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Say one had a function that was itself asymmetric about the abscissa, but one desired to plot it on a symmetric ordinate. How can this be achieved with Plot automatically (without separately computing the limits and providing them to Plot)?

Example The plot of 1/2 serves as a sufficient example.

Table[
  Plot[ 
    1/2, 
    { x, 0, 1 }, 
    Frame -> { False, True }[[i]], 
    Axes -> { True, False }[[i]] 
  ],
  { i, 1, 2 } 
 ]

asymmetric plots

But we might want the PlotRange to go from $+1$ to $-1$ instead, like:

symmetric plot

I did this with the option PlotRange, but short of finding the maximum of the absolute value of the function and setting PlotRange, I don't know how to automate this.

Specifically, is it possible to use Plot in a native manner to do this?

Solution (based on the answer by Szabolcs) Here's a little function that implements the answer.

plotSym[ plot_ ] := 
  plot //
    { 
      #, 
      First[PlotRange /. AbsoluteOptions[#, PlotRange]], 
      Max@Abs@Last[PlotRange /. AbsoluteOptions[#, PlotRange]] 
    } & //
      Show[ #[[1]], PlotRange -> {#[[2]], (1.2*{-#[[3]], #[[3]]})} ] &  

Note: I added the factor of 1.2 as a way to give the plot a little more room. It can be anything nonzero (including the likely more-desirable 1).

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1 Answer 1

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You can try extracting the PlotRange using AbsoluteOptions. You need to use AbsoluteOptions, not simply Options, to make sure All gets translated to a numerical value.

g = Plot[Sin[x]^2, {x, 0, Pi}, PlotRange -> All]

max = Max@Abs@Last[PlotRange /. AbsoluteOptions[g, PlotRange]]

Show[g, PlotRange -> {All, 1.1 {-max, max}}]
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  • $\begingroup$ Nice, thank you Szabolcs. I wrote a little function based on your answer that I'll put in my question. $\endgroup$ Jun 30, 2014 at 20:11

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