如何在Python中触发相同类对象的2个实例中的方法?

Question

我有同一个类(State)的两个实例(st0, st1)，其中有两个方法。当st0当loc_st()被调用时，我希望st1.rem_st()自动触发。我如何用Python OOP实现这个目的?对不起，代码看起来有点幼稚，但它是一个简化的模型，节省了阅读的精力。

import random
class State(object):
def __init__(self) -> None:
self.s = ['S0', 'S1', 'S2']
def loc_sc(self):
# how to trigger a remote method in different class?
print (self.s)
def rem_sc(self):
random.shuffle(self.s)
print (self.s)
if __name__ == '__main__':
st0 = State()
st1 = State()
# how can I trigger st1.rem_sc() automatically 
#  when st0.loc_sc is called  
st0.loc_sc()

Answer 1

如果你想要调用每个实例的方法，你可以在类中添加一个列表，遍历所有实例并调用它们的方法:

import random
class State(object):
instances = []
def __init__(self) -> None:
self.in_trigger = 0
self.ex_trigger = 0
self.s = ['S0', 'S1', 'S2']
State.instances.append(self)

def loc_sc(self):
# how to trigger a remote method in different class?
print (self.s)
for i in State.instances:
i.rem_sc()

def rem_sc(self):
if self.ex_trigger:
random.shuffle(self.s)
print (self.s)

if __name__ == '__main__':
st0 = State()
st1 = State()
st0.loc_sc()

或者您可以向类构造函数中添加实例数组。当方法在一个实例上被调用时，你可以遍历该数组中的所有实例并调用方法:

import random
class State(object):
def __init__(self, instances=[]) -> None:
self.in_trigger = 0
self.ex_trigger = 0
self.s = ['S0', 'S1', 'S2']
self.instances = instances

def loc_sc(self):
# how to trigger a remote method in different class?
print(self.s)
for i in self.instances:
i.rem_sc()

def rem_sc(self):
if self.ex_trigger:
random.shuffle(self.s)
print (self.s)

if __name__ == '__main__':
st1 = State()
st0 = State([st1])
st0.loc_sc()

如果您想先初始化st0，您可以向类添加一个addInstance方法，它将向该实例的实例数组中添加一个实例:

import random
class State(object):
def __init__(self, instances=[]) -> None:
self.in_trigger = 0
self.ex_trigger = 0
self.s = ['S0', 'S1', 'S2']
self.instances = instances

def loc_sc(self):
# how to trigger a remote method in different class?
print(self.s)
for i in self.instances:
i.rem_sc()

def rem_sc(self):
if self.ex_trigger:
random.shuffle(self.s)
print (self.s)
def addInstance(self, instance):
self.instances.append(instance)

if __name__ == '__main__':
st0 = State()
st1 = State()
st0.addInstance(st1)
st0.loc_sc()

或者您也可以向构造函数添加一个引用，以便实例本身被添加到该引用的实例中。

import random
class State(object):
def __init__(self, reference=None) -> None:
self.in_trigger = 0
self.ex_trigger = 0
self.s = ['S0', 'S1', 'S2']
self.instances = []
if reference != None: reference.instances.append(self)

def loc_sc(self):
# how to trigger a remote method in different class?
print(self.s)
for i in self.instances:
i.rem_sc()

def rem_sc(self):
if self.ex_trigger:
random.shuffle(self.s)
print (self.s)

if __name__ == '__main__':
st0 = State()
st1 = State(st0)
st0.loc_sc()

Answer 2

两个实例互不认识。所以这不是推荐的情况，但如果你真的想这么做，请在列表中恢复所有实例，当你想触发其中一个时，只需循环找到它。这就是答案!可能对你有帮助。