Tikz: Calculate position based on node using fit option
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
add a comment |
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
add a comment |
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
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
tikz-pgf fit
asked Jan 18 at 14:20
BenBen
7221419
7221419
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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}
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}
As in any TikZ matrix, every line, including the last one, has to be terminated by \
.
Maybe you want to call attention to the\
before the};
– Ben
Jan 18 at 16:58
add a comment |
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}
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
|
show 1 more comment
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}
I did not knew|-
. This is only available because of thepositioning
library, isn't it? But for your solution one has to know ifa
orb
determines the width ofp
, 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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}
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}
As in any TikZ matrix, every line, including the last one, has to be terminated by \
.
Maybe you want to call attention to the\
before the};
– Ben
Jan 18 at 16:58
add a comment |
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}
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}
As in any TikZ matrix, every line, including the last one, has to be terminated by \
.
Maybe you want to call attention to the\
before the};
– Ben
Jan 18 at 16:58
add a comment |
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}
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}
As in any TikZ matrix, every line, including the last one, has to be terminated by \
.
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}
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}
As in any TikZ matrix, every line, including the last one, has to be terminated by \
.
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
add a comment |
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
add a comment |
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}
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
|
show 1 more comment
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}
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
|
show 1 more comment
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}
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}
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
|
show 1 more comment
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
|
show 1 more comment
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}
I did not knew|-
. This is only available because of thepositioning
library, isn't it? But for your solution one has to know ifa
orb
determines the width ofp
, 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
add a comment |
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}
I did not knew|-
. This is only available because of thepositioning
library, isn't it? But for your solution one has to know ifa
orb
determines the width ofp
, 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
add a comment |
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}
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}
answered Jan 18 at 14:52
ZarkoZarko
123k865161
123k865161
I did not knew|-
. This is only available because of thepositioning
library, isn't it? But for your solution one has to know ifa
orb
determines the width ofp
, 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
add a comment |
I did not knew|-
. This is only available because of thepositioning
library, isn't it? But for your solution one has to know ifa
orb
determines the width ofp
, 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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