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Here's a variation of @sjoerd's answer: data={{1,2},{1,3},{1,4},{2,2},{2,3},{2,4},{2,5}}; ResourceFunction["GroupByList"][data[[All,2]],data[[All,1]],Max] <|1 -> 4, 2 -> 5|> For large datasets, th …

You could give OverBar the HoldFirst attribute so that it doesn't depend on the value of a: SetAttributes[OverBar,HoldFirst] Then: t1 = Table[ p[a, OverBar[a], b, OverBar[b]], {a, 0, 1}, {Ove …

Compiling is usually the best approach for problems like this where an iteration depends on previous state. Here is a compiled function to get the counts (basically a compiled version of the approache …

It's unclear what the Mathematica form of the set is. I will assume that the set elements are just symbols: DeleteDuplicates[ {ab,cd,cea,abc,bcd}, StringContainsQ[SymbolName[#2],SymbolName[#1 …

Based on @Coolwater's fine answer to Replace element in array by checking condition in another list you can use GatherBy: Catenate @ GatherBy[ {{a,b},{f,c},{d,b},{e,c},{a,c}}, Last ][[All, 2; …

If the list consists of positive integers smaller than some reasonable maximum, then the following function should be pretty fast: trunc[a_, b_] := Module[{c = ConstantArray[0, Max[a]]}, c[[Rever …

Since you mentioned you are working with huge arrays, you might consider using Compile. For instance, here are two possible Compile approaches: fc = Compile[{{a, _Integer}, {b, _Integer}}, If[a b …

You could use Nearest for determining intersection1. Here is some sample data: SeedRandom[3]; list1=RandomComplex[1+I, 10^3]; list2=RandomComplex[1+I, 10^4]; Create the Nearest function: nf = Near …

Here is an example of your tables, since you didn't provide one: tb1 = Table[{r,f1[r]}, {r, {r1,r2,r3}}] tb2 = Table[{r,f2[r]}, {r, {r1,r2,r3}}] tb3 = Table[{r,f3[r]}, {r, {r1,r2,r3}}] {{r1, f1[ …

You could "transpose" your lst using Flatten (1), and then take the 4 element: Flatten[lst, {{2}, {1}}][[4]] {d, h} Flatten command: matrix as second argument

In M11.1+ (perhaps earlier as well) you can use Nearest to test multiple elements at once. Here is a comparison with @LeonidShifrin's answer: lrgTest = RandomInteger[100000,{100000}]; r1 = Nearest[l …

Not as fast as some of the other methods, but a TranslationTransform approach is also possible: TranslationTransform[{a,b}][{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}] {{-1 + a, b}, {1 + a, b}, {a, -1 + …

Convert your vector to a diagonal matrix, and then use Dot: data = RandomReal[1,{13000,3}]; data1 = data . DiagonalMatrix[{x, y, z}]; //RepeatedTiming {0.000051, Null} Another very fast approa …

I would use Clip: Clip[{2,3,5,4,6,7,3,7,5,4}, {5, Infinity}, {0, 0}] {0, 0, 5, 0, 6, 7, 0, 7, 5, 0}

This is basically a duplicate of pure functions in nested select. Applying the accepted answer to your question: list = {100, 200, 300, 400, 500}; Array[Function[x, #[[x]]], Length[#]]& @ list { …