Question

给定一个（可能很大的~2 + GB）json文件中的节点之间的事务，具有〜百万个节点和~1000万个事务，每个事务具有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},...
  ]
}

最优雅高效的pythonic方法是将其转换为节点亲和度矩阵，其中亲和度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]

e.g。

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

注意：亲和度矩阵是对称的，因此单独计算下三角就足够了。

值不具代表性***结构示例仅供参考！

node1 | node2 | node3 | node4 | ....
node1 1 .4 .1 .9 ...
node2 .4 1 .6 .3 ...
node3 .1 .6 1 .7 ...... br> node4 .9 .3 .7 1 ... ...

Answer 1

首先，我会清理数据并用整数表示每个节点，并以这样的字典开头

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pythonic但它应该是不言自明的

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)]

查看亲和度矩阵

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 ]]

如何从事务行中有效地构建亲和度矩阵？

1 个答案: