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] |>
6
$\begingroup$
$\endgroup$
7
$\begingroup$
$\endgroup$
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]|>
-
$\begingroup$ 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 toMissing["KeyAbsent", x].Through[Unevaluated[<|a -> f, b -> g|>[x]]]givesAssociation[(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 backThrough[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
