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  • 0 posts edited
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  • 86 votes cast
Jan
27
awarded  Popular Question
Jan
19
awarded  Yearling
Dec
7
awarded  Popular Question
Nov
18
awarded  Notable Question
Nov
18
awarded  Good Question
Jul
20
comment compute the gradient in high dimension
@Jens, for different units (e.g. distance and angle), I added If[ i == angledim, Pi, 1] to the code computeGrad[data_, order_: 2] := Module[{n = ArrayDepth[data]}, MapThread[List, Table[If[i == angledim, Pi, 1]* NDSolve`FiniteDifferenceDerivative[UnitVector[n, i], Range /@ Dimensions[data], "DifferenceOrder" -> order][ data], {i, n}], n] ];
Jul
17
comment Applying a function to a list with fixed intervals and fill the resulting list
@ belisarius, not a good example. It should like this: using {{5, 5, 5}, {5, 5, 5}, {5, 5, 5}} to approximate {{4.6, 4.7, 4.8}, {4.9, 5, 5.1}, {5.2, 5.3, 5.4}}. Something like using the average value to smooth the list, or the first pic shown here : philipbjorge.com/2011/12/02/…
Jul
17
comment Applying a function to a list with fixed intervals and fill the resulting list
@ belisarius, In fact the function f does not matter, I only need to find a subset of a list to approximate the list uniformly(i.e. fixed interval), on the condition that the approximation list has the equal size. It is like using {{5, 5, 5}, {5, 5, 5}, {5, 5, 5}} to approximate {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}. In Chebyshev metric, the elements are 1 element neighbor of 5.
Jul
17
accepted Applying a function to a list with fixed intervals and fill the resulting list
Jul
17
comment Applying a function to a list with fixed intervals and fill the resulting list
@ belisarius, I present a short list for demonstration. The interval is 3, because each element has 2 neighbors. The interval seems unequal at first sight in this specific example, because the resulting list has to match with the original list in length. So I have to apply 'f' to the last element. I was thinking someone had encountered the same problem with me. I think it could be closed.
Jul
16
comment Applying a function to a list with fixed intervals and fill the resulting list
`@ciao, it seems the question is not stated clearly. I added some explanation to my question.
Jul
16
revised Applying a function to a list with fixed intervals and fill the resulting list
added 17 characters in body
Jul
16
revised Applying a function to a list with fixed intervals and fill the resulting list
added 478 characters in body
Jul
16
revised Applying a function to a list with fixed intervals and fill the resulting list
added 89 characters in body
Jul
16
asked Applying a function to a list with fixed intervals and fill the resulting list
Jun
21
awarded  Popular Question
May
27
comment Speed up derivative evaluation
@@Michael E2, the idea of vectorization is great. I get 20x speedup. I am curious about where does the time-saving come from? What if normalVector is also a function of θ?
May
26
comment Speed up derivative evaluation
@@Bichoy, I think Compile fails for functions containing D. The compiled version is slower.
May
26
comment Speed up derivative evaluation
Nice tip for speedup.
May
26
accepted Speed up derivative evaluation