Convert x y coordinates and z data to Matrix

I have three vectors

>> whos x y VAR1
  Name          Size             Bytes  Class     Attributes

  x          21242x1             679744  double              
  y          21242x1             679744  double              
  var1       21242x1             679744  double              

where X and Y represent the location of the corresponding value of var1. What Im trying to do is to convert all this data to a matrix.

What I've done until now is:

a1 = 21242;
m = zeros(a1,a1).*NaN;
indexes = sub2ind([a1,a1],x,y);
m(indexes) = var1

Based on this question but I'm getting an error

Error using sub2ind (line 43)
Out of range subscript.

Which, according to another question about this error, is because I'm trying access elements which does not exist on the matrix.

Any help?

Thanks!

[Dyuman Joshi]

indexes = sub2ind([a1,a1],x,y)

What are the values in x and y?

By the error it seems that values in x and y exceed a1*a1

Also, you can use NaN directly

%example for 2x2
m = NaN(2,2)

m = 2×2

NaN NaN NaN NaN

[user20912]

They are longitude (only negative values) and latitude (only positive values). Could this cause the problem?

Thanks for the comment about NaN, I'll use that!

[Dyuman Joshi]

Yes, it could.

Negetives values are not accepted as indices in MATLAB (even 0 is not accepted for that matter)

i1=sub2ind([3 3],2,2)

i1 = 5

i2=sub2ind([3 3],-2,2)

Error using sub2ind
Out of range subscript.

[user20912]

Thank for the input. However, it's still not working.

I have tried to element by element with a for cycle:

for i1=1:a1
    indexes = sub2ind([a1 a1], abs(x(i1)), y(i1));
    i1
end

and the problem is in the element 3897:

>> x(3896:3898)
   -1.0021
   -0.9993
   -1.0026
   

what's the problem with that numbers? I can not find the reason why keep failing.

[Dyuman Joshi]

Indices should also be integers as well.

Answer 1

[Star Strider]

It would help to have the actual vectors.

It may be possible to use the reshape function to create matrices from the vectors, if the vectors are from a gridded original matrix. The easiest way to determine the dimensions of the arrays is to use the unique function and look at the second output. The elements of the second output should be regularly-spaced, and that difference is one dimension to specify in the reshape call, with the other one left as the empty vector [].

Otherwise, use linspace, ndgrid and griddata to create appropriate matrices from scattered data.

.

[user20912]

The second output from the unique functions gives me a list which is not regularly-spaced, maybe there is the problem?

>> [x1,x2] = unique(x);
x2(1:100)'
Columns 1 through 15

     1     2     3     4     5     6     7     9    10    11     8    14    12    13    20

  Columns 16 through 30

    19    15    16    17    21    22    18    23    26    24    25    34    28    32    29

  Columns 31 through 45

    27    30    33    35    31    36    40    37    38    42    41    39    53    44    46

  Columns 46 through 60

    49    43    51    45    47    54    52    50    48    55    56    57    58    60    61

  Columns 61 through 75

    59    62    63    74    65    68    80    66    71    70    75    64    67    69    78

  Columns 76 through 90

    81    72    77    76    92   104    73    83    82    84    86    79    89   100   109

  Columns 91 through 100

   134    87    85    88    97    98    95    94    91   101
   

Sadly I cannot upload the data.

[Star Strider]

Since you cannot upload the data (that do not appear to be gridded), an appropriate approach is the second option —

x = randn(150,1);
y = randn(150,1);
z = rand(150,1);

figure
stem3(x,y,z, 'Marker','none')
hold on
scatter3(x, y, z, 20, z, 'filled')
hold off
grid on
colormap(turbo)
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Original Data')

xv = linspace(min(x), max(x), numel(x));
yv = linspace(min(y), max(y), numel(y));
[X,Y] = ndgrid(xv, yv);
Z = griddata(x,y,z,X,Y);

figure
surfc(X, Y, Z)
grid on
colormap(turbo)
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Interpolated Surface')

This just uses random data. The approach will be similar for your data.

.

[user20912]

That was a very good aproach. But since the matrix are already really big, this create a huge matrix and make collapse my system. I going to see If I can reduce the size of the matrix to use this method. Thank you!

[Star Strider]

My pleasure!

The lengths of ‘xv’ and ‘yv’ (the third argument to linspace) can be whatever you want them to be. They can be considerably shorter.

Example —

x = randn(150,1);
y = randn(150,1);
z = rand(150,1);

figure
stem3(x,y,z, 'Marker','none')
hold on
scatter3(x, y, z, 20, z, 'filled')
hold off
grid on
colormap(turbo)
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Original Data')

xv = linspace(min(x), max(x), 15);
yv = linspace(min(y), max(y), 15);
[X,Y] = ndgrid(xv, yv);
Z = griddata(x,y,z,X,Y);

figure
surfc(X, Y, Z)
grid on
colormap(turbo)
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Interpolated Surface')

The data are yours to work with, so make whatever accommodations you need to, in order to produce the result you want.

.

[user20912]

Thank you!

This was helpfull to do what I needed to do.

[Star Strider]

As always, my pleasure!

Answer 2

[Robert Daly]

I would use interpolation to convert to a regular grid.

scatteredInterpolant is the funtion to do this with your irregularly spaced data.

Reshape or convert vectors to use it with pcolor - MATLAB Answers - MATLAB Central (mathworks.com)

[X,Y] = meshgrid(linspace(min(x),max(x),500),linspace(min(y),max(y),500));% make 500x500 grid

f = scatteredInterpolant(x,y,var1); % set up the interpolater with your data

Grid_Var1 = f(X,Y); % Do the interpolation onto the grid set up above