pyqtgraph：如何在图形图例上添加单选按钮？

Posted 2023-03-25

【中文标题】pyqtgraph：如何在图形图例上添加单选按钮？

【英文标题】：pyqtgraph: How to add radiobuttons on a figure legend?

【发布时间】：2021-08-01 21:03:46

【问题描述】：

我想通过使用浮动在左上角的单选按钮来控制在图中显示哪些曲线，这也可以用作图例。我该怎么做？

【参考方案1】：

这可以通过继承pg.LegendItem 并覆盖addItem 方法来实现。您需要在其中创建小部件并将它们放入布局中。您可以使用QGraphicsProxyWidget 在QGraphicsScene (PlotWidget) 上显示QRadioButton。

import pyqtgraph as pg
from pyqtgraph.graphicsItems.LegendItem import ItemSample
from PyQt5 import QtCore, QtGui, QtWidgets

class LegendItem(pg.LegendItem):

    clicked = QtCore.pyqtSignal(int)

    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self._group = QtWidgets.QButtonGroup()

    def addItem(self, item, name):
        widget = QtWidgets.QRadioButton(name)
        palette = widget.palette()
        palette.setColor(QtGui.QPalette.Window, QtCore.Qt.transparent)
        palette.setColor(QtGui.QPalette.WindowText, QtCore.Qt.white)
        widget.setPalette(palette)
        self._group.addButton(widget)
        row = self.layout.rowCount()
        widget.clicked.connect(lambda: self.clicked.emit(row))
        proxy = item.scene().addWidget(widget)
        if isinstance(item, ItemSample):
            sample = item
        else:
            sample = ItemSample(item)
        self.layout.addItem(proxy, row, 0)
        self.layout.addItem(sample, row, 1)
        self.items.append((proxy, sample))
        self.updateSize()

if __name__ == '__main__':
    app = QtWidgets.QApplication([])

    plt = pg.PlotWidget()
    plt.setWindowTitle('Legend with radiobuttons')
    
    legend = LegendItem(verSpacing=5, horSpacing=5)
    legend.setParentItem(plt.getViewBox())

    plt.plotItem.legend = legend

    c1 = plt.plot([1,3,2,4], pen='r', symbol='o', symbolPen='r', symbolBrush=0.5, name='red plot') 
    c2 = plt.plot([2,1,4,3], pen='g', fillLevel=0, fillBrush=(255,255,255,30), name='green plot')

    def onClicked(index):
        print(" clicked".format(index))

    legend.clicked.connect(onClicked)

    plt.show()

    app.exec()