Adaptive gridlines | 易学教程

问题

I want to use gridlines to create an effect of millimeter graphing paper on a 2d graph, to show how multi-variable function depends on 1 variable. The scales of different variables differ a lot, so my naive approach (that I have used before) does not seem to work.

Example of what I have at the moment:

<< ErrorBarPlots`
Cmb[x_, y_, ex_, ey_] := {{N[x], N[y]}, ErrorBar[ex, ey]};
SetAttributes[Cmb, Listable];

ELP[x_, y_, ex_, ey_, name_] :=
 ErrorListPlot[
  Cmb[x, y, ex, ey],
  PlotRange -> FromTo[x, y],
  PlotLabel -> name,
  Joined -> True, Frame -> True, GridLines -> GetGrid,
  ImageSize -> {600}
 ]

Both FromTo (I want to leave 5% margin in the frame) and GetGrid do not work exactly as I want them to.

On some axes the variables differs many orders of 10. And I do not want, that one axis has many orders of 10 gridlines more then other. And most importantly I want the gridlines to line up with ticks.

Sample data:

ELP[
  {4124961/25000000, 27573001/100000000, 9162729/25000000, 44635761/
   100000000, 15737089/25000000, 829921/1562500, 4405801/4000000, 
   23068809/25000000, 329386201/100000000, 58079641/100000000},
  {1/10, 1/5, 3/10, 2/5, 3/5, 1/2, 1/2, 1/2, 1/2, 1/2},
  {2031/(250000 Sqrt[10]), 5251/(500000 Sqrt[10]), 3027/(
   250000 Sqrt[10]), 6681/(500000 Sqrt[10]), 3967/(250000 Sqrt[10]), 
   911/(62500 Sqrt[10]), 2099/(100000 Sqrt[10]), 4803/(
   250000 Sqrt[10]), 18149/(500000 Sqrt[10]), 7621/(500000 Sqrt[10])},
  {1/2000, 1/1000, 3/2000, 1/500, 3/1000, 1/400, 1/400, 1/400, 1/400, 
   1/400},
  "T2, m"
]

Would result in:

And my naive GetGrid, that works in some sence:

FromTo[x_, y_] := Module[{dx, dy},
   dx = (Max[x] - Min[x])*0.1;
   dy = (Max[y] - Min[y])*0.1;
   {{Min[x] - dx, Max[x] + dx}, {Min[y] - dy, Max[y] + dy}}];
GetGrid[min_, max_] := Module[{step, i},
  step = (max - min)/100;
  Table[
   {min + i*step,
    If[Equal[Mod[i, 10], 0],
     Directive[Gray, Thick, Opacity[0.5]],
     If[Equal[Mod[i, 5], 0],
      Directive[Gray, Opacity[0.5]],
      Directive[LightGray, Opacity[0.5]]
      ]]},
   {i, 1, 100}]
  ]

Question

How to make GridLines line up with ticks?

edit: With

GetTicks[x_, y_] := Module[{dx, dy},
   dx = (Max[x] - Min[x])*0.1;
   dy = (Max[y] - Min[y])*0.1;
   {
    Min[x] - dx + Table[i*dx*1.2, {i, 1, 9}],
    Min[y] - dy + Table[i*dy*1.2, {i, 1, 9}]
    }];

ELP[x_, y_, ex_, ey_, name_] :=
 ErrorListPlot[
  Cmb[x, y, ex, ey],
  PlotRange -> FromTo[x, y],
  PlotLabel -> name,
  Joined -> True, Frame -> True, GridLines -> GetGrid, 
  FrameTicks -> GetTicks[x, y],
  ImageSize -> {600},
  AspectRatio -> 1
  ]

I can get:

And that is a lot better. But I would like to shift the grid and not the ticks.

edit: @Sjoerd C. de Vries

Your solution does what I wanted to archive and works. I also noticed, that if I take first 5 elements of sample data, then the plot will be (elements are sorted and regression line is added).

Notice the left most element is like off grid.

回答1:

Don't use FrameTicks but shift the grid correctly. This is a first approach. Dinner waits.

getGrid[min_, max_] :=
 Module[{step, i},
  Print[{min, max}];
  step = 1/100;
  Table[
   {
    Floor[min, 0.1] + i*step,
    If[Equal[Mod[i, 10], 0], Directive[Gray, Thick, Opacity[0.5]],
     If[Equal[Mod[i, 5], 0], Directive[Gray, Opacity[0.5]],
      Directive[LightGray, Opacity[0.5]]
      ]
     ]
    },
   {i, 1, (Ceiling[max, 0.1] - Floor[min, 0.1])/step // Round}
   ]
  ]

Use an AspectRatio that's appropriate for the grid (probably the ratio of x and y ranges)

After-dinner update

To make it more robust for different value ranges (per your comment) I generate the ticks that would be chosen by ListPlot and base my steps on that:

getGrid[min_, max_] :=
 Module[{step, i,j},
  i = Cases[(Ticks /. 
       AbsoluteOptions[ListPlot[{{min, min}, {max, max}}], 
        Ticks])[[1]], {a_, ___, {_, AbsoluteThickness[0.25`]}} :> a];
  step = i[[2]] - i[[1]];
  Table[
   {
    i[[1]] + j*step/10,
    If[Equal[Mod[j, 10], 0], Directive[Gray, Thick, Opacity[0.5]],
     If[Equal[Mod[j, 5], 0], Directive[Gray, Opacity[0.5]],
      Directive[LightGray, Opacity[0.5]]
      ]
     ]
    },
   {j, 0, 10 Length[i]}
   ]
  ]

and getting the aspect ratio which yields a square raster

getAspect[{{minX_, maxX_}, {minY_, maxY_}}] :=
 Module[{stepx, stepy, i, rx, ry},
   i = (Ticks /.AbsoluteOptions[ListPlot[{{minX, minY}, {maxX, maxY}}], Ticks]);
   rx = Cases[i[[1]], {a_, ___, {_, AbsoluteThickness[0.25`]}} :> a];
   stepx = rx[[2]] - rx[[1]];
   ry = Cases[i[[2]], {a_, ___, {_, AbsoluteThickness[0.25`]}} :> a];
   stepy = ry[[2]] - ry[[1]];
  ((maxY - minY)/stepy)/((maxX - minX)/stepx)
  ]

Test

ELP[x_, y_, ex_, ey_, name_] := 
 ErrorListPlot[Cmb[x, y, ex, ey], PlotLabel -> name, Joined -> True, 
  Frame -> True, GridLines -> getGrid, ImageSize -> {600}, 
  PlotRangePadding -> 0, AspectRatio -> getAspect[FromTo[x, y]], 
  PlotRange -> FromTo[x, y]]


ELP[{4124961/25000000, 27573001/100000000, 9162729/25000000, 
  44635761/100000000, 15737089/25000000, 829921/1562500, 
  4405801/4000000, 23068809/25000000, 329386201/100000000, 
  58079641/100000000}, {1/10, 1/5, 3/10, 2/5, 3/5, 1/2, 1/2, 1/2, 1/2,
   1/2}, {2031/(250000 Sqrt[10]), 5251/(500000 Sqrt[10]), 
  3027/(250000 Sqrt[10]), 1/100000 6681/(500000 Sqrt[10]), 
  3967/(250000 Sqrt[10]), 911/(62500 Sqrt[10]), 
  2099/(100000 Sqrt[10]), 4803/(250000 Sqrt[10]), 
  18149/(500000 Sqrt[10]), 7621/(500000 Sqrt[10])}, {1/2000, 1/1000, 
  3/2000, 1/500, 3/1000, 1/400, 1/400, 1/400, 1/400, 1/400}, "T2, m"]

Here I divide the y-values by 20 and multiplied the x-values by 10000 to show the grid is still good:

Final update (I hope)

This uses FindDivisions as suggested by belisarius. However, I used the three level line structure standard for milimeter paper as requested by Margus:

getGrid[x_, y_] := 
 FindDivisions[{x, y}, {10, 2, 5}] /. {r_, s_, t_} :> 
   Join[
     {#, Directive[Gray, Thick, Opacity[0.5]]} & /@ r, 
     {#, Directive[Gray, Opacity[0.5]]} & /@ Union[Flatten[s]], 
     {#, Directive[LightGray, Opacity[0.5]]} & /@ Union[Flatten[t]]
   ]

and

getAspect[{{minX_, maxX_}, {minY_, maxY_}}] :=
 Module[{stepx, stepy},
  stepx = (#[[2]] - #[[1]]) &@FindDivisions[{minX, maxX}, 10];
  stepy = (#[[2]] - #[[1]]) &@FindDivisions[{minY, maxY}, 10];
 ((maxY - minY)/stepy)/((maxX - minX)/stepx)
  ]

WARNING!!!

I just noticed that if you have this in MMA:

and you copy it to SO (just ctrl-c ctrl-v), you get this:

(maxY - minY)/stepy/(maxX - minX)/stepx

which is not mathematically equivalent. It should be this:

((maxY - minY)*stepx)/((maxX - minX)*stepy)

I corrected this in the code above, but it has been posted wrong for half a day while working correctly on my computer. Thought that it would be good to mention this.

回答2:

I think FindDivisions[ ] is what you're after:

FindDivisions[{xmin,xmax},n] finds a list of about n "nice" numbers that divide the interval around xmin to xmax into equally spaced parts.

getTicks[x_, y_] := Flatten@FindDivisions[#, {10}] & /@ FromTo[x, y]
getGrid  [x_,y_] := FindDivisions[{x,y},{10,5}]/.
                          {r__,{s__}}:>Join@@{s,{#,{Gray,Thick}}&/@r}

回答3:

If you use the same function for FrameTicks and Gridlines, they'll line up.

See FrameTicks, and GridLines. I think you'll need ImageMargins for the border.

来源：https://stackoverflow.com/questions/5664937/adaptive-gridlines

标签

wolfram-mathematica