Skip to main content

graphics - What does the AspectRatio option actually do?


The Documentation for AspectRatio on its first line under the "Details" section states: "AspectRatio determines the scaling for the final image shape" (emphasis is mine). But the practice shows that is it not true: it seems that this option affects only aspect ratio of the plot range but not the aspect ratio of the whole image (with ImagePadding and ImageMargings added). It is a basic graphics option but we still know a little about it...


What the option AspectRatio actually do? How it interacts with PlotRangePadding, ImagePadding, ImageMargings and ImageSize?



It would be ideal to have a mathematical model of interaction between these options.



Answer



Padding


Without padding of any kind the over-all aspect ratio and element (primitive) aspect ratio are the same and as specified:


g0 =
Graphics[{Opacity[0.5, Red], Rectangle[{0, 0}, {3, 2}]}, AspectRatio -> 2/3,
Background -> GrayLevel[0.8], PlotRangePadding -> 0]

enter image description here


(There is a one pixel discrepancy along the right edge where the background shows through but I believe that is within the margin of error for rasterization in Mathematica. That is to say there are other small discrepancies that would also need to be accounted for before considering this a specific aspect ratio problem.)



g0 // Image // ImageDimensions


{360, 240}

PlotRangePadding is included in the aspect ratio calculation such that the extended plot area has the specified aspect ratio which means that elements have a different aspect ratio unless the padding is such that it exactly matches the aspect ratio.


g1 =
Show[g0, Axes -> True, PlotRangePadding -> {15, 2}, ImagePadding -> 0, ImageMargins -> 0]

enter image description here



The image dimensions are similar to g0 though the Rectangle is clearly distorted.


g1 // Image // ImageDimensions


{360, 240}

If the ratio of the PlotRangePadding matches the numeric AspectRatio the image aspect ratio and the element aspect ratio match:


Show[g0, Axes -> True, PlotRangePadding -> {3, 2}, ImagePadding -> 0, ImageMargins -> 0]

enter image description here



ImagePadding is excluded from the calculation of aspect ratio; it is area added outside the plot area but within the graphic area where e.g. Background applies and where ticks and labels may reside. With PlotRangePadding -> 0 the element aspect ratio is still exactly as specified by AspectRatio.


g2 = 
Show[g0, Axes -> True, PlotRangePadding -> 0,
ImagePadding -> {{70, 70}, {20, 8}}, ImageMargins -> 0]

enter image description here


g2 // Image // ImageDimensions


{360, 175}




ImageMargins is excluded from aspect ratio and image size calculations. It extends the image beyond the specified size with a blank area; it may not contain ticks or labels.


g3 =
Show[g0, Axes -> True, PlotRangePadding -> 0, ImagePadding -> 0,
ImageMargins -> {{30, 30}, {50, 50}}]

enter image description here


The image is larger than the default width-360:


g3 // Image // ImageDimensions



{420, 340}



ImageSize


When an absolute ImageSize is given that does not match the requested ratio the graphic is scaled down to fit entirely within that size and the image area is extended to match the absolute size. The exception is ImageMargins (g3) which as stated before is excluded from ImageSize; it adds padding outside of that bounding box.


Show[#, ImageSize -> {160, 180}] & /@ {g0, g1, g2, g3}

enter image description here


ImageDimensions /@ Image /@ %



{{160, 180}, {160, 180}, {160, 180}, {220, 280}}

Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

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}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.