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I would like to set 

evenFunction[f_][a_, b_, c_, ...] = f[Abs[a], Abs[b], Abs[c], ...]

I have come up with two ways to do this so far.

  1. Use pure functions

    evenFunction = Function[{f}, f[Sequence @@ Abs[{##}]] &]

  2. Use pattern matching

    evenFunction[f_][x__] := f[Sequence @@ (Abs[{x}])]

What is bothering me is that, in both cases, I first have to turn the arguments into a list, and then back to a sequence. Is there a way without this?

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  • 3
    $\begingroup$ evenFunction[f_][x__] := f @@ Abs[{x}] is a bit simpler. $\endgroup$ – Szabolcs Oct 2 '15 at 15:52
  • $\begingroup$ Simply evenFunction[f_][x__] := Abs /@ f[x] would work if Map held its arguments unevaluated $\endgroup$ – user Oct 2 '15 at 16:23
  • $\begingroup$ I like @Szabolcs's answer the best, but you could always force the unevaluation using evenFunction[f_][a__] := ReleaseHold@Map[Abs, Hold@f@a, {2}]. I don't think there's any reason why you would do this instead though. $\endgroup$ – march Oct 2 '15 at 17:12
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Well, the following meets your formal requirements

evenFunction[f_][args__] := f[Abs /@ Unevaluated[args]]

evenFunction[even][a, b, c]

even[Abs[a], Abs[b], Abs[c]]

But is it really better than

evenFunction[f_][args__] := f @@ Abs[{args}]

I, myself, would choose the 2nd version over the 1st.

Update

It is not necessary to set the attribute SequenceHold as I originally did.

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