图树形图不创建根节点 完整代码：

如何解决图树形图不创建根节点 完整代码：

我正在尝试显示以字符串为节点的可绘树图。我在根节点上添加根节点，以遵循列表“ l”内的序列，并以“ david”作为根节点。但是，该图显示根节点还有别的东西。任何帮助将不胜感激。

代码生成图形：

import igraph
from igraph import Graph,EdgeSeq
nr_vertices = 25

import plotly.offline as pyo
pyo.init_notebook_mode()


l=[('david','john'),('david','jenni'),('john','david'),'mavri'),'claire'),]

    
vertices = set()
for line in l:
    vertices.update(line)
vertices = sorted(vertices)

print(len(l))

v_label = list(map(str,range(len(vertices))))
v2 = list(map(str,vertices))
#g = Graph.TupleList(directed=True,edges = l) 
G=Graph()

#G.add_vertices(l)
G.add_vertices(vertices)

# add edges to the graph
G.add_edges(l)


lay = G.layout_reingold_tilford(mode="in",root=0)
print(G)

position = {k: lay[k] for k in range(len(vertices))}

print(position)

Y = [lay[k][1] for k in range(len(vertices))]

M = max(Y)


es = EdgeSeq(G) # sequence of edges
E = [e.tuple for e in G.es] # list of edges
print(E)

L = len(position)

Xn = [position[k][0] for k in range(L)]

Yn = [2*M-position[k][1] for k in range(L)]
Xe = []
Ye = []

for edge in E:
    Xe+=[position[edge[0]][0],position[edge[1]][0],None]
    Ye+=[2*M-position[edge[0]][1],2*M-position[edge[1]][1],None]

labels = v2
print(Xe)

import plotly.graph_objects as go
fig = go.Figure()
fig.add_trace(go.Scatter(x=Xe,y=Ye,mode='lines',line=dict(color='rgb(210,210,210)',width=1),hoverinfo='none'
                   ))
fig.add_trace(go.Scatter(x=Xn,y=Yn,mode='markers',name='bla',marker=dict(symbol='circle-dot',size=30,color='#6175c1',#'#DB4551',line=dict(color='rgb(50,50,50)',width=1)
                                ),text=labels,hoverinfo='text',opacity=0.8
                  ))

def make_annotations(pos,text,font_size=10,font_color='rgb(250,250,250)'):
    L=len(pos)
    if len(text)!=L:
        raise ValueError('The lists pos and text must have the same len')
    annotations = []
    for k in range(L):
        annotations.append(
            dict(
                text=labels[k],# or replace labels with a different list for the text within the circle
                x=pos[k][0],y=2*M-position[k][1],xref='x1',yref='y1',font=dict(color=font_color,size=font_size),showarrow=False)
        )
    return annotations

axis = dict(showline=False,# hide axis line,grid,ticklabels and  title
            zeroline=False,showgrid=False,showticklabels=False,)

fig.update_layout(title= 'Tree with Reingold-Tilford Layout',annotations=make_annotations(position,v_label),font_size=12,showlegend=False,xaxis=axis,yaxis=axis,margin=dict(l=40,r=40,b=85,t=100),hovermode='closest',plot_bgcolor='rgb(248,248,248)'
              )
fig.show()

树输出：

谢谢

解决方法

我不是igraph专家，所以我的建议可能有一些缺陷，但是它使您可以通过简单地更改以下行来将'david'设置为根节点：

lay = G.layout_reingold_tilford(mode="in",root=0)

收件人：

lay = G.layout_reingold_tilford(mode="in",root=1)

这似乎是因为G是使用以下构造的：

G=Graph()

#G.add_vertices(l)
G.add_vertices(vertices)

# add edges to the graph
G.add_edges(l)

顶点是：

['claire','david','jenni','john','mavri']

并且0中的lay = G.layout_reingold_tilford(mode="in",root=0)将'claire'设置为根节点。因此事实证明，您可以通过指定相应的索引将root设置为['claire','mavri']中的任何元素。因此lay = G.layout_reingold_tilford(mode="in",root=1)产生下面的图。然后，您可以通过将1更改为其他内容来验证我的发现。

完整代码：

import igraph
from igraph import Graph,EdgeSeq
nr_vertices = 25

import plotly.offline as pyo
pyo.init_notebook_mode()


l=[('david','john'),('david','jenni'),('john','david'),'mavri'),'claire'),]

    
vertices = set()
for line in l:
    vertices.update(line)
vertices = sorted(vertices)

print(len(l))

v_label = list(map(str,range(len(vertices))))
v2 = list(map(str,vertices))
#g = Graph.TupleList(directed=True,edges = l) 
G=Graph()

#G.add_vertices(l)
G.add_vertices(vertices)

# add edges to the graph
G.add_edges(l)


lay = G.layout_reingold_tilford(mode="in",root=1)
print(G)

position = {k: lay[k] for k in range(len(vertices))}

print(position)

Y = [lay[k][1] for k in range(len(vertices))]

M = max(Y)


es = EdgeSeq(G) # sequence of edges
E = [e.tuple for e in G.es] # list of edges
print(E)

L = len(position)

Xn = [position[k][0] for k in range(L)]

Yn = [2*M-position[k][1] for k in range(L)]
Xe = []
Ye = []

for edge in E:
    Xe+=[position[edge[0]][0],position[edge[1]][0],None]
    Ye+=[2*M-position[edge[0]][1],2*M-position[edge[1]][1],None]

labels = v2
print(Xe)

import plotly.graph_objects as go
fig = go.Figure()
fig.add_trace(go.Scatter(x=Xe,y=Ye,mode='lines',line=dict(color='rgb(210,210,210)',width=1),hoverinfo='none'
                   ))
fig.add_trace(go.Scatter(x=Xn,y=Yn,mode='markers',name='bla',marker=dict(symbol='circle-dot',size=30,color='#6175c1',#'#DB4551',line=dict(color='rgb(50,50,50)',width=1)
                                ),text=labels,hoverinfo='text',opacity=0.8
                  ))

def make_annotations(pos,text,font_size=10,font_color='rgb(250,250,250)'):
    L=len(pos)
    if len(text)!=L:
        raise ValueError('The lists pos and text must have the same len')
    annotations = []
    for k in range(L):
        annotations.append(
            dict(
                text=labels[k],# or replace labels with a different list for the text within the circle
                x=pos[k][0],y=2*M-position[k][1],xref='x1',yref='y1',font=dict(color=font_color,size=font_size),showarrow=False)
        )
    return annotations

axis = dict(showline=False,# hide axis line,grid,ticklabels and  title
            zeroline=False,showgrid=False,showticklabels=False,)

fig.update_layout(title= 'Tree with Reingold-Tilford Layout',annotations=make_annotations(position,v_label),font_size=12,showlegend=False,xaxis=axis,yaxis=axis,margin=dict(l=40,r=40,b=85,t=100),hovermode='closest',plot_bgcolor='rgb(248,248,248)'
              )
fig.show()