Mixing
How to compute the *key* of a pgfkeyssetvalue command?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I have a list of (angles/value) pairs. I'd like to use TikZ to
- draw nodes with those values at these angles (easy),
- then draw an edge from node i to i+1, with the value of node i as a label (not at all easy).
My idea so far: store the values in pgfkeys, with keys the same as the node names. Then, values could be easily accessed. Problem is, I cannot get the pgfkeyssetvalue to evalute the node name first.
Example:
documentclass[tikz]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
pgfkeyssetvalue/nodevalues/pivalue
% to illustrate that it works with explicit node names:
pgfkeyssetvalue/nodevalues/p117
pgfkeyssetvalue/nodevalues/p222
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
This draws the edges for the manually set keys /nodevalues/p1 and /nodevalues/p2 correctly. But pgfkeyssetkeys does not do what I like. Expected result would be that 5->12 arc is label with 5, and 12->17 with 12.
tikz-pgf foreach pgfkeys
edited Aug 21 at 16:50
AndréC
2,872728
asked Aug 21 at 15:46
Holger Karl
354
add a comment |Â
up vote
5
down vote
favorite
I have a list of (angles/value) pairs. I'd like to use TikZ to
- draw nodes with those values at these angles (easy),
- then draw an edge from node i to i+1, with the value of node i as a label (not at all easy).
My idea so far: store the values in pgfkeys, with keys the same as the node names. Then, values could be easily accessed. Problem is, I cannot get the pgfkeyssetvalue to evalute the node name first.
Example:
documentclass[tikz]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
pgfkeyssetvalue/nodevalues/pivalue
% to illustrate that it works with explicit node names:
pgfkeyssetvalue/nodevalues/p117
pgfkeyssetvalue/nodevalues/p222
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
This draws the edges for the manually set keys /nodevalues/p1 and /nodevalues/p2 correctly. But pgfkeyssetkeys does not do what I like. Expected result would be that 5->12 arc is label with 5, and 12->17 with 12.
tikz-pgf foreach pgfkeys
edited Aug 21 at 16:50
AndréC
2,872728
asked Aug 21 at 15:46
Holger Karl
354
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I have a list of (angles/value) pairs. I'd like to use TikZ to
- draw nodes with those values at these angles (easy),
- then draw an edge from node i to i+1, with the value of node i as a label (not at all easy).
My idea so far: store the values in pgfkeys, with keys the same as the node names. Then, values could be easily accessed. Problem is, I cannot get the pgfkeyssetvalue to evalute the node name first.
Example:
documentclass[tikz]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
pgfkeyssetvalue/nodevalues/pivalue
% to illustrate that it works with explicit node names:
pgfkeyssetvalue/nodevalues/p117
pgfkeyssetvalue/nodevalues/p222
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
This draws the edges for the manually set keys /nodevalues/p1 and /nodevalues/p2 correctly. But pgfkeyssetkeys does not do what I like. Expected result would be that 5->12 arc is label with 5, and 12->17 with 12.
tikz-pgf foreach pgfkeys
edited Aug 21 at 16:50
AndréC
2,872728
asked Aug 21 at 15:46
Holger Karl
354
I have a list of (angles/value) pairs. I'd like to use TikZ to
- draw nodes with those values at these angles (easy),
- then draw an edge from node i to i+1, with the value of node i as a label (not at all easy).
My idea so far: store the values in pgfkeys, with keys the same as the node names. Then, values could be easily accessed. Problem is, I cannot get the pgfkeyssetvalue to evalute the node name first.
Example:
documentclass[tikz]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
pgfkeyssetvalue/nodevalues/pivalue
% to illustrate that it works with explicit node names:
pgfkeyssetvalue/nodevalues/p117
pgfkeyssetvalue/nodevalues/p222
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
This draws the edges for the manually set keys /nodevalues/p1 and /nodevalues/p2 correctly. But pgfkeyssetkeys does not do what I like. Expected result would be that 5->12 arc is label with 5, and 12->17 with 12.
tikz-pgf foreach pgfkeys
edited Aug 21 at 16:50
AndréC
2,872728
asked Aug 21 at 15:46
Holger Karl
354
edited Aug 21 at 16:50
AndréC
2,872728
edited Aug 21 at 16:50
AndréC
2,872728
edited Aug 21 at 16:50
AndréC
2,872728
2,872728
asked Aug 21 at 15:46
Holger Karl
354
asked Aug 21 at 15:46
Holger Karl
354
asked Aug 21 at 15:46
Holger Karl
354
354
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01
add a comment |Â
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
Here is how I would do it: build up a list and reuse its entries.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumi=1
xdefLstvalue
else
xdefLstLst,value
fi
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
pgfmathsetmacroXLst[i-1]
typeoutX
draw [->] (pi) to [bend right=45] node[midway,auto,swap]
X (pj) ;
endtikzpicture
enddocument
EDIT: I think this is a nice question which also deserves an explanation why it fails. Essentially the loop does the keys locally, and you need to make them global. This globalization has been achieved by @percusse in this nice answer. Once one employs it, your original approach can also be made work.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
begingroupglobaldefs=1relax
edeftempnoexpandpgfkeyssetvalue/nodevalues/pivalue
temp
endgroup
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
Yet another possibility would be to employ Henri Menke's trick to remember node contents, I'd be happy to spell that out if you ask me to.
answered Aug 21 at 15:57
marmot
55.4k460121
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do inforeachloops and more often in pgfplots. It ensures here that whenpgfkeyssetvaluegets "executed"iandvalueare already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons ofexpanafterandnoexpandstatements. But yes, it is "nicer" if one can do without those. ;-)
– marmot
Aug 22 at 10:11
add a comment |Â
up vote
4
down vote
Just for fun, a single foreach that does what you want.
documentclass[tikz,margin=2mm]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
foreach angle/value [count = i, count=j from 0, remember=value as oldvalue] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumj>0
draw[->] (pj) to [bend right=45] node[midway,auto,swap] oldvalue (pi);
fi
ifnumi=4
draw[->] (pi) to [bend right=45] node[midway,auto,swap] value (p1);
fi
endtikzpicture
enddocument
answered Aug 21 at 16:46
Max
5,89811727
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
@AndréC Thanks! I think yourremember=i as lasti (initially 4)is also very clever, saves you to do theint(mod(...))stuff. I will try to remember that :)
– Max
Aug 22 at 6:19
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
1
down vote
It is possible to solve your problem without using pgf keys but only using the native capabilities of foreach operation.
Here, I used a first loop to draw the nodes and a second to create the arrows between the nodes with the edge operation
documentclass[tikz]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% create the node
foreach angle/value [count=i,remember=angle as lastangle (initially 140),remember=i as lasti (initially 4)] in 0/17, 45/22, 80/5, 140/12
node (i) at (angle:2cm) value;
% draw arrow with edge operation
foreach value [count=i,remember =i as lasti (initially 4),remember=value as lvalue (initially 12)] in 17,22,5,12
path[->](lasti) edge[bend right=45] node[midway,auto,swap]lvalue(i);
endtikzpicture
enddocument
answered Aug 21 at 21:14
AndréC
2,872728
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Here is how I would do it: build up a list and reuse its entries.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumi=1
xdefLstvalue
else
xdefLstLst,value
fi
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
pgfmathsetmacroXLst[i-1]
typeoutX
draw [->] (pi) to [bend right=45] node[midway,auto,swap]
X (pj) ;
endtikzpicture
enddocument
EDIT: I think this is a nice question which also deserves an explanation why it fails. Essentially the loop does the keys locally, and you need to make them global. This globalization has been achieved by @percusse in this nice answer. Once one employs it, your original approach can also be made work.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
begingroupglobaldefs=1relax
edeftempnoexpandpgfkeyssetvalue/nodevalues/pivalue
temp
endgroup
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
Yet another possibility would be to employ Henri Menke's trick to remember node contents, I'd be happy to spell that out if you ask me to.
answered Aug 21 at 15:57
marmot
55.4k460121
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do inforeachloops and more often in pgfplots. It ensures here that whenpgfkeyssetvaluegets "executed"iandvalueare already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons ofexpanafterandnoexpandstatements. But yes, it is "nicer" if one can do without those. ;-)
– marmot
Aug 22 at 10:11
add a comment |Â
up vote
6
down vote
accepted
Here is how I would do it: build up a list and reuse its entries.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumi=1
xdefLstvalue
else
xdefLstLst,value
fi
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
pgfmathsetmacroXLst[i-1]
typeoutX
draw [->] (pi) to [bend right=45] node[midway,auto,swap]
X (pj) ;
endtikzpicture
enddocument
EDIT: I think this is a nice question which also deserves an explanation why it fails. Essentially the loop does the keys locally, and you need to make them global. This globalization has been achieved by @percusse in this nice answer. Once one employs it, your original approach can also be made work.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
begingroupglobaldefs=1relax
edeftempnoexpandpgfkeyssetvalue/nodevalues/pivalue
temp
endgroup
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
Yet another possibility would be to employ Henri Menke's trick to remember node contents, I'd be happy to spell that out if you ask me to.
answered Aug 21 at 15:57
marmot
55.4k460121
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do inforeachloops and more often in pgfplots. It ensures here that whenpgfkeyssetvaluegets "executed"iandvalueare already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons ofexpanafterandnoexpandstatements. But yes, it is "nicer" if one can do without those. ;-)
– marmot
Aug 22 at 10:11
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Here is how I would do it: build up a list and reuse its entries.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumi=1
xdefLstvalue
else
xdefLstLst,value
fi
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
pgfmathsetmacroXLst[i-1]
typeoutX
draw [->] (pi) to [bend right=45] node[midway,auto,swap]
X (pj) ;
endtikzpicture
enddocument
EDIT: I think this is a nice question which also deserves an explanation why it fails. Essentially the loop does the keys locally, and you need to make them global. This globalization has been achieved by @percusse in this nice answer. Once one employs it, your original approach can also be made work.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
begingroupglobaldefs=1relax
edeftempnoexpandpgfkeyssetvalue/nodevalues/pivalue
temp
endgroup
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
Yet another possibility would be to employ Henri Menke's trick to remember node contents, I'd be happy to spell that out if you ask me to.
answered Aug 21 at 15:57
marmot
55.4k460121
Here is how I would do it: build up a list and reuse its entries.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumi=1
xdefLstvalue
else
xdefLstLst,value
fi
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
pgfmathsetmacroXLst[i-1]
typeoutX
draw [->] (pi) to [bend right=45] node[midway,auto,swap]
X (pj) ;
endtikzpicture
enddocument
EDIT: I think this is a nice question which also deserves an explanation why it fails. Essentially the loop does the keys locally, and you need to make them global. This globalization has been achieved by @percusse in this nice answer. Once one employs it, your original approach can also be made work.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% draw nodes on a circle, remembering their value no longer fails:
foreach angle/value [count = i] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
begingroupglobaldefs=1relax
edeftempnoexpandpgfkeyssetvalue/nodevalues/pivalue
temp
endgroup
% edges:
foreach [evaluate = j=int(mod(i, 4)+1)] i in 1,...,4
draw [->] (pi) to [bend right=45] node[midway] pgfkeysvalueof/nodevalues/pi (pj) ;
endtikzpicture
enddocument
Yet another possibility would be to employ Henri Menke's trick to remember node contents, I'd be happy to spell that out if you ask me to.
answered Aug 21 at 15:57
marmot
55.4k460121
edited Aug 21 at 16:33
answered Aug 21 at 15:57
marmot
55.4k460121
answered Aug 21 at 15:57
marmot
55.4k460121
answered Aug 21 at 15:57
marmot
55.4k460121
55.4k460121
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do inforeachloops and more often in pgfplots. It ensures here that whenpgfkeyssetvaluegets "executed"iandvalueare already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons ofexpanafterandnoexpandstatements. But yes, it is "nicer" if one can do without those. ;-)
– marmot
Aug 22 at 10:11
add a comment |Â
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do inforeachloops and more often in pgfplots. It ensures here that whenpgfkeyssetvaluegets "executed"iandvalueare already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons ofexpanafterandnoexpandstatements. But yes, it is "nicer" if one can do without those. ;-)
– marmot
Aug 22 at 10:11
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
Nifty! Thanks a lot, that does it!
– Holger Karl
Aug 21 at 16:01
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
marmot: Your edited answer is enlightening, thanks. I had seen @percusse comment in the other thread, but hadn't understood its relevance here. A bit counterintuitive to have to use noexpand to expand i, though :-). The approach via macro seems quite heavy-weight?
– Holger Karl
Aug 22 at 6:30
@HolgerKarl This expansion thingy is something that one has sometimes to do in
foreach loops and more often in pgfplots. It ensures here that when pgfkeyssetvalue gets "executed" i and value are already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons of expanafter and noexpand statements. But yes, it is "nicer" if one can do without those. ;-)– marmot
Aug 22 at 10:11
@HolgerKarl This expansion thingy is something that one has sometimes to do in
foreach loops and more often in pgfplots. It ensures here that when pgfkeyssetvalue gets "executed" i and value are already expanded. Whether or not this is "heavy weight" is a matter of taste. If you look into the codes in packages, you will often see tons of expanafter and noexpand statements. But yes, it is "nicer" if one can do without those. ;-)– marmot
Aug 22 at 10:11
add a comment |Â
up vote
4
down vote
Just for fun, a single foreach that does what you want.
documentclass[tikz,margin=2mm]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
foreach angle/value [count = i, count=j from 0, remember=value as oldvalue] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumj>0
draw[->] (pj) to [bend right=45] node[midway,auto,swap] oldvalue (pi);
fi
ifnumi=4
draw[->] (pi) to [bend right=45] node[midway,auto,swap] value (p1);
fi
endtikzpicture
enddocument
answered Aug 21 at 16:46
Max
5,89811727
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
@AndréC Thanks! I think yourremember=i as lasti (initially 4)is also very clever, saves you to do theint(mod(...))stuff. I will try to remember that :)
– Max
Aug 22 at 6:19
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
4
down vote
Just for fun, a single foreach that does what you want.
documentclass[tikz,margin=2mm]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
foreach angle/value [count = i, count=j from 0, remember=value as oldvalue] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumj>0
draw[->] (pj) to [bend right=45] node[midway,auto,swap] oldvalue (pi);
fi
ifnumi=4
draw[->] (pi) to [bend right=45] node[midway,auto,swap] value (p1);
fi
endtikzpicture
enddocument
answered Aug 21 at 16:46
Max
5,89811727
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
@AndréC Thanks! I think yourremember=i as lasti (initially 4)is also very clever, saves you to do theint(mod(...))stuff. I will try to remember that :)
– Max
Aug 22 at 6:19
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Just for fun, a single foreach that does what you want.
documentclass[tikz,margin=2mm]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
foreach angle/value [count = i, count=j from 0, remember=value as oldvalue] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumj>0
draw[->] (pj) to [bend right=45] node[midway,auto,swap] oldvalue (pi);
fi
ifnumi=4
draw[->] (pi) to [bend right=45] node[midway,auto,swap] value (p1);
fi
endtikzpicture
enddocument
answered Aug 21 at 16:46
Max
5,89811727
Just for fun, a single foreach that does what you want.
documentclass[tikz,margin=2mm]standalone
usepackagetikz
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
foreach angle/value [count = i, count=j from 0, remember=value as oldvalue] in 0/17, 45/22, 80/5, 140/12
node [circle, draw, fill=white] (pi) at (angle:2cm) value;
ifnumj>0
draw[->] (pj) to [bend right=45] node[midway,auto,swap] oldvalue (pi);
fi
ifnumi=4
draw[->] (pi) to [bend right=45] node[midway,auto,swap] value (p1);
fi
endtikzpicture
enddocument
answered Aug 21 at 16:46
Max
5,89811727
answered Aug 21 at 16:46
Max
5,89811727
answered Aug 21 at 16:46
Max
5,89811727
answered Aug 21 at 16:46
Max
5,89811727
5,89811727
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
@AndréC Thanks! I think yourremember=i as lasti (initially 4)is also very clever, saves you to do theint(mod(...))stuff. I will try to remember that :)
– Max
Aug 22 at 6:19
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
add a comment |Â
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
@AndréC Thanks! I think yourremember=i as lasti (initially 4)is also very clever, saves you to do theint(mod(...))stuff. I will try to remember that :)
– Max
Aug 22 at 6:19
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
1
1
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
Oops, I spent the evening looking to do like you and you were faster. I had not thought at all to use two different counts, one starting at 0, the other at 1, your solution is more beautiful than mine :)
– AndréC
Aug 21 at 21:16
1
1
@AndréC Thanks! I think your
remember=i as lasti (initially 4) is also very clever, saves you to do the int(mod(...)) stuff. I will try to remember that :)– Max
Aug 22 at 6:19
@AndréC Thanks! I think your
remember=i as lasti (initially 4) is also very clever, saves you to do the int(mod(...)) stuff. I will try to remember that :)– Max
Aug 22 at 6:19
1
1
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
Unfortunately, a single loop does not work for me - I am drawing the base graph and then different annotations on top of it (not shown in my question). But yes, that is also nice with the lasti - didnt know about that either :-)
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
1
down vote
It is possible to solve your problem without using pgf keys but only using the native capabilities of foreach operation.
Here, I used a first loop to draw the nodes and a second to create the arrows between the nodes with the edge operation
documentclass[tikz]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% create the node
foreach angle/value [count=i,remember=angle as lastangle (initially 140),remember=i as lasti (initially 4)] in 0/17, 45/22, 80/5, 140/12
node (i) at (angle:2cm) value;
% draw arrow with edge operation
foreach value [count=i,remember =i as lasti (initially 4),remember=value as lvalue (initially 12)] in 17,22,5,12
path[->](lasti) edge[bend right=45] node[midway,auto,swap]lvalue(i);
endtikzpicture
enddocument
answered Aug 21 at 21:14
AndréC
2,872728
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
1
down vote
It is possible to solve your problem without using pgf keys but only using the native capabilities of foreach operation.
Here, I used a first loop to draw the nodes and a second to create the arrows between the nodes with the edge operation
documentclass[tikz]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% create the node
foreach angle/value [count=i,remember=angle as lastangle (initially 140),remember=i as lasti (initially 4)] in 0/17, 45/22, 80/5, 140/12
node (i) at (angle:2cm) value;
% draw arrow with edge operation
foreach value [count=i,remember =i as lasti (initially 4),remember=value as lvalue (initially 12)] in 17,22,5,12
path[->](lasti) edge[bend right=45] node[midway,auto,swap]lvalue(i);
endtikzpicture
enddocument
answered Aug 21 at 21:14
AndréC
2,872728
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is possible to solve your problem without using pgf keys but only using the native capabilities of foreach operation.
Here, I used a first loop to draw the nodes and a second to create the arrows between the nodes with the edge operation
documentclass[tikz]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% create the node
foreach angle/value [count=i,remember=angle as lastangle (initially 140),remember=i as lasti (initially 4)] in 0/17, 45/22, 80/5, 140/12
node (i) at (angle:2cm) value;
% draw arrow with edge operation
foreach value [count=i,remember =i as lasti (initially 4),remember=value as lvalue (initially 12)] in 17,22,5,12
path[->](lasti) edge[bend right=45] node[midway,auto,swap]lvalue(i);
endtikzpicture
enddocument
answered Aug 21 at 21:14
AndréC
2,872728
It is possible to solve your problem without using pgf keys but only using the native capabilities of foreach operation.
Here, I used a first loop to draw the nodes and a second to create the arrows between the nodes with the edge operation
documentclass[tikz]standalone
begindocument
begintikzpicture
draw [dotted] (0,0) circle (2cm);
% create the node
foreach angle/value [count=i,remember=angle as lastangle (initially 140),remember=i as lasti (initially 4)] in 0/17, 45/22, 80/5, 140/12
node (i) at (angle:2cm) value;
% draw arrow with edge operation
foreach value [count=i,remember =i as lasti (initially 4),remember=value as lvalue (initially 12)] in 17,22,5,12
path[->](lasti) edge[bend right=45] node[midway,auto,swap]lvalue(i);
endtikzpicture
enddocument
answered Aug 21 at 21:14
AndréC
2,872728
edited Aug 22 at 8:57
answered Aug 21 at 21:14
AndréC
2,872728
answered Aug 21 at 21:14
AndréC
2,872728
answered Aug 21 at 21:14
AndréC
2,872728
2,872728
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
add a comment |Â
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
1
1
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
See above; answer to @Max
– Holger Karl
Aug 22 at 6:26
add a comment |Â
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f447009%2fhow-to-compute-the-key-of-a-pgfkeyssetvalue-command%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you by chance looking for an array? BTW, the node names can be numbers, so you could just give the nodes the same names as their contents.
– marmot
Aug 21 at 15:50
Numbers or pi makes no difference in MWE. Node names are actually not explicitly used in the keys; could be arbitrary; just the same for convenience. Not sure what you have in mind with array?
– Holger Karl
Aug 21 at 15:52
I use an array in my answer below. BTW Welcome to TeX.SE!
– marmot
Aug 21 at 16:01