edges. if using you code, it said the A list initialize the length as 3. If the set of samples of graph data includes 100 different nodes, that is the node of id will be up to 100. So in the loop because A[triple[0]], so the triple[0] will be 100. So len(A)will be replaced by 100. It is just my opinion. Thanks for your help

for triple in data[i]["graph"]:
        A[triple[0]].append((triple[1], triple[2]))

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