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

Question

chan 2022-6-13

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739410-how-to-insert-a-zero-column-vector-in-a-matrix-according-to-a-particular-position

编辑： Star Strider 2022-6-17

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?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

dpb 2022-6-13

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739410-how-to-insert-a-zero-column-vector-in-a-matrix-according-to-a-particular-position#answer_984630

编辑：dpb 2022-6-14

在 MATLAB Online 中打开

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...

3 个评论
显示 1更早的评论隐藏 1更早的评论

dpb 2022-6-14

编辑：dpb 2022-6-14

在 MATLAB Online 中打开

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 2022-6-16

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

请先登录，再进行评论。

Answer 2

Ayush Singh 2022-6-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739410-how-to-insert-a-zero-column-vector-in-a-matrix-according-to-a-particular-position#answer_985165

在 MATLAB Online 中打开

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

2 个评论
显示无隐藏无

chan 2022-6-15

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

dpb 2022-6-15

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.

请先登录，再进行评论。

Answer 3

Star Strider 2022-6-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1739410-how-to-insert-a-zero-column-vector-in-a-matrix-according-to-a-particular-position#answer_987180

编辑：Star Strider 2022-6-17

在 MATLAB Online 中打开

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