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I was looking for a way to overlay an image with an grid. I found this answer, which allows me to get an evenly sized grid over an arbitrary sized image (see first figure attached). Very nice!

However, if I use a outer bounding box, the grid coordinates are scaled, when moving the image node (see second figure attached).

I found that this effect occurs, as soon as the node is moved. Additionally the grid completely disappears, as soon as the anchor is changed to center (see third figure attached)

Edit: It seems like I missunderstood what scope does. It want to reference the image-node only. Ultimatively I want to overlay an image with an grid. The cells in the grid should all have the same size. I then want to place this image node and grid with respect to some other node

    \documentclass[tikz]{standalone}
    \usetikzlibrary{positioning}
    
    \begin{document}
    
    %%% Fig 1
    % first figure, as found in the answer mentioned
    \begin{tikzpicture}
      % include image as node
      \node[anchor=south west, inner sep=0, minimum height=9, minimum width=12] (image1)  at  (0,0) {\includegraphics[width=0.9\textwidth]{some_image.jpg}};
      
      % grid over the image using scope
      \begin{scope}[x={(image1.south east)},y={(image1.north west)}]
        \draw[thick, red,xstep=.1,ystep=.1] (0,0)  grid (1,1);
      \end{scope}
    
      % Some title
      \node[anchor=south west]  at  (image1.north west) {\Large This works!};
    \end{tikzpicture}
    
    %%% Fig 2
    % outer boundingbox m does funny stuff inside the scope
    \begin{tikzpicture}
      % the boundingbox for the image
      \useasboundingbox[fill=gray!20] (0, 0)  rectangle  (12,9);
    
      % image node as before, but node now at some non-origin coordinates
      \node[anchor=south west, inner sep=0] (image2)  at  (.5, .6) {\includegraphics[width=0.9\textwidth]{some_image.jpg}};
    
      % scopas before
      \begin{scope}[x={(image2.south east)},y={(image2.north west)}]
        \draw[thick,red, xstep=.1,ystep=.1] (0,0)  grid (1,1);
      \end{scope}
    
      % Some title
      \node[anchor=south west]  at  (image2.north west) {\Large This doesn't};
    \end{tikzpicture}
    
    
    %%% Fig 3
    % grid is gone completly
    \begin{tikzpicture}
      % Boundingbox around center (Grid is gone no matter if there is a bounding box or not)
      \useasboundingbox[fill=gray!20] (-6, -4.5)  rectangle  (6, 4.5);
    
      % include image as node
      \node[anchor=center, inner sep=0] (image3)  at  (0,0) {\includegraphics[width=0.9\textwidth]{some_image.jpg}};
      
      % grid over the image using scope
      \begin{scope}[x={(image3.south east)}, y={(image3.north west)}]
        \draw[thick, red,xstep=.1,ystep=.1] (0,0)  grid (1,1);
      \end{scope}
    
      % Some title
      \node[anchor=south west]  at  (image3.north west) {\Large Grid is gone completly};
    \end{tikzpicture}
    
    \end{document}

Output of minimum working example

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1 Answer 1

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UPDATE

I don't think grid is the right way to annotate the image. Cauze grid is the high level command of \pgfpathgrid with which

Conceptually, the origin is part of the grid and the grid is clipped to the rectangle specified by two points. However, no clipping occurs (this command just adds parts to the current path) and the points where the lines enter and leave the “clipping area” are computed and used to add simple lines to the current path.

I define a new coordinate system img cs to help the work. enter image description here

\documentclass[tikz, border=1cm]{standalone}
\usepackage{xparse}
\makeatletter
\newif\ifxtick@rotate
\tikzset{
  img/.cd,
  x/.store in=\img@x,
  y/.store in=\img@y,
  node/.store in=\img@node,
}
\tikzdeclarecoordinatesystem{img}
{%
  \tikzset{img/.cd,#1}%
  \tikz@scan@one@point\pgf@process(\[email protected] west)
  \pgf@xa=\pgf@x
  \pgf@ya=\pgf@y
  \tikz@scan@one@point\pgf@process(\[email protected] east)
  \pgfmathparse{(1-(\img@x))*\pgf@xa+(\img@x)*\pgf@x}
  \pgf@x=\pgfmathresult pt
  \pgfmathparse{(1-(\img@y))*\pgf@ya+(\img@y)*\pgf@y}
  \pgf@y=\pgfmathresult pt
}%
\NewDocumentCommand {\picgrid} {O{10} O{10} D(){} O{}} {
  \pgfkeys{/pgf/number format/.cd,fixed,fixed zerofill,precision=2}
  \foreach \i [evaluate=\i as \x using \i/#1] in {0,...,#1}
  \draw[#4]
    (img cs:node=#3, x=\x, y=0)
    node[rotate=-90, anchor=west] {\pgfmathprintnumber{\x}}
    -- (img cs:node=#3, x=\x, y=1);
  \foreach \i [evaluate=\i as \y using \i/#2] in {0,...,#2}
  \draw[#4]
    (img cs:node=#3, y=\y, x=0)
    node[left] {\pgfmathprintnumber{\y}}
    -- (img cs:node=#3, y=\y, x=1);
}
\makeatother

\begin{document}
\begin{tikzpicture}
  \node[anchor=south west, inner sep=0] (a)  at  (.5, .6) {\includegraphics[width=0.9\textwidth]{example-image}};
  \picgrid[20][20](a)[red, thick]
  \draw[thick, img/node=a] (img cs:x=.1, y=.6) --
    node[sloped, above] {from (.1, .6) to (.85, .95)}
    (img cs:x=.85, y=.95);
\end{tikzpicture}
\end{document}

old answer

<coord-1> grid <coord-2> means draw grid on the rectangle which is specified by <coord-1> and <coord-2>. Let's see where <coord-1> -- (0, 0) and <coord-2> -- (1, 1) actually lay on your second picture: enter image description here

As you see, there is nothing wrong.

So what's wrong with your third picture? Let's draw x unit vector and y unit vector of your scope: Ahha, x = -y, which means any coordinate like (i, i) actually is (0, 0). So (0, 0) grid (0, 0) does nothing. enter image description here

\documentclass[tikz, border=1cm]{standalone}
\usetikzlibrary{positioning, fit}
\usetikzlibrary {arrows.meta}
\tikzset{
  direct/.style={
    line width=1pt,
    arrows={-Stealth[length=3mm, width=3mm, #1]}
  }
}

\begin{document}
\begin{tikzpicture}
  \node[anchor=south west, inner sep=0] (image2)  at  (.5, .6) {\includegraphics[width=0.9\textwidth]{example-image}};

  \begin{scope}[x={(image2.south east)},y={(image2.north west)}]
    \draw[direct, green] (0, 0) -- (1, 1);
    \draw[thick,red, opacity=0.5, xstep=.1,ystep=.1] (0,0)  grid (1,1);
  \end{scope}
\end{tikzpicture}

\begin{tikzpicture}
  \node[anchor=center, inner sep=0] (image3)  at  (0,0) {\includegraphics[width=0.9\textwidth]{example-image}};
  % show x, y vector
  \draw[direct, red] (0, 0) -- (image3.south east) node [red, right] {$y$};
  \draw[direct, blue] (0, 0) -- (image3.north west) node [blue, left] {$x$};
  \begin{scope}[x={(image3.south east)}, y={(image3.north west)}]
    \draw[direct, green] (0, 0) -- (1, 1);
    \draw[thick,red, opacity=0.5, xstep=.1,ystep=.1] (0,0)  grid (1,1);
  \end{scope}
\end{tikzpicture}
\end{document}
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  • Thanks for the answer. Seemingly I did not quite understand what scope is doing. From your example I see that scope references/masks/shadows the whole TikZ image. But how can I scope the image node with the includegraphics? Commented Jul 24, 2020 at 11:07
  • 1
    I don't quite understand what do you want (may be drawing grid on part of image?), could you provide an example image? Commented Jul 24, 2020 at 11:15
  • Considering \node[anchor=south west, inner sep=0] (image2) at (.5, .6) {\includegraphics[width=0.9\textwidth]{example-image}}; If the at (.5, .6) is dropped, so that we end up with \node[anchor=south west, inner sep=0] (image2) {\includegraphics[width=0.9\textwidth]{example-image}}; The arrow starts at the south west and ends at north east of the image node. The coordinates (0,0) and (1,1) within the scope are equal to the corners of the image node. With at (.5, .6) this is not the case any more. I want this, so that I can evenly divide the image with xstep Commented Jul 24, 2020 at 14:05
  • I use another way@rikisa Commented Jul 24, 2020 at 18:39
  • Awesome, exactly what I was looking for! I falsely assumed that scope creates the new coordinate system for me. Thanks again! Commented Jul 27, 2020 at 7:30

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