graphviz - 当graphviz中的rankdir = LR时，如何水平对齐节点

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};
}

它生成了这个:



首先，我希望 I0、I1、I8、I12 水平对齐。它是否必须使用子图来包装某些节点？如果有，怎么做？我试过用子图包装 I0、I1 节点和从 I0 到 I1 的边缘进行测试。但这并没有带来太大的改变。

其次，I5 中的自循环是否可以像 I4 中的自循环一样看起来更优雅，但不是在节点的顶部。我只找到 portpos 选项。如果节点是一个圆，portpos 选项就足够了。但是矩形看起来很奇怪。

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，将 weight=100 添加到边缘。以下是手册中重量的含义:

Weight of edge. In dot, the heavier the weight, the shorter, straighter and more vertical the edge is.

对于第二个问题，没有完美的解决方案。另一种选择是将边缘的控制留给点，但通过选项 labelangle 和 labeldistance 改变标签的位置

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

谢谢，Emden R. Gansner!