Given a sorted list of numbers $S$, I want to create a function that accepts a list of numbers $L$ and for each number $l \in L$ it returns the index of the largest number $s \in S$ such that $s {0.405196, 4.56535, 7.04274, 7.95001, 8.6823} And, here are a couple examples of the argument to the function: {6.67917, 8.33874, 4.61316, 4.83263, 9.52033} {6.0669, 1.22425, 6.13959} Then, I want to create a function such that return: One possibility is to use: But, this approach is quite slow when dealing with large arguments: {1.15025, Null} I'm interested in arguments on the order of 10^6 elements, and ordered sets $S$ on the order 10^4 elements. Is there a faster method?SeedRandom[13];
S = Sort @ RandomReal[10, 5]SeedRandom[10];
list1 = RandomReal[10, 5]
list2 = RandomReal[10, 3]f
:f = findIndices[S];
f[list1]
f[list2](*
{2, 4, 2, 2, 5}
{2, 1, 2}
*)findIndices[s_] := Interpolation[
Thread[{s, Range@Length@s-1}],
InterpolationOrder->0,
"ExtrapolationHandler" -> {Evaluate[Length[s]]&, "WarningMessage" -> False}
]f = findIndices[S];
tst = RandomReal[{.5, 10}, 10^6];
f[tst]; //AbsoluteTiming
Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
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