Python 2.7 Simpy：请求多个（非特定）资源和请求顺序

我正在Python2中开发一个队列模拟模型，该模型中有作业进入系统并请求多个资源。每个到达的作业要求不同数量的资源（而不是特定的资源！），并在不同的时间内使用这些资源

我发现的任何此类示例都需要特定的资源，如res[1]和res[2]。我只需要请求2个资源

此外，我的作业仅在第一个作业完成后运行。我知道我的for循环有问题，但我不确定如何正确修复它。 在这种情况下，因为有两个资源a和b应该能够在时间1运行。但是b要等到a完成。奇怪

如果您能在适当的时间请求多个资源并运行作业，我将不胜感激

以下是我目前的代码：

import simpy


#resource
class SuperComputer:
    def __init__(self, env):
        self.nodes = simpy.Resource(env, capacity = 2)


#users of resource
class Job:
    #enter: time the job enters the system
    #timeout is how long the job occupies a resource for
    #resources is how many resources a job needs in order to run
    def __init__(self, env, name, enter,timeout, resources):
        self.env = env
        self.name = name 
        self.enter = enter
        self.timeout = timeout
        self.resources = resources


#system
def system(env, jobs, super_computer):
    with super_computer.nodes.request() as req:

        for job in jobs:
            print('%s arrives at %s' % (job.name, job.enter)) 
            yield req
            yield env.timeout(job.enter)
            print('%s starts running with %s resources at %s' % (job.name, job.resources, env.now)) 
            yield env.timeout(job.timeout)
            print('%s completed job at %s' % (job.name, env.now)) 




env = simpy.Environment()
super_computer = SuperComputer(env) 

jobs = [
        Job(env, 'a', 1, 4, 1),
        Job(env, 'b', 1, 4, 1),
        Job(env, 'c', 1, 4, 1),
        Job(env, 'd', 1, 4, 1),
    ]


env.process(system(env, jobs, super_computer))    

env.run(50)        

输出：

a arrives at 1
a starts running with 1 resources at 1
a completed job at 5
b arrives at 1
b starts running with 1 resources at 6
b completed job at 10
c arrives at 1
c starts running with 1 resources at 11
c completed job at 15
d arrives at 1
d starts running with 1 resources at 16
d completed job at 20

---

问题是，代码需要先完成一轮for循环，然后再进行下一轮

如果for循环放在流程之外，那么代码就可以工作，而不是为每个作业创建一个新流程，从而允许它们并行运行

import simpy


#resource
class SuperComputer:
        def __init__(self, env):
        self.nodes = simpy.Resource(env, capacity = 2)


#users of resource
class Job:
    #enter: time the job enters the system
    #timeout is how long the job occupies a resource for
    #resources is how many resources a job needs in order to run
    def __init__(self, env, name, enter,timeout, resources):
        self.env = env
        self.name = name 
        self.enter = enter
        self.timeout = timeout
        self.resources = resources


#system
def system(env, job, super_computer):
    with super_computer.nodes.request() as req:
        print('%s arrives at %s' % (job.name, job.enter)) 
        yield req
        yield env.timeout(job.enter)
        print('%s starts running with %s resources at %s' % (job.name, job.resources, env.now)) 
        yield env.timeout(job.timeout)
        print('%s completed job at %s' % (job.name, env.now)) 


env = simpy.Environment()
super_computer = SuperComputer(env) 

jobs = [
        Job(env, 'a', 1, 4, 1),
        Job(env, 'b', 1, 4, 1),
        Job(env, 'c', 1, 4, 1),
        Job(env, 'd', 1, 4, 1),
    ]


for job in jobs:
    env.process(system(env, job, super_computer))    

env.run(50)        

---

非常感谢。这么简单的解决办法。对于如何为不同的工作请求多个资源，您有什么指导吗？目前，如果我有一个作业请求2个资源，我的print函数将打印出所请求的内容，但如果我请求req（job.resources），则req（）不起作用