Is it possible to position ticklabels on negative $y$ axis on its right side? Ticklabels on positive $y$ axis should remain as usual:
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$\begingroup$ Related: mathematica.stackexchange.com/q/2601/121 $\endgroup$– Mr.WizardCommented Jun 4, 2012 at 10:14
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$\begingroup$ @Mr.Wizard: These are Input/Output characteristics of scalar quantization. When these diagrams are small, the lebels and the curves tend to overlap! $\endgroup$– Bhaskar DeyCommented Jun 4, 2012 at 10:16
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3 Answers
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Here's one idea. Notice that the tick-marks are flipped as well.
p = Plot[Round[x], {x, -5, 5}, Exclusions -> None, PlotStyle -> Thick];
ticks = Ticks /. AbsoluteOptions[p];
{yticks, labels} =
Replace[
ticks[[2]],
{a_?Negative, b_, c_, x__} :> {a, Sow@Text[b, {0.3, a}];, -c, x},
1
] // Reap;
Show[p, Graphics[labels], Ticks -> {ticks[[1]], yticks}]
answered Jun 4, 2012 at 11:11
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$\begingroup$ +1 Nice idea! Why do + and - minor $y$-ticks have different lengths? And why does the $y$-tick at +3.5 seem longer than the one at +3.0? Are these rasterization artefacts? $\endgroup$ Commented Jun 4, 2012 at 11:53
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$\begingroup$ @István thanks. I believe it is a rasterization artifact as it appears even at large sizes. $\endgroup$ Commented Jun 4, 2012 at 12:01
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$\begingroup$ @Ajasja the OP asked for the y axis modification. You could do the same thing for only x (
ticks[[2]]->ticks[[1]]&{0.3, a}->{a, 0.3}) or for both in two steps. $\endgroup$ Commented Jun 4, 2012 at 12:29 -
$\begingroup$ Ahh, sorry, sloppy reading, I was just looking at the example image. +1. $\endgroup$– AjasjaCommented Jun 4, 2012 at 12:33
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Lots of magic numbers, no real scalability and rather specific solution for the given case, but it shows how you can build your own tick function easily from scratch. Since I don't think there is any way to finetune the Axes/AxesStyle/Frame options to suit your needs, this is the best I could suggest at the moment. Or you might want to use David Park's Presentations` package or the CustomTicks` package by Mark Caprio to specify aesthetic tick positions for any range (thanks Szabolcs).
{minX, maxX} = {minY, maxY} = {-5, 5}; (* range *)
{xO, yO} = {0, 0}; (* origo *)
d = .2; (* tick length *)
step = 1; (* tick step *)
offset = 1.5; (* tick label offset *)
Plot[IntegerPart@x, {x, minX, maxX}, Exclusions -> None,
PlotStyle -> {Black, Thick}, AspectRatio -> 1,
PlotRange -> {{minX, maxX}, {minY, maxY}}, Axes -> False,
Epilog -> {
Black, Line[{{minX, yO}, {maxX, yO}}], Line[{{xO, minY}, {xO, maxY}}],
Table[{
Line@{{i, yO}, {i, yO + d}},
Text[i, {i, yO + d}, {0, -offset}]
}, {i, minX, xO - 1, step}],(* -x *)
Table[{
Line@{{i, yO}, {i, yO - d}},
Text[i, {i, yO - d}, {0, offset}]
}, {i, xO + 1, maxX, step}],(* +x *)
Table[{
Line@{{xO, i}, {xO + d, i}},
Text[i, {xO + d, i}, {-offset, 0}]
}, {i, minY, yO - 1, step}],(* -y *)
Table[{
Line@{{xO, i}, {xO - d, i}},
Text[i, {xO - d, i}, {offset, 0}]
}, {i, yO + 1, maxY, step}] (* +y *)
}, PlotRangePadding -> [email protected]]
answered Jun 4, 2012 at 10:36
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$\begingroup$ You could use the CustomTicks package to generate tick positions, it'd make the solution a little easier to modify/reuse. $\endgroup$– SzabolcsCommented Jun 4, 2012 at 10:47
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$\begingroup$ @Szabolcs: I never used that package, but it seems very useful. Actually, I think it worths its own answer, and would be a perfect starting point for people willing to learn how to use it! So if you have any knowledge, please post your CustomTicks-based solution! $\endgroup$ Commented Jun 4, 2012 at 10:55
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$\begingroup$ @Szabolcs I agree, your suggestion would make a good answer on it's own. $\endgroup$– AjasjaCommented Jun 4, 2012 at 11:43
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This needs some tweaking, but could be a starting point for a more general solution. The idea is to plot the bottom left quadrant of the plot with a Frame top and right, and the top right quadrant of the plot with a Frame bottom and left. Then Inset those into the complete plot.
bottomleft=Plot[Sin[x],{x,-4,4},Frame->{False,False,True,True},
FrameTicks->All,PlotRange->{{Automatic,0},{Automatic,0}}];
topright=Plot[Sin[x],{x,-4,4},Frame->{True,True,False,False},
FrameTicks->All,PlotRange->{{0,Automatic},{0,Automatic}}];
Plot[Sin[x],{x,-4,4},Ticks->False,
Epilog->{Inset[bottomleft,{0,0},{0,0},Scaled[0.58]],Inset[topright,{0,0},{0,0},Scaled[0.58]]}]
Clearly there is a problem with all the ticks at the origin, and the insets had to be scaled by trial and error, but there might be a way to automate that.
answered Jun 4, 2012 at 11:52
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