Python 正确使用threading.RLock
我正在用Python创建一个实用程序,它在单独的线程上启动时从文件中读取数据,以便可以加载其余的GUI组件。数据存储到Python 正确使用threading.RLock,python,multithreading,Python,Multithreading,我正在用Python创建一个实用程序,它在单独的线程上启动时从文件中读取数据,以便可以加载其余的GUI组件。数据存储到列表中,然后附加到组合框中。如何锁定列表,以便其他方法无法在def read_员工(self,read_文件)使用列表的同时调用列表:方法 这是我能想到的最好的尝试 #left out imports class MyDialog(wx.Frame): def __init__(self, parent, title): self.no_resize
列表中,然后附加到组合框中。如何锁定列表
,以便其他方法无法在def read_员工(self,read_文件)使用列表的同时调用列表:
方法
这是我能想到的最好的尝试
#left out imports
class MyDialog(wx.Frame):
def __init__(self, parent, title):
self.no_resize = wx.DEFAULT_FRAME_STYLE & ~ (wx.RESIZE_BORDER | wx.MAXIMIZE_BOX)
wx.Frame.__init__(self, parent, title=title, size=(500, 450),style = self.no_resize)
self.lock = threading.RLock()
self.empList = []
def read_employees(self, read_file):
with open(read_file) as f_obj:
employees = json.load(f_obj)
with self.lock:
self.empList = [empEmail for empEmail in employees.keys()]
wx.CallAfter(self.emp_selection.Append, self.empList)
def start_read_thread(self):
filename = 'employee.json'
with concurrent.futures.ThreadPoolExecutor(max_workers=1) as executor:
executor.submit(self.read_employees, filename)
app = wx.App(False)
frame = MyDialog(None, "Crystal Rose")
app.MainLoop()
在这里使用RLock
合适吗?我不知道你在应用程序中还做了什么,但我建议你看看wx.CallAfter函数。它是线程安全的,可用于发送消息或发布事件
import wx
from wx.lib.pubsub import Publisher
import json
from threading import Thread
def update_employee_list(read_file):
with open(read_file) as f_obj:
employee_list = json.load(f_obj) # this line should release the GIL so it continues other threads
# next line sends a thread-safe message to the main event thread
wx.CallAfter(Publisher().sendMesage, 'updateEmployeeList', employee_list)
class MyDialog(wx.Frame):
def __init__(self, parent, title):
self.no_resize = wx.DEFAULT_FRAME_STYLE & ~ (wx.RESIZE_BORDER | wx.MAXIMIZE_BOX)
wx.Frame.__init__(self, parent, title=title, size=(500, 450),style = self.no_resize)
self.empList = []
# subscribe our function to be called when 'updateEmployeeList' messages are received
Publisher().subscribe(self.updateDisplay, 'updateEmployeeList')
def updateDisplay(self, employee_list):
# this assignment should be atomic and thread-safe
self.empList = employee_list
# wxPython GUI runs in a single thread, so this is a blocking call
# if you have many many list items, you may want to modify this method
# to add one employee at a time to the list to keep it non-blocking.
self.emp_selection.Append(employee_list)
def start_read_thread(self):
filename = 'employee.json'
t = Thread(target= update_employee_list, args=(filename, ))
t.start() # this starts the thread and immediately continues this thread's execution
更新:
将与ThreadPoolExecutor一起使用会阻塞,因为代码等效于:
executor = ThreadPoolExecutor(max_workers=1)
executor.submit(worker_func, args)
executor.shutdown(wait=True) # <--- wait=True causes Executor to block until all threads complete
executor=ThreadPoolExecutor(最大工作线程数=1)
执行人提交(工人函数,参数)
executor.shutdown(wait=True)#感谢您花时间和精力创建此文件。在start\u read\u线程中,self.read\u employees
应该是update\u employeelist(read\u文件):
和def update\u employeelist(read\u文件):
,updateEmployeeList'
应该是updateDisplay
正确吗?如果我想继续使用R.Lock
,我是否正确使用了它?是的,我已经修复了该错误。在您的示例中,ThreadExecutor将一直阻塞,直到线程完成,所以我不能说它是正确的。至于锁,很难判断你是否正确使用了它,因为我只看到它的一种用法。如果要使用锁,则必须在访问该变量的任何地方使用它。那个看起来不错,虽然有。我需要CallAfter处于锁定区域。您能否更新您的答案以解释threadpoolexecutor
将被阻止的原因?在这里的回答中,有人建议我使用threadpoolexecutor
。文档中说,如果我将与
一起使用,我就不必使用shutdown。它将起作用,就像shutdown
被设置为True一样。不完全如此。请仔细阅读我的更新。如果wait=True
,代码将阻塞,直到所有线程完成
executor = ThreadPoolExecutor(max_workers=1)
executor.submit(worker_func, args)
executor.shutdown(wait=False) # <--- threads will still complete, but execution of this thread continues immediately