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I have an expr which has List at various levels and I want to display them nicely as a column inside a frame but my problem is that the search for List goes inside things like DateObject to ruin their structure.

Replace[
    expr,
    x_List/;!MatchQ[...[...x...],DateObject[x,__]]:>Framed[Column@x]
    All
]

So how should I define such a Rule?

Sample expr:

{1->a,2->b,3->c,4->{5->d,6->e,7->Today},8->Yesterday,9->{10->f,11->{12->Tomorrow,13->g}}}

Current output:

Framed[Column[{1 -> a, 2 -> b, 3 -> c, 4 -> Framed[Column[{5 -> d, 6 -> e, 7 -> DateObject[Framed[Column[{2020, 1, 28}]], "Day", CalendarType -> "Gregorian", DateFormat -> Automatic]}]], 
       8 -> DateObject[Framed[Column[{2020, 1, 27}]], "Day", CalendarType -> "Gregorian", DateFormat -> Automatic], 
       9 -> Framed[Column[{10 -> f, 11 -> Framed[Column[{12 -> DateObject[Framed[Column[{2020, 1, 29}]], "Day", CalendarType -> "Gregorian", DateFormat -> Automatic], 13 -> g}]]}]]}]]

Desired output:

Framed[Column[{1 -> a, 2 -> b, 3 -> c, 4 -> Framed[Column[{5 -> d, 6 -> e, 7 -> DateObject[{2020, 1, 28}, "Day", CalendarType -> "Gregorian", DateFormat -> Automatic]}]], 
       8 -> DateObject[{2020, 1, 27}, "Day", CalendarType -> "Gregorian", DateFormat -> Automatic], 
       9 -> Framed[Column[{10 -> f, 11 -> Framed[Column[{12 -> DateObject[{2020, 1, 29}, "Day", CalendarType -> "Gregorian", DateFormat -> Automatic], 13 -> g}]]}]]}]]
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5
  • $\begingroup$ Can you supply a sample expr? It's hard to provide help if other people don't have anything they can evaluate. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jan 28, 2020 at 17:16
  • $\begingroup$ @J.M. sample expr added. $\endgroup$
    – user13892
    Commented Jan 28, 2020 at 17:38
  • $\begingroup$ I'm a little bit confused about what exactly you want the output to look like. Are you trying to extract every expression that has List as a Head and then put those in a column? Or are you trying to extract just the rules at level 1, or what? Can you give a (rough) example of your desired output? $\endgroup$
    – march
    Commented Jan 28, 2020 at 17:59
  • $\begingroup$ @march There is no extraction, it is a replacement problem at any level. See the current vs desired output. $\endgroup$
    – user13892
    Commented Jan 28, 2020 at 18:13
  • $\begingroup$ Related: (25538) $\endgroup$
    – Mr.Wizard
    Commented Jan 30, 2020 at 3:51

3 Answers 3

Reset to default
4
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Another option is Developer`ReplaceAllUnheld.

expr = {1 -> a, 2 -> b, 3 -> c, 4 -> {5 -> d, 6 -> e, 7 -> Today}, 8 -> Yesterday, 
   9 -> {10 -> f, 11 -> {12 -> Tomorrow, 13 -> g}}};

Block[{DateObject},
 Attributes[DateObject] = HoldAll; 
 Developer`ReplaceAllUnheld[expr, List -> Framed@*Column@*List]
]

enter image description here

Also see What is the potential usage of ReplaceAllUnheld in Developer Utilities Package?

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2
  • 1
    $\begingroup$ Thank you all for the wonderful answers and I have upvoted all of them but after careful consideration, I am accepting this one since it is very easy to extend and I don't have to define additional rules. In my application, I had to skip multiple Heads and just increasing Block variables and a line like SetAttributes[{DateObject,Entity,Style,Missing},HoldAll]; seemed far convenient than defining separate rules. $\endgroup$
    – user13892
    Commented Feb 10, 2020 at 16:08
  • $\begingroup$ @user13892 I am glad my answer was useful, and thanks for the Accept. $\endgroup$
    – Mr.Wizard
    Commented Feb 11, 2020 at 2:28
15
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You can use the fact that once an object is replaced, it doesn't get replaced by any further rules. So, just add a rule that replaces a DateObject with itself:

expr = {1 -> a, 2 -> b, 3 -> c, 4 -> {5 -> d, 6 -> e, 7 -> Today}, 8 -> Yesterday, 9 -> {10 -> f, 11 -> {12 -> Tomorrow, 13 -> g}}}l
expr /. {a_DateObject :> a, List -> Framed@*Column@*List}

enter image description here

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3
  • $\begingroup$ Please explain the reason for changing x_List:>Framed[Column@x] to List->Framed@*Column@*List and why the former doesn't work in your trick? $\endgroup$
    – user13892
    Commented Jan 28, 2020 at 18:42
  • $\begingroup$ Also, can you provide a solution that works with Replace[#,{rules},All]& rather than ReplaceAll[#,{rules}]&. $\endgroup$
    – user13892
    Commented Jan 28, 2020 at 18:43
  • 2
    $\begingroup$ @user13892 you won't be able to get this to work well with Replace[#, {rules}, All]& since that traverses the expression tree from the lowest level upwards. You really do need ReplaceAll. And List->Framed@*Column@*List is replacing the Head List with (in essence) the function (Framed@Column@List[##]&) which just means that you never need to bind x $\endgroup$
    – b3m2a1
    Commented Jan 28, 2020 at 19:26
5
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We can get the desired result using Replace in several ways:

1. We can temporarily change the behavior of Framed inside expressions with head DateObject (using Block and TagSetDelayed):

Block[{Framed}, 
 Framed /: DateObject[Framed[a_], b___] := DateObject[First @ a, b]; 
 Replace[expr, a_List :> Framed[Column @ a], All]]

enter image description here

Alternatively,

Block[{Framed}, 
 Framed /: DateObject[Framed[Column[a_]], b___] := DateObject[a, b];
 Replace[expr, a_List :> Framed[Column@a], All]]

same picture

2. Alternatively, we can use two replacement rules where the second rule undoes the replacements inside DateObjects effected by the first rule:

Replace[expr, {a_List :> Framed[Column@a], d_DateObject :> (d /. Framed -> First)}, All]

enter image description here

Alternatively,

Replace[expr, {List -> Framed@*Column@*List, 
  DateObject -> (DateObject[##] /. Framed -> First &)}, All, Heads -> True]

same picture

Note the need to use the option Heads -> True in the second approach.

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