我试图从我使用experta库构建的规则引擎中提取规则。
下面是代码
from experta import *
import pydot
class Iris(Fact):
"""IRIS"""
sepal_length = Field(float)
sepal_width = Field(float)
petal_length = Field(float)
petal_width = Field(float)
class Species(Fact):
"""Species characteristics"""
pass
class IrisExpert(KnowledgeEngine):
def __init__(self):
super().__init__()
self.status = None
@Rule(Iris(sepal_length=P(lambda x: x > 5), sepal_width=P(lambda x: x < 3)))
def rule1(self):
self.status = 'Iris Setosa'
return self.declare(Species(flower='Iris Setosa'))
@Rule(AND(
Iris(sepal_length=P(lambda x: x > 5.5)),
Iris(sepal_width=P(lambda x: x > 2.5))
))
def rule2(self):
self.status = 'Iris Versicolor'
return self.declare(Species(flower='Iris Versicolor'))
@Rule(AND(
Iris(petal_length=P(lambda x: x > 4.8)),
Iris(petal_width=P(lambda x: x > 1.8))
))
def rule3(self):
self.status = 'Iris Virginica'
return self.declare(Species(flower='Iris Virginica'))
def execute_fuc(key):
if key == 'rule1':
print('executing rule-1')
engine.rule1()
elif key == 'rule2':
print('executing rule-2')
engine.rule2()
else:
print('executing rule-3')
engine.rule3()
# Create a new knowledge engine instance
engine = IrisExpert()
engine.reset()
# Create a Pydot graph
graph = pydot.Dot(graph_type='digraph', comment='IRIS Expert Rules')
# Add nodes for each rule
for rule_instance in engine.get_rules():
quality = rule_instance()
qual_node = pydot.Node(repr(quality))
graph.add_node(qual_node)
for rule in rule_instance:
print('rule_format: ',repr(rule))
我得到了以下格式
rule_format: Iris(petal_width=P(<function IrisExpert.<lambda> at 0x7fe3fd4788c8>,))
但我想找的是
Iris(petal_width=P(<function IrisExpert. x > 5)) (for function "rule1")
1条答案
按热度按时间kadbb4591#
你显然是想重写lambda函数的
repr
方法。This answer展示了如何做到这一点。应用到你的案例(两种< and >表示),你可以做以下事情:输出: