Python,SimPy:在函数内部使用yield

lam*_*mba 1 python simulation simpy

Helo,我正在SimPy中构建一个相对复杂的离散事件仿真模型.

当我尝试将yield语句放在函数中时,我的程序似乎不起作用.下面显示了一个示例.

import SimPy.SimulationTrace as Sim
import random

## Model components ##
class Customer(Sim.Process):
    def visit(self):
        yield Sim.hold, self, 2.0
        if random.random()<0.5:
            self.holdLong()
        else:
            self.holdShort()

    def holdLong(self):
        yield Sim.hold, self, 1.0
        # more yeild statements to follow

    def holdShort(self):
        yield Sim.hold, self, 0.5
        # more yeild statements to follow

## Experiment data ##
maxTime = 10.0 #minutes

## Model/Experiment ##
#random.seed(12345)
Sim.initialize()
c = Customer(name = "Klaus") #customer object
Sim.activate(c, c.visit(), at = 1.0)
Sim.simulate(until=maxTime)
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我从运行中获得的输出是:

0 activate <Klaus > at time: 1.0 prior: False
1.0 hold  < Klaus >  delay: 2.0
3.0 <Klaus > terminated
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holdLong()和holdShort方法似乎根本不起作用.我怎样才能解决这个问题?提前致谢.

Sve*_*ach 6

调用生成器函数会返回可以迭代的生成器对象.你只是忽略了这个返回值,所以没有任何反应.相反,您应该迭代生成器并重新生成所有值:

class Customer(Sim.Process):
    def visit(self):
        yield Sim.hold, self, 2.0
        if random.random()<0.5:
            for x in self.holdLong():
                yield x
        else:
            for x in self.holdShort():
                yield x
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  • 我相信Python 3.3的新'yield from'语法在这里很有用. (2认同)