自适应网格线 [英] Adaptive gridlines
问题描述
我想使用网格线在二维图形上创建.但是，我按照 Margus 的要求使用了毫米纸的三级线结构标准:
getGrid[x_, y_] :=FindDivisions[{x, y}, {10, 2, 5}]/.{r_，s_，t_}:＞加入[{#，指令[灰色，厚，不透明度[0.5]]} &/@r,{#, Directive[Gray, Opacity[0.5]]} &/@联合[展平[s]]，{#, Directive[LightGray, Opacity[0.5]]} &/@联合[展平[t]]]和
getAspect[{{minX_, maxX_}, {minY_, maxY_}}] :=模块[{stepx, stepy},stepx = (#[[2]] - #[[1]]) &@FindDivisions[{minX, maxX}, 10];stepy = (#[[2]] - #[[1]]) &@FindDivisions[{minY, maxY}, 10];((maxY - minY)/stepy)/((maxX - minX)/stepx)]<小时>
警告！！！
我刚刚注意到，如果你在 MMA 中有这个:
然后将其复制到 SO(只需 ctrl-c ctrl-v)，您会得到:
(maxY - minY)/stepy/(maxX - minX)/stepx这在数学上不等价.应该是这样的:
((maxY - minY)*stepx)/((maxX - minX)*stepy)我在上面的代码中更正了这个问题，但是在我的电脑上正常工作时它已经发布了半天.觉得提一下就好了.
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.
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.
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