Re: Controling the color of line segments in a Log plot
- To: mathgroup at smc.vnet.net
- Subject: [mg90157] Re: Controling the color of line segments in a Log plot
- From: "David Park" <djmpark at comcast.net>
- Date: Tue, 1 Jul 2008 06:57:53 -0400 (EDT)
- References: <g4a6r8$8s5$1@smc.vnet.net>
Using your sample point list:
yValues = Table[10^n, {n, 10}]
xy = Transpose[{Range[Length[yValues]], yValues}]
One might have hoped that the following construction would work:
ListLogPlot[xy, Joined -> True,
ColorFunction -> Function[{x, y}, If[3 <= x <= 6, Red, Black]]]
But it doesn't. Perhaps someone will know how to make that approach work.
So I go to a more detailed approach using the Presentations package. This
parallels your second approach but without all the level jumping. First, we
need a format function for the y tick labels.
Needs["Presentations`Master`"]
nformat[n_] := Switch[n,
1, 1,
10, 10,
_, Superscript[10, Log[10, n]]]
Then we construct custom ticks and grids. I'm going to use a Frame plot with
a light grid.
yticks = CustomTicks[Log, {1, 10, {1}, {2, 3, 4, 5, 6, 7, 8, 9}},
CTNumberFunction -> (nformat[#] &)];
ygrid = CustomGridLines[Log, {1, 10, Range[9]}, {LightGray}];
xgrid = CustomGridLines[Identity, {1, 10, 1}, {LightGray}];
Then we can draw the custom plot, using separate draw statements for the
regions of the curve:
Draw2D[
{ListLogDraw[Take[xy, {1, 3}], Joined -> True],
ListLogDraw[Take[xy, {6, 10}], Joined -> True],
Red,
ListLogDraw[Take[xy, {3, 6}], Joined -> True]},
AspectRatio -> 1,
PlotRange -> {{1, 10}, {Log[10], Log[10^10]}},
Frame -> True,
FrameTicks -> {{yticks, yticks // NoTickLabels}, {Automatic,
Automatic}},
GridLines -> {xgrid, ygrid},
ImageSize -> 500]
I could have used PlotRange -> {{1, 10}, All} but I gave explicit values to
the y plot range just to show that, in fact, Mathematica DOES use the
natural log for the underlying plot values. Also notice that I used natural
log in the CustomTicks and CustomGrid functions as the routine to get from
the label value to the underlying plot value.
--
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
"Donald DuBois" <donabc at comcast.net> wrote in message
news:g4a6r8$8s5$1 at smc.vnet.net...
>I would like to control the color of line segments in a Log plot. I have
>three Questions (A,B,C) regarding this problem.
>
> A simple example: I start with the data yValues = Table[10^n, {n,10}] -
> which should also be the Y-axis tick values - and use ListLogPlot to graph
> it.
>
> yValues = Table[10^n, {n, 10}]
> xy = Transpose[{Range[Length[yValues]], yValues}]
> ListLogPlot[xy, Joined -> True]
>
>
> Question A . Does anyone find it peculiar that ListLogPlot seems to be
> taking
> Log to base 10 as a default as in the above plot when Log by itself
> defaults to base E? This is FINE for me, as I want it to be to the base
> 10. But still,
> does anyone know why this is so or am I misconstruing something?
>
>
> Question B.
> I would like to change the color to Red of the line segment with X values
> going from
> 3 to 6. (In general, I would like to change the segment color to Red for
> any number of segments of the same Log plot.) Is there a way to do this at
> the ListLogPlot level (i.e.giving directives directly to ListLogPlot)?
> That would be the simplest solution.
>
> I can change the color of this part of the graph by going down to the
> Graphics element level using Line as in the following, but this is
> tedious.
>
> logY = Map[N[ Log[10, #]] &, yValues]
> xyLog = Transpose[{Range[Length[logY]], logY}]
> l2 = Graphics[{Red, Thick, Line[Take[xyLog, {3, 6}]]}];
> l1 = Graphics[{Line[Take[xyLog, {1, 3}]]}];
> l3 = Graphics[{Line[Take[xyLog, {6, 10}]]}];
> Show[l1, l2, l3, Axes -> True, AxesOrigin -> {0, 0}]
>
> Question C. The above graph loses the original Y axis tick values which
> are contained in the yValues list defined above. Is there any way to
> solve this? I want the graph to look like the first graph above with the
> same Y tick values (using ListLogPlot) except the Line segment with X
> values from 3 to 6 should be Red.
>
> Thank you in advance for any help you can give me.
>
> Don
>