Tikz: Calculate position based on node using fit option












3















I want the nodes c and d be inside q and the position of q to be determined by p.



A second problem is, that the distance between p and q isn't correct even with on grid=false (I guess the width of p isn't available because of using fit?).



documentclass[tikz]{standalone}
usetikzlibrary{positioning,fit}

begin{document}

begin{tikzpicture}
node (a) [draw=black] {a};
node (b) [draw=black,on grid,below=of a] {long text};
node (p) [draw=black,fit={(a) (b)}] {};

node (c) [draw=red] {c};
node (d) [draw=red,below=of c] {d};
node (q) [right=of p,draw=red,fit={(c) (d)}] {};
end{tikzpicture}
end{document}









share|improve this question



























    3















    I want the nodes c and d be inside q and the position of q to be determined by p.



    A second problem is, that the distance between p and q isn't correct even with on grid=false (I guess the width of p isn't available because of using fit?).



    documentclass[tikz]{standalone}
    usetikzlibrary{positioning,fit}

    begin{document}

    begin{tikzpicture}
    node (a) [draw=black] {a};
    node (b) [draw=black,on grid,below=of a] {long text};
    node (p) [draw=black,fit={(a) (b)}] {};

    node (c) [draw=red] {c};
    node (d) [draw=red,below=of c] {d};
    node (q) [right=of p,draw=red,fit={(c) (d)}] {};
    end{tikzpicture}
    end{document}









    share|improve this question

























      3












      3








      3








      I want the nodes c and d be inside q and the position of q to be determined by p.



      A second problem is, that the distance between p and q isn't correct even with on grid=false (I guess the width of p isn't available because of using fit?).



      documentclass[tikz]{standalone}
      usetikzlibrary{positioning,fit}

      begin{document}

      begin{tikzpicture}
      node (a) [draw=black] {a};
      node (b) [draw=black,on grid,below=of a] {long text};
      node (p) [draw=black,fit={(a) (b)}] {};

      node (c) [draw=red] {c};
      node (d) [draw=red,below=of c] {d};
      node (q) [right=of p,draw=red,fit={(c) (d)}] {};
      end{tikzpicture}
      end{document}









      share|improve this question














      I want the nodes c and d be inside q and the position of q to be determined by p.



      A second problem is, that the distance between p and q isn't correct even with on grid=false (I guess the width of p isn't available because of using fit?).



      documentclass[tikz]{standalone}
      usetikzlibrary{positioning,fit}

      begin{document}

      begin{tikzpicture}
      node (a) [draw=black] {a};
      node (b) [draw=black,on grid,below=of a] {long text};
      node (p) [draw=black,fit={(a) (b)}] {};

      node (c) [draw=red] {c};
      node (d) [draw=red,below=of c] {d};
      node (q) [right=of p,draw=red,fit={(c) (d)}] {};
      end{tikzpicture}
      end{document}






      tikz-pgf fit






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 18 at 14:20









      BenBen

      7221419




      7221419






















          3 Answers
          3






          active

          oldest

          votes


















          2














          How about using a matrix?



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm] {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black] {long text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b)}] {};
          node (q) [draw=red,fit={(c) (d)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          The advantage of using one matrix (instead of two) is that it also looks good if the vertical dimensions of the nodes vary.



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm,anchor=center] (mat) {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black,align=center] {long\ text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b) (b|-mat.south)}] {};
          node (q) [draw=red,fit={(c) (d) (d|-mat.south)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          As in any TikZ matrix, every line, including the last one, has to be terminated by \.






          share|improve this answer


























          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58



















          3














          marmot proposes to use a matrix, but I think two matrices are better:



          documentclass[tikz]{standalone}
          usetikzlibrary{positioning,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[draw, matrix of nodes, nodes=draw, row sep=3mm] (a) {
          a\
          long text\
          };

          matrix[draw=red, matrix of nodes, nodes={draw=red},
          right=3mm of a, row sep=3mm] (b) {
          c\
          dvphantom{g}\
          };
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

            – marmot
            Jan 18 at 16:52











          • What are the advantages of using two matrices?

            – Ben
            Jan 18 at 16:53











          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58











          • @marmot It's true, but you can always define the desired dimensions or inner nodes.

            – Ignasi
            Jan 18 at 17:15











          • Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

            – marmot
            Jan 18 at 17:18



















          1














          documentclass[tikz,margin=3mm]{standalone}
          usetikzlibrary{fit, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 8mm and 4 mm,
          every node/.style = {inner sep=1mm, minimum height=1.5em}
          ]
          node (a) [draw=black] {a};
          node (b) [draw=black,below=of a] {long text};
          node (p) [draw=black,fit={(a) (b)}] {};

          node (c) [draw=red,right=of b.east |- a] {c};
          node (d) [draw=red,below=of c] {d};
          node (q) [draw=red,fit={(c) (d)}] {};%right=of p,
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

            – Ben
            Jan 18 at 15:19






          • 3





            @Ben No, -| and |- have nothing to do with the positioning library.

            – Torbjørn T.
            Jan 18 at 15:37











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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          How about using a matrix?



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm] {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black] {long text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b)}] {};
          node (q) [draw=red,fit={(c) (d)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          The advantage of using one matrix (instead of two) is that it also looks good if the vertical dimensions of the nodes vary.



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm,anchor=center] (mat) {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black,align=center] {long\ text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b) (b|-mat.south)}] {};
          node (q) [draw=red,fit={(c) (d) (d|-mat.south)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          As in any TikZ matrix, every line, including the last one, has to be terminated by \.






          share|improve this answer


























          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58
















          2














          How about using a matrix?



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm] {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black] {long text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b)}] {};
          node (q) [draw=red,fit={(c) (d)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          The advantage of using one matrix (instead of two) is that it also looks good if the vertical dimensions of the nodes vary.



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm,anchor=center] (mat) {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black,align=center] {long\ text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b) (b|-mat.south)}] {};
          node (q) [draw=red,fit={(c) (d) (d|-mat.south)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          As in any TikZ matrix, every line, including the last one, has to be terminated by \.






          share|improve this answer


























          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58














          2












          2








          2







          How about using a matrix?



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm] {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black] {long text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b)}] {};
          node (q) [draw=red,fit={(c) (d)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          The advantage of using one matrix (instead of two) is that it also looks good if the vertical dimensions of the nodes vary.



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm,anchor=center] (mat) {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black,align=center] {long\ text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b) (b|-mat.south)}] {};
          node (q) [draw=red,fit={(c) (d) (d|-mat.south)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          As in any TikZ matrix, every line, including the last one, has to be terminated by \.






          share|improve this answer















          How about using a matrix?



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm] {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black] {long text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b)}] {};
          node (q) [draw=red,fit={(c) (d)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          The advantage of using one matrix (instead of two) is that it also looks good if the vertical dimensions of the nodes vary.



          documentclass[tikz]{standalone}
          usetikzlibrary{fit,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[column sep=8mm,row sep=3mm,anchor=center] (mat) {
          node (a) [draw=black] {a};
          &
          node (c) [draw=red] {c};
          \
          node (b) [draw=black,align=center] {long\ text};
          &
          node (d) [draw=red] {d};
          \
          };
          node (p) [draw=black,fit={(a) (b) (b|-mat.south)}] {};
          node (q) [draw=red,fit={(c) (d) (d|-mat.south)}] {};
          end{tikzpicture}
          end{document}


          enter image description here



          As in any TikZ matrix, every line, including the last one, has to be terminated by \.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jan 18 at 17:00

























          answered Jan 18 at 16:31









          marmotmarmot

          94.6k4109209




          94.6k4109209













          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58



















          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58

















          Maybe you want to call attention to the \ before the };

          – Ben
          Jan 18 at 16:58





          Maybe you want to call attention to the \ before the };

          – Ben
          Jan 18 at 16:58











          3














          marmot proposes to use a matrix, but I think two matrices are better:



          documentclass[tikz]{standalone}
          usetikzlibrary{positioning,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[draw, matrix of nodes, nodes=draw, row sep=3mm] (a) {
          a\
          long text\
          };

          matrix[draw=red, matrix of nodes, nodes={draw=red},
          right=3mm of a, row sep=3mm] (b) {
          c\
          dvphantom{g}\
          };
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

            – marmot
            Jan 18 at 16:52











          • What are the advantages of using two matrices?

            – Ben
            Jan 18 at 16:53











          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58











          • @marmot It's true, but you can always define the desired dimensions or inner nodes.

            – Ignasi
            Jan 18 at 17:15











          • Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

            – marmot
            Jan 18 at 17:18
















          3














          marmot proposes to use a matrix, but I think two matrices are better:



          documentclass[tikz]{standalone}
          usetikzlibrary{positioning,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[draw, matrix of nodes, nodes=draw, row sep=3mm] (a) {
          a\
          long text\
          };

          matrix[draw=red, matrix of nodes, nodes={draw=red},
          right=3mm of a, row sep=3mm] (b) {
          c\
          dvphantom{g}\
          };
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

            – marmot
            Jan 18 at 16:52











          • What are the advantages of using two matrices?

            – Ben
            Jan 18 at 16:53











          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58











          • @marmot It's true, but you can always define the desired dimensions or inner nodes.

            – Ignasi
            Jan 18 at 17:15











          • Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

            – marmot
            Jan 18 at 17:18














          3












          3








          3







          marmot proposes to use a matrix, but I think two matrices are better:



          documentclass[tikz]{standalone}
          usetikzlibrary{positioning,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[draw, matrix of nodes, nodes=draw, row sep=3mm] (a) {
          a\
          long text\
          };

          matrix[draw=red, matrix of nodes, nodes={draw=red},
          right=3mm of a, row sep=3mm] (b) {
          c\
          dvphantom{g}\
          };
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer













          marmot proposes to use a matrix, but I think two matrices are better:



          documentclass[tikz]{standalone}
          usetikzlibrary{positioning,matrix}

          begin{document}

          begin{tikzpicture}
          matrix[draw, matrix of nodes, nodes=draw, row sep=3mm] (a) {
          a\
          long text\
          };

          matrix[draw=red, matrix of nodes, nodes={draw=red},
          right=3mm of a, row sep=3mm] (b) {
          c\
          dvphantom{g}\
          };
          end{tikzpicture}
          end{document}


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 18 at 16:47









          IgnasiIgnasi

          92.6k4166307




          92.6k4166307













          • I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

            – marmot
            Jan 18 at 16:52











          • What are the advantages of using two matrices?

            – Ben
            Jan 18 at 16:53











          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58











          • @marmot It's true, but you can always define the desired dimensions or inner nodes.

            – Ignasi
            Jan 18 at 17:15











          • Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

            – marmot
            Jan 18 at 17:18



















          • I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

            – marmot
            Jan 18 at 16:52











          • What are the advantages of using two matrices?

            – Ben
            Jan 18 at 16:53











          • Maybe you want to call attention to the \ before the };

            – Ben
            Jan 18 at 16:58











          • @marmot It's true, but you can always define the desired dimensions or inner nodes.

            – Ignasi
            Jan 18 at 17:15











          • Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

            – marmot
            Jan 18 at 17:18

















          I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

          – marmot
          Jan 18 at 16:52





          I thought about this as well but I was afraid that, if the nodes have different vertical dimensions, this will no longer work.

          – marmot
          Jan 18 at 16:52













          What are the advantages of using two matrices?

          – Ben
          Jan 18 at 16:53





          What are the advantages of using two matrices?

          – Ben
          Jan 18 at 16:53













          Maybe you want to call attention to the \ before the };

          – Ben
          Jan 18 at 16:58





          Maybe you want to call attention to the \ before the };

          – Ben
          Jan 18 at 16:58













          @marmot It's true, but you can always define the desired dimensions or inner nodes.

          – Ignasi
          Jan 18 at 17:15





          @marmot It's true, but you can always define the desired dimensions or inner nodes.

          – Ignasi
          Jan 18 at 17:15













          Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

          – marmot
          Jan 18 at 17:18





          Yes, sure. I just wanted to say that I do not necessarily think that " two matrices are better". Either of our two options may be advantageous depending on what you precisely intent to do.

          – marmot
          Jan 18 at 17:18











          1














          documentclass[tikz,margin=3mm]{standalone}
          usetikzlibrary{fit, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 8mm and 4 mm,
          every node/.style = {inner sep=1mm, minimum height=1.5em}
          ]
          node (a) [draw=black] {a};
          node (b) [draw=black,below=of a] {long text};
          node (p) [draw=black,fit={(a) (b)}] {};

          node (c) [draw=red,right=of b.east |- a] {c};
          node (d) [draw=red,below=of c] {d};
          node (q) [draw=red,fit={(c) (d)}] {};%right=of p,
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

            – Ben
            Jan 18 at 15:19






          • 3





            @Ben No, -| and |- have nothing to do with the positioning library.

            – Torbjørn T.
            Jan 18 at 15:37
















          1














          documentclass[tikz,margin=3mm]{standalone}
          usetikzlibrary{fit, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 8mm and 4 mm,
          every node/.style = {inner sep=1mm, minimum height=1.5em}
          ]
          node (a) [draw=black] {a};
          node (b) [draw=black,below=of a] {long text};
          node (p) [draw=black,fit={(a) (b)}] {};

          node (c) [draw=red,right=of b.east |- a] {c};
          node (d) [draw=red,below=of c] {d};
          node (q) [draw=red,fit={(c) (d)}] {};%right=of p,
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer
























          • I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

            – Ben
            Jan 18 at 15:19






          • 3





            @Ben No, -| and |- have nothing to do with the positioning library.

            – Torbjørn T.
            Jan 18 at 15:37














          1












          1








          1







          documentclass[tikz,margin=3mm]{standalone}
          usetikzlibrary{fit, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 8mm and 4 mm,
          every node/.style = {inner sep=1mm, minimum height=1.5em}
          ]
          node (a) [draw=black] {a};
          node (b) [draw=black,below=of a] {long text};
          node (p) [draw=black,fit={(a) (b)}] {};

          node (c) [draw=red,right=of b.east |- a] {c};
          node (d) [draw=red,below=of c] {d};
          node (q) [draw=red,fit={(c) (d)}] {};%right=of p,
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer













          documentclass[tikz,margin=3mm]{standalone}
          usetikzlibrary{fit, positioning}

          begin{document}

          begin{tikzpicture}[
          node distance = 8mm and 4 mm,
          every node/.style = {inner sep=1mm, minimum height=1.5em}
          ]
          node (a) [draw=black] {a};
          node (b) [draw=black,below=of a] {long text};
          node (p) [draw=black,fit={(a) (b)}] {};

          node (c) [draw=red,right=of b.east |- a] {c};
          node (d) [draw=red,below=of c] {d};
          node (q) [draw=red,fit={(c) (d)}] {};%right=of p,
          end{tikzpicture}
          end{document}


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 18 at 14:52









          ZarkoZarko

          123k865161




          123k865161













          • I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

            – Ben
            Jan 18 at 15:19






          • 3





            @Ben No, -| and |- have nothing to do with the positioning library.

            – Torbjørn T.
            Jan 18 at 15:37



















          • I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

            – Ben
            Jan 18 at 15:19






          • 3





            @Ben No, -| and |- have nothing to do with the positioning library.

            – Torbjørn T.
            Jan 18 at 15:37

















          I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

          – Ben
          Jan 18 at 15:19





          I did not knew |-. This is only available because of the positioning library, isn't it? But for your solution one has to know if a or b determines the width of p, which isn't perfectly easy, but should work for me

          – Ben
          Jan 18 at 15:19




          3




          3





          @Ben No, -| and |- have nothing to do with the positioning library.

          – Torbjørn T.
          Jan 18 at 15:37





          @Ben No, -| and |- have nothing to do with the positioning library.

          – Torbjørn T.
          Jan 18 at 15:37


















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