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Re: plot variance(s)


>> Why doesn't p2 directly overlay the dots in p1?

ListPlot assumes the x-coordinates (when you don't specify them) are 1, ..., Length@data.

To get the right results, specify the x-coordinates:

data = Table[{x, Sin[x]}, {x, 0, 2*Pi}];
p1 = ListPlot[data, PlotStyle ->
      {Red, PointSize[0.03]}, DisplayFunction ->
      Identity];
p2 = Plot[Sin[x], {x, 0, 2*Pi},
     DisplayFunction -> Identity];
Show[p1, p2, DisplayFunction ->
     $DisplayFunction];

or

data = Table[Sin[x], {x, 0, 2*Pi}];
p1 = ListPlot[Transpose[{Range[0, 2*Pi], data}],
     PlotStyle -> {Red, PointSize[0.03]},
     DisplayFunction -> Identity];
p2 = Plot[Sin[x], {x, 0, 2*Pi},
     DisplayFunction -> Identity];
Show[p1, p2, DisplayFunction -> $DisplayFunction];

or just

DisplayTogether[Graphics[{PointSize[0.03], Red,
      Table[Point@{x, Sin[x]}, {x, 0, 2*Pi}]}],
    Plot[Sin[x], {x, 0, 2*Pi}]];

I don't think ListPlot is very useful, in general.

Bobby

On Sun, 31 Oct 2004 01:17:31 -0500 (EST), Steven Shippee <slshippee at comcast.net> wrote:

> Input1:  data = Table[Sin[x], {x, 0, 2\[Pi]}];
>
> Input 2:  p1 = ListPlot[data,
> PlotStyle -> {Red, PointSize[0.03]}, DisplayFunction -> Identity];
>
> Input 3:  p2 = Plot[Sin[x], {x, 0, 2\[Pi]}, DisplayFunction -> Identity];
>
> Input 4:  Show[p1, p2, DisplayFunction -> $DisplayFunction];
>
> Why doesn't p2 directly overlay the dots in p1?  Thanks in advance,
>
> Steven Shippee
>
> slshippee at comcast.net
>
> 360-493-8353
>
>
>
>
>
>



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