我正在开发一个项目,在该项目中需要创建nx.Graph()的预览,该预览允许更改使用鼠标拖动节点的节点的位置。如果在特定节点上单击鼠标,则我的当前代码能够在每次鼠标移动后立即重绘整个图形。但是,这大大增加了等待时间。如何仅更新所需的演出者,即单击的节点,其标签文本和相邻边缘,而不是刷新每个plt.subplots()的演出者?我至少可以得到所有需要搬迁的艺术家的引用吗?
我从在networkx中显示图形的标准方法开始:

import networkx as nx
import matplotlib.pyplot as plt
import numpy as np
import scipy.spatial

def refresh(G):
    plt.axis((-4, 4, -1, 3))
    nx.draw_networkx_labels(G, pos = nx.get_node_attributes(G, 'pos'),
                                bbox = dict(fc="lightgreen", ec="black", boxstyle="square", lw=3))
    nx.draw_networkx_edges(G, pos = nx.get_node_attributes(G, 'pos'), width=1.0, alpha=0.5)
    plt.show()

nodes = np.array(['A', 'B', 'C', 'D', 'E', 'F', 'G'])
edges = np.array([['A', 'B'], ['A', 'C'], ['B', 'D'], ['B', 'E'], ['C', 'F'], ['C', 'G']])
pos = np.array([[0, 0], [-2, 1], [2, 1], [-3, 2], [-1, 2], [1, 2], [3, 2]])

G = nx.Graph()
# IG = InteractiveGraph(G) #>>>>> add this line in the next step
G.add_nodes_from(nodes)
G.add_edges_from(edges)
nx.set_node_attributes(G, dict(zip(G.nodes(), pos.astype(float))), 'pos')

fig, ax = plt.subplots()
# fig.canvas.mpl_connect('button_press_event', lambda event: IG.on_press(event))
# fig.canvas.mpl_connect('motion_notify_event', lambda event: IG.on_motion(event))
# fig.canvas.mpl_connect('button_release_event', lambda event: IG.on_release(event))
refresh(G) # >>>>> replace it with IG.refresh() in the next step
python - 绘制网络x.Graph : how to change node position instead of resetting every node?-LMLPHP
在下一步中,我更改了上一个脚本的5行(取消注释4行,并替换了1行),并使用InteractiveGraph实例使其具有交互性:
class InteractiveGraph:
    def __init__(self, G, node_pressed=None, xydata=None):
        self.G = G
        self.node_pressed = node_pressed
        self.xydata = xydata

    def refresh(self, show=True):
        plt.clf()
        nx.draw_networkx_labels(self.G, pos = nx.get_node_attributes(self.G, 'pos'),
                                bbox = dict(fc="lightgreen", ec="black", boxstyle="square", lw=3))
        nx.draw_networkx_edges(self.G, pos = nx.get_node_attributes(self.G, 'pos'), width=1.0, alpha=0.5)
        plt.axis('off')
        plt.axis((-4, 4, -1, 3))
        fig.patch.set_facecolor('white')
        if show:
            plt.show()

    def on_press(self, event):
        if event.inaxes is not None and len(self.G.nodes()) > 0:
            nodelist, coords = zip(*nx.get_node_attributes(self.G, 'pos').items())
            kdtree = scipy.spatial.KDTree(coords)
            self.xydata = np.array([event.xdata, event.ydata])
            close_idx = kdtree.query_ball_point(self.xydata, np.sqrt(0.1))
            i = close_idx[0]
            self.node_pressed = nodelist[i]

    def on_motion(self, event):
        if event.inaxes is not None and self.node_pressed:
            new_xydata = np.array([event.xdata, event.ydata])
            self.xydata += new_xydata - self.xydata
            #print(d_xy, self.G.nodes[self.node_pressed])
            self.G.nodes[self.node_pressed]['pos'] = self.xydata
            self.refresh(show=False)
            event.canvas.draw()

    def on_release(self, event):
        self.node_pressed = None
python - 绘制网络x.Graph : how to change node position instead of resetting every node?-LMLPHP
相关资料:
  • Event handling
  • Optimized removal of closest node
  • 最佳答案

    为了扩展我上面的评论,可以在 netgraph 中复制您的示例

    import numpy as np
    import matplotlib.pyplot as plt; plt.ion()
    import networkx as nx
    import netgraph
    
    nodes = np.array(['A', 'B', 'C', 'D', 'E', 'F', 'G'])
    edges = np.array([['A', 'B'], ['A', 'C'], ['B', 'D'], ['B', 'E'], ['C', 'F'], ['C', 'G']])
    pos = np.array([[0, 0], [-2, 1], [2, 1], [-3, 2], [-1, 2], [1, 2], [3, 2]])
    
    G = nx.Graph()
    G.add_nodes_from(nodes)
    G.add_edges_from(edges)
    
    I = netgraph.InteractiveGraph(G,
                                  node_positions=dict(zip(nodes, pos)),
                                  node_labels=dict(zip(nodes,nodes)),
                                  node_label_bbox=dict(fc="lightgreen", ec="black", boxstyle="square", lw=3),
                                  node_size=12,
    )
    
    # move stuff with mouse
    
    python - 绘制网络x.Graph : how to change node position instead of resetting every node?-LMLPHP

    关于您编写​​的代码,如果您拥有所有艺术家的句柄,则不需要kd树。通常,matplotlib艺术家具有contains方法,这样,当您记录按钮按下事件时,您可以简单地检查artist.contains(event)来查找按钮按下是否发生在艺术家身上。当然,如果使用networkx进行绘制,则无法以一种可查询的好形式获得句柄(ax.get_children()都不是),所以这是不可能的。

    09-30 16:05
    查看更多