# Is it possible to change the direction of axis in pgfplots?

Is it possible to change the direction of one (or both) axis?

I could achieve the visual result by using tikz and compute the coordinate by myself but this is obviously too long for a complex picture.

In the following there is the point X = (1,1) in a,b coordinates (the angle between a and b is 2*pi/3). drawing using the following code

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\draw[->] (0,0) -- (-1,1.732050808) node[anchor=south] {$b$}; % (2*cos(2*pi/3),2*sin(2*pi/3))
\draw[->] (0,0) -- (2,0) node[anchor=west] {$a$};
\draw[dashed] (-0.5,0.866025404) -- (0.5,0.866025404);
\draw[dashed] (1,0) -- (0.5,0.866025404) node[anchor=west] {$X$};
\node[anchor=north] at (1,0) {$1$};
\node[anchor=east] at (-0.5,0.866025404) {$1$};
\end{tikzpicture}
\end{document}


I want the same result using pgfplots and using (1,1) as coordinates.

## Edit

Both solutions posted by percusse has the same problem, that is that TikZ takes as argument the angle in radians, not in degree. But that it's not a big problem.

After a check, this is the code:

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[y={(-0.5, 0.866025404)} ]
\draw[->] (0,0) -- (0,2) node[anchor=south] {$b$};
\draw[->] (0,0) -- (2,0) node[anchor=west] {$a$};
\draw[dashed] (1,0) -- (1,1);
\draw[dashed] (0,1) -- (1,1) node[anchor=west] {$X$};
\node[anchor=north] at (1,0) {$1$};
\node[anchor=east] at (0,1) {$1$};
\end{tikzpicture}
\end{document}


and this is the result: that is the visual result that I want. In his second solution instead the scale between the two axis is wrong.

If I try to implement the first solution with pgfplots I obtain nothing. This is the code:

\documentclass{standalone}

\usepackage{pgfplots}
\pgfplotsset{compat=1.15}
\begin{document}
\begin{tikzpicture}[y={(-0.5, 0.866025404)}]
\begin{axis}[axis lines=middle]
\draw[dashed] (1,0) -- (1,1);
\draw[dashed] (0,1) -- (1,1) node[anchor=west] {$X$};
\end{axis}
\end{tikzpicture}
\end{document}


And this is the result: There are two direct possibilities together with other hacks

The first being, changing the unit vector and use dimensionless coordinates x=<...>, y=<...> as an option to the tikzpicture environment.

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[y={({cos(120)}, {sin(120)})}]
\draw[->] (0,0) -- (-1,1) node[anchor=south] {$b$};
\draw[->] (0,0) -- (2,0) node[anchor=west] {$a$};
\draw[dashed] (-0.5,0.5) -- (0.5,0.5);
\draw[dashed] (1,0) -- (0.5,0.5) node[anchor=west] {$X$};
\node[anchor=north] at (1,0) {$1$};
\node[anchor=east] at (-0.5,0.5) {$1$};
\end{tikzpicture}
\end{document}


The other is applying a transformation to the coordinate system and use just like you would code in a cartesian CS.

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[cm={1,0,{0.5*cos(120)},1,(0,0)}]
\draw[->] (0,0) -- (-1,1) node[anchor=south] {$b$};
\draw[->] (0,0) -- (2,0) node[anchor=west] {$a$};
\draw[dashed] (-0.5,0.5) -- (0.5,0.5);
\draw[dashed] (1,0) -- (0.5,0.5) node[anchor=west] {$X$};
\node[anchor=north] at (1,0) {$1$};
\node[anchor=east] at (-0.5,0.5) {$1$};
\end{tikzpicture}
\end{document} It looks like I've made a mistake about the degree but you get the idea. Look at the manual for the explanation for each entry of cm key.

• There is no problem if the degree are wrong, I can fix it by myself. However can you please explain better the first possibility? I didn't understand it. – gvgramazio May 24 '18 at 13:53
• @gvgramazio Added another example for it. – percusse May 24 '18 at 13:56
• The mistake about degree is due to the fact that trigonometric functions in TikZ use radians, so it should be cos(2*pi/3). However there is still a point missing, the plot is not in scale now. – gvgramazio May 24 '18 at 13:57
• I updated the question. Your first solution works fine with tikz but I'm not able to implement it with pgfplots. Your second solution change the scale of between the two axis instead. – gvgramazio May 24 '18 at 16:51
• @gvgramazio In pgfplots you shouldn't be needing these. It has its own tools to handle these projections – percusse May 24 '18 at 17:07