Skip to main content

matrix - Square bracketed matrices in Mathematica - precision markers?


I'm trying to display Matrices with square brackets in Mathematica.


I've found this post from a mailing list in 2009 http://forums.wolfram.com/mathgroup/archive/2009/Aug/msg00458.html and it seems to work fine for whole numbers and symbols.


NotebookWrite[InputNotebook[], 
TemplateBox[{GridBox[{{a, b}, {c, d}}]}, "Identity",

DisplayFunction -> (RowBox[{StyleBox["[",
SpanMaxSize -> \[Infinity]], #1,
StyleBox["]", SpanMaxSize -> \[Infinity]]}] &)]]

(* Outputs: *)
Identity[{
{a, b},
{c, d}
}]


(* Which displays correctly as a square matrix. *)

But as soon as I try to input floating point numbers, I get floating point precision markers displayed. For example:


NotebookWrite[InputNotebook[], 
TemplateBox[{GridBox[{{2.1, 1}, {1, 1.2}}]}, "Identity",
DisplayFunction -> (RowBox[{StyleBox["[",
SpanMaxSize -> \[Infinity]], #1,
StyleBox["]", SpanMaxSize -> \[Infinity]]}] &)]]

(* Outputs: *)

Identity[{
{2.1000000000000001`, 1},
{1, 1.2`}
}]

(* Which displays correctly as a square matrix. *)

I don't see any function in there that would force the numbers to be evaluated into machine precision form. I think using HoldForm or others could solve this issue but I'm not too sure where that can be placed since GridBox needs a list and RowBox needs a box - evaluating them individually displays decimal numbers just fine.


For clarification, I'm looking for something like this:


(* In the square bracketed matrix display form of course *)

Identity[{
{2.1, 1},
{1, 1.2}
}]

Any help for a beginner? Thank you!



Answer



makeBrackMat[mat_?MatrixQ] := 
DisplayForm[
RowBox[{StyleBox["[", SpanMaxSize -> \[Infinity]],

GridBox[mat],
StyleBox["]", SpanMaxSize -> \[Infinity]]}
]
];

Exact numbers:


mat1 = Partition[Range[12], 3];
makeBrackMat[mat1]

enter image description here



Machine-precision numbers:


mat2 = {{1.3, 2.9}, {9.5, 8.4}, {7.6, 0.2}};
makeBrackMat[mat2]

enter image description here


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.