Unexpected behaviour of Nothing inside List inside Association












7












$begingroup$


When evaluating the following input:



<|a -> {c, d, e}|> /. d -> Nothing


I get (expression 1):



<|a -> {c, Nothing, e}|>


However I would expect to get (expression 2):



<|a -> {c, e}|>


Nevertheless, if i place the cursor on the unexpected output (expression 1) and evaluate it by pressing Shift + Enter, I get the expected output (expression 2), which is even more puzzling.



Mathematica version is 11.0.0.0



Should this behaviour be expected or is it a bug? Why?






list-manipulation replacement associations







share|improve this question









$endgroup$








  • 1




    $begingroup$
    closely related: Pattern matching to a function evaluation inside an Association
    $endgroup$
    – kglr
    2 days ago
















7












$begingroup$


When evaluating the following input:



<|a -> {c, d, e}|> /. d -> Nothing


I get (expression 1):



<|a -> {c, Nothing, e}|>


However I would expect to get (expression 2):



<|a -> {c, e}|>


Nevertheless, if i place the cursor on the unexpected output (expression 1) and evaluate it by pressing Shift + Enter, I get the expected output (expression 2), which is even more puzzling.



Mathematica version is 11.0.0.0



Should this behaviour be expected or is it a bug? Why?






list-manipulation replacement associations







share|improve this question









$endgroup$








  • 1




    $begingroup$
    closely related: Pattern matching to a function evaluation inside an Association
    $endgroup$
    – kglr
    2 days ago














7












7








7


1



$begingroup$


When evaluating the following input:



<|a -> {c, d, e}|> /. d -> Nothing


I get (expression 1):



<|a -> {c, Nothing, e}|>


However I would expect to get (expression 2):



<|a -> {c, e}|>


Nevertheless, if i place the cursor on the unexpected output (expression 1) and evaluate it by pressing Shift + Enter, I get the expected output (expression 2), which is even more puzzling.



Mathematica version is 11.0.0.0



Should this behaviour be expected or is it a bug? Why?






list-manipulation replacement associations







share|improve this question









$endgroup$




When evaluating the following input:



<|a -> {c, d, e}|> /. d -> Nothing


I get (expression 1):



<|a -> {c, Nothing, e}|>


However I would expect to get (expression 2):



<|a -> {c, e}|>


Nevertheless, if i place the cursor on the unexpected output (expression 1) and evaluate it by pressing Shift + Enter, I get the expected output (expression 2), which is even more puzzling.



Mathematica version is 11.0.0.0



Should this behaviour be expected or is it a bug? Why?






list-manipulation replacement associations




list-manipulation replacement associations






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 days ago









darmualdarmual

413




413








  • 1




    $begingroup$
    closely related: Pattern matching to a function evaluation inside an Association
    $endgroup$
    – kglr
    2 days ago














  • 1




    $begingroup$
    closely related: Pattern matching to a function evaluation inside an Association
    $endgroup$
    – kglr
    2 days ago








1




1




$begingroup$
closely related: Pattern matching to a function evaluation inside an Association
$endgroup$
– kglr
2 days ago




$begingroup$
closely related: Pattern matching to a function evaluation inside an Association
$endgroup$
– kglr
2 days ago










5 Answers
5






active

oldest

votes


















8












$begingroup$

The observed puzzle is resolved by the following two observations:



From Leonid's answer in the linked q/a:






  • Association is HoldAllComplete. Once it is created, its parts will then normally be held unevaluated.




From Nothing>> Details





  • Nothing is removed as part of the standard evaluation process. It is not removed in expressions that are held or inactive.




You can Map ReplaceAll on the association to force removal of Nothings:



ReplaceAll[d -> Nothing] /@ assoc (* or *)

Map[# /. d -> Nothing &]@<|a -> {c, d, e}|>



<|a -> {c, e}|>







share|improve this answer











$endgroup$





















    5












    $begingroup$

    Use Replace with level spec All rather than ReplaceAll:



    Replace[<|a -> {c, d, e}|>, d -> Nothing, All]



    <|a -> {c, e}|>




    Related: Is there a difference between Replace with parameter "All" and ReplaceAll






    share|improve this answer









    $endgroup$





















      3












      $begingroup$

      Pattern matching to a function evaluation inside an Association answers your question about bug vs expected result.



      This will not do what your input suggests but let me anticipate your needs:



      <|a -> {b, c, d}, e :> {d, Print[1], d}, f -> d|> //. 
      
  {a___, d, b___} :> {a, b} /. (*this way Print is not evaluated*)
      
  d -> Nothing 
      




      <|a -> {b, c}, e :> {Print[1]}, f -> Nothing|>






      share|improve this answer









      $endgroup$





















        3












        $begingroup$

        For completeness sakes (I believe Coolwater's answer is best):



        assoc = Association[ a → {c,d,e} ];


        Query



        assoc // Query[ All, ReplaceAll[ d → Nothing ] ]
        

        
(* <|a -> {c, e}|> *)


        Note: This will need repeated use of All for nested associations.



        DeleteCases



        Before there was Nothing there was DeleteCases and the like:



        assoc // DeleteCases[ #, d, Infinity ]&
        

        
(* <|a -> {c, e}|> *)





        share|improve this answer











        $endgroup$





















          1












          $begingroup$

          You may use Query to evaluate the result of the replace in an Association. Query gives direct access to the key's value which then evaluated in Query.



          Query[All, # /. d -> Nothing &]@<|a -> {c, d, e}|>
          




          <|a -> {c, e}|>



          Hope this helps.






          share|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8












            $begingroup$

            The observed puzzle is resolved by the following two observations:



            From Leonid's answer in the linked q/a:






            • Association is HoldAllComplete. Once it is created, its parts will then normally be held unevaluated.




            From Nothing>> Details





            • Nothing is removed as part of the standard evaluation process. It is not removed in expressions that are held or inactive.




            You can Map ReplaceAll on the association to force removal of Nothings:



            ReplaceAll[d -> Nothing] /@ assoc (* or *)
            
Map[# /. d -> Nothing &]@<|a -> {c, d, e}|>



            <|a -> {c, e}|>







            share|improve this answer











            $endgroup$


















              8












              $begingroup$

              The observed puzzle is resolved by the following two observations:



              From Leonid's answer in the linked q/a:






              • Association is HoldAllComplete. Once it is created, its parts will then normally be held unevaluated.




              From Nothing>> Details





              • Nothing is removed as part of the standard evaluation process. It is not removed in expressions that are held or inactive.




              You can Map ReplaceAll on the association to force removal of Nothings:



              ReplaceAll[d -> Nothing] /@ assoc (* or *)
              
Map[# /. d -> Nothing &]@<|a -> {c, d, e}|>



              <|a -> {c, e}|>







              share|improve this answer











              $endgroup$
















                8












                8








                8





                $begingroup$

                The observed puzzle is resolved by the following two observations:



                From Leonid's answer in the linked q/a:






                • Association is HoldAllComplete. Once it is created, its parts will then normally be held unevaluated.




                From Nothing>> Details





                • Nothing is removed as part of the standard evaluation process. It is not removed in expressions that are held or inactive.




                You can Map ReplaceAll on the association to force removal of Nothings:



                ReplaceAll[d -> Nothing] /@ assoc (* or *)
                
Map[# /. d -> Nothing &]@<|a -> {c, d, e}|>



                <|a -> {c, e}|>







                share|improve this answer











                $endgroup$



                The observed puzzle is resolved by the following two observations:



                From Leonid's answer in the linked q/a:






                • Association is HoldAllComplete. Once it is created, its parts will then normally be held unevaluated.




                From Nothing>> Details





                • Nothing is removed as part of the standard evaluation process. It is not removed in expressions that are held or inactive.




                You can Map ReplaceAll on the association to force removal of Nothings:



                ReplaceAll[d -> Nothing] /@ assoc (* or *)
                
Map[# /. d -> Nothing &]@<|a -> {c, d, e}|>



                <|a -> {c, e}|>








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered 2 days ago









                kglrkglr

                179k9198410




                179k9198410























                    5












                    $begingroup$

                    Use Replace with level spec All rather than ReplaceAll:



                    Replace[<|a -> {c, d, e}|>, d -> Nothing, All]



                    <|a -> {c, e}|>




                    Related: Is there a difference between Replace with parameter "All" and ReplaceAll






                    share|improve this answer









                    $endgroup$


















                      5












                      $begingroup$

                      Use Replace with level spec All rather than ReplaceAll:



                      Replace[<|a -> {c, d, e}|>, d -> Nothing, All]



                      <|a -> {c, e}|>




                      Related: Is there a difference between Replace with parameter "All" and ReplaceAll






                      share|improve this answer









                      $endgroup$
















                        5












                        5








                        5





                        $begingroup$

                        Use Replace with level spec All rather than ReplaceAll:



                        Replace[<|a -> {c, d, e}|>, d -> Nothing, All]



                        <|a -> {c, e}|>




                        Related: Is there a difference between Replace with parameter "All" and ReplaceAll






                        share|improve this answer









                        $endgroup$



                        Use Replace with level spec All rather than ReplaceAll:



                        Replace[<|a -> {c, d, e}|>, d -> Nothing, All]



                        <|a -> {c, e}|>




                        Related: Is there a difference between Replace with parameter "All" and ReplaceAll







                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered 2 days ago









                        CoolwaterCoolwater

                        14.8k32553




                        14.8k32553























                            3












                            $begingroup$

                            Pattern matching to a function evaluation inside an Association answers your question about bug vs expected result.



                            This will not do what your input suggests but let me anticipate your needs:



                            <|a -> {b, c, d}, e :> {d, Print[1], d}, f -> d|> //. 
                            
  {a___, d, b___} :> {a, b} /. (*this way Print is not evaluated*)
                            
  d -> Nothing 
                            




                            <|a -> {b, c}, e :> {Print[1]}, f -> Nothing|>






                            share|improve this answer









                            $endgroup$


















                              3












                              $begingroup$

                              Pattern matching to a function evaluation inside an Association answers your question about bug vs expected result.



                              This will not do what your input suggests but let me anticipate your needs:



                              <|a -> {b, c, d}, e :> {d, Print[1], d}, f -> d|> //. 
                              
  {a___, d, b___} :> {a, b} /. (*this way Print is not evaluated*)
                              
  d -> Nothing 
                              




                              <|a -> {b, c}, e :> {Print[1]}, f -> Nothing|>






                              share|improve this answer









                              $endgroup$
















                                3












                                3








                                3





                                $begingroup$

                                Pattern matching to a function evaluation inside an Association answers your question about bug vs expected result.



                                This will not do what your input suggests but let me anticipate your needs:



                                <|a -> {b, c, d}, e :> {d, Print[1], d}, f -> d|> //. 
                                
  {a___, d, b___} :> {a, b} /. (*this way Print is not evaluated*)
                                
  d -> Nothing 
                                




                                <|a -> {b, c}, e :> {Print[1]}, f -> Nothing|>






                                share|improve this answer









                                $endgroup$



                                Pattern matching to a function evaluation inside an Association answers your question about bug vs expected result.



                                This will not do what your input suggests but let me anticipate your needs:



                                <|a -> {b, c, d}, e :> {d, Print[1], d}, f -> d|> //. 
                                
  {a___, d, b___} :> {a, b} /. (*this way Print is not evaluated*)
                                
  d -> Nothing 
                                




                                <|a -> {b, c}, e :> {Print[1]}, f -> Nothing|>







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered 2 days ago









                                Kuba♦Kuba

                                104k12201520




                                104k12201520























                                    3












                                    $begingroup$

                                    For completeness sakes (I believe Coolwater's answer is best):



                                    assoc = Association[ a → {c,d,e} ];


                                    Query



                                    assoc // Query[ All, ReplaceAll[ d → Nothing ] ]
                                    

                                    
(* <|a -> {c, e}|> *)


                                    Note: This will need repeated use of All for nested associations.



                                    DeleteCases



                                    Before there was Nothing there was DeleteCases and the like:



                                    assoc // DeleteCases[ #, d, Infinity ]&
                                    

                                    
(* <|a -> {c, e}|> *)





                                    share|improve this answer











                                    $endgroup$


















                                      3












                                      $begingroup$

                                      For completeness sakes (I believe Coolwater's answer is best):



                                      assoc = Association[ a → {c,d,e} ];


                                      Query



                                      assoc // Query[ All, ReplaceAll[ d → Nothing ] ]
                                      

                                      
(* <|a -> {c, e}|> *)


                                      Note: This will need repeated use of All for nested associations.



                                      DeleteCases



                                      Before there was Nothing there was DeleteCases and the like:



                                      assoc // DeleteCases[ #, d, Infinity ]&
                                      

                                      
(* <|a -> {c, e}|> *)





                                      share|improve this answer











                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$

                                        For completeness sakes (I believe Coolwater's answer is best):



                                        assoc = Association[ a → {c,d,e} ];


                                        Query



                                        assoc // Query[ All, ReplaceAll[ d → Nothing ] ]
                                        

                                        
(* <|a -> {c, e}|> *)


                                        Note: This will need repeated use of All for nested associations.



                                        DeleteCases



                                        Before there was Nothing there was DeleteCases and the like:



                                        assoc // DeleteCases[ #, d, Infinity ]&
                                        

                                        
(* <|a -> {c, e}|> *)





                                        share|improve this answer











                                        $endgroup$



                                        For completeness sakes (I believe Coolwater's answer is best):



                                        assoc = Association[ a → {c,d,e} ];


                                        Query



                                        assoc // Query[ All, ReplaceAll[ d → Nothing ] ]
                                        

                                        
(* <|a -> {c, e}|> *)


                                        Note: This will need repeated use of All for nested associations.



                                        DeleteCases



                                        Before there was Nothing there was DeleteCases and the like:



                                        assoc // DeleteCases[ #, d, Infinity ]&
                                        

                                        
(* <|a -> {c, e}|> *)






                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited 2 days ago

























                                        answered 2 days ago









                                        gwrgwr

                                        7,73822558




                                        7,73822558























                                            1












                                            $begingroup$

                                            You may use Query to evaluate the result of the replace in an Association. Query gives direct access to the key's value which then evaluated in Query.



                                            Query[All, # /. d -> Nothing &]@<|a -> {c, d, e}|>
                                            




                                            <|a -> {c, e}|>



                                            Hope this helps.






                                            share|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              You may use Query to evaluate the result of the replace in an Association. Query gives direct access to the key's value which then evaluated in Query.



                                              Query[All, # /. d -> Nothing &]@<|a -> {c, d, e}|>
                                              




                                              <|a -> {c, e}|>



                                              Hope this helps.






                                              share|improve this answer









                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$

                                                You may use Query to evaluate the result of the replace in an Association. Query gives direct access to the key's value which then evaluated in Query.



                                                Query[All, # /. d -> Nothing &]@<|a -> {c, d, e}|>
                                                




                                                <|a -> {c, e}|>



                                                Hope this helps.






                                                share|improve this answer









                                                $endgroup$



                                                You may use Query to evaluate the result of the replace in an Association. Query gives direct access to the key's value which then evaluated in Query.



                                                Query[All, # /. d -> Nothing &]@<|a -> {c, d, e}|>
                                                




                                                <|a -> {c, e}|>



                                                Hope this helps.







                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered 2 days ago









                                                EdmundEdmund

                                                26.2k330100




                                                26.2k330100






























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