python - 如何制作具有多个根树的圆形树

Question

Radial Positions algorithm使用从this question转换为python的C ，我已经成功地创建了一个基于根节点的径向图，一直到每个节点的最后一个子节点。现在我有一个新问题，我有多个根节点，需要它们以找到的第一个根为中心，或者至少是一个中心点。

我发现的最接近的例子是这张图：

到目前为止，我所能想到的只是对于找到的每个根节点，将其乘以它的索引，然后将 x 位置与 ay 半径相加。到目前为止，它的效果并不好，因为我的子节点不遵循它。我已经为此困扰了几天。

def RadialPositions(node, id):

    children_in_node = len(timelinedatta.neighbors(id, mode="out"))

    def rotate_node(x, y, nangle):
        nx = x * math.cos(nangle) - y * math.sin(nangle)
        ny = x * math.sin(nangle) + y * math.cos(nangle)
        return nx, ny

    def get_depth(id):
        count = 0
        for v in timelinedatta.bfsiter(id, mode="in", advanced=True):
            count = count + 1
        return count - 1

    if len(timelinedatta.neighbors(id, mode="out")) > 0 and len(timelinedatta.neighbors(id, mode="in")) == 0:
        node["positions"] = (0, 0)

    node["depth"] = get_depth(id)
    
    node_children_list = []

    for child in timelinedatta.neighbors(id, mode="out"):
        node_children_list.append((child, timelinedatta.vs[child]))

    for idx, (child_node_id, child_node) in enumerate(node_children_list, start=0):


        centerAdjust = 0

        if timelinedatta.neighbors(id, mode="in"):
            centerAdjust = (-node["angleRange"] + node["angleRange"] / children_in_node) / 2

        child_node["depth"] = get_depth(child_node_id)
        child_depth = child_node["depth"]



        child_node["nodeAngle"] = node["nodeAngle"] + node["angleRange"] / children_in_node * idx + centerAdjust
        child_node["angleRange"] = node["angleRange"] / children_in_node

        nx = rotate_node(40 * child_depth, 0, child_node["nodeAngle"])[0]
        ny = rotate_node(40 * child_depth, 0, child_node["nodeAngle"])[1]
        
        child_node["positions"] = [2 * nx, 2 * ny]


        RadialPositions(child_node, child_node_id)

我在pastebin 上也有我的图表示例

标签： pythonmathtreenodesigraph

Answer 1

您可以将graphviz（如@thoku 所建议的）与neato布局引擎（man neato）一起使用：

neato使用弹簧模型绘制无向图并减少相关能量（参见Kamada 和 Kawai，Information Processing Letters 31:1，April 1989）。

例如，示例

from graphviz import Digraph

tree = Digraph(engine='neato')

tree.node('root1',"R1")
tree.node('root2',"R2")
tree.node('root3',"R3")
tree.node('root4',"R4")
tree.edge('root1','root2')
tree.edge('root1','root3')
tree.edge('root3','root4')

tree.node('child11',"C11")
tree.node('child12',"C12")
tree.node('child13',"C13")
tree.edge('root1','child11')
tree.edge('root1','child12')
tree.edge('root1','child13')

tree.node('child21',"C21")
tree.node('child22',"C22")
tree.edge('root2','child21')
tree.edge('root2','child22')

tree.node('child31',"C31")
tree.node('child32',"C32")
tree.edge('root3','child31')
tree.edge('root3','child32')

tree.node('child41',"C41")
tree.node('child42',"C42")
tree.node('child43',"C43")
tree.edge('root4','child41')
tree.edge('root4','child42')
tree.edge('root4','child43')

tree.render("tree")

生成以下输出：