Replacing elements of one table with the elements of another table for common grid

Question

Consider two tables:

tab1=Flatten[Table[{i,j,0},{i,0.5,0.7,0.1},{j,0.1,0.3,0.1}],{1,2}];
tab2=Flatten[Table[{i,j,i*j},{i,0.5,0.7,0.2},{j,0.1,0.3,0.2}],{1,2}];

These tables have a common subset of the first and second elements of their rows: e.g., {0.5,0.1}, and {0.7,0.3}. How to make a table from tab1 where zeros (the third element) in the rows from the common subset are replaced with the corresponding values from tab2?

Answer 1

Something along these lines might work:

tab2Rules = Thread[ReplacePart[3 -> _] /@ tab2 -> tab2];
tab1 /. tab2Rules

{{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0}, {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07}, {0.7, 0.2, 0}, {0.7, 0.3, 0.21}}

Answer 2

Clear["Global`*"]
tab1 = Flatten[
   Table[{i, j, 0}, {i, 0.5, 0.7, 0.1}, {j, 0.1, 0.3, 0.1}], {1, 2}];
tab2 = Flatten[
   Table[{i, j, i*j}, {i, 0.5, 0.7, 0.2}, {j, 0.1, 0.3, 0.2}], {1, 
    2}];

Create rules for a Lookup:

r1 = Sequence @@@ ({Most@# -> Last@#} & /@ tab2)

{{0.5, 0.1} -> 0.05, {0.5, 0.3} -> 0.15, {0.7, 0.1} -> 0.07, {0.7, 0.3} -> 0.21}

Sequence @@@ {Most@#, Lookup[r1, {Most@#}, Last@#]} & /@ tab1 

{{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0}, {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07}, {0.7, 0.2, 0}, {0.7, 0.3, 0.21}}

Answer 3

Using Association - overwriting

Join @@ Values @ Association[GroupBy[Most] @ tab1, GroupBy[Most] @ tab2]

{{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0}, {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07}, {0.7, 0.2, 0}, {0.7, 0.3, 0.21}}

Answer 4

Replace[tab1,(MapThread[{#1,#2,0}->{##1}&,Transpose@tab2]),2]

(* {
    {0.5,0.1,0.05},
    {0.5,0.2,0},
    {0.5,0.3,0.15},
    {0.6,0.1,0},
    {0.6,0.2,0},
    {0.6,0.3,0},
    {0.7,0.1,0.07},
    {0.7,0.2,0},
    {0.7,0.3,0.21}
   } *)

Answer 5

Using Merge and MapApply (new in 13.1)

res = Merge[Last] @ MapApply[{#1, #2} -> #3 &] @ Join[tab1, tab2]

<|{0.5, 0.1} -> 0.05, {0.5, 0.2} -> 0, {0.5, 0.3} -> 0.15, {0.6, 0.1} -> 0, {0.6, 0.2} -> 0, {0.6, 0.3} -> 0, {0.7, 0.1} -> 0.07, {0.7, 0.2} -> 0, {0.7, 0.3} -> 0.21|>

Convert to lists

KeyValueMap[Flatten @* List] @ res

{{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0}, {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07}, {0.7, 0.2, 0}, {0.7, 0.3, 0.21}}

Answer 6

My attempt to do this, defining ReplaceByCoords:

 ReplaceByCoords[tab1_, tab2_] := 
 Module[{assis, res}, 
 assis[elem_] := 
 Module[{match}, match = SelectFirst[tab2, Most[#] == Most[elem] &];
 If[MissingQ[match], elem, match]]; es = assis /@ tab1; res]

Testing ReplaceByCoords:

res = {{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0}, 
      {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07}, {0.7, 0.2, 0}, 
      {0.7, 0.3, 0.21}};

ReplaceByCoords[tab1, tab2] === res

(*True*)

Answer 7

Values @ AssociationThread[Most /@ #, #] & @ Join[tab1, tab2]

{{0.5, 0.1, 0.05}, {0.5, 0.2, 0}, {0.5, 0.3, 0.15}, {0.6, 0.1, 0},   
 {0.6, 0.2, 0}, {0.6, 0.3, 0}, {0.7, 0.1, 0.07},   
 {0.7, 0.2, 0}, {0.7, 0.3, 0.21}}

Replacements highlighted:

% /. a : Alternatives @@ tab2 :> Highlighted[a]