如何有效地从事务行构建亲和矩阵？

Question

给定json文件中节点之间的事务（可能大到2 GBs），其中约有100万个节点和约1000万个事务，每个事务有10-1000个节点，例如

{"transactions":
 [
  {"transaction 1": ["node1","node2","node7"], "weight":0.41},
  {"transaction 2": ["node4","node2","node1","node3","node10","node7","node9"], "weight":0.67},
  {"transaction 3": ["node3","node10","node11","node2","node1"], "weight":0.33},...
  ]
}

将其转换为节点亲和力矩阵的最优雅和有效的pythonic方法是什么，其中亲和力是节点之间加权事务的总和。

affinity [i,j] = weighted transaction count between nodes[i] and nodes[j] = affinity [j,i]

例如

affinity[node1, node7] = [0.41 (transaction1) + 0.67 (transaction2)] / 2 = affinity[node7, node1]

注意：亲和矩阵将是对称的，因此仅计算下三角形就足够了。

数值不具有代表性***仅结构示例！

Answer 1

首先，我会清理数据，用一个整数表示每个节点，并从这样一个字典开始

data=[{'transaction': [1, 2, 7], 'weight': 0.41},
      {'transaction': [4, 2, 1, 3, 10, 7, 9], 'weight': 0.67},
      {'transaction': [3, 10, 11, 2, 1], 'weight': 0.33}]

不确定这是否足够Python化，但它应该是不言而喻的

def weight(i,j,data_item):
    return data_item["weight"] if i in data_item["transaction"] and j in data_item["transaction"] else 0

def affinity(i,j):
    if j<i: # matrix is symmetric
        return affinity(j,i)
    else:
        weights = [weight(i,j,data_item) for data_item in data if weight(i,j,data_item)!=0]
        if len(weights)==0:
            return 0
        else:
            return sum(weights) / float(len(weights))

ln = 10 # number of nodes
A = [[affinity(i,j) for j in range(1,ln+1)] for i in range(1,ln+1)]

要查看关联性矩阵

import numpy as np
print(np.array(A))

    [[ 0.47  0.47  0.5   0.67  0.    0.    0.54  0.    0.67  0.5 ]
     [ 0.47  0.47  0.5   0.67  0.    0.    0.54  0.    0.67  0.5 ]
     [ 0.5   0.5   0.5   0.67  0.    0.    0.67  0.    0.67  0.5 ]
     [ 0.67  0.67  0.67  0.67  0.    0.    0.67  0.    0.67  0.67]
     [ 0.    0.    0.    0.    0.    0.    0.    0.    0.    0.  ]
     [ 0.    0.    0.    0.    0.    0.    0.    0.    0.    0.  ]
     [ 0.54  0.54  0.67  0.67  0.    0.    0.54  0.    0.67  0.67]
     [ 0.    0.    0.    0.    0.    0.    0.    0.    0.    0.  ]
     [ 0.67  0.67  0.67  0.67  0.    0.    0.67  0.    0.67  0.67]
     [ 0.5   0.5   0.5   0.67  0.    0.    0.67  0.    0.67  0.5 ]]