Question

当在onclick函数中使用fig.canvas.draw（）时，它不会等待onclick事件，而是来自函数。如何使其连续工作，以便每次点击饼图时都可以显示标签。

import matplotlib.pyplot as plt
labels = ['Beans', 'Squash', 'Corn']
def main():
    # Make an example pie plot
    fig = plt.figure()
    ax = fig.add_subplot(111)

    #labels = ['Beans', 'Squash', 'Corn']
    wedges, plt_labels = ax.pie([20, 40, 60], labels=labels)
    ax.axis('equal')

    make_picker(fig, wedges)
    plt.show()

def make_picker(fig, wedges):

    def onclick(event):
        print event.__class__
        wedge = event.artist
        label = wedge.get_label()
        print label
        fig.canvas.figure.clf() 
        ax=fig.add_subplot(111)
        wedges, plt_labels = ax.pie([50, 100, 60],labels=labels)
        fig.canvas.draw()

    # Make wedges selectable
    for wedge in wedges:
        wedge.set_picker(True)

    fig.canvas.mpl_connect('pick_event', onclick)

if __name__ == '__main__':
    main()

Answer 1

您的问题出在onclick函数中。

def onclick(event):
    print event.__class__
    wedge = event.artist
    label = wedge.get_label()
    print label
    fig.canvas.figure.clf() 
    ax=fig.add_subplot(111)
    wedges, plt_labels = ax.pie([50, 100, 60],labels=labels)
    fig.canvas.draw()

在这里，您要创建新的wedges（覆盖旧的wedges个实例）并且您没有将它们设置为可选。一个快速补丁是将onclick更改为：

def onclick(event):
    print event.__class__
    wedge = event.artist
    label = wedge.get_label()
    print label
    fig.canvas.figure.clf()
    ax=fig.add_subplot(111)
    wedges, plt_labels = ax.pie([50, 100, 60],labels=labels)
    fig.canvas.draw()
    for wedge in wedges:
        wedge.set_picker(True)

matplotlib中pick_event的异常行为

1 个答案: