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I want to check if this is a bug before I email WR about it.

I can't get ListPlot[] to make scatter plots over a very small area:

n = 40;
c = I + RandomReal[1, n]*(2^-48)*Exp[I*Pi*2*RandomReal[1, n]];
ListPlot[Transpose[{Re[c], Im[c]}]]

BadScatterPlot

Providing an explicit PlotRange returns the exact same plot:

ListPlot[Transpose[{Re[c], Im[c]}], PlotRange -> {{Min[Re[c]], Max[Re[c]]}, {Min[Im[c]], Max[Im[c]]}}]

Subtracting I from c as a workaround yields a plot, but I shouldn't have to do this:

ListPlot[Transpose[{Re[c], Im[c - I]}]]

cMinusI

Is this a bug, or is there some quirk of ListPlot[] that I'm not aware of? Thanks.

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  • $\begingroup$ At least when I try this, Im[c] evaluates to 1.0 for all c. Are you sure you're not in some sort of machine-precision-limited situation here? $\endgroup$
    – flip
    Jan 22, 2016 at 1:03
  • $\begingroup$ @flip Is it just displaying as 1? Try subtracting 1 from Im[c], and try Im[c-i]. This this was just a precision error, my second plot would fail. $\endgroup$
    – Jerry Guern
    Jan 22, 2016 at 1:08
  • $\begingroup$ In short, Im[c] - I is a complex number, however you may probably want to plot this: Im[c-I]. $\endgroup$
    – luyuwuli
    Jan 22, 2016 at 6:40
  • $\begingroup$ +1. Except for the typo of the input, it's a valuable question. Since 2^-48 > 2^-53, there is no reason for the lost of the precision. The plot range issue also happens to Plot etc.. Quite weird. $\endgroup$
    – luyuwuli
    Jan 22, 2016 at 7:34
  • $\begingroup$ @luyuwuli Thank you. But what typo in the input? Did you fix it? $\endgroup$
    – Jerry Guern
    Jan 22, 2016 at 7:38

1 Answer 1

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This is a feature of ListPlot. It tries to avoid generating graphics where all the ticks would look the same ("1", in this case.)

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