当使用graphviz中的rankdir=LR时，我该如何使节点水平对齐？

Question

我的代码是

digraph{
rankdir=LR;
ratio=auto
node[shape=rectangle];

i0[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="2">
                I<sub>0</sub>:
            </TD>
            <TD align="left">
                S'→.bexpr<BR ALIGN="LEFT"/>
            </TD>
        </TR>
        <TR>
            <TD align="left" bgcolor="#aaaaaa">
                bexpr→.bexpr or bterm <BR ALIGN="LEFT"/>
                bexpr→.bterm <BR ALIGN="LEFT"/>
                bterm→ .bterm and bfactor <BR ALIGN="LEFT"/>
                bterm→.bfactor <BR ALIGN="LEFT"/>
                bfactor→.not bfactor <BR ALIGN="LEFT"/>
                bfactor→.(bexpr) <BR ALIGN="LEFT"/>
                bfactor→.true <BR ALIGN="LEFT"/>
                bfactor→.false <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i1[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top">
                I<sub>1</sub>:
            </TD>
            <TD align="left">
                S'→bexpr. <BR ALIGN="LEFT"/>
                bexpr→bexpr .or bterm <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i2[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>2</sub>:
            </TD>
            <TD align="left">
                bexpr→bterm. <BR ALIGN="LEFT"/>
                bterm→bterm .and bfactor <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i3[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>3</sub>:
            </TD>
            <TD align="left">
                bterm→bfactor.<BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i4[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="2">
                I<sub>4</sub>:
            </TD>
            <TD align="left">
                bfactor→not .bfactor <BR ALIGN="LEFT"/>
            </TD>
        </TR>
        <TR>
            <TD align="left" bgcolor="#aaaaaa">
                bfactor→.not bfactor <BR ALIGN="LEFT"/>
                bfactor→.(bexpr) <BR ALIGN="LEFT"/>
                bfactor→.true <BR ALIGN="LEFT"/>
                bfactor→.false <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];
i5[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="2">
                I<sub>5</sub>:
            </TD>
            <TD align="left">
                bfactor→(.bexpr) <BR ALIGN="LEFT"/>
            </TD>
        </TR>
        <TR>
            <TD align="left" bgcolor="#aaaaaa">
                bexpr→.bexpr or bterm <BR ALIGN="LEFT"/>
                bexpr→.bterm <BR ALIGN="LEFT"/>
                bterm→.bterm and bfactor <BR ALIGN="LEFT"/>
                bterm→.bfactor <BR ALIGN="LEFT"/>
                bfactor→.not bfactor <BR ALIGN="LEFT"/>
                bfactor→.(bexpr) <BR ALIGN="LEFT"/>
                bfactor→.true <BR ALIGN="LEFT"/>
                bfactor→.false <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i6[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>6</sub>:
            </TD>
            <TD align="left">
                 bfactor→true.<BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i7[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>7</sub>:
            </TD>
            <TD align="left">
                bfactor→false. <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i8[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="2">
                I<sub>8</sub>:
            </TD>
            <TD align="left">
                bexpr→bexpr or .bterm  <BR ALIGN="LEFT"/>
            </TD>
        </TR>
        <TR>
            <TD align="left" bgcolor="#aaaaaa">
                bterm→ .bterm and bfactor  <BR ALIGN="LEFT"/>
                bterm→.bfactor  <BR ALIGN="LEFT"/>
                bfactor→.not bfactor  <BR ALIGN="LEFT"/>
                bfactor→.(bexpr)  <BR ALIGN="LEFT"/>
                bfactor→.true  <BR ALIGN="LEFT"/>
                bfactor→.false  <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i9[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="2">
                I<sub>9</sub>:
            </TD>
            <TD align="left">
                bterm→ bterm and .bfactor <BR ALIGN="LEFT"/>
            </TD>
        </TR>
        <TR>
            <TD align="left" bgcolor="#aaaaaa">
                bfactor→.not bfactor <BR ALIGN="LEFT"/>
                bfactor→.(bexpr) <BR ALIGN="LEFT"/>
                bfactor→.true <BR ALIGN="LEFT"/>
                bfactor→.false <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i10[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>10</sub>:
            </TD>
            <TD align="left">
                bfactor→not bfactor. <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i11[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>11</sub>:
            </TD>
            <TD align="left">
                bfactor→(bexpr.) <BR ALIGN="LEFT"/>
                bexpr→bexpr .or bterm <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i12[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>12</sub>:
            </TD>
            <TD align="left">
                    bexpr→bexpr or bterm. <BR ALIGN="LEFT"/>
                    bterm→ bterm .and bfactor <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i13[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>13</sub>:
            </TD>
            <TD align="left">
                bterm→ bterm and bfactor. <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

i14[label=<
    <TABLE border="0">
        <TR>
            <TD valign="top" rowspan="1">
                I<sub>14</sub>:
            </TD>
            <TD align="left">
                bfactor→(bexpr). <BR ALIGN="LEFT"/>
            </TD>
        </TR>
    </TABLE>
>];

node[width=0.15,shape=none,fixedsize=false];
i2_1[label=<I<sub>2</sub>>];
i3_1[label=<I<sub>3</sub>>];
i3_2[label=<I<sub>3</sub>>];
i4_2[label=<I<sub>4</sub>>];
i4_3[label=<I<sub>4</sub>>];
i4_3[label=<I<sub>4</sub>>];
i5_1[label=<I<sub>5</sub>>];
i5_3[label=<I<sub>5</sub>>];
i5_4[label=<I<sub>5</sub>>];
i6_1[label=<I<sub>6</sub>>];
i6_2[label=<I<sub>6</sub>>];
i6_3[label=<I<sub>6</sub>>];
i6_4[label=<I<sub>6</sub>>];
i7_1[label=<I<sub>7</sub>>];
i7_2[label=<I<sub>7</sub>>];
i7_3[label=<I<sub>7</sub>>];
i7_4[label=<I<sub>7</sub>>];
i8_1[label=<I<sub>8</sub>>];
i9_1[label=<I<sub>9</sub>>];

i0 -> i1 [label ="bexpr"];
i0 -> i2 [label = "bterm"];
i0 -> i3 [label = "bfactor"];
i0 -> i4 [label = "not"];
i0 -> i5 [label = "("];
i0 -> i6 [label = "true"];
i0 -> i7 [label = "false"];
i1 -> i8 [label = "or"];
i2 -> i9 [label = "and"];
i4 -> i10 [label = "bfactor"];
i4 -> i4 [label = "not",weight=1];
i4 -> i5_1 [label = "("];
i4 -> i6_1 [label = "true"];
i4 -> i7_1 [label = "false"];
i5 -> i11 [label = "bexpr"];
i5 -> i2_1 [label ="bterm"];
i5 -> i3_1 [label = "bfactor"];
i5 -> i4_2 [label = "not"];
i5:sw -> i5:_ [label = "("];
i5 -> i6_2 [label = "true"];
i5 -> i7_2 [label = "false"];
i8 -> i12 [label = "bterm"];
i8 -> i3_2 [label = "bfactor"];
i8 -> i4_3 [label = "not"];
i8 -> i5_3 [label = "("];
i8 -> i6_3 [label = "true"];
i8 -> i7_3 [label = "false"];
i9 -> i13 [label = "bfactor"];
i9 -> i4_4 [label = "not"];
i9 -> i5_4 [label = "("];
i9 -> i6_4 [label = "true"];
i9 -> i7_4 [label = "false"];
i11 -> i14 [label = ")"];
i11 -> i8_1 [label = "or"];
i12 -> i9_1[label = "and"];
{rank = same;i0,i5};
}

它生成以下内容：

首先，我想让I0、I1、I8和I12水平对齐。需要使用子图将一些节点包裹起来吗？如果需要，如何操作？我已经尝试使用子图包装I0、I1节点和从I0到I1的边缘进行测试，但变化不大。

其次，I5中的自环能否像I4中的自环一样更优雅，而不是在节点的顶部。我只找到了portpos选项。如果节点是圆形，则portpos选项就足够了。但是矩形看起来很奇怪。

- Sisyphus

Answer 1

由于没有人回答，我在graphviz邮件列表中问了同样的问题并得到了答案。

对于第一个问题，在边缘中添加weight=100。以下是手册中weight的含义：

边缘的权重。在dot中，权重越大，边缘就越短、直接和垂直。

对于第二个问题，没有完美的解决方案。一种替代方案是将边缘控制留给dot，但通过labelangle和labeldistance选项更改标签位置。

i5 -> i5 [labelangle=20 labeldistance=2.5 taillabel = "("];

感谢Emden R. Gansner！

- Sisyphus

Answer 2

我可以确认将weight=100设置为节点居中在集群中起作用。我正在寻找graphviz中的设置，以便将神经网络单元居中可视化。在http://www.graphviz.org/doc/info/attrs.html中没有一个属性看起来足够明显，但是在边缘上设置weight=100确实像魔法一样奏效。

- Chaobin