Filling in Matrix (Interpolation
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If I have a (10,3) matrix like so,
      10    20    1
       0     0    0
       0     0    0
      40    80    4
       0     0    0
      60   120    6
       0     0    0
       0     0    0
       0     0    0
      80   200   10
Is there any way I can interpolate the data to get it to turn like so which displays the average values in between the endpoints? Using interp1 was not clear to me and resulted in NaN everywhere.
    10    20    1
    20    40    2
    30    60    3
    40    80    4
    50   100    5
    60   120    6 
    65   140    7
    70   160    8
    75   180    9
    80   200   10
Any and all help/guidance is very much appreciated.
0 Comments
Accepted Answer
  Stephen23
      
      
 on 10 Feb 2017
        
      Edited: Stephen23
      
      
 on 10 Feb 2017
  
      >> M = [10,20,1;0,0,0;0,0,0;40,80,4;0,0,0;60,120,6;0,0,0;0,0,0;0,0,0;80,200,10];
>> X = ~all(M==0,2);
>> N = interp1(find(X),M(X,:),1:size(M,1))
N =
    10    20     1
    20    40     2
    30    60     3
    40    80     4
    50   100     5
    60   120     6
    65   140     7
    70   160     8
    75   180     9
    80   200    10
2 Comments
  D. Plotnick
      
 on 11 Feb 2017
				
      Edited: D. Plotnick
      
 on 11 Feb 2017
  
			Very nice, hadn't thought about using the logical indexing. Also, I hadn't realized interp1 could handle multiple rows like that. Has that always been true? And, more importantly, does that work on pages as well? i.e., interp1 will always assume it is working along the first dimension of any N-D array?
Scratch that, I just tried it and it works. Amazing!
Try
 M = repmat(M,1,1,5);
 X = all(~all(M==0,2),3);
 N = interp1(find(X),M(X,:,:),1:size(M,1))
  Stephen23
      
      
 on 11 Feb 2017
				It works along columns, exactly as the documentation says: "...you can pass v as an array. Each column of array v contains a different set of 1-D sample values".
More Answers (2)
  John BG
      
 on 10 Feb 2017
        Hi Suki
A=[ 10    20    1
         0     0    0
         0     0    0
        40    80    4
         0     0    0
        60   120    6
         0     0    0
         0     0    0
         0     0    0
        80   200   10]  
    B=A;
  [sz1 sz2]=size(B);
  for s=1:1:sz2
     L=B(:,s);
     [nx ny val]=find(L)
     for k=2:1:numel(nx) 
        L([nx(k-1):1:nx(k)])=linspace(val(k-1),val(k),nx(k)-nx(k-1)+1); 
     end
     B(:,s)=L;
  end
    A 
    =
        10    20     1
         0     0     0
         0     0     0
        40    80     4
         0     0     0
        60   120     6
         0     0     0
         0     0     0
         0     0     0
        80   200    10
     B
     =
        10    20     1
        20    40     2
        30    60     3
        40    80     4
        50   100     5
        60   120     6
        65   140     7
        70   160     8
        75   180     9
        80   200    10
Suki
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
0 Comments
  D. Plotnick
      
 on 10 Feb 2017
        
      Edited: D. Plotnick
      
 on 10 Feb 2017
  
      You can either break up the columns, or do it in one go. Example below.
In future, please include how you ran interp1 or any other code, since that will make helping you easier.
 % This returns the right answer
 x = [1,4,6,10]; %indexes of known points
 y1 = [1,4,6,10]; % y values at those points
 Xq = 1:11; % add an extra point so you can see how extrap works.
 Yq = interp1(x,y1,Xq,'linear','extrap');
 % This also returns the right answer
 y2 = [20,80,120,200]; 
 Yq2 = interp1(x,y2,Xq,'linear','extrap');
 % Also the right answer
 y3 = [10 40 60 80];
 Yq3 = interp1(x,y3,Xq,'linear','extrap');
 % In one swell-foop
 % We can do this in one go. The meshgrids are to make sure you have matrices of  % the right size.
 z = 1:3; % Index for columns.
 Y = cat(1,y3,y2,y1); 
 [Xq,Zq] = meshgrid(Xq,z);
 % Extrap does not work with interp2. 
 YqAll = interp2(x,z,Y,Xq,Zq,'linear')'
 % you could also add an extrapval after the 'linear', which sets those
 % NaNs to that number
2 Comments
  Stephen23
      
      
 on 10 Feb 2017
				I guess that is one way to avoid the fowl/foul/fell confusion: http://www.quickanddirtytips.com/education/grammar/one-fell-swoop
  D. Plotnick
      
 on 11 Feb 2017
				Hah, I always assumed it was a Shakespeareanism of some kind but didn't know the specific source. I have to give credit for 'swell-foop' to Piers 'Xanth' Anthony.
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