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使用变换坐标计算 TikZ 角度并进行旋转

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使用变换坐标计算 TikZ 角度并进行旋转

请考虑以下 MWE：

\documentclass[]{article}
\usepackage{tikz}

\begin{document}

\typeout{Using internal functions}
\begin{tikzpicture}[
    color=green!50!black]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) to (2,2) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \pgfmathanglebetweenpoints{(\tikztostart)}{(\tikztotarget)}\xdef\aaa{\pgfmathresult}};
    \node [draw] at (1,1) {\aaa}; % shouldn't be 45?
\end{tikzpicture}
\quad
\typeout{Using coordinates}
\begin{tikzpicture}[
    color=blue]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \pgfmathanglebetweenpoints{\pgfpointanchor{a}{center}}{\pgfpointanchor{b}{center}}\xdef\aaa{\pgfmathresult}};
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Ok, now it's 45
\end{tikzpicture}
\quad
\typeout{Using coordinates and scaling}
\begin{tikzpicture}[
    color=red, xscale=2
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \pgfmathanglebetweenpoints{\pgfpointanchor{a}{center}}{\pgfpointanchor{b}{center}}\xdef\aaa{\pgfmathresult}};
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Why?
\end{tikzpicture}
\quad
\typeout{Using coordinates and rotating}
\begin{tikzpicture}[
    color=black, rotate=-45
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \pgfmathanglebetweenpoints{\pgfpointanchor{a}{center}}{\pgfpointanchor{b}{center}}\xdef\aaa{\pgfmathresult}};
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Why?
\end{tikzpicture}
\end{document}

结果是：

在此处输入图片描述

并且在日志中我有（预期）：

Using internal functions
0.0pt,0.0pt--- 2,2
Using coordinates
0.0pt,0.0pt--- 2,2
Using coordinates and scaling
0.0pt,0.0pt--- 2,2
Using coordinates and rotating
0.0pt,0.0pt--- 2,2

因此，第一个（绿色）不起作用（我尝试添加括号，但(\tikztostart)仍然……）。

蓝色技巧有效，我可以正确获取角度并绘制它。

但是，如果我以非对称方式缩放（或使用负比例因子）并进行全局旋转，它就会失败。我理解为什么计算出的角度仍然是 45 度（我们处于旋转/缩放的坐标系中），但似乎rotate节点的是在绝对坐标中。

我不想使用transform shape，因为结果是（预期的，正确的）：

在此处输入图片描述

是否可以计算路径角度以便它适用于红色和黑色的情况而不使用该transform shape选项？

（是的，这与...有关circuitikz）

我想法我找到了一种方法，但我不确定我在这里做的是否正确，所以我发布它以征求意见。看（我删除了@AndréC 评论的第一件事）：

\documentclass[]{article}
\usepackage{tikz}

\makeatletter
\newdimen{\@@xa}
\newdimen{\@@ya}
\newdimen{\@@xb}
\newdimen{\@@yb}
\def\computerot{%
        \pgfgettransformentries\a\b\c\d\temp\temp
        \pgfextractx{\@@xa}{\pgfpointanchor{a}{center}}
        \pgfextracty{\@@ya}{\pgfpointanchor{a}{center}}
        \pgfextractx{\@@xb}{\pgfpointanchor{b}{center}}
        \pgfextracty{\@@yb}{\pgfpointanchor{b}{center}}
        \typeout{XA:\space\the\@@xa\space YA:\space\the\@@ya\space XB:\space\the\@@xb\space YB:\space\the\@@yb}
        \typeout{MATRIX\space A:\a\space B:\b\space C:\c\space D:\d}
        \pgfmathsetmacro{\newx}{\a*(\@@xb-\@@xa)+\b*(\@@yb-\@@ya)}
        \pgfmathsetmacro{\newy}{\c*(\@@xb-\@@xa)+\d*(\@@yb-\@@ya)}
        \typeout{NEWX:\space\newx\space NEWY:\space\newy}
        \pgfmathsetmacro{\rot}{atan2(\c*(\@@xb-\@@xa)+\d*(\@@yb-\@@ya), \a*(\@@xb-\@@xa)+\b*(\@@yb-\@@ya) )}
        \typeout{ROT\space\rot}
}
\makeatother
\begin{document}

\typeout{Using coordinates}
\begin{tikzpicture}[
    color=blue]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \computerot
        \xdef\aaa{\rot}
    };
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Ok, now it's 45
\end{tikzpicture}
\quad
\typeout{Using coordinates and scaling}
\begin{tikzpicture}[
    color=red, xscale=2,
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \computerot
        \xdef\aaa{\rot}
    };
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Why?
\end{tikzpicture}
\quad
\typeout{Using coordinates and rotating}
\begin{tikzpicture}[
    color=black, rotate=30,
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b) \pgfextra{%
            \typeout{\tikztostart --- \tikztotarget}
        \computerot
        \xdef\aaa{\rot}
    };
    \node [draw, rotate=\aaa] at (1,1) {\aaa}; % Why?
\end{tikzpicture}
\end{document}

这似乎适用于具有不同缩放比例的情况（其中角度大约 26 度是正确的），但不适用于旋转：我有 15 度，而我应该有 75 度或大约这个数字......

在此处输入图片描述

输出为：

Using coordinates
0.0pt,0.0pt--- 2,2
XA: 0.0pt YA: 0.0pt XB: 56.90549pt YB: 56.90549pt
MATRIX A:1.0 B:0.0 C:0.0 D:1.0
NEWX: 56.90549 NEWY: 56.90549
ROT 45.0
Using coordinates and scaling
0.0pt,0.0pt--- 2,2
XA: 0.0pt YA: 0.0pt XB: 56.90549pt YB: 56.90549pt
MATRIX A:2.0 B:0.0 C:0.0 D:1.0
NEWX: 113.81097 NEWY: 56.90549
ROT 26.56505
Using coordinates and rotating
0.0pt,0.0pt--- 2,2
XA: 0.0pt YA: 0.0pt XB: 56.9057pt YB: 56.90572pt
MATRIX A:0.86603 B:0.5 C:-0.5 D:0.86603
NEWX: 77.73476 NEWY: 20.82907
ROT 15.00056

并且看起来，为了进行旋转，X 和 Y 坐标有所交换。

答案1

买得好，罗马诺！

\documentclass[]{article}
\usepackage{tikz}
\tikzset{good bye/.code=\pgftransformreset}
\begin{document}
\typeout{Using coordinates and scaling}
\begin{tikzpicture}[
    color=red, xscale=2,
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b);
    \path[good bye] (a) -- (b) node[midway,sloped,draw] {pft};
\end{tikzpicture}
\quad
\typeout{Using coordinates and rotating}
\begin{tikzpicture}[
    color=black, rotate=-45
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b);
    \path[good bye] (a) -- (b) node[midway,sloped,draw] {pft};
\end{tikzpicture}
\end{document}

在此处输入图片描述

让您更快乐。我强调，这没有必要那么复杂，也就是说，对于所有目的，我都能想象有更简单的解决方案。

\documentclass[]{article}
\usepackage{tikz}
\makeatletter
\newdimen{\@@xa}
\newdimen{\@@ya}
\newdimen{\@@xb}
\newdimen{\@@yb}
\def\computerot{%
        \pgfgettransformentries\a\b\c\d\temp\temp
        \pgfextractx{\@@xa}{\pgfpointanchor{a}{center}}
        \pgfextracty{\@@ya}{\pgfpointanchor{a}{center}}
        \pgfextractx{\@@xb}{\pgfpointanchor{b}{center}}
        \pgfextracty{\@@yb}{\pgfpointanchor{b}{center}}
        \typeout{XA:\space\the\@@xa\space YA:\space\the\@@ya\space XB:\space\the\@@xb\space YB:\space\the\@@yb}
        \typeout{MATRIX\space A:\a\space B:\b\space C:\c\space D:\d}
        \pgfmathsetmacro{\newx}{\a*(\@@xb-\@@xa)+\b*(\@@yb-\@@ya)}
        \pgfmathsetmacro{\newy}{\c*(\@@xb-\@@xa)+\d*(\@@yb-\@@ya)}
        \typeout{NEWX:\space\newx\space NEWY:\space\newy}
        \pgfmathsetmacro{\rot}{atan2(\c*(\@@xb-\@@xa)+\d*(\@@yb-\@@ya), \a*(\@@xb-\@@xa)+\b*(\@@yb-\@@ya) )}
        \typeout{ROT\space\rot}
}
\makeatother
\tikzset{good bye/.code=\pgftransformreset\computerot%
\xdef\aaa{\rot}%
}
\begin{document}
\typeout{Using coordinates and scaling}
\begin{tikzpicture}[
    color=red, xscale=2,
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b)
    [good bye] (a) -- (b) node[midway,rotate=\aaa,draw] {pft};
\end{tikzpicture}
\quad
\typeout{Using coordinates and rotating}
\begin{tikzpicture}[
    color=black, rotate=-45
    ]
    \draw[thin, dotted] (0,0) grid (2,2);
    \draw (0,0) coordinate(a) to (2,2) coordinate(b)
    [good bye] (a) -- (b) node[midway,rotate=\aaa,draw] {pft};
\end{tikzpicture}
\end{document}

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