Python 向PyQt5中的线程函数发送信号会导致装饰槽的类型错误

我目前正在开发一个Gui，它会一直计数，直到达到给定的输入数字。应该可以在计数期间停止while循环，并使用按钮重新启动它，而无需结束主Gui线程。这已经起作用了，但仅当目标计数数在线程工作函数中是固定的，例如n=10->counts to 10，并且在不关闭Gui并更改主代码中的数字的情况下，不能更改为20或任何其他数字。我想对数字输入使用行编辑，并将所需的目标数字发送到线程，以便它在我开始计数时计数到给定的数字。但是，当启动Gui（在实现了通常需要发送和获取输入值的信号之后）在行编辑中键入数字并按下开始按钮时，我得到一个TypeError，指出“装饰的插槽没有与started（）兼容的签名”（“started”应该连接线程和工作函数）。是否有人碰巧知道我在哪里犯了错误或必须更改某些内容和/或是否有解决此错误的方法

我非常绝望（我已经尝试解决这个问题很长一段时间了，在网上找不到任何提示…），非常感谢您的帮助！致以最良好的问候和感谢

这是带有注释的完整代码：

import sys
import time
from PyQt5.QtWidgets import QPushButton, QMainWindow, QApplication, QLineEdit
from PyQt5.QtCore import QObject, QThread, pyqtSignal, pyqtSlot

class Worker(QObject):

    finished = pyqtSignal()  # signal out to main thread to alert it that work is completed

    def __init__(self):
        super(Worker, self).__init__()
        self.working = True  # flag to control our loop

    @pyqtSlot(int) # should take the sended integer n from signal...
def work(self, n):

        s = 0

        while self.working:

            if s != n: # count until input integer is reached (doesn't work so far)
                print(s)
                s += 1
                time.sleep(0.5)

        self.finished.emit() # alert gui that the loop stopped

class Window(QMainWindow):

    sendnumber = pyqtSignal(int) # send signal (this is how it is usually done...?)

    def __init__(self):

        super(Window, self).__init__()
        self.setGeometry(50, 50, 200, 250)
        self.setWindowTitle("Program")

        self.inputnumber=QLineEdit(self, placeholderText="number")
        self.inputnumber.resize(self.inputnumber.minimumSizeHint())
        self.inputnumber.move(50, 50)

        self.startbtn = QPushButton("Start", self)
        self.startbtn.resize(self.startbtn.minimumSizeHint())
        self.startbtn.move(50, 100)

        self.stopbtn = QPushButton("Stop", self)
        self.stopbtn.resize(self.stopbtn.minimumSizeHint())
        self.stopbtn.move(50, 150)

        self.thread = None
        self.worker = None

        self.startbtn.clicked.connect(self.start_loop)  

    def start_loop(self):

        self.thread = QThread()  # a new thread to run the background tasks in
        self.worker = Worker()  # a new worker to perform those tasks
        self.worker.moveToThread(self.thread)  # move the worker into the thread, do this first before connecting the signals

        self.thread.started.connect(self.worker.work)  # begin worker object loop when the thread starts running
        self.sendnumber.connect(self.worker.work) # connect input number to worker in thread
        # this doesn't work so far and gives a TypeError!

        self.stopbtn.clicked.connect(self.stop_loop)  # stop the loop on the stop button click
        self.worker.finished.connect(self.loop_finished)  # do something in the gui when the worker loop ends
        self.worker.finished.connect(self.thread.quit)  # tell the thread it's time to stop running
        self.worker.finished.connect(self.worker.deleteLater)  # have worker mark itself for deletion
        self.thread.finished.connect(self.thread.deleteLater)  # have thread mark itself for deletion
        # make sure those last two are connected to themselves or you will get random crashes

        self.thread.start()

    def stop_loop(self):

        self.worker.working = False
        # when ready to stop the loop, set the working flag to false

    @pyqtSlot() # as far as I know you need this Slot to send the input to Slot in thread?
    def getnumber(self):

        try:
            n = int(self.inputnumber.text()) # input for the work function in thread
            print('Trying number...')
        except:
            print('Use an integer!')
            return

        self.sendnumber.emit(n) # emit integer signal
        print('Emitting signal to worker...')

    def loop_finished(self):
        # received a callback from the thread that it's completed
        print('Loop finished.')

if __name__ == '__main__':
    def run():

        app = QApplication(sys.argv)
        gui = Window()
        gui.show()
        app.exec_()

    run() 

下面是显示错误的控制台输出：

Traceback (most recent call last):

  File "C:/Users/***/WhileLoopInterruptTest2.py", line 63, in start_loop
    self.thread.started.connect(self.worker.work)

TypeError: decorated slot has no signature compatible with started()

这就是所有的错误，Gui必须关闭。 我使用的是Python 3.6.1、Spyder 3.3.1和PyQt5.9。

替换这一行：

self.thread.started.connect(self.worker.work)

为此：

self.thread.started.connect(lambda: self.worker.work(10))

您可以用文本框中的值替换

10

。请记住首先将其转换为

int

---

查看更详细的解释。

有趣的是，我通常使用

partial

来代替lambdas<代码>从functools导入部分

self.thread.started.connect（部分（self.worker.work，input=10））

@Guimoute我最近发现了部分。我一直在使用lambda表达式，使用partial的解决方案对我有效！不知何故，lambda没有完成这项工作（Gui在为输入设置新值后停止工作）。我还删除了getnumber函数和所有pyqtSlot的东西（不再需要了），因为我现在可以通过文本框控制目标编号。我将self.thread.started.connect（partial（self.worker.work，input=10））更改为self.thread.started.connect（partial（self.worker.work，n=int（self.inputnumber.text（））。太好了，很高兴它有帮助<当您想强制某些特定的命名输入，而将其他输入留空时，code>partial非常方便。装饰器和默认值可能会填充其他输入。请参阅示例，其中显示了处理线程的3种不同且简单的方法。