Creating new window using keyboard keys [duplicate] - pyqt

Question

I am programming a simple GUI, that will open a opencv window at a specific point. This window has some very basic keyEvents to control it. I want to advance this with a few functions. Since my QtGui is my Controller, I thought doing it with the KeyPressedEvent is a good way. My Problem is, that I cannot fire the KeyEvent, if I am active on the opencv window.
So How do I fire the KeyEvent, if my Gui is out of Focus?
Do I really need to use GrabKeyboard?
The following code reproduces my Problem:
import sys
from PyQt5.QtWidgets import (QApplication, QWidget)
from PyQt5.Qt import Qt
import cv2
class MainWindow(QWidget):
def __init__(self):
super().__init__()
self.first = True
def openselect(self):
im = cv2.imread(str('.\\images\\Steine\\0a5c8e512e.jpg'))
self.r = cv2.selectROI("Image", im)
def keyPressEvent(self, event):
if event.key() == Qt.Key_Space and self.first:
self.openselect()
self.first = False
print('Key Pressed!')
if __name__ == '__main__':
app = QApplication(sys.argv)
win = MainWindow()
win.show()
sys.exit(app.exec_())

Answer 1

The keyPressEvent method is only invoked if the widget has the focus so if the focus has another application then it will not be notified, so if you want to detect keyboard events then you must handle the OS libraries, but in python they already exist libraries that report those changes as pyinput(python -m pip install pyinput):
import sys
from PyQt5 import QtCore, QtWidgets
from pynput.keyboard import Key, Listener, KeyCode
class KeyMonitor(QtCore.QObject):
keyPressed = QtCore.pyqtSignal(KeyCode)
def __init__(self, parent=None):
super().__init__(parent)
self.listener = Listener(on_release=self.on_release)
def on_release(self, key):
self.keyPressed.emit(key)
def stop_monitoring(self):
self.listener.stop()
def start_monitoring(self):
self.listener.start()
class MainWindow(QtWidgets.QWidget):
pass
if __name__ == "__main__":
app = QtWidgets.QApplication(sys.argv)
monitor = KeyMonitor()
monitor.keyPressed.connect(print)
monitor.start_monitoring()
window = MainWindow()
window.show()
sys.exit(app.exec_())