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如何計算旋轉的邊界框？

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如何計算旋轉的邊界框？

編輯：正如打擊的答案所建議的，我可以使用transform shape選項。我的第一個問題現在已經解決了。但是，儘管第二個邊界框現在已局部對齊，但它並不正確。事實上，local bounding box總是計算一個全球對齊邊界框（綠點）並fit包圍該邊界框的north、east、south和錨點。west

我現在的問題是：如何計算local bounding box 本地對齊？

新範例（右側的藍色矩形不適合紅色路徑）：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{fit}
\tikzset{
  pt/.style={circle,minimum size=3pt,fill=#1,inner sep=0},
  red pt/.style={pt=red},
  green pt/.style={pt=green},
  every picture/.style={line width=1pt,inner sep=0pt},
}
\begin{document}
\begin{tikzpicture}[rotate=20]
  \draw[gray,line width=.4pt] (0,0) grid (6.5,2.5);
  % first case: fitting some nodes
  \node[red pt] (a) at (.5,.5){};
  \node[red pt] (b) at (.5,2){};
  \node[red pt] (c) at (3,.5){};
  \node[red pt] (d) at (2,2.2){};
  \begin{scope}[transform shape]
    \node[fit=(a)(b)(c)(d),draw=blue]{};
  \end{scope}

  % second case: fitting arbitrary path
  \begin{scope}[local bounding box=bb]
    \draw[red] (4,1) to[bend right] (6,1) -- (5,2);
  \end{scope}
  \node[green pt] at (bb.north west){};
  \node[green pt] at (bb.north east){};
  \node[green pt] at (bb.south west){};
  \node[green pt] at (bb.south east){};
  % how to find correct bounding box locally aligned ?
  \begin{scope}[transform shape]
    \node[fit=(bb),draw=blue]{};
  \end{scope}
\end{tikzpicture}
\end{document}

在此輸入影像描述

原問題：

我想計算一些邊界框。這是我的兩個案例：

  1. 為了適應一些節點（或座標），我可以使用fit函式庫。
  2. 為了適應任意路徑，我可以使用帶有local bounding box=bb.

下面的程式碼展示了這兩種情況：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{fit}
\tikzset{
  red pt/.style={circle,minimum size=3pt,fill=red,inner sep=0},
  every picture/.style={line width=1pt,inner sep=0pt},
}
\begin{document}
\begin{tikzpicture}
  \draw[gray,line width=.4pt] (0,0) grid (6.5,2.5);
  % first case: fitting some nodes
  \node[red pt] (a) at (.5,.5){};
  \node[red pt] (b) at (.5,2){};
  \node[red pt] (c) at (3,.5){};
  \node[red pt] (d) at (2,2.2){};
  \node[fit=(a)(b)(c)(d),draw=blue]{};

  % second case: fitting arbitrary path
  \begin{scope}[local bounding box=bb]
    \draw[red] (4,1) to[bend right] (6,1) -- (5,2);
  \end{scope}
  \node[fit=(bb),draw=blue]{};
\end{tikzpicture}

在此輸入影像描述

現在，我想做同樣的事情在旋轉圖片內！

我的兩個問題：

  1. 在我的第一種情況（某些節點）中，我必須rotate為我的擬合節點添加選項（因為rotate選項不會旋轉節點）。如何自動找到此選項的正確值（例如：目前座標系和畫布座標系之間的角度）？

  2. 在我的第二種情況（任意路徑）中，我可以計算本地邊界框但全域對齊。如何計算任意路徑局部對齊的局部邊界框？

這是我的嘗試：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{fit}
\tikzset{
  red pt/.style={circle,minimum size=3pt,fill=red,inner sep=0},
  every picture/.style={line width=1pt,inner sep=0pt},
}
\begin{document}
\begin{tikzpicture}[rotate=20]
  \draw[gray,line width=.4pt] (0,0) grid (6.5,2.5);
  % first case: fitting some nodes
  \node[red pt] (a) at (.5,.5){};
  \node[red pt] (b) at (.5,2){};
  \node[red pt] (c) at (3,.5){};
  \node[red pt] (d) at (2,2.2){};
  % how to find the good value for rotate (here 20)?
  \node[rotate=20,fit=(a)(b)(c)(d),draw=blue]{};

  % second case: fitting arbitrary path
  \begin{scope}[local bounding box=bb]
    \draw[red] (4,1) to[bend right] (6,1) -- (5,2);
  \end{scope}
  % how to find bounding box locally aligned ?
  \node[rotate=20,rotate fit=-20,fit=(bb),draw=blue]{};
\end{tikzpicture}
\end{document}

在此輸入影像描述

答案1

我錯過重點了嗎？transform shape重置旋轉似乎是一個解決方案。

編輯：我希望這次我明白你的意思。如果沒有，我真的很感激用旋轉的矩形和形狀來解釋，而不是當地的和全球的這是關於旋轉的 Tikz 圖片環境的相關術語。

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{fit}

\tikzset{
  pt/.style={circle,minimum size=3pt,fill=#1,inner sep=0},
  red pt/.style={pt=red},
  green pt/.style={pt=green},
  every picture/.style={line width=1pt,inner sep=0pt},
}
\begin{document}
\begin{tikzpicture}[rotate=20]
  \draw[gray,line width=.4pt] (0,0) grid (6.5,2.5);
  % first case: fitting some nodes
  \node[red pt] (a) at (.5,.5){};
  \node[red pt] (b) at (.5,2){};
  \node[red pt] (c) at (3,.5){};
  \node[red pt] (d) at (2,2.2){};
  \begin{scope}[transform shape]
    \node[fit=(a)(b)(c)(d),draw=blue]{};
  \end{scope}
\pgfgettransform{\currtrafo}     %Save the current trafo 

  % second case: fitting arbitrary path
\begin{scope}[local bounding box=bb]
\pgftransformresetnontranslations % Now there is no rotation and it doesn't know 
                                  % things are going to be rotated
\begin{scope}                     % We open a new scope and restore the outer trafo
\pgfsettransform{\currtrafo}      % inside the scope

\draw[red] (4,1) to[bend right] (6,1) -- (5,2);  % Draw anything
\end{scope}                                      % Now the trafo is reset again

\node[fit=(bb),draw=blue]{};     % Externally it doesn't know the content is
                                 % rotated or not

  \end{scope}                    % Back to original trafo.
\end{tikzpicture}
\end{document}

在此輸入影像描述

答案2

筆記：我終於自己找到了解決方案......但我不能給我自己的賞金。 ;-)

我定義了三種風格：

  1. memoize points命名路徑的每個點（在memoizepoints計數器的幫助下）並將這些名稱累積到全域巨集（其參數）。

  2. cont memoize points與 相同memoize points，但不重置全域巨集（其參數）。

  3. init memoize points重置全域宏（其參數）和memoizepoints計數器。

這是序言：

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{fit,intersections,decorations.pathreplacing,decorations.markings}

櫃檯memoizepoints及三種款式：

\newcounter{memoizepoints}
\tikzset{
  init memoize points/.code={\xdef#1{}\setcounter{memoizepoints}{0}},  
  cont memoize points/.style={postaction={
      decorate,decoration={show path construction,
        moveto code={},
        lineto code={
          \foreach \coord in {\tikzinputsegmentfirst,\tikzinputsegmentlast}{
            \addtocounter{memoizepoints}{1}
            \coordinate(memoizepoints-\arabic{memoizepoints}) at (\coord);
            \xdef#1{#1 (memoizepoints-\arabic{memoizepoints})}
          }
        },
        curveto code={
          \foreach \coord in {\tikzinputsegmentfirst,\tikzinputsegmentsupporta,%
            \tikzinputsegmentsupportb,\tikzinputsegmentlast}{
            \addtocounter{memoizepoints}{1}
            \coordinate(memoizepoints-\arabic{memoizepoints}) at (\coord);
            \xdef#1{#1 (memoizepoints-\arabic{memoizepoints})}
          }
        },
        closepath code={
          \foreach \coord in {\tikzinputsegmentfirst,\tikzinputsegmentlast}{
            \addtocounter{memoizepoints}{1}
            \coordinate(memoizepoints-\arabic{memoizepoints}) at (\coord);
            \xdef#1{#1 (memoizepoints-\arabic{memoizepoints})}
          }
        },
      },
    },
  },
  memoize points/.style={init memoize points=#1,cont memoize points=#1},
}

然後是一個範例及其程式碼：

在此輸入影像描述

\begin{document}
\begin{tikzpicture}[rotate=30,inner sep=0pt,line width=1pt]
  \tikzset{
    pt/.style={circle,minimum size=3pt,fill=#1,inner sep=0},
    red pt/.style={pt=red},
  }
  \draw[gray,line width=.4pt] (0,0) grid (9.5,5.5);
  % first case: fitting some nodes
  \node[red pt] (a) at (.5,.5){};
  \node[red pt] (b) at (.5,2){};
  \node[red pt] (c) at (3,.5){};
  \node[red pt] (d) at (2,2.2){};
  \begin{scope}[transform shape]
    \node[fit=(a)(b)(c)(d),draw=blue]{};
  \end{scope}

  % second case: fitting arbitrary path
  \draw[red,memoize points=\allpoints] (4,1) to[bend right] (6,1) -- (5,2);
  \begin{scope}[transform shape]
    \node[fit=\allpoints,draw=blue]{};
  \end{scope}

  % another example of fitting arbitrary path
  \draw[red,memoize points=\allpoints] (8,1.2) circle ();
  \begin{scope}[transform shape]
    \node[fit=\allpoints,draw=blue]{};
  \end{scope}

  % another example of fitting arbitrary paths
  \begin{scope}[yshift=2cm]
    \draw[red,memoize points=\allpoints]
    plot[domain=4:8,samples=100] (\x,{2+sin(3 * \x r)});
    \draw[red,cont memoize points=\allpoints]
    plot[domain=4:8,samples=100] (\x,{2.1+cos(3 * \x r)});
  \end{scope}
  \begin{scope}[transform shape]
    \node[fit=\allpoints,draw=blue]{};
  \end{scope}
\end{tikzpicture}

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