Replacing the Zeros of One List by the (i-1)th Element of Another List












5












$begingroup$


My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if



list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}


and 



list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}


the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.



My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1. 



zeroslist1 = Flatten[Position[list1, 0]]

 DeleteCases[

 Flatten[Reap@

   Do[Sow[ReplacePart[vtest1, 

      i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]


The results of the output are:



{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.



Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.






list-manipulation functions function-construction







share|improve this question









$endgroup$

















    5












    $begingroup$


    My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if



    list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}


    and 



    list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}


    the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.



    My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1. 



    zeroslist1 = Flatten[Position[list1, 0]]
    
 DeleteCases[
    
 Flatten[Reap@
    
   Do[Sow[ReplacePart[vtest1, 
    
      i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]


    The results of the output are:



    {{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.



    Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.






    list-manipulation functions function-construction







    share|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if



      list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}


      and 



      list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}


      the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.



      My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1. 



      zeroslist1 = Flatten[Position[list1, 0]]
      
 DeleteCases[
      
 Flatten[Reap@
      
   Do[Sow[ReplacePart[vtest1, 
      
      i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]


      The results of the output are:



      {{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.



      Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.






      list-manipulation functions function-construction







      share|improve this question









      $endgroup$




      My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if



      list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}


      and 



      list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}


      the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.



      My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i in$zeroslist1. 



      zeroslist1 = Flatten[Position[list1, 0]]
      
 DeleteCases[
      
 Flatten[Reap@
      
   Do[Sow[ReplacePart[vtest1, 
      
      i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]


      The results of the output are:



      {{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.



      Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.






      list-manipulation functions function-construction




      list-manipulation functions function-construction






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 11 hours ago









      WillWill

      3347




      3347






















          5 Answers
          5






          active

          oldest

          votes


















          5












          $begingroup$

          You can use a multiplication instead of looping or conditionals:



          list1 + (1 - list1) Prepend[Most[list2], 0]



          {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




          The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.






          share|improve this answer











          $endgroup$













          • $begingroup$
            What a simple way to do this. Thank you so much.
            $endgroup$
            – Will
            10 hours ago



















          4












          $begingroup$

          idx = Random`Private`PositionsOf[Rest[list1], 0];
          
result = list1;
          
result[[idx + 1]] = list2[[idx]];
          
result



          {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







          share|improve this answer









          $endgroup$





















            3












            $begingroup$

            Here's another possibility:



            Module[{tmp=list1},
            
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
            
        tmp[[i+1]]=list2[[i]]
            
    ];
            
    tmp
            
]



            {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







            share|improve this answer









            $endgroup$





















              2












              $begingroup$

              Also MapIndexed works fine



              (# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]


              The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.






              share|improve this answer









              $endgroup$





















                0












                $begingroup$

                MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most



                {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                Your answer from above




                {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                share|improve this answer









                $endgroup$













                  Your Answer





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                  5 Answers
                  5






                  active

                  oldest

                  votes








                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  5












                  $begingroup$

                  You can use a multiplication instead of looping or conditionals:



                  list1 + (1 - list1) Prepend[Most[list2], 0]



                  {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                  The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    What a simple way to do this. Thank you so much.
                    $endgroup$
                    – Will
                    10 hours ago
















                  5












                  $begingroup$

                  You can use a multiplication instead of looping or conditionals:



                  list1 + (1 - list1) Prepend[Most[list2], 0]



                  {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                  The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    What a simple way to do this. Thank you so much.
                    $endgroup$
                    – Will
                    10 hours ago














                  5












                  5








                  5





                  $begingroup$

                  You can use a multiplication instead of looping or conditionals:



                  list1 + (1 - list1) Prepend[Most[list2], 0]



                  {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                  The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.






                  share|improve this answer











                  $endgroup$



                  You can use a multiplication instead of looping or conditionals:



                  list1 + (1 - list1) Prepend[Most[list2], 0]



                  {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                  The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 10 hours ago

























                  answered 10 hours ago









                  RomanRoman

                  1,147511




                  1,147511












                  • $begingroup$
                    What a simple way to do this. Thank you so much.
                    $endgroup$
                    – Will
                    10 hours ago


















                  • $begingroup$
                    What a simple way to do this. Thank you so much.
                    $endgroup$
                    – Will
                    10 hours ago
















                  $begingroup$
                  What a simple way to do this. Thank you so much.
                  $endgroup$
                  – Will
                  10 hours ago




                  $begingroup$
                  What a simple way to do this. Thank you so much.
                  $endgroup$
                  – Will
                  10 hours ago











                  4












                  $begingroup$

                  idx = Random`Private`PositionsOf[Rest[list1], 0];
                  
result = list1;
                  
result[[idx + 1]] = list2[[idx]];
                  
result



                  {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                  share|improve this answer









                  $endgroup$


















                    4












                    $begingroup$

                    idx = Random`Private`PositionsOf[Rest[list1], 0];
                    
result = list1;
                    
result[[idx + 1]] = list2[[idx]];
                    
result



                    {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                    share|improve this answer









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      idx = Random`Private`PositionsOf[Rest[list1], 0];
                      
result = list1;
                      
result[[idx + 1]] = list2[[idx]];
                      
result



                      {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                      share|improve this answer









                      $endgroup$



                      idx = Random`Private`PositionsOf[Rest[list1], 0];
                      
result = list1;
                      
result[[idx + 1]] = list2[[idx]];
                      
result



                      {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}








                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 11 hours ago









                      Henrik SchumacherHenrik Schumacher

                      52.9k471148




                      52.9k471148























                          3












                          $begingroup$

                          Here's another possibility:



                          Module[{tmp=list1},
                          
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
                          
        tmp[[i+1]]=list2[[i]]
                          
    ];
                          
    tmp
                          
]



                          {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                          share|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            Here's another possibility:



                            Module[{tmp=list1},
                            
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
                            
        tmp[[i+1]]=list2[[i]]
                            
    ];
                            
    tmp
                            
]



                            {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                            share|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Here's another possibility:



                              Module[{tmp=list1},
                              
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
                              
        tmp[[i+1]]=list2[[i]]
                              
    ];
                              
    tmp
                              
]



                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                              share|improve this answer









                              $endgroup$



                              Here's another possibility:



                              Module[{tmp=list1},
                              
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
                              
        tmp[[i+1]]=list2[[i]]
                              
    ];
                              
    tmp
                              
]



                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}








                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 11 hours ago









                              Carl WollCarl Woll

                              67.9k389176




                              67.9k389176























                                  2












                                  $begingroup$

                                  Also MapIndexed works fine



                                  (# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]


                                  The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.






                                  share|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    Also MapIndexed works fine



                                    (# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]


                                    The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.






                                    share|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Also MapIndexed works fine



                                      (# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]


                                      The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.






                                      share|improve this answer









                                      $endgroup$



                                      Also MapIndexed works fine



                                      (# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]


                                      The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 11 hours ago









                                      Mane.andreaMane.andrea

                                      1366




                                      1366























                                          0












                                          $begingroup$

                                          MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most



                                          {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                                          Your answer from above




                                          {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                                          share|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most



                                            {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                                            Your answer from above




                                            {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                                            share|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most



                                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                                              Your answer from above




                                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}







                                              share|improve this answer









                                              $endgroup$



                                              MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most



                                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}




                                              Your answer from above




                                              {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}








                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered 11 hours ago









                                              MikeYMikeY

                                              2,697412




                                              2,697412






























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