# Tag Info

25

f = {1, 5, 9, 14}; v = {-1, 1, 3, 4, 6, 9, 10, 13, 14, 15}; BinLists[v, {Join[{-Infinity}, f, {Infinity}]}] {{-1}, {1, 3, 4}, {6}, {9, 10, 13}, {14, 15}}

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Summary We can look at the code of DeleteDuplicatesBy and it turns out it uses GroupBy. The test cases proposed by Mr.Wizard are all handled by some part of the code of DeleteDuplicatesBy. Other parts of this code also seem to have some issues. Most of the members of the *By family of functions seem to have side effects. How DeleteDuplicatesBy works It ...

22

First let me note that I didn't write PositionIndex, so I can't speak to its internals without doing a bit of digging (which at the moment I do not have time to do). I agree performance could be improved in the case where there are many collisions. Let's quantify how bad the situation is, especially since complexity was mentioned! We'll use the ...

21

test = {5, 6, 9, 3, 2, 6, 7, 8, 1, 1, 4, 7} MaxFilter[test, 1] (* {6, 9, 9, 9, 6, 7, 8, 8, 8, 4, 7, 7} *) You can also use Max /@ Transpose[{Rest[Append[#, 0]], #, Most[Prepend[#, 0]]}] &[yourList] which is competitive with the MM MaxFilter, but will allow you to change the 'slide' (e.g.pad with zeroes, or other arbitrary 'start').

19

There was an update for Array, not done to the end. The method below does not work for earlier versions even though that Array is New in 1 | Last modified in 4 Moreover WRI forgot to update docs for error messages: Array::plen - the first example gives no error in V9. V9 Array[# &, n, {start, stop}] Array[# &, 10, {-1, 1}] {-1, ...

17

Not sure if this will work for you, but... There is a cool blog by Roman Osipov in Russian (use Google Translate to translate): Study of arbitrary functions by methods of mathematical analysis in the system Mathematica I will give 2 functions from there (see the blog for more tricks). The domain of the function DefinitionDomain[expr_, variable_: x] := ...

16

Maximize[{y, x^2 + y^2 == (2 x^2 + 2 y^2 - x)^2}, {x, y}] {(3 Sqrt[3])/8, {x -> 3/8, y -> (3 Sqrt[3])/8}}

16

I see no mention of the new-in-10 PositionIndex in the other answers, which takes a list (or association) of values and returns a 'reverse lookup' that maps from values in the list to the positions where they occur: In[1]:= index = PositionIndex[{a, b, c, a, c, a}] Out[1]= <|a -> {1, 4, 6}, b -> {2}, c -> {3, 5}|> It doesn't take a level ...

16

Attempting to analyze the performance of this function in the manner that Taliesin Beynon did for PositionIndex I shall use the same tools. The old method that will be compared in all cases below: myDeDupeBy[x_, f_] := GatherBy[x, f][[All, 1]] Speed A BenchmarkPlot of DeleteDuplicatesBy versus myDeDupeBy: Needs["GeneralUtilities"] BenchmarkPlot[ ...

15

May be the most compact approach: Hold[i1++, i1--, i2++, i2--][[RandomInteger[{1, 4}]]]

15

The best I have is manual RHS holding and Join, after which an arbitrary head could be Applied: Join @@ Cases[expr, x : _Times :> Hold[x], 3] Hold[2/2, 8/4, 1/0] This could be done automatically as follows: makeHeld[(L_ -> R_) | (L_ :> R_)] := L :> HoldComplete[R]; makeHeld[pat_] := x : pat :> HoldComplete[x]; heldCases[expr_, rule_, ...

15

There are nice trigonometric formulas δ = 0.01; trg[x_] := 1 - 2 ArcCos[(1 - δ) Sin[2 π x]]/π; sqr[x_] := 2 ArcTan[Sin[2 π x]/δ]/π; swt[x_] := (1 + trg[(2 x - 1)/4] sqr[x/2])/2; Plot[{TriangleWave[x], trg[x]}, {x, -2, 2}, PlotRange -> All] Plot[{SquareWave[x], sqr[x]}, {x, -2, 2}, PlotRange -> All, Exclusions -> None] Plot[{SawtoothWave[x], ...

15

Looking at the Trace of one which does work: x = Sin[Pi/5] (* Sqrt[5/8 - Sqrt[5]/8] *) Trace[ArcSin[x], TraceInternal -> True] It appears that Mathematica computes the ArcSin numerically and then recognises the result, 0.628319 as possibly equal to Pi/5. To check it computes Sin[Pi/5], and subtracts it from the original argument to see if it gets ...

15

I'm guessing you're coming from a programming language where every expression must evaluate to a value, and if it didn't evaluate to something (like 5[Cos+Sin]), it's a syntax error. To me, Mathematica started to make a lot more sense, once I stopped thinking about functions and values, and started to think of every expression as evaluating to an "expression ...

14

Short story $$\vartheta(x) = \arg \left[(\operatorname{Bi}x+i \operatorname{Ai}x)e^{-\frac{2}{3} i (-x)^{3/2}}\right]+\frac{2}{3} \operatorname{Re}\left[(-x)^{3/2}\right]$$ Update: I see that you want use only real functions, so you can expand this as  \vartheta(x) = \begin{cases} \arctan\frac{\cos \left(\frac{2}{3} (-x)^{3/2}\right) ...

13

You can give the function one of the Hold Attributes. SetAttributes[fun, HoldFirst] Then as Leonid suggested fun[var_] := SymbolName[Unevaluated@var] Without the hold attribute, this will not work.

13

I don't know why no one mentioned this, but all you have to do is to use a special form of OptionsPattern: pfunc[x0_, plotopts : OptionsPattern[{Plot, pfunc}]] := your-code where inside OptionsPattern go all sub-functions you need, in a list. Now everything is fine and dandy. This has been explained already in this answer of Mr. Wizard, so this answer ...

12

The question made me wonder about zero-order interpolations. It's hardly clear which is the right way. When I tried to figure out why ListLinePlot would use a different Interpolation, I noticed it didn't seem to use an Interpolation for orders 0 or 1 at all, but did it the simple way which you might use by hand: connect the dots. This was probably done ...

12

Maybe : Reduce[Exists[{x}, x^2 + y^2 == 1 ], Reals] -1<=y<=1

12

To avoid those scoping constructs being recognized as such and having their variables renamed, I like wrapping their heads with Identity. In your case, func[opt_] := Identity[With][{a = True}, "x" /. opt]

12

r[n_] := Symbol["r" <> ToString[n]] Then: r[1] gives r1

12

Yes, there is. Group your extra arguments in a list, and address them by their positions in the function under Fold. For your particular example: FoldList[#1 (1 + First@#2) - Last@#2 &, 1000, Transpose@{{.01, .02, .03}, {100, 200, 300}}] (* {1000, 910., 728.2, 450.046} *)

11

The "unnecessary" complication is needed for those cases where you specify deeper levels than the first: MapIndexed[f, {{a}, {b}}, {2}] (* {{f[a, {1, 1}]}, {f[b, {2, 1}]}} *) The following code produces what you want: myMapIndexed[f_, l_] := Inner[f, l, Range[Length[l]], List]; myMapIndexed[f, {a, b, c, d}] (* {f[a, 1], f[b, 2], f[c, 3], f[d, 4]} *)

11

You have many possible solution satisfying: Reduce[{f[-6] == -4, f[-5] == -5, f[1] == 3, f[2] == 2}, {a, b, c, d}] b == -10 a && c == -a && d == -2 a && a != 0 To find a few of those: FindInstance[{f[-6] == -4, f[-5] == -5, f[1] == 3, f[2] == 2}, {a, b, c, d}, Reals, 5] {{a -> -235, b -> 2350, c -> 235, d -> ...

11

Using the fourth and fifth arguments of Partition gives you exactly what you want lis = {5, 6, 9, 3, 2, 6, 7, 8, 1, 1, 4, 7} Max @@@ Partition[lis, 3, 1, {2, 2}, {}] Gives: {6, 9, 9, 9, 6, 7, 8, 8, 8, 4, 7, 7} Update As Simon Wood suggested in the comment below (I also know this but on my system the difference isn't that much), Maping Max instead ...

11

Another option is DeveloperPartitionMap. In RunnyKine's solution we first partition the list and sweep through it to add Max to every element. With Developer`PartitionMap we can do both at the same time, which is faster. Here's a table for reference. My first table was incorrect and I apologize for that, it was an honest mistake which I am not sure how it ...

11

It gives those errors because you explicitly specified that pfunc only has "test" as an option. OptionValue is finicky and will complain if it sees options that it doesn't recognize. There are a couple of alternatives that I can think of: 1: Use FilterRules everywhere instead of OptionValue ClearAll@pfunc2 pfunc2[x0_, plotopts : OptionsPattern[]] := ...

11

This function will be rewritten in C for 10.0.2 and should come down to average-case complexity of $O(n)$ from its current $O(n \log(n))$. Note that the version most users will be bothered to write (and the way we advertized this before in the docpage for DeleteDuplicates) is $O(n^2)$, so most users are probably already winning. In the meantime, my advice ...

10

If we define f[x] e.g. like this: f[x_] := Abs[x] the following returns interesting points: Reduce[ Limit[(f[x + h] - f[x])/h, h -> 0, Assumptions -> x ∈ Reals, Direction -> -1] != Limit[(f[x + h] - f[x])/h, h -> 0, Assumptions -> x ∈ Reals, Direction -> 1], x] x == 0 Let's try another function defined with Piecewise, ...

10

A new solution I realized that comparing each and every value in the sections might be inefficient, especially in cases where the sections are long. Instead we need only the relative ordering of these elements from which we can compute the number of Less pairs. Here is my solution: Edit: Ray Koopman provided a greatly improved counting method (applied to ...

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