5
$\begingroup$

I have a 11x11 matrix and a 23x23 one. The 23x23 is a null matrix and I need to insert the 11x11 matrix in the null matrix for a specific position of the element 6,6 in the first matrix like this:

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

aaaaaaaaaaa000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

00000000000000000000000

So element 6,6 at the first matrix is inserted at the position 6,6 at the second matrix. The elements of the first matrix are not all the same, it is a generic set of data. I need to be able to put the element 6,6 in any position of the second matrix, assuming that it won't overflow the limit of 23x23. How can I do that?

$\endgroup$
4
$\begingroup$

Here is one way:

n=ConstantArray[0,{23,23}];
m=RandomInteger[1,{11,11}];

With[{i=10,j=10},
  ReplacePart[n,Select[
      Flatten@MapIndexed[(#2-{6-1,6-1}+{i-1,j-1})->#1&,m,{2}],
      AllTrue[First@#,Between[#,{1,23}]&]&
    ]
  ]
]//MatrixForm

It puts the element $6,6$ in position $i=10,j=10$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, Kenny. I ended up doing like this: newM = Sum[ Sum[SparseArray[{Band[{(3*j + 1), (3*i + 1)}] -> M}, {n, n}], {i, 0, (d - 1)}], {j, 0, (d - 1)}];, so I can also sum all the combinations I need at the same time :p $\endgroup$ – Rodrigo May 13 '17 at 23:33
8
$\begingroup$

Assuming that you want to insert the 11x11 into the 23x23 at position {6,6}

n = ConstantArray[0, {23, 23}];
m = RandomInteger[1, {11, 11}];

First make a copy of n if you want to keep it (you could work directly with n if it was not needed).

newMatrix = n;

Next equate specific elements of newMatrix with m.

newMatrix[[6 ;; 16, 6 ;; 16]] = m;

Here is what it looks like

newMatrix // MatrixForm

Mathematica graphics

| improve this answer | |
$\endgroup$
7
$\begingroup$

Since the matrix into which you wish to insert is all zeros you can simply PadRight the input matrix. I'll use dimensions 5x5 and 11x11 to keep the examples manageable.

SeedRandom[0]

input = RandomInteger[9, {5, 5}]

$\left( \begin{array}{ccccc} 7 & 0 & 8 & 2 & 1 \\ 5 & 8 & 0 & 6 & 7 \\ 2 & 1 & 0 & 6 & 1 \\ 2 & 8 & 6 & 5 & 5 \\ 8 & 4 & 5 & 9 & 0 \\ \end{array} \right)$

This embeds into an 11x11 array starting three rows down and one column in:

PadRight[input, {11, 11}, 0, {3, 1}]

$\left( \begin{array}{ccccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 0 & 8 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 8 & 0 & 6 & 7 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 6 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 8 & 6 & 5 & 5 & 0 & 0 & 0 & 0 & 0 \\ 0 & 8 & 4 & 5 & 9 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$

You can also use ArrayPad but you'll need to calculate the dimensions:

ArrayPad[input, {{3, 3}, {1, 5}}]

(* same output *)
| improve this answer | |
$\endgroup$
3
$\begingroup$

The reason to post this answer is that although it is longer than a dedicated answer, its code shows the systematic use of the package RSparseMatrix for these kind of tasks.

Import["https://raw.githubusercontent.com/antononcube/\
MathematicaForPrediction/master/Misc/RSparseMatrix.m"]

mat0 = ConstantArray[0, {23, 23}];
mat1 = RandomInteger[1, {11, 11}];

rmat0 = ToRSparseMatrix[SparseArray[mat0], 
   "RowNames" -> Map[ToString, Range[Dimensions[mat0][[1]]]], 
   "ColumnNames" -> Map[ToString, Range[Dimensions[mat0][[2]]]]];

rmat1 = ToRSparseMatrix[SparseArray[mat1], 
   "RowNames" -> Map[ToString, Range[Dimensions[mat1][[1]]]], 
   "ColumnNames" -> Map[ToString, Range[Dimensions[mat1][[2]]]]];

SetRowNames[rmat1, 
  Map[ToString, (6 - 1) + Range[Dimensions[mat1][[1]]]]];
SetColumnNames[rmat1, 
  Map[ToString, (6 - 1) + Range[Dimensions[mat1][[2]]]]];

rmat1 // MatrixForm

enter image description here

rmat2 = ImposeColumnNames[
          ImposeRowNames[rmat1, RowNames[rmat0]], 
          ColumnNames[rmat0]];
rmat2 // MatrixForm

enter image description here

Also see these answers:

  1. https://mathematica.stackexchange.com/a/145785/34008 ,
  2. https://mathematica.stackexchange.com/a/96351/34008 .
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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