获取相对于图像限制的鼠标位置

如何解决获取相对于图像限制的鼠标位置

我在自定义 QLabel 中显示了一张图片，并获得了对该标签的点击。我只对图像内的点击感兴趣，并且位置由 0 到 1 之间的数字表示，0 是最左侧或最顶部的像素，1 是最右侧或最底部的像素，无论图像实际大小如何。

我无法使用图像矩形来计算位置。当我调用 self.pixmap.rect() 时，宽度和高度是原始图像尺寸，而不是缩放以适合标签的图像尺寸。

我做错了什么？

from PyQt5.QtCore import Qt,pyqtSignal
from PyQt5.QtWidgets import QApplication,QWidget,QLabel,QGridLayout
from PyQt5.QtGui import QPixmap

class Window(QWidget):
    def __init__(self):
        super().__init__()
        image_path = '092.jpg'
        image_area = Image_Area(QPixmap(image_path))
        image_area.left_click.connect(self.image_clicked)

        layout = QGridLayout()
        layout.addWidget(image_area,0)
        self.setLayout(layout)

    # Left click on image
    def image_clicked(self,x,y):
        print(f'{x:.3f},{y:.3f}')


class Image_Area(QLabel):

    left_click = pyqtSignal(float,float)

    def __init__(self,pixmap):
        super().__init__()
        self.pixmap = pixmap
        self.setPixmap(pixmap)
        self.setScaledContents(False)
        self.setMinimumSize(100,100)

    def resizeEvent(self,e):
        if self.pixmap != None:
            width,height = int(self.width()),int(self.height())
            scaled_pixmap = self.pixmap.scaled(width,height,Qt.KeepAspectRatio)
            self.setPixmap(scaled_pixmap)

        return super().resizeEvent(e)

    def mousePressEvent(self,e):
        x,y,= e.x(),e.y()

        if self.pixmap:
            # Get pixmap rectangle
            r = self.pixmap.rect()
            x0,y0 = r.x(),r.y()
            x1,y1 = x0+r.width(),y0+r.height()

            # Check we clicked on the pixmap
            if x >= x0 and x < x1 and y >= y0 and y < y1:

                # emit position relative to pixmap bottom-left corner
                x_relative = (x - x0) / (x1 - x0)
                y_relative = (y - y0) / (y1 - y0)
                self.left_click.emit(x_relative,y_relative)

        super().mousePressEvent(e)


app = QApplication([])
window = Window()
window.show()
app.exec()

更新：在提供答案后，我意识到相对于图像原点的点击位置的计算是错误的，以下是感兴趣的人的准确版本：

def mousePressEvent(self,e):
        # Mouse position is in label coordinates
        x,e.y()

        if self.pixmap():
            # pixmap is not a widget,we don't have its location,# so we assume the pixmap is centered in the label and
            # compute the location with respect to label
            label_size = self.size()
            pixmap_size = self.pixmap().size()
            width = pixmap_size.width()
            height = pixmap_size.height()

            x0 = int((label_size.width() - width) / 2)
            y0 = int((label_size.height() - height) / 2)

            # Check we clicked on the pixmap
            if (x >= x0 and x < (x0 + width) and
                y >= y0 and y < (y0 + height)):

                # emit position relative to pixmap top-left corner
                x_relative = (x - x0) / width
                y_relative = (y - y0) / height
                self.left_click.emit(x_relative,y_relative)

        super().mousePressEvent(e)

解决方法

必须使用QLabel中建立的QPixmap，可以通过pixmap()方法获取。问题是您正在混淆对该方法的访问，因为您有一个具有相似名称的属性。所以解决方案是使用 pixmap() 并重命名属性 pixmap。

class Image_Area(QLabel):

    left_click = pyqtSignal(float,float)

    def __init__(self,pixmap):
        super().__init__()
        self._pixmap = pixmap
        self.setPixmap(pixmap)
        self.setScaledContents(False)
        self.setMinimumSize(100,100)

    def resizeEvent(self,e):
        if self._pixmap is not None:
            scaled_pixmap = self._pixmap.scaled(self.size(),Qt.KeepAspectRatio)
            self.setPixmap(scaled_pixmap)
        return super().resizeEvent(e)

    def mousePressEvent(self,e):
        x,y,= (
            e.x(),e.y(),)
        if self.pixmap:
            r = self.pixmap().rect()
            x0,y0 = r.x(),r.y()
            x1,y1 = x0 + r.width(),y0 + r.height()
            if x >= x0 and x < x1 and y >= y0 and y < y1:
                x_relative = (x - x0) / (x1 - x0)
                y_relative = (y - y0) / (y1 - y0)
                self.left_click.emit(x_relative,y_relative)
        super().mousePressEvent(e)