Python:具有短生命周期键的线程安全字典，这是正确的吗？

Question

import threading
import weakref
_mainlock = threading.RLock()
_job_locks = weakref.WeakValueDictionary()
def do_thing(job_id):
    _mainlock.acquire() #Dictionary modification lock acquire
    _job_locks.setdefault(job_id, threading.RLock()) #Possibly modifies the dictionary
    _mainlock.release()
    _job_locks[job_id].acquire()
    try:
        one_time_init(job_id)
    finally:
        _job_locks[job_id].release()
    #On function return, the weakref.WeakValueDictionary should cause the key to evaporate

假设 do_thing() 在 ID 编号可能相同或不同的许多线程上被多次调用(例如，4 次 ID 3，各 1 次 ID 不同)，这个线程安全吗？ one_time_init() 是否会一次针对特定作业 ID 运行多次？ (PS:one_time_init 保存的是每个 ID 都运行过一次的状态，所以如果已经运行完成就调用它是空操作)

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更新代码(感谢 THC4k):

import threading
import weakref
_mainlock = threading.RLock()
_job_locks = weakref.WeakValueDictionary()

def do_thing(job_id):
    with _mainlock:
        jl = _job_locks.setdefault(job_id, threading.RLock())
    with jl:
        one_time_init(job_id)

Answer 1

似乎非常安全。为什么你甚至需要 _job_locks 如果 one_time_init 再次检查它是否运行？你可以在那里加锁。为什么是RLock而不是Lock(函数似乎永远不会重入)？

无论如何，with 语句看起来更好:

import threading
import weakref
_mainlock = threading.RLock()
_job_locks = weakref.WeakValueDictionary()

def do_thing(job_id):
    with _mainlock:
        _job_locks.setdefault(job_id, threading.RLock())
    with _job_locks[job_id]:
        one_time_init(job_id)