在graphviz python包中难以分离等级

Question

我正在尝试使用 graphviz Python 包创建神经网络图，但在尝试防止排名重叠时遇到了问题。我在这篇文章中专门询问如何防止排名重叠。

这是我目前的代码，

from graphviz import Graph

d = Graph(name="Network Graph",graph_attr={'rankdir':'TD'},engine='dot')

with d.subgraph() as s:
    s.attr(rank='same')
    s.node('density','density')
    s.node('te','Elec. Temp.')
    s.node('tr','Rad. Temp.')
    s.node('alpha','alpha')
    
layer0 = ['density','te','tr','alpha']
layer1 = []
layer2 = []
layer3 = []
layer4 = []
layer5 = []
allLayers =[ layer0, layer1, layer2, layer3, layer4, layer5 ]

with d.subgraph() as s1:
    s1.attr(rank='same')
    for i in range(1,7):
        layer1.append('a1%i'%i)
        s1.node('a1%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph() as s2:
    s2.attr(rank='same')
    for i in range(1,10):
        layer2.append('a1%i'%i)
        s2.node('a1%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph() as s3:
    s3.attr(rank='same')
    for i in range(1,3):
        layer3.append('a1%i'%i)
        s3.node('a1%i'%i,'',tailclip='true',headclip='true')

with d.subgraph() as s4:
    s4.attr(rank='same')
    for i in range(1,7):
        layer4.append('a1%i'%i)
        s4.node('a1%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph() as s5:
    s5.attr(rank='same')
    for i in range(1,14):
        layer5.append('a1%i'%i)
        s5.node('a1%i'%i,'',tailclip='true',headclip='true')

edges = []
d.attr(rankdir="LR",splines='line',ranksep='2.0',overlap='false')
for n in range(0,len(allLayers)-1):
    # d.attr(rankdir="LR",splines='line',ranksep='2.0',overlap='false')
    for i in allLayers[n]:
        for j in allLayers[n+1]:
            # edges.append((i,j))
            d.edge(i,j,directed="False")

print(d.source)

d.render('network.pdf',view=True)

这会产生这是图像

我不知道为什么我能够将等级/层 0 和 1 分开，但随后所有后续层都堆叠在 layer1 之上。出于某种原因，d.attr(rankdir="LR",splines='line',ranksep='2.0',overlap='false') 行似乎只适用一次，但是当我将该行放在 for 循环内时，它的行为是相同的。

graphviz 包的文档还有很多不足之处，我能找到的关于漂浮在互联网上的 Python Graphviz 的帮助很少。此外，显然不能在创建Graph() 对象后立即放置d.attr(rankdir='LR')。它必须在所有节点创建之后放置，因此顺序显然很重要。

仅供参考：我还尝试在不同的地方添加参数overlap='false'，但这并没有改变我使用它的方式。

如何划分等级？使用 Graphviz 是否还有其他一般提示，特别是在如何正确排序命令和设置属性方面？

Answer 1

很高兴看到您解决了问题。但您可能会发现以下更改很有帮助。看看他们。

• 为每个子图命名
• 为您的节点使用不同的名称（您已经意识到）
• 改用有向图
• 实际上，您可以在实例化时设置所有图形属性

from graphviz import Digraph

d = Digraph(name="Network Graph",
          graph_attr={'rankdir':'LR', 'splines':'line','ranksep':'2.0','overlap':'false'},
          engine='dot')

with d.subgraph(name='s') as s:
    s.node('density','density')
    s.node('te','Elec. Temp.')
    s.node('tr','Rad. Temp.')
    s.node('alpha','alpha')
    
layer0 = ['density','te','tr','alpha']
layer1 = []
layer2 = []
layer3 = []
layer4 = []
layer5 = []
allLayers =[ layer0, layer1, layer2, layer3, layer4, layer5 ]

with d.subgraph(name='s1') as s1:
    s1.attr(rank='same')
    for i in range(1,7):
        layer1.append('a1%i'%i)
        s1.node('a1%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph(name='s2') as s2:
    s2.attr(rank='same')
    for i in range(1,10):
        layer2.append('a2%i'%i)
        s2.node('a2%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph(name='s3') as s3:
    s3.attr(rank='same')
    for i in range(1,3):
        layer3.append('a3%i'%i)
        s3.node('a3%i'%i,'',tailclip='true',headclip='true')

with d.subgraph(name='s4') as s4:
    s4.attr(rank='same')
    for i in range(1,7):
        layer4.append('a4%i'%i)
        s4.node('a4%i'%i,'',tailclip='true',headclip='true')
        
with d.subgraph(name='s5') as s5:
    s5.attr(rank='same')
    for i in range(1,14):
        layer5.append('a5%i'%i)
        s5.node('a5%i'%i,'',tailclip='true',headclip='true')

for n in range(0,len(allLayers)-1):
    for i in allLayers[n]:
        for j in allLayers[n+1]:
            d.edge(i,j,directed="False")

# print(d.source)

# d.render('network.pdf',view=True)

您可能还会发现以下链接很有帮助

https://logicatcore.github.io/scratchpad/machine%20learning/jupyter/graphviz/2020/12/26/Graphviz-Neural-Networks-visualisation.html

https://github.com/martisak/dotnets

Answer 2

我发现的问题是由于复制意大利面，我未能更改节点名称。我之前在每一层的生成中都有a1%i'%i 的标签。当然，解决方法是将每个后续层中的1 更改为与层号对应的整数。