How can I insert a row in the middle of a matrix/vector?

My question is also about this subject, but appears a bit more complicated. I also need to insert new rows into a matrix (it's a vector column actually), but in different positions. Let's see: I have this column vector for instance:

[1;2;3;8;9;10;14;16;17]

what I need is to insert new elements in different places. As you can see, the values 4 5 6 7, 11 12 13 and 15 are missing in specific positions. In position 4 I need to insert 4 values (4 5 6 7), in position 7 I need to insert (11 12 13) and in position 8 I need to insert (15).

I used the function diff in order to calculate the difference in the "leaps" and I got somthing like this (11511421) I thought of using a loop for ----- for diff~=1 but after that I got stuck I didn't know to follow.

I would really thank to any who would give me an answer back. I'm waiting for any light!!

[Jan]

Acoording to Imaga Analyst's answer: This sounds like a (too) complicated method to create the vector (1:17).' . Perhaps you simplified the problem too much.

Answer 1

[Roger Stafford]

Suppose you want to insert the column vector y in the column vector x starting at index ix.

   x = [x(1:ix-1);y;x(ix,end)];

[Warning: If you plan to make other insertions afterwards on the same vector, its indexing will be different after the point ix.]

Answer 2

[ConsistencyPLS]

Say you have the matrix A given by

A = [1,  1;
     2,  4;
     3,  9;
     4, 16;
     5, 25; 
     6, 36;
     7, 49];

and say that want to add the rows [1.5, 2.25] after row 1, [5.5, 30.25] after row 5 and [6.5, 42.25] after row 6. Lets put this information in the matrix B and vector ind,

B = [1.5,  2.25;
     5.5, 30.25;
     6.5, 42.25];
ind = [1, 5, 6];

Here is my solution for adding these rows in,

% Preallocate output
>> Anew = zeros(size(A,1)+size(B,1),size(A,2));
% Find indices for old data
>> addRows = ismember(1:size(A,1), ind)

addRows =

    1×7 logical array

     1   0   0   0   1   1   0

>> oldDataInd = (1:size(A,1)) + cumsum([0, addRows(1:end-1)])

oldDataInd =

     1     3     4     5     6     8    10

% Add in old data
>> Anew(oldDataInd,:) = A(:,:)

Anew =

       1     1
       0     0
       2     4
       3     9
       4    16
       5    25
       0     0
       6    36
       0     0
       7    49

% Find indices for new data
>> newDataInd = (1:length(ind)) + ind

newDataInd =

       2     7     9

% Add in new data
>> Anew(newDataInd,:) = B(:,:);

% Verify output
>> display(Anew)

Anew =

      1.0000    1.0000
      1.5000    2.2500
      2.0000    4.0000
      3.0000    9.0000
      4.0000   16.0000
      5.0000   25.0000
      5.5000   30.2500
      6.0000   36.0000
      6.5000   42.2500
      7.0000   49.0000

% Pass!

Hopefully this explanation is easy enough to follow and adapt.

[Maryam S]

You're a great teacher! Thanks

[Scott Prince]

This was a great find for me; thanks so much for posting this solution!

[Dana Galili]

Thank you for this useful solution and detailed explaination!

Answer 3

[Kuifeng]

 %   Here is a walkaround for this problem
%   1. Define new vector Y based on existing vector x,
Y = ones(length(x) + No.Of.New.Values);
Y(:) = nan;
%2. Insert New.Values into respective cells of Y, for example
Y(4:7) = 4:7; %and others
% 3. after all new values inserted, find locations of NaN in Y;
locations = find(isnan(Y));
% 4. Replace the found locations of Y with existing x values;
Y(locations) = x;

[Jan]

Logical indexing is faster, so omit the find().

[Kuifeng]

Good idea. Thank you.

[Image Analyst]

Seem way more complicated than it needs to be. Also, not sure why Y is sometimes a 2D matrix and sometimes a 1D vector.

[Kuifeng]

thank you for pointing this out.

Answer 4

[Image Analyst]

If you simply need consecutive numbers, do this:

vec = (1:max(vec))'

Answer 5

[Suresh Babu Annamraju]

A=[1;2;3;4;5]

add rows 2 and 5 to A to create B=[1;0;2;3;4;0;5]

A_temp=A;

index_rows=[2;5]

for i=1:size(index_rows,1)

mat1=A_temp(1:index_rows(i)-1);

mat2=A_temp(index_rows(i):end);

A_temp=[mat1;0;mat2];

end

B=A_temp;