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Given a table of values, I know how to find find maximum values, but how do I find the position of the maximum value in each column.

For example, given

{ {0.803279, 0.958913, 0.600443, 0.928255, 0.425632, 0.165858}, 
  {0.550107, 0.929972, 0.990928, 0.110509, 0.803279, 0.939139}, 
  {0.693203, 0.823982, 0.645499, 0.617851, 0.461366, 0.252978}, 
  {0.277155, 0.321569, 0.796915, 0.976772, 0.462962, 0.944314} }

what so I apply to get

{{1, 1}, {1, 2}, {2, 3}, {4, 4}, {2, 5}, {4, 6}}

especially in cases where the maximum may not be unique (e.g. two columns could have the same maximum value).

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  • $\begingroup$ Related: 18660 $\endgroup$
    – Michael E2
    Nov 14, 2014 at 19:49

6 Answers 6

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You can use this (where t is your dataset):

Ordering[#, -1] & /@ Transpose[t]

which produces

{{1}, {1}, {2}, {4}, {2}, {2}}

Incidentally, the list of positions you gave in your question is wrong (the 4th element should be {4,4}, and the 6th element should be {2,6}).

The above method omits the first coordinates in your expected output, since they are redundant. If you want to have them anyways, do this:

{Flatten[Ordering[#, -1] & /@ Transpose[t]], Range[6]}\[Transpose]

which gives

{{1, 1}, {1, 2}, {2, 3}, {4, 4}, {2, 5}, {4, 6}}
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  • 2
    $\begingroup$ +1, Nice solution for Ordering[#, -1] & /@ Transpose[t] $\endgroup$
    – xyz
    Nov 15, 2014 at 2:31
  • 1
    $\begingroup$ @DumpsterDoofus this is a neat answer +1, it does not deal with repeated maxima in column, however.:) $\endgroup$
    – ubpdqn
    Nov 15, 2014 at 11:12
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Dealing with repeated maxima and to reproduce the expressed desired output:

func[list_] := 
 Join @@ MapIndexed[Thread[{First /@ Position[#, Max@#], First@#2}] &,
    Transpose[list]]

Testing

test = {{0.803279, 0.958913, 0.600443, 0.928255, 0.425632, 
   0.165858}, {0.550107, 0.929972, 0.990928, 0.110509, 0.803279, 
   0.939139}, {0.693203, 0.823982, 0.645499, 0.617851, 0.461366, 
   0.252978}, {0.277155, 0.321569, 0.796915, 0.976772, 0.462962, 
   0.944314}};
func[test]

yields:

(*{{1, 1}, {1, 2}, {2, 3}, {4, 4}, {2, 5}, {4, 6}}*)

Example with repeated values:

func[{{1, 2, 3}, {3, 4, 3}, {2, 5, 2}, {3, 5, 1}}]

yields:

(*{{2, 1}, {4, 1}, {3, 2}, {4, 2}, {1, 3}, {2, 3}}*)
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The solution by @DumpsterDoofus is no doubt the simplest and most elegant Mathematica solution. Just for fun I wrote a more direct construction, that gives both the maxima and their positions.

mat={{0.803279,0.958913,0.600443,0.928255,0.425632,0.165858},
  {0.550107,0.929972,0.990928,0.110509,0.803279,0.939139},
  {0.693203,0.823982,0.645499,0.617851,0.461366,0.252978},
  {0.277155,0.321569,0.796915,0.976772,0.462962,0.944314}};

maxvalues=mat[[1]];
maxpositions=Table[1, {Length[mat[[1]]]}];
Do[
  maxpositions=MapThread[Max, {maxpositions,  n  Sign[mat[[n]] - maxvalues]}];
   maxvalues=MapThread[Max,{maxvalues, mat[[n]]}],
  {n, 2, Length[mat]}];
maxvalues
maxpositions

(* {0.803279,0.958913,0.990928,0.976772,0.803279,0.944314} *)

(* {1,1,2,4,2,4} *)

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My solution:

data =
 {{0.803279, 0.958913, 0.600443, 0.928255, 0.425632, 0.165858}, 
  {0.550107, 0.929972, 0.990928, 0.110509, 0.803279, 0.939139}, 
  {0.693203, 0.823982, 0.645499, 0.617851, 0.461366, 0.252978},
  {0.277155, 0.321569, 0.796915, 0.976772, 0.462962, 0.944314}}

SortBy[
 Union@
   Flatten[Position[data, #] & /@ (Max /@ Transpose@data), 1], Last]
{{1, 1}, {1, 2}, {2, 3}, {4, 4}, {2, 5}, {4, 6}}
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list =
  {{0.803279, 0.958913, 0.600443, 0.928255, 0.425632, 0.165858},
   {0.550107, 0.929972, 0.990928, 0.110509, 0.803279, 0.939139},
   {0.693203, 0.823982, 0.645499, 0.617851, 0.461366, 0.252978},
   {0.277155, 0.321569, 0.796915, 0.976772, 0.462962, 0.944314}};

1.

Using TakeLargest (new in 10.1) and Splice (new in 12.1)

pos = Splice @ TakeLargest[# -> "Index", 1] & /@ Transpose[list]

To get the desired indexed result:

Flatten /@ MapIndexed[List] @ pos

{{1, 1}, {1, 2}, {2, 3}, {4, 4}, {2, 5}, {4, 6}}

To also get the elements:

Splice @ TakeLargest[# -> {"Element", "Index"}, 1] & /@ Transpose[list]

{{0.803279, 1}, {0.958913, 1}, {0.990928, 2}, {0.976772, 4}, {0.803279, 2}, {0.944314, 4}}

2.

Using PositionLargest (new in 13.2)

Splice @* PositionLargest /@ Transpose[list]

{1, 1, 2, 4, 2, 4}

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Using Cases and Position:

list = {{0.803279, 0.958913, 0.600443, 0.928255, 0.425632, 0.165858}, 
        {0.550107, 0.929972, 0.990928, 0.110509, 0.803279, 0.939139}, 
        {0.693203, 0.823982, 0.645499, 0.617851, 0.461366, 0.252978}, 
        {0.277155, 0.321569, 0.796915, 0.976772, 0.462962, 0.944314}};

Position[#, Alternatives @@ Cases[Transpose@#, v_ :> Max[v]] &@#] &@list

(*{{1, 1}, {1, 2}, {2, 3}, {2, 5}, {4, 4}, {4, 6}}*)
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