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(* combine[expr_, l_] := With[{matches = Select[l, First[#] == First[expr] &]}, If[Length[matches] == 0, {Append[expr, Missing[]]}, Map[Join[expr, Rest[#]] &, match …

To find minima, run the same command with your data negated. To find the times and values, do Transpose[{data[[PeakDetect[data]], 1]], FindPeaks[data[[All, 2]]]}]

For the particular question: If[Or @@ (# > 1& /@ a), 1, 0] but there is a problem you should be aware of: Or returns immediately after a True clause, so if your list is $10^{10}$ long, the version …

Assuming I understand the question correctly, Permute[xp, Cycles[{{1, 21}}] will do the needful. Notice that this works to effect an arbitrary permutation of the elements.

This looks like a bug, but presumably a workaround is to use a hash table. That is, in @Szabolcs's code, change the first line to: Table[arr[i] = RandomInteger[10, 5].x^Range[5], {i, 1, 1000000}]; …

mrandomperms[n_, m_]:= Table[PermutationList[RandomPermutation[n]], {m}] Was my initial answer, but, as pointed out by Mr. Wizard, PermutationList should be given $n$ as a second argument, since othe …

I am not sure what it is you are trying to do, but presumably list = foo /@ RotateRight[list] is something like what you want?! The most general solution involves the function ListConvolve[] whic …

I don't understand the question: Mathematica likes to represent permutations as products of cycles (do a help on "Permutations"), so it seems natural that it would represent the symmetry group of the …

Well, here is the simplest I can think of: foo[l_, x_] := Module[{ll = l[[-1, -1]]}, If[ll + 1 == x, Append[l[[;; -2]], Append[l[[-1]], x]], Append[l, {x}]]] z[l_] := Fold[foo, {{l[[1]]}}, Rest …

This will certainly be less efficient, but is slightly shorter: ReactionPower = Tr[RF1.Transpose[RF1vel] + RF2.Transpose[RF2vel]];

Suppose I have a list of data, and a list of quantiles ({0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9} is a perennial favorite), and I want to assign to each element the quantile it is in (so for Range …

To answer the specific question, I think that instead of Apply[Dot, l], Fold[Dot, l] will not run into any problems. Of course, all the other comments are valid.

Suppose first that I want to generate a matrix whose elements are independent and identically distributed (i.i.d.) with distribution dist. This is easy: randMat[dist_, n_, m_] := RandomVariate[dist, …

I am not sure what you are trying to do exactly, but this works fine: ContourPlot[w[x, y, z] /. z -> int[x, y], {x, 0, 8}, {y, 0, 8}]