如何从sparql查询中过滤任何定义概念？

如何解决如何从sparql查询中过滤任何定义概念？

我将使用从DBpedia中提取的SPARQL查询来构建主题层次结构树。 如何从sparql查询中过滤掉没有定义的概念？

我当前的代码如下：

from SPARQLWrapper import SPARQLWrapper,N3
from rdflib.extras.external_graph_libs import rdflib_to_networkx_graph
from pprint import pprint
from rdflib import Graph
import networkx as nx
import copy

# Main concept 
concept = '<http://dbpedia.org/resource/Category:Machine_learning>'

sparql = SPARQLWrapper("http://dbpedia.org/sparql")

sparql.setQuery(f"""
CONSTRUCT {{ ?child skos:broader {concept} .}} 
WHERE {{ ?child skos:broader {concept} .}}
""")

sparql.setReturnFormat(N3)
results = sparql.query().convert()
g_child = Graph()
g_child.parse(data=results,format="n3")

sparql.setQuery(f"""
CONSTRUCT {{ ?child skos:broader {concept} .
             ?grandchild skos:broader ?child .}} 
WHERE {{ ?child skos:broader {concept} .
         ?grandchild skos:broader ?child .}}
""")

sparql.setReturnFormat(N3)
results = sparql.query().convert()
g_grandchild = Graph()
g_grandchild.parse(data=results,format="n3")

我曾经尝试：

sparql.setQuery(f"""
CONSTRUCT {{ ?child skos:broader {concept} .}} 
WHERE {{ ?child skos:broader {concept} .
        FILTER(LANG(?deFinition) = 'en')}}
""")

sparql.setReturnFormat(N3)
results = sparql.query().convert()
g_child = Graph()
g_child.parse(data=results,format="n3")

和

sparql.setQuery(f"""
CONSTRUCT {{ ?child skos:broader {concept} .
             ?grandchild skos:broader ?child .}} 
WHERE {{ ?child skos:broader {concept} .
         ?grandchild skos:broader ?child .
         FILTER(LANG(?deFinition) = 'en') }}
""")

sparql.setReturnFormat(N3)
results = sparql.query().convert()
g_grandchild = Graph()
g_grandchild.parse(data=results,format="n3")

但是，图形变为空。

提前谢谢！