0
$\begingroup$

The following code works for matrices without row/column headings and I like to carry out the aggregation of the selected rows and columns by keeping row/column headings.

ClearAll[mat2, aggregationSUM, aggmat];

aggregationSUM =(*IO matrix aggregation with headings*)
  Fold[Module[{m = Transpose@#, x = #2}, 
    ReplacePart[m, 
      Join[#[[1]] -> Plus @@ m[[#]] & /@ 
        x, {Alternatives @@ Flatten[x[[All, 2 ;;]]] -> 
         Sequence[]}]]] &, #, {##2}] &;    (developed by @kglr)

mat2 = {
{0, c1s1, c1s2, c1s3, c2s1, c2s2, c2s3, c3s1, c3s2, c3s3, 
c1d1, c1d2, c2d1, c2d2, c2d3, c3d1, c3d2}, {c1s1, 1, 3, 0, 2, 2, 
2, 3, 2, 0, 3, 1, 2, 0, 3, 3, 0}, {c1s2, 2, 2, 0, 0, 3, 0, 0, 0, 
0, 0, 0, 3, 0, 1, 2, 0}, {c1s3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0}, {c2s1, 2, 1, 0, 0, 2, 0, 2, 3, 0, 2, 3, 1, 0, 2, 
2, 1}, {c2s2, 1, 3, 0, 0, 2, 1, 3, 0, 0, 3, 3, 2, 0, 1, 2, 
3}, {c2s3, 3, 3, 0, 0, 2, 2, 3, 1, 0, 2, 2, 2, 0, 0, 2, 3}, {c3s1,
1, 2, 0, 0, 2, 0, 3, 0, 0, 1, 0, 1, 0, 2, 3, 1}, {c3s2, 2, 0, 0, 
2, 2, 0, 3, 0, 0, 3, 1, 1, 0, 1, 3, 3}, {c3s3, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {c1t, 3, 1, 0, 0, 0, 3, 1, 0, 0, 0,
1, 2, 0, 2, 0, 0}, {c2t, 2, 1, 0, 0, 3, 3, 3, 0, 0, 2, 1, 0, 0, 
3, 1, 2}, {c3t, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0}, {va, 3, 2, 0, 0, 1, 3, 0, 2, 0, 1, 3, 2, 0, 3, 3, 1}};

aggmat = aggregationSUM[mat2,
         {{2, 3}, {6, 7, 8}},   (*rows*)
         {{2, 3}, {6, 7, 8}}    (*columns*)]

labRow={r1, r2};   (*new labels for the aggregated rows*) 
labCol={c1, c2};   (*new labels for the aggregated columns*)

It will convenient if these labels are created automatically by linking them to aggregationSUM[].

$\endgroup$

1 Answer 1

2
$\begingroup$
ClearAll[aggregationSumWithHeaders]
aggregationSumWithHeaders = Fold[Module[{m = Transpose @ #, x = 1 + #2}, 
      ReplacePart[m, Join[#[[1, 1]] -> 
           ReplacePart[Total[m[[#[[1]]]]], 1 -> #[[2]] - 1] & /@ x, 
        {Alternatives @@ Flatten[x[[All, 1, 2 ;;]]] -> Sequence[]}]]] &, #, {##2}] &;

labRow = Highlighted /@ {r1, r2};
labCol = Highlighted /@ {c1, c2};


aggregationSumWithHeaders[mat2, 
  Thread[{{{2, 3}, {6, 7, 8}}, Highlighted /@ labCol}], 
  Thread[{{{2, 3}, {6, 7, 8}}, Highlighted /@ labRow}] ] // MatrixForm

enter image description here

$\endgroup$
2
  • $\begingroup$ When I run aggregationSumWithHeaders[ mat2, {{2, 3}, {6, 7, 8}}, {{2, 3}, {6, 7, 8}} ], I receive the error Part specification {3,4}[[1,1]] is longer than depth of object. Is there anything I do wrong here? $\endgroup$ Feb 5, 2021 at 1:40
  • 1
    $\begingroup$ @Tugrul, to avoid hard-coded new labels we need to provide Indices as well as labels for each list of rows/columns to be combined. So we need to use aggregationSumWithHeaders[mat2, Thread[{{{2, 3}, {6, 7, 8}}, labCol}], Thread[{{{2, 3}, {6, 7, 8}}, labRow}]] or aggregationSumWithHeaders[mat2, {{{2, 3}, c1}, {{6, 7, 8}, c2}}, {{{2, 3}, r1}, {{6, 7, 8}, r2}}] $\endgroup$
    – kglr
    Feb 5, 2021 at 1:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.