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I have had this problem way too often and still have not found a solution, yet. One acually used code is:`

VectorQuantity[perms_List, dim_Integer] :=
 Module[
  {elementList := Range[1, dim],
   collectionList := {},
   v = {}},
  If[collectionList = Union[collectionList, v]; 
     MemberQ[collectionList, #],
     Unevaluated[Sequence[]],
     v = PermutationReplace[#, perms]] & /@ elementList]

perms is a list of cycles. During the mapping over elementList I find elements over which I do not want to map anymore. Right now, they are put into a list, and I use If and MemberQ. I know that there are more issues with the code like the ; within the condition, but those do not matter to me, right now. What I am asking for is whether there is a good way to control the part behind the /@

I apologize for unclearness in the unedited question. I am rather new to Mathematica, and my English is terrible ;)

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  • $\begingroup$ MapAt[f, {1, 2, 3, 4, 5}, {{2}, {5}}] ({1, f[2], 3, 4, f[5]}) ?? $\endgroup$
    – Basheer Algohi
    Feb 5, 2015 at 13:33
  • $\begingroup$ I'm sorry. I have edited the question. Please inform me, if it is still unclear. $\endgroup$
    – Fred
    Feb 5, 2015 at 14:47
  • $\begingroup$ I must admit I still have trouble understanding this. Assuming your code works as you intend what is wrong with using MemberQ and then effectively dropping the element with Unevaluated[Sequence[]]? (Which by the way you can replace with ## &[].) Conceptually how else would you do this? $\endgroup$
    – Mr.Wizard
    Feb 5, 2015 at 16:51
  • $\begingroup$ The problem here is that this code will be performed for very huge dim. But in general, I am really curious about whether there is a way to implement this more efficiently. Now, a friend of mine has replaced it with a "reap-nestwhile-saw"-function which is actually better. (I do not have the code here, right now. If you want, I can add it soon) $\endgroup$
    – Fred
    Feb 13, 2015 at 14:18

2 Answers 2

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Answer rewritten after changes to question

[See the edit history for the previous version.]

As I understand the revised example code, and I'm not sure if this is correct or even if it actually has any utility, the function $f$ should be mapped such that the result $y_i = f(x_i)$ is not calculated if the argument $x_i$ appears in the prior results, $y_1, y_2, ... y_{i-1}$.

I think the following is a function that generalises this:

strangelyMap[f_, elementList_List] := Block[
  {collectionList = {f[First[elementList]]}, g},
  g[x_] := (collectionList = Join[collectionList, {f[x]}]) /; ! MemberQ[collectionList, x];
  g /@ Rest[elementList];
  collectionList]

(Replace Join by Union if you require sorted results with no repeats.)

A somewhat contrived example:

strangelyMap[Mod[#, 10] &, Reverse[Range[20]]]

{0, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0}

And the way to use it with the incomplete example in the question:

strangelyMap[PermutationReplace[#, perms] &, Range[dim]]
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  • $\begingroup$ That appears to be about what I had in mind. Bu I see, I had to show an actual example^^ Nevertheless, thank you! :) $\endgroup$
    – Fred
    Feb 5, 2015 at 14:51
  • $\begingroup$ (after your edit) That function does exactly what I meant. But my actual question is: Is there a more efficient way to implement strangelyMap? Still, thank you very much! :) $\endgroup$
    – Fred
    Feb 13, 2015 at 14:27
  • $\begingroup$ @Fred As you said "I have had this problem way too often" I thought a generalisation would be helpful. I'm not sure there can be a better method unless you know the results in advance. $\endgroup$
    – MikeLimaOscar
    Feb 13, 2015 at 14:39
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$\begingroup$ 
Take[Range[50],25]

???????????????????????????????

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