How to insert a zero column vector in a matrix according to a particular position?

I have a matrix 4*4. i want to insert a zero column vector based on a particular position. position can be 1 or 2 or 3 or 4 or 5. How can we implement this?

Answer 1

[dpb]

ixInsert=N;            % set the index
Z=zeros(size(M,1));    % the insert vector (don't hardcode magic numbers into code)
if ixInsert==1         % insert new column in front
  M=[Z M];  
elseif ixInsert==(size(M,2)+1)  % or at end
  M=[M Z];  
else                    % somewhere in the middle
  M=[M(:,1:ixInsert-1) Z M(:,ixInsert:end)];
end

Add error checking to heart's content...

[chan]

I want to insert only 1 single column of zeros in the matrix M . It should be according to the users choice of defining the position where to insert . Here in your code you are putting the whole Z into the matrix M which i dnt prefer in my problem

[dpb]

Read more carefully... Z is a single column vector the length of which matches the number of rows of the original M so it does precisely what you asked for...I simply made the code generic for any matrix/array, M, rather than, again, hardocding in a magic number for the specific instance.

Ideally, package the above code in a function and call it...

function M=insertColumnN(M,C)
% insertColumn(M,N) places a column of zeros in input Array at column C

Z=zeros(size(M,1));    % the insert vector (don't hardcode magic numbers into code)
if C==1
  M=[Z M];  
elseif C==(size(M,2)+1)
  M=[M Z];  
else
  M=[M(:,1:C-1) Z M(:,C:end)];
end

[chan]

Thank you. I got some brief idea from your code.

Answer 2

[Ayush Singh]

Hi chan,

From your statement I understand that you want to insert a zero column vector at a given position in a 4*4 matrix.

Now suppose you have a N*N matrix and you want to insert the zero column vector at nth position For that

Create a N*1 zero column vector first.

zero_column_vector=zeros(4,1)  % to create N row and 1 column zero vector (Here N is 4)

zero_column_vector = 4×1

0 0 0 0

Now that you have your zero column vector you would need the position where you want to place the vector in the matrix.

Lets say the position is 'n'. So now concatenate the zero column vector in the given position by

[A(:,1:n) zero_column_vector A(:,n+1:end)]    % A is the N*N matrix here

[chan]

Thank you. This gives me some idea how to solve my problem

[dpb]

Exactly what my solution above does except I chose to insert before the input column number instead of after -- that's simply a "count adjust by one" difference.

Answer 3

[Star Strider]

Another approach —

M = randn(4)                                            % Original Matrix

M = 4×4

2.0100 0.5562 0.0693 -1.9815 1.0954 -1.7139 -1.1686 1.0233 -0.1497 0.6061 -0.2158 -0.4245 0.7403 -1.1888 -0.1240 -0.3121

[rows,cols] = size(M);
A = zeros(rows,cols+1);                                 % Augmented Matrix

zeroCol = 3;                                            % Insert Zero Column At #3
idx = setdiff(1:size(A,2), zeroCol);
A(:,idx) = M

A = 4×5

2.0100 0.5562 0 0.0693 -1.9815 1.0954 -1.7139 0 -1.1686 1.0233 -0.1497 0.6061 0 -0.2158 -0.4245 0.7403 -1.1888 0 -0.1240 -0.3121

A = zeros(rows,cols+1);
zeroCol = 5;                                            % Insert Zero Column At #5
idx = setdiff(1:size(A,2), zeroCol)

idx = 1×4

1 2 3 4

A(:,idx) = M

A = 4×5

2.0100 0.5562 0.0693 -1.9815 0 1.0954 -1.7139 -1.1686 1.0233 0 -0.1497 0.6061 -0.2158 -0.4245 0 0.7403 -1.1888 -0.1240 -0.3121 0

EDIT — (17 Jun 2022 at 10:59)

This can easily be coded to a function —

M = randn(4)

M = 4×4

0.2606 1.2419 -0.0297 -0.1365 0.2304 1.6802 0.0920 0.8995 0.1835 0.3499 -1.6399 -0.0510 1.9357 -0.2835 1.1602 -0.0601

for k = 1:size(M,2)+1
    fprintf(repmat('—',1,50))
    InsertZerosColumn = k
    Result = InsertZeroCol(M,k)
end

——————————————————————————————————————————————————

InsertZerosColumn = 1

Result = 4×5

0 0.2606 1.2419 -0.0297 -0.1365 0 0.2304 1.6802 0.0920 0.8995 0 0.1835 0.3499 -1.6399 -0.0510 0 1.9357 -0.2835 1.1602 -0.0601

——————————————————————————————————————————————————

InsertZerosColumn = 2

Result = 4×5

0.2606 0 1.2419 -0.0297 -0.1365 0.2304 0 1.6802 0.0920 0.8995 0.1835 0 0.3499 -1.6399 -0.0510 1.9357 0 -0.2835 1.1602 -0.0601

——————————————————————————————————————————————————

InsertZerosColumn = 3

Result = 4×5

0.2606 1.2419 0 -0.0297 -0.1365 0.2304 1.6802 0 0.0920 0.8995 0.1835 0.3499 0 -1.6399 -0.0510 1.9357 -0.2835 0 1.1602 -0.0601

——————————————————————————————————————————————————

InsertZerosColumn = 4

Result = 4×5

0.2606 1.2419 -0.0297 0 -0.1365 0.2304 1.6802 0.0920 0 0.8995 0.1835 0.3499 -1.6399 0 -0.0510 1.9357 -0.2835 1.1602 0 -0.0601

——————————————————————————————————————————————————

InsertZerosColumn = 5

Result = 4×5

0.2606 1.2419 -0.0297 -0.1365 0 0.2304 1.6802 0.0920 0.8995 0 0.1835 0.3499 -1.6399 -0.0510 0 1.9357 -0.2835 1.1602 -0.0601 0

function NewMtx = InsertZeroCol(OldMtx, InsertZerosCol)
    [rows,cols] = size(OldMtx);                             % Get 'OldMtx' Information
    NewMtx = zeros(rows,cols+1);                            % Augmented Matrix
    idx = setdiff(1:cols+1, InsertZerosCol);                % 'NewMtx' Column Index Vector
    NewMtx(:,idx) = OldMtx;                                 % New Matrix With Inserted Zeros Column
end

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