Adding a column every nth column without replacing existing one
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Hey all,
I am looking for a solution to the problem I mentioned in the title.
FOr example, I have matrix A
A = [1 2 3 4; 5 6 7 8; 9 10 11 12];
I now want to add every 2nd (and later 3th) column a new column containing zeros.
The result should be like this:
result = [1 0 2 0 3 0 4 0; 5 0 6 0 7 0 8 0; 9 0 10 0 11 0 12 0];
I know how to create a new column containg only zeros
B = zeros(number_of_rows,1);
Previously, I used
C = [A(:,1:2) B A(:,2+1:end)];
But this only works for a specific column and not every nth column.
Thanks for your answers in advance!
Respuesta aceptada
Más respuestas (2)
Max Murphy
el 11 de Dic. de 2019
Test Data
A = [1 2 3 4; 5 6 7 8; 9 10 11 12];
Function
function A = addDataNthColumn(A,B,n)
% ADDDATANTHCOLUMN Add data in column vector or matrix B to every nth
% column of A
if nargin < 3
n = 3;
end
if size(A,1) ~= size(B,1)
error('A and B must have same number of rows');
end
if size(A,2) >= n
% Use recursion to iterate on A
A = [A(:,1:(n-1)), B, addDataNthColumn(A(:,n:end),B,n)];
else
A = [A, B]; % Last "set" just concatenate them
end
end
Usage
>> test = addDataNthColumn(A,zeros(3,2),2)
test =
1 0 0 2 0 0 3 0 0 4 0 0
5 0 0 6 0 0 7 0 0 8 0 0
9 0 0 10 0 0 11 0 0 12 0 0
5 comentarios
Chuguang Pan
el 11 de Dic. de 2019
Editada: Chuguang Pan
el 11 de Dic. de 2019
rowA=size(A,1);
colA=size(A,2);
B=zeros(rowA,colA*2);
B(:,1:2:2*colA)=A;
It is every 2th column.
Max Murphy
el 11 de Dic. de 2019
Nice profile picture. Is that a well isolated neuron I see? :)
Haha it's from a sparse extracellular array, so I don't know about well-isolated, but otherwise yes :)
Adam Danz
el 11 de Dic. de 2019
Well there must have been good thresholding on your instrumentation, then. It's such a good feeling to work with an isolatable single neuron but mult-iunit activity is also fun. Now you have me all nostalgic.
Enthusiasten
el 12 de Dic. de 2019
Chuguang Pan
el 12 de Dic. de 2019
It is every 2th column. Now I put it in Answer this question
rowA=size(A,1);
colA=size(A,2);
B=zeros(rowA,colA*2);
B(:,1:2:2*colA)=A;
1 comentario
Enthusiasten
el 12 de Dic. de 2019
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