plotting - Controlling the 2D $x–y$ aspect ratio of a 3D plot

I have some data representing a nested list, each element being composed of 3 real numbers like {{652.112, 0, 0.111838}, {664.096, 29.134, 0.0000485323}, {713.531, 12.1382, 0.805022} ...}. I would like to make an illustration for a journal article out of this list. The first idea is to use ListPlot3D function. The points in my list span in space from 600 to 1300 along $x$ (e.g. about 700 units along $x$ ) and from -100 to 100 along $y$ (e.g. about 200 units along $y$ ).

It is principally important for me that the illustration gives a correct feeling about the $x/y$ ratio. However, when I build it the image looks about equal along $x$ and along $y$ . Changing AspectRatio->0.2 to 0.5 only helps partially, but disturbs another aspects of the image.

Is there a way to control the $x\text{–}y$ aspect ratio of a 3D image?

Answer

You need BoxRatios, it works also in ListPlot3D.

GraphicsRow[ Table[ Graphics3D[ Sphere[], BoxRatios -> a],
                    {a, {{1, 1, 1}, {2, 1, 1}, {1, 2, 1}, {1, 1, 2}}}]]

enter image description here

You can use it in ListPlot3D, e.g.

GraphicsRow[ Table[ ListPlot3D[Table[Sin[x y], {x, 0, 3, 0.1}, {y, 0, 3, 0.1}], 
                                                       BoxRatios -> {1, 1, 1/k}],                 
                   {k, 3}] ]

enter image description here

Comments

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}]

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