图列阵LR但堆积TB

Question

digraph G
{
    rankdir=LR;

    Array1 [ shape = record, label = "{ <a1>A(1) | <a2>A(2) | <a3>A(...) | <an>A(n)}"] ; 
    Array2 [ shape = record, label = "{ <b1>B(1) | <b2>B(2) | <b3>B(...) | <bn1>B(n+1)}"] ;  
    Array1:a1 -> Array2:b1 [style=solid];
    Array1:an -> Array2:bn1 [style=solid];
    {rank=same; Array1; Array2;}
}

我试图从左到右绘制两个数组，但Array1位于顶部，Array2位于底部。

用rank=same输出

正如您在图片中看到的，没有绘制箭头，但是布局应该是这样的。我也从点中得到了一些错误，但是我找不到任何有用的信息：

dot -Tpng algoA.dot -o algoA.png                                         :(
Warning: flat edge between adjacent nodes one of which has a record shape - replace records with HTML-like labels
  Edge Array2 -> Array1
Error: lost Array1 Array2 edge
Error: lost Array1 Array2 edge

digraph G
{
    rankdir=LR;

    Array1 [ shape = record, label = "{ <a1>A(1) | <a2>A(2) | <a3>A(...) | <an>A(n)}"] ; 
    Array2 [ shape = record, label = "{ <b1>B(1) | <b2>B(2) | <b3>B(...) | <bn1>B(n+1)}"] ;  
    Array1:a1 -> Array2:b1 [style=solid];
    Array1:an -> Array2:bn1 [style=solid];
    //{rank=same; Array1; Array2;}
}

移除rank=same的输出

如果代码中没有rank=same，我就会得到我想要的箭头，但不像我想要的那样--顶部是Array1，底部是Array2。错误信息消失了。我做错了什么？我尝试了几种方法，但它们都不适用于我，我总是得到一个错误的布局，或者我必须添加其他几个节点来绘制它，而那些不可见的箭头也不会被绘制。

Answer 1

键在错误消息中：replace records with HTML-like labels。详细信息可以找到这里。

根据您的代码以及我所理解的您想要实现的目标，这可以通过创建两个类似HTML的标签来实现。代码中的注释。

digraph so 
{
    # plaintext is being used to create HTML-like labels
    node [shape=plaintext]

    # no border for the table, a single border for each cell and
    # no distance between individual cells looks like record shape

    # PORTs being defined as addresses of individual cells
    # that edges can be directed to

    array1 [label=< 
        <TABLE BORDER="0" CELLBORDER="1" CELLSPACING="0"> <TR>
            <TD PORT="a1">A(1)</TD>
            <TD PORT="a2">A(2)</TD>
            <TD PORT="ax">A(...)</TD>
            <TD PORT="an">A(n)</TD>
        </TR> </TABLE>>];

    array2 [label=< 
        <TABLE BORDER="0" CELLBORDER="1" CELLSPACING="0"> <TR>
            <TD PORT="b1">B(1)</TD>
            <TD PORT="b2">B(2)</TD>
            <TD PORT="bx">B(...)</TD>
            <TD PORT="bn">B(n+1)</TD>
    </TR> </TABLE>>]; 

    array1:a1 -> array2:b1;
    array1:an -> array2:bn;
}

这产生了(我认为)你想要的东西：

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