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Check if a point is right or left of another one in TikZ


Extract x, y coordinate of an arbitrary point in TikZRotate a node but not its content: the case of the ellipse decorationNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingReferencing locations within a text boxLine up nested tikz enviroments or how to get rid of themTikz-qtree : graphs in the nodesHow to draw more than 8 edges to one node in tikz-er2Using tikz Calc package to add cordinatesIn Tikz is there a way to make Tikz images begin “exactly” at the leftmost point in the page?













3















Trying to implement a handy way to check if a TikZ point is right or left of another one, I came up with the following MWE.



documentclass[border=1mm, tikz]standalone

makeatletter
% prints 1 if #1.center is right than #2.center, 0 otherwise
newcommandisRight[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center
pgfmathparsegreater(pgf@x,0)pgfmathresult


% 1 if #1.center is left than #2.center, 0 otherwise -> result in pgfmathresult
newcommandcheckIfLeft[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center%
pgfmathparseless(pgf@x,0)

makeatother

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;

checkIfLeftAB
ifnumpgfmathresult=1
node at (0.5,1)A is left of B;
fi

%ifnumisRightBA=1
% node at (0.5,-1)B is right of A;
%fi
endtikzpicture
enddocument


but I am quite unsatisfied with it. What I would like to have is something that may be used in a ifnum ... fi construct, as in the commented code (and I would like not to use packages beyond what pgf loads).



Here my questions:



  1. In the commented code, what is exactly happening in the ifnum expansion, which makes the compilation fail?

  2. How can I fix the isRight command in order to be able to use it together with ifnum?

  3. If question 2 is tricky, how can I affect the ifnum expansion to achieve what I wish?

  4. Is there in pgf a smarter/more straightforward way to check if a TikZ point is left/right of another point? My final goal would be to draw something only if a point is right/left of another.

Bonus question:



  1. How (and where) is pdfstrcmp implemented, since it is harmless to use it in a ifnum ... fi construct?









share|improve this question






















  • your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

    – Ulrike Fischer
    Apr 2 at 12:46












  • Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

    – Axel Krypton
    Apr 2 at 13:14











  • You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

    – Ulrike Fischer
    Apr 2 at 13:24















3















Trying to implement a handy way to check if a TikZ point is right or left of another one, I came up with the following MWE.



documentclass[border=1mm, tikz]standalone

makeatletter
% prints 1 if #1.center is right than #2.center, 0 otherwise
newcommandisRight[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center
pgfmathparsegreater(pgf@x,0)pgfmathresult


% 1 if #1.center is left than #2.center, 0 otherwise -> result in pgfmathresult
newcommandcheckIfLeft[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center%
pgfmathparseless(pgf@x,0)

makeatother

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;

checkIfLeftAB
ifnumpgfmathresult=1
node at (0.5,1)A is left of B;
fi

%ifnumisRightBA=1
% node at (0.5,-1)B is right of A;
%fi
endtikzpicture
enddocument


but I am quite unsatisfied with it. What I would like to have is something that may be used in a ifnum ... fi construct, as in the commented code (and I would like not to use packages beyond what pgf loads).



Here my questions:



  1. In the commented code, what is exactly happening in the ifnum expansion, which makes the compilation fail?

  2. How can I fix the isRight command in order to be able to use it together with ifnum?

  3. If question 2 is tricky, how can I affect the ifnum expansion to achieve what I wish?

  4. Is there in pgf a smarter/more straightforward way to check if a TikZ point is left/right of another point? My final goal would be to draw something only if a point is right/left of another.

Bonus question:



  1. How (and where) is pdfstrcmp implemented, since it is harmless to use it in a ifnum ... fi construct?









share|improve this question






















  • your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

    – Ulrike Fischer
    Apr 2 at 12:46












  • Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

    – Axel Krypton
    Apr 2 at 13:14











  • You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

    – Ulrike Fischer
    Apr 2 at 13:24













3












3








3








Trying to implement a handy way to check if a TikZ point is right or left of another one, I came up with the following MWE.



documentclass[border=1mm, tikz]standalone

makeatletter
% prints 1 if #1.center is right than #2.center, 0 otherwise
newcommandisRight[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center
pgfmathparsegreater(pgf@x,0)pgfmathresult


% 1 if #1.center is left than #2.center, 0 otherwise -> result in pgfmathresult
newcommandcheckIfLeft[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center%
pgfmathparseless(pgf@x,0)

makeatother

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;

checkIfLeftAB
ifnumpgfmathresult=1
node at (0.5,1)A is left of B;
fi

%ifnumisRightBA=1
% node at (0.5,-1)B is right of A;
%fi
endtikzpicture
enddocument


but I am quite unsatisfied with it. What I would like to have is something that may be used in a ifnum ... fi construct, as in the commented code (and I would like not to use packages beyond what pgf loads).



Here my questions:



  1. In the commented code, what is exactly happening in the ifnum expansion, which makes the compilation fail?

  2. How can I fix the isRight command in order to be able to use it together with ifnum?

  3. If question 2 is tricky, how can I affect the ifnum expansion to achieve what I wish?

  4. Is there in pgf a smarter/more straightforward way to check if a TikZ point is left/right of another point? My final goal would be to draw something only if a point is right/left of another.

Bonus question:



  1. How (and where) is pdfstrcmp implemented, since it is harmless to use it in a ifnum ... fi construct?









share|improve this question














Trying to implement a handy way to check if a TikZ point is right or left of another one, I came up with the following MWE.



documentclass[border=1mm, tikz]standalone

makeatletter
% prints 1 if #1.center is right than #2.center, 0 otherwise
newcommandisRight[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center
pgfmathparsegreater(pgf@x,0)pgfmathresult


% 1 if #1.center is left than #2.center, 0 otherwise -> result in pgfmathresult
newcommandcheckIfLeft[2]
%pgfpointdiffab gives b-a
pgfpointdiffpgfpointanchor#2centerpgfpointanchor#1center%
pgfmathparseless(pgf@x,0)

makeatother

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;

checkIfLeftAB
ifnumpgfmathresult=1
node at (0.5,1)A is left of B;
fi

%ifnumisRightBA=1
% node at (0.5,-1)B is right of A;
%fi
endtikzpicture
enddocument


but I am quite unsatisfied with it. What I would like to have is something that may be used in a ifnum ... fi construct, as in the commented code (and I would like not to use packages beyond what pgf loads).



Here my questions:



  1. In the commented code, what is exactly happening in the ifnum expansion, which makes the compilation fail?

  2. How can I fix the isRight command in order to be able to use it together with ifnum?

  3. If question 2 is tricky, how can I affect the ifnum expansion to achieve what I wish?

  4. Is there in pgf a smarter/more straightforward way to check if a TikZ point is left/right of another point? My final goal would be to draw something only if a point is right/left of another.

Bonus question:



  1. How (and where) is pdfstrcmp implemented, since it is harmless to use it in a ifnum ... fi construct?






tikz-pgf expansion






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 2 at 12:01









Axel KryptonAxel Krypton

477211




477211












  • your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

    – Ulrike Fischer
    Apr 2 at 12:46












  • Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

    – Axel Krypton
    Apr 2 at 13:14











  • You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

    – Ulrike Fischer
    Apr 2 at 13:24

















  • your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

    – Ulrike Fischer
    Apr 2 at 12:46












  • Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

    – Axel Krypton
    Apr 2 at 13:14











  • You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

    – Ulrike Fischer
    Apr 2 at 13:24
















your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

– Ulrike Fischer
Apr 2 at 12:46






your commands are not expandable (they assign a number to pgfmathresult) and will never work in a ifnum. pdfstrcmp is a primitive.

– Ulrike Fischer
Apr 2 at 12:46














Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

– Axel Krypton
Apr 2 at 13:14





Thanks for the comment. I find it a bit cryptic, though. Do you want to say that I should give up and what I got so far is the best I can have? The fact that pdfstrcmp is a primitive clarifies why I was not finding its implementation... I should have had a look to the pdftex manual before.

– Axel Krypton
Apr 2 at 13:14













You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

– Ulrike Fischer
Apr 2 at 13:24





You will have to give up one part - either ifnum or pgfmath commands. You could implement something working with ifnum by using e.g. zrefsavepos (needs two compilations) but not with pgfmathparse.

– Ulrike Fischer
Apr 2 at 13:24










3 Answers
3






active

oldest

votes


















6














Here is some low level code based on this answer that do not use calc.



documentclass[tikz,border=7pt]standalone
newififleft
makeatletter
defisleft(#1)of(#2)?%
tikz@scan@one@pointpgfutil@firstofone(#1)relax%
pgf@xa=pgf@x%
tikz@scan@one@pointpgfutil@firstofone(#2)relax%
ifdimpgf@xa<pgf@xrelaxlefttrueelseleftfalsefi

makeatother

begindocument
begintikzpicture[nodes=circle]
node[draw] (A) at (0,0) A;
node[draw] (B) at (2,1) right of A;
isleft(A)of(B)?
path (A) ifleft edge[latex-] (B) else edge[-latex] (B) fi ;
foreach~in1,...,70
path[ultra thin] (180*rand:1) coordinate(N)
pgfextraisleft(N)of(A)?
ifleft [red] else [blue] fi node[scale=2].
(N) ifleft edge[latex-] (A) else edge[-latex] (A) fi;
endtikzpicture
enddocument


enter image description here






share|improve this answer

























  • I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

    – Axel Krypton
    Apr 3 at 8:55


















6














How about



documentclass[border=1mm, tikz]standalone
usetikzlibrarycalc
tikzsetif left/.style n args=4insert path=%
let p1=($#1-#2$) in ifdimx1<0pt
#3
else
#4
fi

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;
path[if left=(A)(B)(0.5,1) node A is left of B];
path[if left=(B)(A)(0.5,-1) node B is left of A];
endtikzpicture
enddocument


enter image description here






share|improve this answer























  • This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

    – Axel Krypton
    Apr 3 at 8:42











  • @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

    – marmot
    Apr 3 at 13:46











  • @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

    – marmot
    2 days ago











  • In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

    – Axel Krypton
    2 days ago











  • And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

    – Axel Krypton
    2 days ago


















0














Just for future record, I want to share a tiny elaboration of what Kpym suggested in his answer. Just to avoid having the newif, it is possible to supply two more arguments to the macro. Of course, having the ifleft gives versatility in the usage (as nicely demonstrated in the Kpym's answer), but maybe this is not really needed (like in my case).



documentclass[border=1mm, tikz]standalone

makeatletter
longdefIfLeft(#1)of(#2)#3#4%
tikz@scan@one@pointpgfutil@firstofone(#1)relax%
pgf@xa=pgf@x%
tikz@scan@one@pointpgfutil@firstofone(#2)relax%
ifdimpgf@xa<pgf@xrelax#3else#4fi

makeatother

begindocument

begintikzpicture
node[draw] (A) at (0,0) A;
node[draw] (B) at (1,0) B;
node[draw] (C) at (2,0) C;

IfLeft(A)of(B)%
foreach n in 10,20,...,90
fill[red!n!yellow] (0.5,0) ++(n/100,0.5) circle(1mm);

%
foreach n in 10,20,...,90
fill[black!n!gray] (0.5,0) ++(n/100,-0.5) circle(1mm);



IfLeft(C)of(B)%
foreach n in 10,20,...,90
fill[black!n!gray] (0.5,0) ++(n/100,0.5) circle(1mm);

%
foreach n in 10,20,...,90
fill[blue!n!magenta] (0.5,0) ++(n/100,-0.5) circle(1mm);


endtikzpicture
enddocument



 



    enter image description here






share|improve this answer























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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    Here is some low level code based on this answer that do not use calc.



    documentclass[tikz,border=7pt]standalone
    newififleft
    makeatletter
    defisleft(#1)of(#2)?%
    tikz@scan@one@pointpgfutil@firstofone(#1)relax%
    pgf@xa=pgf@x%
    tikz@scan@one@pointpgfutil@firstofone(#2)relax%
    ifdimpgf@xa<pgf@xrelaxlefttrueelseleftfalsefi

    makeatother

    begindocument
    begintikzpicture[nodes=circle]
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (2,1) right of A;
    isleft(A)of(B)?
    path (A) ifleft edge[latex-] (B) else edge[-latex] (B) fi ;
    foreach~in1,...,70
    path[ultra thin] (180*rand:1) coordinate(N)
    pgfextraisleft(N)of(A)?
    ifleft [red] else [blue] fi node[scale=2].
    (N) ifleft edge[latex-] (A) else edge[-latex] (A) fi;
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer

























    • I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

      – Axel Krypton
      Apr 3 at 8:55















    6














    Here is some low level code based on this answer that do not use calc.



    documentclass[tikz,border=7pt]standalone
    newififleft
    makeatletter
    defisleft(#1)of(#2)?%
    tikz@scan@one@pointpgfutil@firstofone(#1)relax%
    pgf@xa=pgf@x%
    tikz@scan@one@pointpgfutil@firstofone(#2)relax%
    ifdimpgf@xa<pgf@xrelaxlefttrueelseleftfalsefi

    makeatother

    begindocument
    begintikzpicture[nodes=circle]
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (2,1) right of A;
    isleft(A)of(B)?
    path (A) ifleft edge[latex-] (B) else edge[-latex] (B) fi ;
    foreach~in1,...,70
    path[ultra thin] (180*rand:1) coordinate(N)
    pgfextraisleft(N)of(A)?
    ifleft [red] else [blue] fi node[scale=2].
    (N) ifleft edge[latex-] (A) else edge[-latex] (A) fi;
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer

























    • I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

      – Axel Krypton
      Apr 3 at 8:55













    6












    6








    6







    Here is some low level code based on this answer that do not use calc.



    documentclass[tikz,border=7pt]standalone
    newififleft
    makeatletter
    defisleft(#1)of(#2)?%
    tikz@scan@one@pointpgfutil@firstofone(#1)relax%
    pgf@xa=pgf@x%
    tikz@scan@one@pointpgfutil@firstofone(#2)relax%
    ifdimpgf@xa<pgf@xrelaxlefttrueelseleftfalsefi

    makeatother

    begindocument
    begintikzpicture[nodes=circle]
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (2,1) right of A;
    isleft(A)of(B)?
    path (A) ifleft edge[latex-] (B) else edge[-latex] (B) fi ;
    foreach~in1,...,70
    path[ultra thin] (180*rand:1) coordinate(N)
    pgfextraisleft(N)of(A)?
    ifleft [red] else [blue] fi node[scale=2].
    (N) ifleft edge[latex-] (A) else edge[-latex] (A) fi;
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer















    Here is some low level code based on this answer that do not use calc.



    documentclass[tikz,border=7pt]standalone
    newififleft
    makeatletter
    defisleft(#1)of(#2)?%
    tikz@scan@one@pointpgfutil@firstofone(#1)relax%
    pgf@xa=pgf@x%
    tikz@scan@one@pointpgfutil@firstofone(#2)relax%
    ifdimpgf@xa<pgf@xrelaxlefttrueelseleftfalsefi

    makeatother

    begindocument
    begintikzpicture[nodes=circle]
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (2,1) right of A;
    isleft(A)of(B)?
    path (A) ifleft edge[latex-] (B) else edge[-latex] (B) fi ;
    foreach~in1,...,70
    path[ultra thin] (180*rand:1) coordinate(N)
    pgfextraisleft(N)of(A)?
    ifleft [red] else [blue] fi node[scale=2].
    (N) ifleft edge[latex-] (A) else edge[-latex] (A) fi;
    endtikzpicture
    enddocument


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Apr 2 at 19:33

























    answered Apr 2 at 18:00









    KpymKpym

    17.8k24191




    17.8k24191












    • I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

      – Axel Krypton
      Apr 3 at 8:55

















    • I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

      – Axel Krypton
      Apr 3 at 8:55
















    I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

    – Axel Krypton
    Apr 3 at 8:55





    I like your idea a lot! Indeed it is perfect for my needs, up to a tiny adjustment. The comments to the question clarify the other points.

    – Axel Krypton
    Apr 3 at 8:55











    6














    How about



    documentclass[border=1mm, tikz]standalone
    usetikzlibrarycalc
    tikzsetif left/.style n args=4insert path=%
    let p1=($#1-#2$) in ifdimx1<0pt
    #3
    else
    #4
    fi

    begindocument

    begintikzpicture
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (1,0) B;
    path[if left=(A)(B)(0.5,1) node A is left of B];
    path[if left=(B)(A)(0.5,-1) node B is left of A];
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer























    • This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

      – Axel Krypton
      Apr 3 at 8:42











    • @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

      – marmot
      Apr 3 at 13:46











    • @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

      – marmot
      2 days ago











    • In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

      – Axel Krypton
      2 days ago











    • And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

      – Axel Krypton
      2 days ago















    6














    How about



    documentclass[border=1mm, tikz]standalone
    usetikzlibrarycalc
    tikzsetif left/.style n args=4insert path=%
    let p1=($#1-#2$) in ifdimx1<0pt
    #3
    else
    #4
    fi

    begindocument

    begintikzpicture
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (1,0) B;
    path[if left=(A)(B)(0.5,1) node A is left of B];
    path[if left=(B)(A)(0.5,-1) node B is left of A];
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer























    • This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

      – Axel Krypton
      Apr 3 at 8:42











    • @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

      – marmot
      Apr 3 at 13:46











    • @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

      – marmot
      2 days ago











    • In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

      – Axel Krypton
      2 days ago











    • And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

      – Axel Krypton
      2 days ago













    6












    6








    6







    How about



    documentclass[border=1mm, tikz]standalone
    usetikzlibrarycalc
    tikzsetif left/.style n args=4insert path=%
    let p1=($#1-#2$) in ifdimx1<0pt
    #3
    else
    #4
    fi

    begindocument

    begintikzpicture
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (1,0) B;
    path[if left=(A)(B)(0.5,1) node A is left of B];
    path[if left=(B)(A)(0.5,-1) node B is left of A];
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer













    How about



    documentclass[border=1mm, tikz]standalone
    usetikzlibrarycalc
    tikzsetif left/.style n args=4insert path=%
    let p1=($#1-#2$) in ifdimx1<0pt
    #3
    else
    #4
    fi

    begindocument

    begintikzpicture
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (1,0) B;
    path[if left=(A)(B)(0.5,1) node A is left of B];
    path[if left=(B)(A)(0.5,-1) node B is left of A];
    endtikzpicture
    enddocument


    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Apr 2 at 13:24









    marmotmarmot

    115k5146277




    115k5146277












    • This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

      – Axel Krypton
      Apr 3 at 8:42











    • @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

      – marmot
      Apr 3 at 13:46











    • @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

      – marmot
      2 days ago











    • In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

      – Axel Krypton
      2 days ago











    • And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

      – Axel Krypton
      2 days ago

















    • This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

      – Axel Krypton
      Apr 3 at 8:42











    • @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

      – marmot
      Apr 3 at 13:46











    • @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

      – marmot
      2 days ago











    • In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

      – Axel Krypton
      2 days ago











    • And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

      – Axel Krypton
      2 days ago
















    This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

    – Axel Krypton
    Apr 3 at 8:42





    This is for sure a good idea and I indeed thought about something in this direction at the beginning. However, if I have to draw a lot (nodes, paths, for-loops, etc.) in the if-else branches, then this approach is not handy.

    – Axel Krypton
    Apr 3 at 8:42













    @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

    – marmot
    Apr 3 at 13:46





    @AxelKrypton I actually beg to disagree. You can definitely use this in loops and so on. Can you please tell me a scenario in which you think this has disadvantages?

    – marmot
    Apr 3 at 13:46













    @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

    – marmot
    2 days ago





    @AxelKrypton Just add path foreach X in 1,...,100 (rnd*360:1) coordinate(aux) [if left=(aux)(A) (A) edge[red,-latex] (aux)(A) edge[blue,-latex] (aux)]; to my code. I did not have to adjust anything. You can pick whatever answer you like. But making statements about my code that do not apply just to make Kpym happy is IMHO not useful for other users who have a similar problem and want a simple solution without any makeatletter and so on stuff. All the foreach's of your own post can be done in the very same way.

    – marmot
    2 days ago













    In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

    – Axel Krypton
    2 days ago





    In my comment I actually recognized that your approach is good, I do not know why you think my comment is against you in favour of Kpym. I simply said that if I have to draw a lot, like many nodes, paths, for loops, whatever then your tikzstyle is not as handy as having an if-clause.

    – Axel Krypton
    2 days ago













    And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

    – Axel Krypton
    2 days ago





    And this does not mean that your solution would not work, just that having something very similar to a if ... fi block makes the conditional code stand out and this in my opinion also increases readability. It was simply better in my use case.

    – Axel Krypton
    2 days ago











    0














    Just for future record, I want to share a tiny elaboration of what Kpym suggested in his answer. Just to avoid having the newif, it is possible to supply two more arguments to the macro. Of course, having the ifleft gives versatility in the usage (as nicely demonstrated in the Kpym's answer), but maybe this is not really needed (like in my case).



    documentclass[border=1mm, tikz]standalone

    makeatletter
    longdefIfLeft(#1)of(#2)#3#4%
    tikz@scan@one@pointpgfutil@firstofone(#1)relax%
    pgf@xa=pgf@x%
    tikz@scan@one@pointpgfutil@firstofone(#2)relax%
    ifdimpgf@xa<pgf@xrelax#3else#4fi

    makeatother

    begindocument

    begintikzpicture
    node[draw] (A) at (0,0) A;
    node[draw] (B) at (1,0) B;
    node[draw] (C) at (2,0) C;

    IfLeft(A)of(B)%
    foreach n in 10,20,...,90
    fill[red!n!yellow] (0.5,0) ++(n/100,0.5) circle(1mm);

    %
    foreach n in 10,20,...,90
    fill[black!n!gray] (0.5,0) ++(n/100,-0.5) circle(1mm);



    IfLeft(C)of(B)%
    foreach n in 10,20,...,90
    fill[black!n!gray] (0.5,0) ++(n/100,0.5) circle(1mm);

    %
    foreach n in 10,20,...,90
    fill[blue!n!magenta] (0.5,0) ++(n/100,-0.5) circle(1mm);


    endtikzpicture
    enddocument



     



        enter image description here






    share|improve this answer



























      0














      Just for future record, I want to share a tiny elaboration of what Kpym suggested in his answer. Just to avoid having the newif, it is possible to supply two more arguments to the macro. Of course, having the ifleft gives versatility in the usage (as nicely demonstrated in the Kpym's answer), but maybe this is not really needed (like in my case).



      documentclass[border=1mm, tikz]standalone

      makeatletter
      longdefIfLeft(#1)of(#2)#3#4%
      tikz@scan@one@pointpgfutil@firstofone(#1)relax%
      pgf@xa=pgf@x%
      tikz@scan@one@pointpgfutil@firstofone(#2)relax%
      ifdimpgf@xa<pgf@xrelax#3else#4fi

      makeatother

      begindocument

      begintikzpicture
      node[draw] (A) at (0,0) A;
      node[draw] (B) at (1,0) B;
      node[draw] (C) at (2,0) C;

      IfLeft(A)of(B)%
      foreach n in 10,20,...,90
      fill[red!n!yellow] (0.5,0) ++(n/100,0.5) circle(1mm);

      %
      foreach n in 10,20,...,90
      fill[black!n!gray] (0.5,0) ++(n/100,-0.5) circle(1mm);



      IfLeft(C)of(B)%
      foreach n in 10,20,...,90
      fill[black!n!gray] (0.5,0) ++(n/100,0.5) circle(1mm);

      %
      foreach n in 10,20,...,90
      fill[blue!n!magenta] (0.5,0) ++(n/100,-0.5) circle(1mm);


      endtikzpicture
      enddocument



       



          enter image description here






      share|improve this answer

























        0












        0








        0







        Just for future record, I want to share a tiny elaboration of what Kpym suggested in his answer. Just to avoid having the newif, it is possible to supply two more arguments to the macro. Of course, having the ifleft gives versatility in the usage (as nicely demonstrated in the Kpym's answer), but maybe this is not really needed (like in my case).



        documentclass[border=1mm, tikz]standalone

        makeatletter
        longdefIfLeft(#1)of(#2)#3#4%
        tikz@scan@one@pointpgfutil@firstofone(#1)relax%
        pgf@xa=pgf@x%
        tikz@scan@one@pointpgfutil@firstofone(#2)relax%
        ifdimpgf@xa<pgf@xrelax#3else#4fi

        makeatother

        begindocument

        begintikzpicture
        node[draw] (A) at (0,0) A;
        node[draw] (B) at (1,0) B;
        node[draw] (C) at (2,0) C;

        IfLeft(A)of(B)%
        foreach n in 10,20,...,90
        fill[red!n!yellow] (0.5,0) ++(n/100,0.5) circle(1mm);

        %
        foreach n in 10,20,...,90
        fill[black!n!gray] (0.5,0) ++(n/100,-0.5) circle(1mm);



        IfLeft(C)of(B)%
        foreach n in 10,20,...,90
        fill[black!n!gray] (0.5,0) ++(n/100,0.5) circle(1mm);

        %
        foreach n in 10,20,...,90
        fill[blue!n!magenta] (0.5,0) ++(n/100,-0.5) circle(1mm);


        endtikzpicture
        enddocument



         



            enter image description here






        share|improve this answer













        Just for future record, I want to share a tiny elaboration of what Kpym suggested in his answer. Just to avoid having the newif, it is possible to supply two more arguments to the macro. Of course, having the ifleft gives versatility in the usage (as nicely demonstrated in the Kpym's answer), but maybe this is not really needed (like in my case).



        documentclass[border=1mm, tikz]standalone

        makeatletter
        longdefIfLeft(#1)of(#2)#3#4%
        tikz@scan@one@pointpgfutil@firstofone(#1)relax%
        pgf@xa=pgf@x%
        tikz@scan@one@pointpgfutil@firstofone(#2)relax%
        ifdimpgf@xa<pgf@xrelax#3else#4fi

        makeatother

        begindocument

        begintikzpicture
        node[draw] (A) at (0,0) A;
        node[draw] (B) at (1,0) B;
        node[draw] (C) at (2,0) C;

        IfLeft(A)of(B)%
        foreach n in 10,20,...,90
        fill[red!n!yellow] (0.5,0) ++(n/100,0.5) circle(1mm);

        %
        foreach n in 10,20,...,90
        fill[black!n!gray] (0.5,0) ++(n/100,-0.5) circle(1mm);



        IfLeft(C)of(B)%
        foreach n in 10,20,...,90
        fill[black!n!gray] (0.5,0) ++(n/100,0.5) circle(1mm);

        %
        foreach n in 10,20,...,90
        fill[blue!n!magenta] (0.5,0) ++(n/100,-0.5) circle(1mm);


        endtikzpicture
        enddocument



         



            enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 3 at 8:53









        Axel KryptonAxel Krypton

        477211




        477211



























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