Interpolation of an array of values

Question

Pranjal Pathak 2013-11-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/105000-interpolation-of-an-array-of-values

评论： G A 2013-11-13

Hi,

I have an array of values which is not a square one(different rows have different number of elements and they are placed in such a way that the first element (i.e. 1) of second row is placed in the mid of first two elements (i.e. 3 & 5) of the first row and so on). I want to interpolate (bi-linear) this matrix into a square matrix of dimension 36x36.

 A = [3 5 8 9 1 4;
       1 6 9 2 5;
      6 7 8 9 10 12;
       2 5 1 3 4;
      1 2 4 6 8 9;
       3 1 4 5 8];

Can anybody please help me in doing this in Matlab.

Thanking You!

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Jan 2013-11-5

This is not a matrix. Matrices have the same number of elements in each row by definition. Therefore the shown definition of A is not clear. In consequence we cannot guess how your input is represented and this does not allow to guess which kind of procedure you are looking for.

Please edit the question and show us, how your input matrix A is defined in a way, Matlab does understand.

Pranjal Pathak 2013-11-6

Dear Simon, You are right,it is not a matrix. But my set of values are of this type and dimension. In this case we can assume, each of the rows as a different row matrix and interpolate each of them separately upto a diemnsion of 1x36 along horizontal direction. After that, it can be interpolated along vertical direction separately upto a dimension of 36x1. Finally concatenate them to form a square matrix of dimension 36x36. Please help me in doing this.

Thanking You!

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Answer 1

G A 2013-11-6

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/105000-interpolation-of-an-array-of-values#answer_114398

在 MATLAB Online 中打开

Do you mean something like this?

   A = {[3 5 8 9 1 4];
         [1 6 9 2 5];
        [6 7 8 9 10 12];
         [2 5 1 3 4];
        [1 2 4 6 8 9];
         [3 1 4 5 8]};
  dim1=36;
  dim2=36;
  B=zeros(length(A),dim2); 
     for k=1:length(A)
  x1=1:length(A{k});
  x2=linspace(1,length(x1),dim2);
  B(k,:)=interp1(x1,A{k},x2,'linear');
     end
  x3=1:length(A);
  x4=linspace(1,length(x3),dim1);
  C=zeros(dim1,dim2);  
     for k=1:dim2
  C(:,k)=interp1(x3,B(:,k),x4,'linear');
     end