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A long comment...I'm using V13.1, Mac M1 Pro, which might matter... First, I would point out the "About" section of the profile of user @WReach. Toward the end, the question "Why don't you like to ans …

Compiled, parallelized ordering plus a more functional approach. The switch between DistanceMatrix and Outer (for dm) may be system dependent. ordC = Compile[{{x, _Real, 1}}, First@Ordering[x, 1], …

Best on packed arrays... PadLeft[Transpose@{#2}, {Automatic, 2}, {#1}] &[ n0, {n1, n2, n3, n4, n5}] (* {{n0, n1}, {n0, n2}, {n0, n3}, {n0, n4}, {n0, n5}} *) PadLeft[ArrayReshape[#2, {Length@#2, …

Let's examine the problem as given. I will consider it as a general code rewriting problem. We have an expression of the form h[.., f[ Evaluate[g[x]] ],..] (* or more generally... *) h[.., f[ …

Here's a slight refactoring of the OP's solution. Expressions like #[[5]] are hard to read and therefore hard to debug. One can use a function with symbolic names for arguments to make the code ea …

I sometimes want to apply a linear combination of functions to an argument. This can be more difficult if the linear combinations are generated by code, so that rewriting them some form to make Throu …

This gets the desired result: (MyFunction @@ #1) @@ #2 & @@ {{1, 2, 3}, {4, 5, 6}} (* MyFunction[1, 2, 3][4, 5, 6] *)

This addresses the underlying problem, which implies a really fast way is not likely to be found, and shows yet another way to sum a list, Tr. One reason the OP's sum is slow is that the sum is bigge …

Another way: (* ad hoc def. of Mat[] *) ClearAll[Mat]; Mat[x_, y_] := { {2 y, x, 1}, {x + y, 0, -x}, {x, 2, y} }; Block[{x = 1.0}, First@ Differences@ MinMax[y /. NSolve[Det[Mat[x, y]] = …

Maybe the following. Simplifying a left-over constant term along with the other coefficients seems hard to comprise in a single, simple function. forms = {_f, _e, _g, _h}; funcs = {simpF, simpE, sim …

Here's another way that's quite a bit faster on David Stork's example: #.Y.# &[A.X] SeedRandom[0]; A = Table[RandomReal[], {3000}]; X = Table[RandomReal[], {3000}, {8000}]; Y = Table[RandomReal[], {8 …

Another way: a = RandomReal[{0, 1}, {2, 8}]; b = RandomReal[{0, 1}, {2, 8, 2}]; m = First@Outer[Dot, {a}, b, 1]; Compare with Sjoerd's: s = a.# & /@ b; (* Sjoerd's *) s == m (* True *)

If you have a large amount of calibData, then preserving packed arrays can greatly improve speed. Convert it all to Reals and then apply logarithm. calibData = 10 RandomInteger[{1, 20}, {10^6, 2}]; …

Here's a way using NestWhile: f[lst_?(VectorQ[#, NumericQ] &), n_Integer?Positive, p_?Positive] := If[Length @ # > n, #[[n + 1]], None] &@ NestWhile[Rest, lst, Length @ # > n && Abs[Mea …

I'm afraid my comment was too obscure to be noticed. Further, I disagree with one premise somewhere in the commentary, and I wish to make a fuller explanation to see if I understand correctly or inco …