# Tag Info

10

Slightly expanding my comment. It is a partial answer, explaining most but not all of the observed timing increase. First I show some representative timings. Note that I use 2^16-1 and 2^21-1 below, for reasons that will be explained. AbsoluteTiming[RandomInteger[2^16 - 1, 100000000];] AbsoluteTiming[RandomInteger[2^21 - 1, 100000000];] (* Out[37]= ...

8

m = 5; r = 3000; Table[pt[i] = RandomReal[{-2000,2000},2], {i, m}]; circles = Table[Circle[pt[i], r], {i, m}]; disks = circles /. Circle -> Disk; reg1 = RegionIntersection[disks]; area1 = RegionMeasure[reg1]; reg2 = RegionUnion[disks]; area2 = RegionMeasure[reg2]; g10 = Show[Graphics[{RandomColor[], #}& /@ circles, Axes-> True, ...

8

Dirichelet random variates are constructed from gamma random variates. Let $$v_j \stackrel{\text{iid}}{\sim} \textsf{Gamma}(\alpha_j,1)$$ for $j = 1, \ldots, n$ and set $$x_j = \frac{v_j}{\sum_{j=1}^n v_j}.$$ Then $x \sim \textsf{Dirichlet}(\alpha)$, where $x = (x_1,\ldots,x_n)$ and $\alpha = (\alpha_1,\ldots,\alpha_n)$. Gamma random variates ...

7

@kglr beat me to posting and has a much nicer answer, but since I interpreted the question slightly differently, I thought it might be helpful to post my answer too. (Edit: Updated based on comment by @JackLaVigne to accelerate by getting rid of empty intersections, see module below for timing.) Since OP says that "some of the circles may overlap", I ...

4

Following up on @Karsten 7.'s approach, with a more convenient parameterization of PearsonDistribution (using pieces from PearsonDistribution >> Applications): ClearAll[pearsonD, dis, tdisn, tdisp] pearsonD[μ_, σ_, γ_, κ_] := PearsonDistribution[2 (9 + 6 γ^2 - 5 κ), -12 μ γ^2 - σ γ (3 + κ) + 2 μ (-9 + 5 κ), 6 + 3 γ^2 - 2 κ, -6 μ γ^2 + 4 μ ...

4

TL;DR: There is no PearsonDistribution that matches all characteristics exactly, but there is an infinite number of PearsonDistributions that resemble the given characteristics quite well. The difference between them is their standard deviation. In my personal opinion the information given in the question and its foundation are insufficient to make any ...

3

Extract the path from the TemporalData object, then isolate the $y$ values and produce a random permutation using RandomSample; finally add back the $t$ values: ts5List = First@ts5["Paths"]; scrambled = Transpose@{ts5List[[All, 1]], RandomSample[ts5List[[All, 2]], Length[ts5List]]} ListPlot[scrambled]

3

Variant 1 define distribution_1 (losses) NSolve[Kurtosis[LogNormalDistribution[0, x]] == 12.2, x, Reals] {{x -> -0.579872}, {x -> 0.579872}} LogNormalDistribution[0, 0.5798723392706395] // Kurtosis 12.2 LogNormalDistribution[0, 0.5798723392706395] // Skewness 2.14932 dist1 = TruncatedDistribution[ {-\[Infinity], 0}, ...

2

Letting the 5-vector $x = \{a^2, b^2, c^2, d^2, e^2 \} = \{x_1, x_2, x_3, x_4, x_5\}$ have DirichletDistribution[{1,1,1,1,1}], RandomVariate[DirichletDistribution[{1,1,1,1,1}]] gives a random 4-vector that satisfies $0 \leq x_i \leq 1$ and $\sum_{i=1}^4 x_i <1$. The 5th component of $x$ is determined by the condition $x_5= 1 - x_1 - x_2 - x_3 - x_4$. So ...

2

To settle this: Append[Normalize[RandomReal[1, 5]], RandomReal[3]] produces a set of numbers that satisfies the OP's request. What distribution this tuple follows is a different kettle of fish.

2

Working from what @MarcoB suggested, I came up with: ts5randomized = RandomSample[ts5["Values"]]; ts5randomizeSeries = TimeSeries[ts5randomized, {0, 999, 1}] ListLinePlot[ts5randomizeSeries] This appears to be what I needed. Thanks again @MarcoB!

2

Is the implementation of the classifier or method wrong or are these results ok? We cannot conclude from the posted table of classification accuracy results that "the classifier or method is wrong". I can come up with at least several explanations. (And it would be nice if data is provided.) The simplest explanation is that none of your variables is ...

1

UPDATE: You request no duplication in the rows, but duplication is possible in the columns. We can achieve that using RandomChoice instead of RandomSample to generate the permutations. After generation of a new list, we check that each row is free of duplicates; if not, we generate a new one until we get an appropriate new list. The following uses your ...

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