3
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Consider for example the nested association:

myA = <|"d" -> <|"c" -> <|"a" -> 1, "b" -> 2|>|>|>

It is easy to map a function to a key on the first level:

MapAt["c" /. # &, myA, "d"]

enter image description here

but MapAt doesn't work with levels higher than 1

MapAt["a" /. # &, myA, "c"]

enter image description here

So I tried a workaround, trasforming all the associations to lists:

  Replace[
    myA /. Association -> ass
    ,
    Rule["c", x_] -> Rule["c", ("a" + "b") /. x], Infinity
  ] /. ass -> Association 

getting the correct result, but with some errors that I can't fix

enter image description here

I've tried with Hold, but it looks like it works at level 1 only.

Thank you for your help!

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2
  • $\begingroup$ Do you need to extract a value or is this only an example of a function "c" /. # &? p.s. it would help to have desired results presented as it is not clear what is the ultimate goal. $\endgroup$
    – Kuba
    Feb 10, 2020 at 11:49
  • $\begingroup$ MapAt["a" /. # &, myA, {All, "c"}]? $\endgroup$
    – kglr
    Feb 10, 2020 at 12:34

1 Answer 1

6
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Here is a function that finds all keys of a given name anywhere in your data structure and maps a function onto those positions:

mapAtKeyEveryWhere[assoc_, fun_, key_] := MapAt[
  fun,
  assoc,
  Append[Key[key]] /@ Position[
    assoc,
    KeyValuePattern[key -> _]
  ]
];

Test it:

myA = <|"d" -> <|"c" -> <|"a" -> 1, "b" -> 2|>|>|>;
mapAtKeyEveryWhere[myA, ("a" + "b") /. # &, "c"]
(* <|"d" -> <|"c" -> 3|>|> *)

Is that what you have in mind?

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1
  • $\begingroup$ Perfect, thanks Sjoerd! $\endgroup$ Feb 10, 2020 at 13:05

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