编辑：
Juan 的方法很棒，因为在绘制整个图片之前就可以知道坐标。但有时存在具有相对位置的复杂路径，可能需要在路径中间创建坐标。
原始问题：
我怎样才能隐藏或不显示pic但仍让它计算内部坐标/节点？
这里我画了从原点到 的每个内部坐标的蓝线。我当前的解决方案是使用图片范围的pic命令：opacity=0
但是，当任何内部绘图的pic不透明度发生变化时，它就不成立，因为它会覆盖opacity=0命令：
\fill[fill=yellow,opacity=0.5] (0,0) rectangle (-1,-1) coordinate (-point 2);
\documentclass[margin=1cm]{standalone}
\usepackage{tikz}
\tikzset
{
draw it/.store in=\drawit,
draw it=1,
random symbols/.pic=
{
\begin{scope}[opacity=\drawit]
\draw (1,-1) -- +(0,1) coordinate (-point 1);
\fill[fill=yellow] (0,0) rectangle (-1,-1) coordinate (-point 2);
\node[draw,minimum size=1cm] (-point 3) at (-1,1) {a};
\path (0,0)
-- ++(1,1)
edge[red] coordinate[pos=1] (-point 4) +(-1,0);
\end{scope}
}
}
\begin{document}
\begin{tikzpicture}[x=2cm,y=2cm]
\path[draw it=1] pic (thepic) at (0,0) {random symbols};
\draw[blue,very thick] (0,0)
edge (thepic-point 1)
edge (thepic-point 2)
edge (thepic-point 3)
edge (thepic-point 4);
\end{tikzpicture}
\end{document}
答案1
如果我理解正确的话，您可以先在代码中定义坐标pic，然后在命令pic内绘制（或不绘制）\ifnum。
就像是：
\documentclass[margin=1cm]{standalone}
\usepackage{tikz}
\tikzset
{
draw it/.store in=\drawit,
draw it=1,
random symbols/.pic=
{ % coordinates (and one node)
\node[minimum size=1cm] (-aux) at (-1, 1) {};
\coordinate (-point 1) at ( 1, 0);
\coordinate (-point 2) at (-1,-1);
\coordinate (-point 3) at (-aux.south east);
\coordinate (-point 4) at ( 0, 1);
% optional drawing
\ifnum\drawit = 1
\draw (1,-1) -- (-point 1);
\fill[fill=yellow,opacity=0.5] (0,0) rectangle (-point 2);
\node[draw,minimum size=1cm] at (-aux) {a};
\path (0,0) -- (1,1) edge[red] (-point 4);
\fi
}
}
\begin{document}
\begin{tikzpicture}[x=2cm,y=2cm]
\foreach\i in {0,1}
{
\path pic[draw it=\i] (thepic\i) at (3*\i,0) {random symbols};
\draw[blue,very thick] (3*\i,0)
edge (thepic\i-point 1)
edge (thepic\i-point 2)
edge (thepic\i-point 3)
edge (thepic\i-point 4);
}
\end{tikzpicture}
\end{document}
生成结果：
