Skip to main content

graphics - What is wrong with triangle PlotMarkers in v.10.0.0?



Bug introduced in 10.0.0 and fixed in 10.0.1




In Mathematica 10.0.0 we have built-in graphical triangle PlotMarkers. Let us look closer on them:


ListLinePlot[{{Missing[]}, {{0, 0}}}, PlotTheme -> "Monochrome", 
ImageSize -> 20, Ticks -> False, AxesOrigin -> {0, 0},
BaseStyle -> {Magnification -> 10, Thickness -> Tiny}]

ListLinePlot[{{Missing[]}, {{0, 0}}},
PlotTheme -> {"OpenMarkersThick", "LargeLabels"}, ImageSize -> 20,
Ticks -> False, AxesOrigin -> {0, 0},

BaseStyle -> {Magnification -> 10, Thickness -> Tiny}]


plot1


plot2



It is clear that there is something wrong with the triangles. Is this functionality implemented correctly?



Answer



The triangle plot markers


It is natural to expect that the triangle marker is placed in such a way that its center of mass (center of circumcircle) coincides with the point it marks. That's how it is implemented in all major scientific plotting software, for example Origin:



plot markers in Origin 7.5


Some time ago I published my own implementation of triangle-based plot markers. Let us check how the new markers are implemented:


ListLinePlot[{{Missing[]}, {{0, 0}}}, PlotTheme -> "Monochrome", 
ImageSize -> 10, Ticks -> False, AxesOrigin -> {0, 0},
BaseStyle -> {Magnification -> 10, Thickness -> Tiny}]
%[[1, 2, 2, 2, -1]] // InputForm


screenshot


GeometricTransformation[Inset[Graphics[{<...>

Line[{Offset[{0., 2.7625}],
Offset[{-2.7625, -2.022290355909023}],
Offset[{2.7625, -2.022290355909023}],
Offset[{0., 2.7625}]}]}], {0., 0.}],
{{{0., 0.}}, {{0., 0.}}}]

Apart of the fact that the curve is not closed, the triangle is positioned in a strange way: the "center" is placed on the


2.022290355909023/(2.7625 + 2.022290355909023)



0.4226497308103742



part of the height of the triangle instead of expected 1/3 (the center of circumcircle). So current implementation is clearly wrong and leads to producing incorrect plots. Here is an example of correct implementation:


Graphics[{AbsoluteThickness[1], JoinedCurve[
Line[{Offset[{0, 2}], Offset[{Sqrt[3], -1}],
Offset[{-Sqrt[3], -1}]}], CurveClosed -> True]},
ImageSize -> 10, Axes -> True, Ticks -> False, AxesOrigin -> {0, 0},
BaseStyle -> {Magnification -> 10, Thickness -> Tiny}]



screenshot



The following is correct implementation of both empty and filled triangle plot markers of strictly identical sizes with consistent explicit control over their sizes and thickness:


emptyUpTriangle = 
Graphics[{AbsoluteThickness[absoluteThickness],
JoinedCurve[Line[{Offset[size {0, 2}], Offset[size {Sqrt[3], -1}],
Offset[size {-Sqrt[3], -1}]}], CurveClosed -> True]},
AlignmentPoint -> {0, 0}];
filledUpTriangle =
Graphics[{Triangle[{Offset[size {0, 2} + absoluteThickness {0, 1}],

Offset[size {Sqrt[3], -1} + absoluteThickness {Sqrt[3/4], -1/2}],
Offset[size {-Sqrt[3], -1} + absoluteThickness {-Sqrt[3/4], -1/2}]}]},
AlignmentPoint -> {0, 0}];
{emptyLeftTriangle, filledLeftTriangle, emptyDownTriangle,
filledDownTriangle, emptyRightTriangle, filledRightTriangle} =
Flatten[{emptyUpTriangle, filledUpTriangle} /. {x_?NumericQ, y_?NumericQ} :>
RotationTransform[#][{x, y}] & /@ {Pi/2, Pi/3, -Pi/2}];

SeedRandom[12]
ListLinePlot[Accumulate /@ RandomReal[3, {8, 10}] Range[8],

PlotMarkers -> {emptyUpTriangle, filledUpTriangle, emptyLeftTriangle,
filledLeftTriangle, emptyDownTriangle, filledDownTriangle,
emptyRightTriangle, filledRightTriangle}, AspectRatio -> 1]

plot


And here is an extended version which includes open triangles with white filling:


size = 4; absoluteThickness = 2;

triangle[Up, Empty] =
Graphics[{AbsoluteThickness[absoluteThickness],

JoinedCurve[Line[{Offset[size {0, 2}], Offset[size {Sqrt[3], -1}],
Offset[size {-Sqrt[3], -1}]}], CurveClosed -> True]},
AlignmentPoint -> {0, 0}];
triangle[Up, Filled] =
Graphics[{Triangle[{Offset[size {0, 2} + absoluteThickness {0, 1}],
Offset[size {Sqrt[3], -1} + absoluteThickness {Sqrt[3/4], -1/2}],
Offset[size {-Sqrt[3], -1} + absoluteThickness {-Sqrt[3/4], -1/2}]}]},
AlignmentPoint -> {0, 0}];
triangle[Up, Open] =
Graphics[{{White, Triangle[{Offset[size {0, 2}], Offset[size {Sqrt[3], -1}],

Offset[size {-Sqrt[3], -1}]}]}, {AbsoluteThickness[absoluteThickness],
JoinedCurve[Line[{Offset[size {0, 2}], Offset[size {Sqrt[3], -1}],
Offset[size {-Sqrt[3], -1}]}], CurveClosed -> True]}},
AlignmentPoint -> {0, 0}];
triangle[dir_: {Up, Right, Down, Left}, fill_: {Empty, Filled, Open}] :=
triangle[Up, fill] /. {x_?NumericQ, y_?NumericQ} :>
RotationTransform[dir /. {Right -> -Pi/2, Down -> Pi/3, Left -> Pi/2}][{x, y}]

pl = ListPlot[Flatten[Table[{{n, y}}, {y, Range[2]}, {n, 6}], 1],
PlotMarkers ->

Flatten@Table[triangle[dir, fill],
{dir, {Up, Right, Down, Left}}, {fill, {Empty, Filled, Open}}],
GridLines -> {Range[6], Range[2]},
PlotRange -> {{0, 7}, {0, 3}}, Axes -> False, Frame -> True]

plot2




Other plot markers


Not only triangle plot markers have problems:


ListLinePlot[{{#, 0}} & /@ Range[5], PlotTheme -> "Monochrome", 

ImageSize -> 70, Ticks -> False, Axes -> False, Frame -> True,
BaseStyle -> {Magnification -> 15, Thickness -> Tiny},
PlotRange -> {{0.5, 5.5}, All}, AspectRatio -> 1/10,
FrameTicks -> None, BaselinePosition -> Center,
GridLines -> {Range[5], {0}},
GridLinesStyle -> Directive[{Dashing[None], Gray}],
Method -> {"GridLinesInFront" -> True}]
Cases[%, g_Graphics :>
Show[g, ImageSize -> 8, BaseStyle -> {Magnification -> 16},
BaselinePosition -> Center], Infinity]

Cases[%[[4]], _Line, Infinity]


screenshot


{Line[{Offset[{-2.5, -2.5}],   Offset[{2.125, -2.125}], 
Offset[{2.125, 2.125}], Offset[{-2.125, 2.125}],
Offset[{-2.125, -2.125}]}]}

As one can see, the square starts from the point {-2.5, -2.5} and ends in {-2.125, -2.125}!


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...