我正在绘制一个带有节点的图形。我使用fill=white和在节点标签后面放置了一个白色透明框，opacity=0.8然后尝试使用，text opacity=1.0以便文本仍然可读。问题是，在 for 循环中创建的节点标签最终得到的1.0不是正确的数字。我得到的图表和我想要的图表如下图所示：
我找不到任何与相关的帖子text opacity。pgfmathresult这是我拥有的代码的 MWE：
\usepackage{xifthen}
\usetikzlibrary{backgrounds}
\begin{document}
\begin{tikzpicture}
[
every label/.style={inner sep=0pt, opacity=0.8, text opacity=1.0, fill=white},
cnode/.style={draw=black,fill=#1,minimum width=3mm,circle},
]
\node[cnode=red,label=0:{$\hat y_1$}] (s1) at (6,-1) {};
\node[cnode=red,label=0:{$\hat y_2$}] (s2) at (6,-2) {};
\node at (6,-3) {$\vdots$};
\node[cnode=red,label=0:{$\hat y_K$}] (sK) at (6,-4) {};
\foreach \x in {0,...,4}
{
\pgfmathparse{\x<4 ? \x : "q-1"}
\ifthenelse{\x>0}
{
\node[cnode=gray,label={90:$z_{\pgfmathresult}$}] (h-\x) at (3,{-\x-div(\x,4)+.5}) {};
}
{
\node[cnode=gray,label=90:$1$] (h-0) at (3,0.5) {};
}
\begin{scope}[on background layer]
\draw (h-\x) -- (s1);
\draw (h-\x) -- (s2);
\draw (h-\x) -- (sK);
\end{scope}
}
\node at (3,-3.5) {$\vdots$};
\node[cnode=gray,label=90:$z_{M}$] (h-q) at (3,-5.5) {};
\begin{scope}[on background layer]
\draw (h-q) -- (s1);
\draw (h-q) -- (s2);
\draw (h-q) -- (sK);
\end{scope}
\end{tikzpicture}
\end{document}
先谢谢了。
答案1
问题是\pgfmathresult会被覆盖，因此您可能希望使用\pgfmathsetmacro来将计算结果存储在宏中。（我也摆脱了它，xifthen因为整数比较不需要它，您可以使用简单的\ifnum。）
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{backgrounds}
\begin{document}
\begin{tikzpicture}
[
every label/.style={inner sep=0pt, opacity=0.8, text opacity=1.0, fill=white},
cnode/.style={draw=black,fill=#1,minimum width=3mm,circle},
]
\node[cnode=red,label=0:{$\hat y_1$}] (s1) at (6,-1) {};
\node[cnode=red,label=0:{$\hat y_2$}] (s2) at (6,-2) {};
\node at (6,-3) {$\vdots$};
\node[cnode=red,label=0:{$\hat y_K$}] (sK) at (6,-4) {};
\foreach \x in {0,...,4}
{
\pgfmathsetmacro{\myindex}{\x<4 ? \x : "q-1"}
\ifnum\x>0
\node[cnode=gray,label={90:$z_{\myindex}$}] (h-\x) at (3,{-\x-div(\x,4)+.5}) {};
\else
\node[cnode=gray,label=90:$1$] (h-0) at (3,0.5) {};
\fi
\begin{scope}[on background layer]
\draw (h-\x) -- (s1);
\draw (h-\x) -- (s2);
\draw (h-\x) -- (sK);
\end{scope}
}
\node at (3,-3.5) {$\vdots$};
\node[cnode=gray,label=90:$z_{M}$] (h-q) at (3,-5.5) {};
\begin{scope}[on background layer]
\draw (h-q) -- (s1);
\draw (h-q) -- (s2);
\draw (h-q) -- (sK);
\end{scope}
\end{tikzpicture}
\end{document}
答案2
另一种方法无法解决您的问题pgfmathparse。略有不同的是，节点的定位计算省略了\pgfmathparse节点位置。对于定位，仅使用chains、fit和positioning库：
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{backgrounds,
chains,
fit}
\begin{document}
\begin{tikzpicture}[
node distance = 6mm and 30mm,
start chain = going below,
every label/.style = {label distance=2pt, inner sep=0.5pt,
fill=white, fill opacity=0.8, text opacity=1},
cnode/.style = {minimum size=3mm, on chain,
append after command = {\pgfextra{% this is used for correct positioning of nodes
\node[circle, draw, fill=#1, inner sep=0pt,
fit=(\tikzlastnode)] {};}
}% end of append after command
}
]
\node (s1) [cnode=red,label=0:$\hat{y}_1$] {};
\node (s2) [cnode=red,label=0:$\hat{y}_2$] {};
\node (s3) [inner ysep=0pt, on chain] {$\vdots$};
\node (s4) [cnode=red,label=0:$\hat{y}_K$] {};
%
\node (h1) [cnode=green,label=$1$,
above left = of s1] {};
\foreach \i in {1,2,4}
{
\scoped[on background layer]
\draw (h1) -- (s\i);
}
\foreach \i [count=\ii from 2] in {2,3,4,q-1,M}
{
\ifnum\ii=4
\node (h\ii) [inner ysep=0pt,on chain] {$\vdots$};
\else
\node (h\ii) [cnode=green,label=$z_{\i}$] {};
\begin{scope}[on background layer]
\draw (h\ii) -- (s1);
\draw (h\ii) -- (s2);
\draw (h\ii) -- (s4);
\end{scope}
\fi
}
\end{tikzpicture}
\end{document}
