Python RQ-当其他多个作业完成时，如何触发工作?多职位依赖工作工作

Question

我的python redis队列中有一个嵌套的作业结构。首先执行了RNCopy作业。完成此操作后，接下来会有3个依赖的注册工作。当所有这3个作业的计算完成时，我想触发作业，以将Websocket发送到我的前端。

我当前的尝试：

    rncopy = redisqueue.enqueue(raw_nifti_copymachine, patientid, imagepath, timeout=6000)
    t1c_reg = redisqueue.enqueue(modality_registrator, patientid, "t1c", timeout=6000, depends_on=rncopy)
    t2_reg = redisqueue.enqueue(modality_registrator, patientid, "t2", timeout=6000, depends_on=rncopy)
    fla_reg = redisqueue.enqueue(modality_registrator, patientid, "fla", timeout=6000, depends_on=rncopy)
    notify = redisqueue.enqueue(print, patient_finished, patientid, timeout=6000, depends_on=(t1c_reg, t2_reg, fla_reg))

不幸的是，似乎从未将多职位依赖性功能合并到主人中。我看到目前在Git上有两个拉的请求。有能力可以使用的解决方法吗？

很抱歉未能提供可再现的例子。

Answer 1

新版本(rq＆gt; = 1.8(

您可以简单地使用depends_on参数，通过列表或元组。

rncopy = redisqueue.enqueue(raw_nifti_copymachine, patientid, imagepath, timeout=6000)
t1c_reg = redisqueue.enqueue(modality_registrator, patientid, "t1c", timeout=6000, depends_on=rncopy)
t2_reg = redisqueue.enqueue(modality_registrator, patientid, "t2", timeout=6000, depends_on=rncopy)
fla_reg = redisqueue.enqueue(modality_registrator, patientid, "fla", timeout=6000, depends_on=rncopy)
notify = redisqueue.enqueue(first_wrapper, patient_finished, patientid,t2_reg.id,fla_reg.id, timeout=6000, depends_on=(t1c_reg, t2_reg, fla_reg))
# you can also use a list instead of a tuple:
# notify = redisqueue.enqueue(first_wrapper, patient_finished, patientid,t2_reg.id,fla_reg.id, timeout=6000, depends_on=[t1c_reg, t2_reg, fla_reg])

旧版本(RQ＆LT; 1.8(

我使用此解决方法：如果依赖项为 n ，我创建了真实功能的 n-1 包装器：每个包装器都取决于其他作业。

此解决方案有点偏僻，但起作用。

rncopy = redisqueue.enqueue(raw_nifti_copymachine, patientid, imagepath, timeout=6000)
t1c_reg = redisqueue.enqueue(modality_registrator, patientid, "t1c", timeout=6000, depends_on=rncopy)
t2_reg = redisqueue.enqueue(modality_registrator, patientid, "t2", timeout=6000, depends_on=rncopy)
fla_reg = redisqueue.enqueue(modality_registrator, patientid, "fla", timeout=6000, depends_on=rncopy)
notify = redisqueue.enqueue(first_wrapper, patient_finished, patientid,t2_reg.id,fla_reg.id, timeout=6000, depends_on=t1c_reg)
def first_wrapper(patient_finished, patientid,t2_reg_id,fla_reg_id):
    queue = Queue('YOUR-QUEUE-NAME'))
    queue.enqueue(second_wrapper, patient_finished, patientid, fla_reg_id, timeout=6000, depends_on=t2_reg_id)
def second_wrapper(patient_finished, patientid,fla_reg_id):
    queue = Queue('YOUR-QUEUE-NAME'))
    queue.enqueue(print, patient_finished, patientid, timeout=6000, depends_on=fla_reg_id)

一些警告：

• 我不会将队列对象传递给包装器，因为出现了一些序列化问题。因此，队列必须通过名称恢复...

• 出于相同的原因，我将Job.ID(而不是作业对象(传递给包装器。

Answer 2

我创建了'an＆quot'rq-manager;解决与依赖性等多重和树的类似问题：https://github.com/crispydyne/rq-manager

具有多个依赖性的项目结构看起来像这样。

def simpleTask(x):
    return 2*x

project = {'jobs':[
            {
                'blocking':True, # this job must finished before moving on.
                'func':simpleTask,'args': 0
            },
            {
                'blocking':True, # this job, and its child jobs, must finished before moving on.
                'jobs':[ # these child jobs will run in parallel
                    {'func':simpleTask,'args': 1},
                    {'func':simpleTask,'args': 2},
                    {'func':simpleTask,'args': 3}],
            },
            { # this job will only run when the blocking jobs above finish.
                'func':simpleTask,'args': 4
            }
        ]}

然后将其传递给经理完成。

from rq_manager import manager, getProjectResults
managerJob = q.enqueue(manager,project)
projectResults = getProjectResults(managerJob)

返回

projectResults = [0, [2, 4, 6], 8]

---

当抚养工作需要父母的结果时。我创建了执行第一个作业的函数，然后在项目中添加了其他作业。因此，就您的示例：

def firstTask(patientid,imagepath):
    raw_nifti_result  = raw_nifti_copymachine(patientid,imagepath)
    moreTasks = {'jobs':[
        {'func':modality_registrator,'args':(patientid, "t1c", raw_nifti_result)},
        {'func':modality_registrator,'args':(patientid, "t2", raw_nifti_result)},
        {'func':modality_registrator,'args':(patientid, "fla", raw_nifti_result)},
    ]}
    # returning a dictionary with an "addJobs" will add those tasks to the project. 
    return {'result':raw_nifti_result, 'addJobs':moreTasks}

该项目看起来像这样：

project = {'jobs':[
            {'blocking':True, # this job, and its child jobs, must finished before moving on.
             'jobs':[
                {
                    'func':firstTask, 'args':(patientid, imagepath)
                    'blocking':True, # this job must finished before moving on.
                },
                # "moreTasks" will be added here
                ]
            }
            { # this job will only run when the blocking jobs above finish.
                'func':print,'args': (patient_finished, patientid)
            }
        ]}

---

如果最终作业需要上一个作业的结果，请设置"上议院"。旗帜。＆quot" finaljob"将收到以前结果的数组，并带有其子作业结果的嵌套数组。

def finalJob(previousResults):
    # previousResults = [ 
    #     raw_nifti_copymachine(patientid,imagepath),
    #     [
    #         modality_registrator(patientid, "t1c", raw_nifti_result),
    #         modality_registrator(patientid, "t2", raw_nifti_result),
    #         modality_registrator(patientid, "fla", raw_nifti_result),
    #     ]
    # ]
    return doSomethingWith(previousResults)

那么该项目看起来像这样

project = {'jobs':[
            {
             #'blocking':True, # Blocking not needed.
             'jobs':[
                {
                    'func':firstTask, 'args':(patientid, imagepath)
                    'blocking':True, # this job must finished before moving on.
                },
                # "moreTasks" will be added here
                ]
            }
            { # This job will wait, since it needs the previous job's results. 
                'func':finalJob, 'previousJobArgs': True # it gets all the previous jobs results
            }
        ]}

---

希望https://github.com/rq/rq/rq/issues/260可以实现，我的解决方案将被淘汰！