Using surf function with data from excel table

I keep getting the "Z must be a matrix, not a scalar or vector" error using the following code, and it is currently reshaped into a matrix using the reshape function. I'm not sure how to convert it to a form the surf function can use:

The ismatrix function produces a result of 1, so Z does identify as a matrix. Does anyone have insight on this?

[Jackson Pohl]

As an extra note, the scatter plot does produce the desired result, I was advised to use it as a tool to determine if the data was being manipulated properly and it seems that it is undergoing the correct changes to provide the wanted outcome.

[Stephen23]

Do NOT change the MATLAB search path just to access data files. Calling ADDPATH and REHASH PATH like that is inefficient. All MATLAB functions which import/export data files accept absolute/relative filenames, so you should just provide the data import/export function with the path. FULLFILE is often useful for that.

Answer 1

[Voss]

It would help to have the data, in order to know whether the x and y are gridded or scattered.

https://www.mathworks.com/help/matlab/math/overview-of-interpolation-techniques.html

If they are gridded

[x,y] = ndgrid(linspace(5.4,1,5),linspace(4,10,8));
x = x(:);
y = y(:);
z = rand(numel(x),1);
T = table(x,y,z);
disp(T)

     x       y           z    
    ___    ______    _________

    5.4         4      0.20813
    4.3         4      0.12496
    3.2         4      0.39786
    2.1         4      0.71341
      1         4       0.8041
    5.4    4.8571      0.50956
    4.3    4.8571      0.31989
    3.2    4.8571    0.0053885
    2.1    4.8571      0.75143
      1    4.8571      0.86634
    5.4    5.7143      0.52135
    4.3    5.7143      0.17024
    3.2    5.7143      0.30531
    2.1    5.7143      0.42971
      1    5.7143      0.41343
    5.4    6.5714      0.01241
    4.3    6.5714       0.3412
    3.2    6.5714      0.73762
    2.1    6.5714        0.181
      1    6.5714      0.70824
    5.4    7.4286      0.89059
    4.3    7.4286      0.48137
    3.2    7.4286      0.29383
    2.1    7.4286     0.027363
      1    7.4286      0.63855
    5.4    8.2857      0.26877
    4.3    8.2857      0.44245
    3.2    8.2857      0.71812
    2.1    8.2857      0.98614
      1    8.2857      0.69864
    5.4    9.1429      0.39224
    4.3    9.1429      0.24801
    3.2    9.1429      0.88306
    2.1    9.1429       0.3135
      1    9.1429     0.047909
    5.4        10     0.028884
    4.3        10      0.23169
    3.2        10      0.59952
    2.1        10       0.6228
      1        10      0.33481

then you can determine the number of grids in the x and y directions and reshape z accordingly before plotting the surface

x = T{:,1};
y = T{:,2};
z = T{:,3};
X = unique(x,'stable');
Y = unique(y,'stable');
Z = reshape(z,numel(X),numel(Y)).'

Z = 8x5

0.2081 0.1250 0.3979 0.7134 0.8041 0.5096 0.3199 0.0054 0.7514 0.8663 0.5214 0.1702 0.3053 0.4297 0.4134 0.0124 0.3412 0.7376 0.1810 0.7082 0.8906 0.4814 0.2938 0.0274 0.6385 0.2688 0.4424 0.7181 0.9861 0.6986 0.3922 0.2480 0.8831 0.3135 0.0479 0.0289 0.2317 0.5995 0.6228 0.3348

figure()
surf(X,Y,Z)
hold on
scatter3(x,y,z,32,'r','o','filled')

In this case the scatter points lie exactly on the surface.

On the other hand, if they are scattered

x = 1+4.4*rand(40,1);
y = 4+6*rand(40,1);
z = rand(numel(x),1);
T = table(x,y,z);
disp(T)

      x         y          z    
    ______    ______    ________

    2.9591    7.1576     0.31834
    4.7758    9.3023    0.065207
    4.6927    8.1744     0.66528
    3.1441    7.8177     0.79885
    3.8019    4.0164     0.55441
     5.363    4.4674     0.13931
    4.8018    9.0908    0.054529
     5.226    9.1488     0.48932
    2.3734    7.6413     0.39042
    3.5804    4.7519     0.35436
    3.7212    7.6837     0.58757
    1.0912    5.0552      0.3594
     1.427    8.9829     0.59153
    2.7388    4.0643     0.57094
     2.758    4.2733     0.39638
    1.7745    4.1681     0.81471
    4.2311    7.0143     0.54371
    4.6068    8.5291     0.97366
    3.4731    6.1257     0.15647
    1.2959    9.2176    0.013094
    3.7287    6.3124     0.73943
     1.069    8.3555     0.60209
    2.0307    9.1535     0.91088
    3.3958    4.9858      0.6511
    4.3794     6.109      0.8781
    2.7535    7.2607     0.35717
    1.8027     5.773     0.30448
    1.6564    7.2397     0.49787
    4.9911    8.6641     0.14537
    3.8204    7.4689     0.72007
    1.0116    5.1652     0.89224
    3.8024    7.7002     0.59089
    2.7738     5.607      0.1959
    4.8534    9.7478     0.13545
    1.8173    6.1251     0.19459
    2.8249    9.3558      0.9205
    2.8219    7.0614     0.96915
    5.2179    4.6159     0.24682
    4.1396    9.2608     0.82085
    3.2687    7.3872     0.93174

then you'll have to do something else, e.g., use a scatteredInterpolant and define a grid on which the Z should be calculated

x = T{:,1};
y = T{:,2};
z = T{:,3};
I = scatteredInterpolant(x,y,z);
X = linspace(min(x),max(x),5);
Y = linspace(min(y),max(y),8);
[X,Y] = meshgrid(X,Y);
Z = I(X,Y)

Z = 8x5

0.6661 0.8074 0.5805 0.4482 0.0005 0.1862 0.5507 0.5549 0.3658 0.1641 0.8704 0.2928 0.2902 0.6973 0.2275 0.8163 0.4408 0.4783 0.7391 0.3024 0.7404 0.4393 0.7430 0.5872 0.3888 0.6426 0.4838 0.8146 0.7551 0.4867 0.0721 0.8413 0.8673 0.8583 0.5960 -1.2803 -0.5411 -0.1017 0.1129 0.2303

figure()
surf(X,Y,Z)
hold on
scatter3(x,y,z,32,'r','o','filled')

In this case the scatter points do not lie on the surface because the surface was calculated by interpolation/extrapolation of the data onto a grid.

[Jackson Pohl]

Unfortunately I cannot give exact data because the information is potentially proprietary, but I can say that the information is a 3-column excel sheet of data points. Another thing to consider is the sheer size of the data pool - there are 126,000 rows of data per column.

[Jackson Pohl]

The scatter plot produces a result like this:

Which I would say is gridded.

[Jackson Pohl]

When attempting to use the gridded method for mapping the data that you provided, I get an error using reshape, as the number of elements attempts to change when using the number of elements of x and y.

[Voss]

Please share the code you are running.

[Jackson Pohl]

I actually ended up figuring it out! Although now I am having trouble establishing a gradient but that's for a different post...