6
$\begingroup$

I have an association

<| a->f, b->g |>

where f and g are pure functions. Is there a nice way to apply arguments to both f and g? E.g.

<| a->f, b->g |>[x] --> <| a->f[x], b->g[x] |>
$\endgroup$
7
$\begingroup$

As Kuba provided in a comment we can Map a Function that applies its argument to a specific expression, e.g. x:

#[x] & /@ <|a -> f, b -> g|>
<|a -> f[x], b -> g[x]|>
$\endgroup$
  • $\begingroup$ Still, as did OP, I'd expect Through to work (does it?) $\endgroup$ – LLlAMnYP Sep 26 '17 at 8:16
  • $\begingroup$ @LLlAMnYP you mean like Through[<|a -> f, b -> g|>[x]] ? No, it does not; the body evaluates to Missing["KeyAbsent", x]. Through[Unevaluated[<|a -> f, b -> g|>[x]]] gives Association[(a -> f)[x], (b -> g)[x]] but I don't find that helpful. With[{asc = <|a -> f, b -> g|>}, Through[Unevaluated[asc[x]]]] just kicks back Through[Unevaluated[<|a -> f, b -> g|>[x]]]. $\endgroup$ – Mr.Wizard Sep 26 '17 at 8:37
  • $\begingroup$ Thanks, I do not mind a non cw answer next time :) $\endgroup$ – Kuba Sep 26 '17 at 9:42

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