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Is it possible to shorten the length of a (delayed) rule (i.e. lhs :> rhs) when there are expressions which repeat several times? Those expressions involve values of the patterns matched in lhs, so ideas that come to my mind, such as creating a pure function that takes the repeated elements as arguments, do not work.

A simple example of what I refer to can be:

Replace[ list , a_ :> {IntegerDigits[a],Length@IntegerDigits[a],IntegerDigits[a][[1]]} ]

Is there a way to write in such case IntegerDigits[a] only once instead of three in there?

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    $\begingroup$ With[{rd = RealDigits[a]}, {a, Length@a, First@a}] $\endgroup$
    – Michael E2
    Jun 2, 2020 at 17:28
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    $\begingroup$ Through[{Identity, Length, First}[RealDigits[a]]] $\endgroup$
    – LouisB
    Jun 2, 2020 at 17:43
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    $\begingroup$ Are you sure that's what you want though? Length@RealDigits[anything] will always be $2$, because RealDigits always returns a list of two elements. Was that supposed to be Length@First@RealDigits[a] instead? $\endgroup$
    – MarcoB
    Jun 2, 2020 at 18:42
  • $\begingroup$ I've edited it and changed the function for IntegerDigits. The function doesn't directly relate to my problem, it was just a simple example to clarify a bit what I meant. $\endgroup$
    – abcd
    Jun 2, 2020 at 19:33

1 Answer 1

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You can define the thing you want to use repeatedly using a variable (instead of a pure function):

Replace[list, a_ :> {b = IntegerDigits[a], Length@b, b[[1]]}]
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