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is there an elegant way of replacing the RHS of a rule with the value of some function of it. Say I have {{a -> 1.2}, {a -> 2.3}} and I want to replace 1.2 with Sqrt[1.2] and 2.3 with Sqrt[2.3]? I know it can be done by pulling out the values, applying the function and building a new list of rules but this doesn't look very nice.

Thanks, Pete

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    $\begingroup$ MapAt[Sqrt, {{a -> 1.2}, {a -> 2.3}}, {;; , 1, 2}] $\endgroup$
    – Kuba
    Commented Nov 22, 2014 at 13:25

2 Answers 2

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You can use ReplaceAll.

{{a -> 1.2}, {a -> 2.3}} /. ({lhs_ -> rhs_} -> {lhs -> Sqrt[rhs]})
(* {{a -> 1.09545}, {a -> 1.51658}} *)

You can get fancier on the selection rule by changing or conditioning the pattern.

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  • $\begingroup$ great, thank you so much. $\endgroup$
    – Pete
    Commented Nov 22, 2014 at 13:22
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    $\begingroup$ I'd use :>. You never know what lives in Global` :) $\endgroup$
    – Kuba
    Commented Nov 22, 2014 at 15:54
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lst = {{a -> 1.2}, {a -> 2.3}} 
Normal /@ Sqrt /@ Association /@ lst

{{a -> 1.095445}, {a -> 1.516575}}

More generally,

Normal /@ Map[foo] /@ Association /@ lst

{{a -> foo[1.2]}, {a -> foo[2.3]}}

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