Skip to main content

plotting - improving sequence dot operator for matrix rows and plot of Norm-imaginary


I have written a code to Dot a row (conjugate of that) of mm to the next one:


mm={{1,2,I},{I,I,3},{2,6,I},{1,6,4},{1,4,5}}

fIdD = ConstantArray[0, {Dimensions[mm][[1]] - 1, 2}];

Do[

fIdD[[j]][[1]] = j;
fIdD[[j]][[2]] = Dot[mm[[j]]\[Conjugate], mm[[j + 1]]]

, {j, 1, Dimensions[mm][[1]] - 1}];

Is there any possibility to use of mathematica syntax instead of this code for this aim(dot a conjugated row to the next one).


fIdD={{1, 0}, {2, -5 I}, {3, 38 - 4 I}, {4, 45}}


And if I am going to plot the {{1,Norm[0]},{2,Norm[-5I]},{3,Norm[38-4I]},{4,Norm[45]}} with the ListPlot. same as this: however it is wrong: ListPlot[{#1, Norm[ #2]}, & fIdD]?!



Answer



I would simply do


Dot[Conjugate[#1], #2] & @@@ Partition[mm, 2, 1]

and do away with the sequential numbers, as those are easy enough to generate when needed any way (MapIndexed?) ListPlot does not need these indices.


But instead of jumping straight to the solution, let's take your code and improve it step by step.




  1. Instead of arr[[i]][[j]] you can write arr[[i,j]]. This is simpler and more readable.





  2. You are pre-initializing an array of known dimensions and then compute each element. This is never necessary in Mathematica. This is what Table is for.


    At this point we have


    Table[
    {j, Conjugate[mm[[j]]].mm[[j + 1]]},
    {j, 1, Dimensions[mm][[1]] - 1}
    ]

    Much simper than the original code.



    Additionally, Dimensions[mm][[1]] is simpler and clearer as Length[mm].




  3. When you use the iterator in Table to index an array, there are usually simpler ways. Instead of Table[ f[ arr[[i]] ], {i, 1, Length[arr]}] we can always write Table[f[elem], {elem, arr}].


    But here you need to iterate through consecutive pairs of elements, not each element. Partition can make a list of pairs (or tuples) from a simple list. Partition[{a,b,c,d}, 2] would give {{a,b}, {c,d}} while Partition[{a,b,c,d}, 2, 1] will give {{a,b}, {b,c}, {c,d}. This latter is what you need.


    This changes our code to


    Table[
    Conjugate[elem[[1]]].elem[[2]],
    {elem, Partition[mm, 2, 1]}
    ]


    Instead of elem[[1]] people often write First[elem], which might be more readable, depending on context. Since we have a list of pairs, in this particular case elem[[2]] would be Last[elem].


    You might rightly argue that this change is not really a simplification, and it also loses the integer index form the result. But it leads up to the next step.




  4. When we have a list of the form list = {{x1,y1}, {x2,y2}, ..., {xn,yn}}, and we need to transform it to {f[x1,y1], f[x2,y2], ..., f[xn,yn]}, one pattern I like to use is


    f @@@ list

    This is short for Apply[f, list, {1}], please look up Apply. If we already have a pre-defined function f, this makes for very simple and clear code. If we don't, we can construct f on the fly as a pure function.


    This gives us the final version:



    result = Dot[Conjugate[#1], #2] & @@@ Partition[mm, 2, 1]

    One reason why I like this style is that I find #1 much more readable than elem[[1]]. Too many brackets give me a headache :-)




To re-insert the integer indices, you can use


Transpose[{Range@Length[result], result}]

But in applications that need the index of each element, MapIndexed can be useful too. MapIndexed[{First[#2], #1} &, result] is another way to get the result, though I'd integrate MapIndexed with the step that ultimately uses these indices.


ListPlot does not need these indices, it automatically assumes x coordinates to be 1, 2, ... if they're not given. So plotting is as simple as



ListPlot[ Norm /@ result ]

f /@ list is short for Map[f, list], please look up Map.




As you can see, Mathematica has many ways to do the same thing, and it's up to you to choose one. I hope you found this little introduction educational.


Comments

Popular posts from this blog

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 - 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 - 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....