[Python知识库]python 如何实现依据依存关系构造邻接矩阵（有向图）

如何实现依据依存关系构造邻接矩阵（有向图）

#!/usr/bin/env python
# -*- encoding: utf-8 -*-
'''
@Filename :depency_matix.py
@Description :
@Datatime :2021/10/08 11:03:42
@Author :qtxu
@Version :v1.0
@function:通过依存关系生成邻接矩阵
'''


import spacy
from tqdm import tqdm
import numpy as np

def adj_dependcy_tree(argments, max_length):
    # nlp = spacy.load('en')
    nlp = spacy.load("en_core_web_sm")
    depend = []
    depend1 = []
    doc = nlp(str(argments)) # doc表示当前的sentence
    print(doc)
    d = {}
    i = 0
    for(_, token) in enumerate(doc):
        if str(token) in d.keys():
            continue
        d[str(token)] = i
        i = i+1
    
    for token in doc:
        depend.append((token.text, token.head.text)) #将关系对加入depend[]
        # print("token.text: ",token.text)
        # print("token.head.text:", token.head.text)
        depend1.append((d[str(token)], d[str(token.head)]))

    ze = np.identity(max_length) # np.identity(n) 创建n*n的矩阵
    for(i,j) in depend1:
        if i >= max_length or j >= max_length:
            continue
        ze[i][j] = 1
    
    # D = np.array(np.sum(ze, axis=1))
    # D = np.matrix(np.diag(D))
    # DSN = np.sqrt(D** -1)
    # DN = D ** -1
    # return ze, DN, DSN
    return ze

if __name__ == "__main__":
    cur_length = 5  # 长度大于等于句子长度即可
    argments = "I love eating apple."
    print(adj_dependcy_tree(argments,max_length=cur_length) )

程序结果：

[[1. 1. 0. 0. 0.]
 [0. 1. 0. 0. 0.]
 [0. 1. 1. 0. 0.]
 [0. 0. 1. 1. 0.]
 [0. 1. 0. 0. 1.]]