创建自定义符号：钻石/菱形

Question

我读了这篇小文章：http://www.elfsoft2000.com/projects/bipole.pdf以及本文提到的另外两个。我想创建一个符号用于我的论文和项目中。在我的国家，其中一些与欧洲或美国的不同，所以我决定自己制作它们。

我设法得到了两个罗马尼亚电路符号，如下所示：github.com/PopAdi/circuitikz-romanian-symbols

现在我必须做类似的事情：不是圆形，而是菱形。我该怎么做？我希望它们看起来像这里展示的那样：Circuitikz 美国控制电压标志

在我的代码中，对于带有圆圈的代码，我有：

\pgfpathellipse{\pgfpointorigin}{\pgfpoint{0cm}{\ResUp}}{\pgfpoint{\ResRight}{0cm}}

我设法用这个制作了一个正方形：

\pgfpathrectanglecorners{\southwest}{\northeast}

但我真的不知道如何旋转那个正方形或用菱形替换它。你能帮我解决这个问题吗？谢谢！

编辑：我的代码如下所示：

\documentclass{article}  

\usepackage{tikz}
\usepackage{circuitikz}
\usetikzlibrary{shapes,arrows,positioning}
\usetikzlibrary{decorations.markings}

\makeatletter
\pgf@circ@Rlen = \pgfkeysvalueof{/tikz/circuitikz/bipoles/length}
\def\TikzBipolePath#1#2{\pgf@circ@bipole@path{#1}{#2}}
\makeatother

\newlength{\ResUp} 
\newlength{\ResRight}

\ctikzset{bipoles/romanianCCS/height/.initial=.60}
\ctikzset{bipoles/romanianCCS/width/.initial=.60}


\pgfcircdeclarebipole{} 
    {\ctikzvalof{bipoles/romanianCCS/height}}
    {romanianCCS}
    {\ctikzvalof{bipoles/romanianCCS/height}}
    {\ctikzvalof{bipoles/romanianCCS/width}}
    {
        \pgfsetlinewidth{\pgfkeysvalueof{/tikz/circuitikz/bipoles/thickness}\pgfstartlinewidth}

        \pgfextracty{\ResUp}{\northeast}
        \pgfextractx{\ResRight}{\southwest}

        %Desenam cerculetul
        \pgfpathrectanglecorners{\southwest}{\northeast}

        %First little arrow
        \pgfmoveto{\pgfpoint{1.0\ResRight}{0.0\ResUp}}
        \pgflineto{\pgfpoint{0.1\ResRight}{0.0\ResUp}} 
        \pgflineto{\pgfpoint{0.3\ResRight}{-0.25\ResUp}} 
        \pgfmoveto{\pgfpoint{0.1\ResRight}{0.0\ResUp}}   
        \pgflineto{\pgfpoint{0.3\ResRight}{0.25\ResUp}}  

        %Second arrow
        \pgfmoveto{\pgfpoint{-0.2\ResRight}{0.0\ResUp}}
        \pgflineto{\pgfpoint{-1.0\ResRight}{0.0\ResUp}}
        \pgfmoveto{\pgfpoint{0.0\ResRight}{0.25\ResUp}}
        \pgflineto{\pgfpoint{-0.2\ResRight}{0.0\ResUp}}
        \pgflineto{\pgfpoint{0.0\ResRight}{-0.25\ResUp}}

        \pgfusepath{draw}
    }

\def\romanianCCS#1{\TikzBipolePath{romanianCCS}{#1}}
\tikzset{romanianCCS/.style = {\circuitikzbasekey, /tikz/to path=\romanianCCS, l=#1}}


\begin{document}

\begin{center}
\begin{circuitikz} 
\draw (0, 0) 
      to[romanianCCS, l=${j_1 = 4A}$, *-*] (4, 0);
\end{circuitikz}
\end{center}

\end{document}

将会得到以下结果：

我找到了解决方案！我这样做了：

\pgftransformrotate{-45}
\pgfpathrectanglecorners{\southwest}{\northeast}
\pgftransformrotate{45}

而且效果非常好！

Answer 1

如果您想要菱形而不是菱形，您可以使用：

\documentclass{article}  

\usepackage{tikz}
\usepackage{circuitikz}
\usetikzlibrary{shapes,arrows,positioning}
\usetikzlibrary{decorations.markings}

\makeatletter
\pgf@circ@Rlen = \pgfkeysvalueof{/tikz/circuitikz/bipoles/length}
\def\TikzBipolePath#1#2{\pgf@circ@bipole@path{#1}{#2}}
\makeatother

\newlength{\ResUp} 
\newlength{\ResRight}

\ctikzset{bipoles/romanianCCS/height/.initial=.60}
\ctikzset{bipoles/romanianCCS/width/.initial=.60}


\pgfcircdeclarebipole{} 
    {\ctikzvalof{bipoles/romanianCCS/height}}
    {romanianCCS}
    {\ctikzvalof{bipoles/romanianCCS/height}}
    {\ctikzvalof{bipoles/romanianCCS/width}}
    {
        \pgfsetlinewidth{\pgfkeysvalueof{/tikz/circuitikz/bipoles/thickness}\pgfstartlinewidth}

        \pgfextracty{\ResUp}{\northeast}
        \pgfextractx{\ResRight}{\southwest}

        %Desenam cerculetul
        \pgfpathmoveto{\pgfpoint{-\ResRight}{0pt}}
        \pgfpathlineto{\pgfpoint{0pt}{-\ResUp}}
        \pgfpathlineto{\pgfpoint{\ResRight}{0pt}}
        \pgfpathlineto{\pgfpoint{0pt}{\ResUp}}
        \pgfpathclose

        %First little arrow
        \pgfpathmoveto{\pgfpoint{1.0\ResRight}{0.0\ResUp}}
        \pgfpathlineto{\pgfpoint{0.1\ResRight}{0.0\ResUp}} 
        \pgfpathlineto{\pgfpoint{0.3\ResRight}{-0.25\ResUp}} 
        \pgfpathmoveto{\pgfpoint{0.1\ResRight}{0.0\ResUp}}   
        \pgfpathlineto{\pgfpoint{0.3\ResRight}{0.25\ResUp}}  

        %Second arrow
        \pgfpathmoveto{\pgfpoint{-0.2\ResRight}{0.0\ResUp}}
        \pgfpathlineto{\pgfpoint{-1.0\ResRight}{0.0\ResUp}}
        \pgfpathmoveto{\pgfpoint{0.0\ResRight}{0.25\ResUp}}
        \pgfpathlineto{\pgfpoint{-0.2\ResRight}{0.0\ResUp}}
        \pgfpathlineto{\pgfpoint{0.0\ResRight}{-0.25\ResUp}}

        \pgfusepath{draw}
    }

\def\romanianCCS#1{\TikzBipolePath{romanianCCS}{#1}}
\tikzset{romanianCCS/.style = {\circuitikzbasekey, /tikz/to path=\romanianCCS, l=#1}}


\begin{document}

\begin{center}
\begin{circuitikz} 
\draw (0, 0) 
      to[romanianCCS, l=${j_1 = 4A}$, *-*] (4, 0);
\end{circuitikz}
\end{center}

\end{document}