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I want to create an array that must be symmetric with respect to the centre of it, as in the following example:

0 1 2 3 4 3 2 1

but, instead of having such simple numbers in the array, I need to fill the array by extracting the i-th number from a Gaussian distribution with mean 0 and variance equal to a function of the index i. I have tried to use Piecewise, at least to fill the first half of the array, but I don't know how to fill the other half:

KfieldREAL[k_] := Piecewise[{{0, k == 0},
     {RandomVariate[NormalDistribution[mu, Exp[-(2*Pi*k*sigma/L)^2]]], 
     0 < k <= n/2}, (*missing part here*)

How can I create an array like that?

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3 Answers 3

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How about creating the left half and using its image for the right half.

sigma = .25; L = 300; mu = 1; n = 1000;

left = Table[{k,RandomVariate[NormalDistribution[mu, Exp[-(2*Pi*k*sigma/L)^2]]]}, {k, n/2}];

right = Reverse[left] /. {x_, y_} -> {n - x, y}; (*imaging*)
full = Join[Most[left], right];
ListPlot[full, Frame -> True]

enter image description here

Adding extra elements

Let say you want to put the element x1 [say, (0,0)] at the beginning or end. You can simply put it during the Join.

x1={0,0.0}; x2={n+1,0.0}; (*first and last element in (k,f(k)) format*)
full=Join[{x1},Most[left], right,{x2}];

Note that when using join, you don't use a single element x1, rather a list containing a single element {x1}.

for n=8 it looks like

$ \begin{array}{l l} 0 & 0.00000 \\ 1 & 1.47757 \\ 2 & 2.29881 \\ 3 & 1.46615 \\ 4 & 1.34819 \\ 5 & 1.46615 \\ 6 & 2.29881 \\ 7 & 1.47757 \\ 9 & 0.00000 \\ \end{array} $

And your list now has n+1 element. This is because the axis of reflection goes through 4. If you want even number of element then use n+1 for right and omit Most in final.

right = Reverse[left] /. {x_, y_} -> {n + 1 - x, y};
full = Join[left, right]; 

And the result for n=8 after adding the zeros

x1 = {0, 0.0}; x2 = {n + 1, 0.0};
full = Join[{x1}, left, right, {x2}];

$ \begin{array}{ll} 0 & 0.00000 \\ 1 & 0.380134 \\ 2 & 1.48306 \\ 3 & 0.509712 \\ 4 & 1.76137 \\ \hline 5 & 1.76137 \\ 6 & 0.509712 \\ 7 & 1.48306 \\ 8 & 0.380134 \\ 9 & 0.00000 \\ \end{array} $

Now 4 and 5 has same element and number of elements in your final list iss n+2.

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  • $\begingroup$ You could also use full = ArrayPad[left, n/2-1, "Reflected"]. $\endgroup$
    – C. E.
    Apr 13, 2015 at 10:14
  • $\begingroup$ Thanks @Pickett. I didn't know about that syntax. $\endgroup$
    – Sumit
    Apr 13, 2015 at 10:21
  • $\begingroup$ Thanks to all of you. The method described in @Sumit 's answer works very well, I just need to add the first element, which is 0 (because that's what the theory imposes). How can I do that? I have tried using Join, but I did not succeed in adding a first element, set to 0. $\endgroup$
    – johnhenry
    Apr 13, 2015 at 10:52
  • $\begingroup$ Just a wild guess for adding the leading 0: Join[{0},(a = RandomVariate[NormalDistribution[0, 1], i]), Drop[Reverse[a], 1]] Note the curly braces around the zero in the Join function. $\endgroup$
    – LLlAMnYP
    Apr 13, 2015 at 11:15
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myMatrix[i_, j_] := 
 Table[Join[(a = RandomVariate[NormalDistribution[0, 1], i]), 
   Drop[Reverse[a], 1]], {j}]

 ListPlot[myMatrix[10, 10], Joined -> True]

enter image description here

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ClearAll[kfR]
kfR[mu_: 0, sigma_: 1, L_: 100] := With[{m = Array[
     RandomVariate[NormalDistribution[mu, Exp[-(2 Pi # sigma/L)^2]]] &, Ceiling[#/2]]},
   Prepend[m[[MapThread[Min, {Range[#], Reverse@Range[#]}]]], 0]] &

MatrixPlot[{kfR[][10]}, AspectRatio -> 1/10, ImageSize -> 800,
 ColorFunction -> "TemperatureMap",
 FrameTicks -> {{None, None}, {{#, #-1} & /@ Range[11], {#, #-1} & /@ Range[11]}}]

enter image description here

Using MultinormalDistribution instead of NormalDistribution:

ClearAll[kfR2]
kfR2[mu_: 0, sigma_: 1, L_: 100] := With[{m = ConstantArray[mu, Ceiling[#/2]], 
  s = DiagonalMatrix[Exp[-(2*Pi*Range[Ceiling[#/2]]*sigma/L)^2]], 
  ind = MapThread[Min, {Range[#], Reverse@Range[#]}]}, 
 Prepend[RandomVariate[MultinormalDistribution[m, s]][[ind]], 0]] &;

Panel[Column[MatrixPlot[{aa = (kfR2[][#])}, AspectRatio -> 1/10, 
     ImageSize -> 800, ColorFunction -> "TemperatureMap", 
     FrameTicks -> {{None, None},
       {rng = ({#, # - 1} & /@ Range[# + 1]), rng},
     Epilog -> (MapIndexed[Style[Text[Round[#, .2], {First[#2] - 1/2, 1/2}],
         "Panel", 18, Bold, Background -> Transparent] &, aa])] & /@ 
    {5, 6, 11, 12}]]

enter image description here

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  • $\begingroup$ thanks for your answer, but unfortunately this is not what I need, because you are changing only the variance of the distribution from which the numbers are extracted in the second half of the array. I need the numbers to be an exact duplicate of the first half, with the symmetry described in my question before $\endgroup$
    – johnhenry
    Apr 13, 2015 at 9:21
  • $\begingroup$ @user, i see.. I will post a fixed version, $\endgroup$
    – kglr
    Apr 13, 2015 at 9:30

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