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Use Clip: Clip[{0.61975, 10562., 0.43274, 0.15068, 0.13703, 0.057564}, {-Infinity, 1}] {0.61975, 1, 0.43274, 0.15068, 0.13703, 0.057564}

I assume you want to find an exact match, instead of just that individual members of the subset are found in the larger list. This can be quite easily be accomplished with the following: ClearAll[sub …

Do you absolutely need a built-in function? At least if performance is not a concern, this is the simplest way to do it: {0,1,0,0,2,0,0} /. {x___, 0 ..} :> {x} {0, 1, 0, 0, 2} This can be writ …

updatedm = Flatten /@ Tuples[{m, l}] List-of-lists variant of Tuples does this directly, you just need to flatten (or Apply@Append) entries of l for each element of output to achieve your intended r …

The question is in my opinion more about math than Mathematica and so is my answer (unless the goal is to create these lists through specific constructs), but here it is nonetheless: Table[(1 + x - A …

Use PatternSequence, Condition (/;) and a form of Repeated that allows matching zero or one times: SeedRandom[6]; With[{alist = RandomInteger[{-2, 2}, 200]}, SequenceReplace[alist, k : {PatternSe …

I quite often run into a situation where I want to treat a list as a circular (repeating) list, and want to take a specific sublist of it, such as... One past the end of the list: Append[#, First@#] …

It is sometimes useful to remove exactly the amount of specific entries on one list from another. Consider: CountingComplement[{1, 2, 1, 3, 1, 3, 4, 5}, {6, 1, 3, 1, 5}] {2, 1, 3, 4} Two 1s a …

My solution is ugly, but task-specific. It builds a bitmap out of machine-sized integers in imperative fashion and uses Compile. This works reasonably in memory usage for ranges that have at least cou …

Turning my comment into an answer: ClearAll[MapAccumulate]; MapAccumulate[f_, list_List] := f@Take[list, #] & /@ Range@Length@list; Now you can do stuff like this: MapAccumulate[Mean, mylist]

It's far from pretty, using pattern matching (OrderlessPatternSequence): ReplaceList[#1, Module[{names = Unique[] & /@ #1, partitions}, partitions = Internal`PartitionRagged[names, #2]; Ac …

RegionEqual, like many region functions, is able to compute symbolic results as long as arguments are fully specified. This allows more efficient constructions of the following kind - where the symbol …

If we ignore random sampling, this is a simple question of Boolean satisfiability: Module[{m, n, greens, matrix, sol}, m = 8; n = 5; greens = 20; matrix = Array[c, {m, n}]; sol = matrix /. …

Assuming polygons follow the same (clockwise or counterclockwise) vertex order, find all good quality two line segment rigid mappings between polygons without overlap with each other (at least much ov …

As people on the comments pointed out this may be a problematic in various ways, but if all y-values of your points are primes, this can be accomplished with ScalingFunctions option. PrimePi is an inv …