Skip to main content

plotting - How can I make a grid of plots with no spacing, and have the plots line up perfectly?


I am making a bunch of two dimensional plots and I would like to have them arranged in a grid with no space in between them. That is, I want their frames touching each other.


My problem is twofold. First I don't know how to get the plots to have a consistent size. When I try to give the plots the same size via ImageSize, it applies the size to the overall image, including the tick marks and labels. But when I have two different plots, one with tick labels and one without, giving the same value to ImageSize doesn't give the result I'm looking for.


Secondly, I want to get rid of any white space around the images such that when I give Grid the option Spacings ->{0,0} there really is no visible space between the plots.



In the example code I use the CustomTicks package.


(*an example of what I might plot, simple sum of two dimensional \
Lorentzians*)
lzn[x_, w_, g_] := (x - w + I g)^-1;
exampledata2D =
Table[Re[lzn[w1, .2, .1] lzn[w2, .3, .2] +
lzn[w1, .4, .1] lzn[w2, .8, .1] +
lzn[w1, .8, .2] lzn[w2, .4, .2]], {w1, 0, 1, .01}, {w2, 0,
1, .01}];
<< "CustomTicks`";

standardticks = LinTicks[0, 1, .2, 4];
(*I don't want the last tick label for one plot overlapping with the \
first tick label of the plot right next to it.*)
ticks2 = LinTicks[0, 1, .2, 4, ShowLast -> False];
(*Custom plotting function so I don't have to reprint the same \
options over and over *)
plottingfunction[list_, plotopts : OptionsPattern[]] :=
ListContourPlot[list,
Evaluate[FilterRules[{plotopts}, Options[ListContourPlot]]],
Contours -> 30, PlotRange -> All, DataRange -> {{0, 1}, {0, 1}},

BaseStyle -> 18,
ColorFunction -> (Blend[{Red, Orange, Yellow, Green, Blue,
Darker[Blue]}, #] &)];

bottomleft =
plottingfunction[exampledata2D,
FrameTicks -> {{ticks2, StripTickLabels[standardticks]}, {ticks2,
StripTickLabels[standardticks]}}];
bottomright =
plottingfunction[exampledata2D,

FrameTicks -> {{StripTickLabels[standardticks],
StripTickLabels[standardticks]}, {standardticks,
StripTickLabels[standardticks]}}];
topleft =
plottingfunction[exampledata2D,
FrameTicks -> {{ticks2,
StripTickLabels[standardticks]}, {StripTickLabels[
standardticks], StripTickLabels[standardticks]}}];
topright =
plottingfunction[exampledata2D,

FrameTicks -> {{StripTickLabels[standardticks],
StripTickLabels[standardticks]}, {StripTickLabels[
standardticks], StripTickLabels[standardticks]}}];

Grid[{{topleft, topright}, {bottomleft, bottomright}},
BaseStyle -> ImageSizeMultipliers -> 1, Spacings -> {0, 0}]

enter image description here



Answer



Thanks to Kuba and Szabolcs for pointing out many related posts. I recognize that LevelScheme is probably the best way to go here, but at the moment I don't have the time to learn everything I need in order to use it.



I am going to use Jens's solution from this page, but I have to remove the certain tick labels so that they don't overlap each other.


(*an example of what I might plot, simple sum of two dimensional \
Lorentzians*)
lzn[x_, w_, g_] := (x - w + I g)^-1;
exampledata2D =
Table[Re[lzn[w1, .2, .1] lzn[w2, .3, .2] +
lzn[w1, .4, .1] lzn[w2, .8, .1] +
lzn[w1, .8, .2] lzn[w2, .4, .2]], {w1, 0, 1, .01}, {w2, 0,
1, .01}];
<< "CustomTicks`";

tickbottom = LinTicks[-1, 1, .5, 5, ShowLast -> False];
ticktop = LinTicks[-1, 1, .5, 5, ShowFirst -> False];
tickmiddle =
LinTicks[-1, 1, .5, 5, ShowFirst -> False, ShowLast -> False];
standardticks = LinTicks[-1, 1, .5, 5];
plottingfunction[list_, horizontalposition_, verticalposition_,
plotopts : OptionsPattern[]] :=
ListContourPlot[list,
Evaluate[FilterRules[{plotopts}, Options[ListContourPlot]]],
Contours -> 30, DataRange -> {{-1, 1}, {-1, 1}}, BaseStyle -> 18,

ColorFunction -> (Blend[{Red, Orange, Yellow, Green, Blue,
Darker[Blue]}, #] &), PlotRangePadding -> None,
PlotRange -> All,
FrameTicks -> {{Which[verticalposition == "Top", ticktop,
verticalposition == "Middle", tickmiddle,
verticalposition == "Bottom", tickbottom],
StripTickLabels[standardticks]}, {If[
verticalposition == "Bottom",
Which[horizontalposition == "Right", ticktop,
horizontalposition == "Middle", tickmiddle,

horizontalposition == "Left", tickbottom],
StripTickLabels[standardticks]],
StripTickLabels[standardticks]}}];

Options[plotGrid] = {ImagePadding -> 40};
plotGrid[l_List, w_, h_, opts : OptionsPattern[]] :=
Module[{nx, ny, sidePadding = OptionValue[plotGrid, ImagePadding],
topPadding = 0, widths, heights, dimensions, positions,
frameOptions =
FilterRules[{opts},

FilterRules[Options[Graphics],
Except[{ImagePadding, Frame, FrameTicks}]]]}, {ny, nx} =
Dimensions[l];
widths = (w - 2 sidePadding)/nx Table[1, {nx}];
widths[[1]] = widths[[1]] + sidePadding;
widths[[-1]] = widths[[-1]] + sidePadding;
heights = (h - 2 sidePadding)/ny Table[1, {ny}];
heights[[1]] = heights[[1]] + sidePadding;
heights[[-1]] = heights[[-1]] + sidePadding;
positions =

Transpose@
Partition[
Tuples[Prepend[Accumulate[Most[#]], 0] & /@ {widths, heights}],
ny];
Graphics[
Table[Inset[
Show[l[[ny - j + 1, i]],
ImagePadding -> {{If[i == 1, sidePadding, 0],
If[i == nx, sidePadding, 0]}, {If[j == 1, sidePadding, 0],
If[j == ny, sidePadding, topPadding]}}, AspectRatio -> Full],

positions[[j, i]], {Left, Bottom}, {widths[[i]],
heights[[j]]}], {i, 1, nx}, {j, 1, ny}],
PlotRange -> {{0, w}, {0, h}}, ImageSize -> {w, h},
Evaluate@Apply[Sequence, frameOptions]]]

plotGrid[Table[
plottingfunction[exampledata2D, h,
v], {v, {"Top", "Middle", "Bottom"}}, {h, {"Left", "Middle",
"Right"}}], 1000, 1000]


enter image description here


I'm not 100% happy with the way it looks, with the missing tick labels, but it is better than having the labels overlap or get cut off.


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]