Keeping columns names but replacing value for iteration

Hi all. I'm new at MATLAB and in coding in general. I'm practicing with some modified K-means clustering and I encountered this problem when iterating. I can't seem to figure it out. I uploaded the sample data named 'CCP.xlsx'. This is my code:

A = readtable('CCP.xlsx');
A = sortrows(A,'d','descend');
C = 64;
num_r = length(A.Customer);
num_K = round(sum(A.d / C));
iter = 10;

%getting initial centroid K coordinates
K = A(1:num_K,2:3);

for h = [1:iter]

    %finding Euc.distance from K to r
    for t = [1:num_K]
        for i = [1:num_r]
            dist(i,t) = sqrt((K.x(t) - A.x(i))^2 + (K.y(t) - A.y(i))^2);
        end
    end
    dist_T = array2table(dist)
    A = [A dist_T]
    
    
    % unrelated = unrelated code that changes the value of column 'dist1', 'dist2', 'dist3' every iteration     
    % ......
    % ......
    % ......
    
end    

When I ran this, it outputs this for the 1st iteration:

A =

  10×7 table

    Customer    x     y     d     dist1     dist2     dist3 
    ________    __    __    __    ______    ______    ______

        4        2    19    30         0    16.279    3.1623
        2       18    16    30    16.279         0        17
        9        1    16    22    3.1623        17         0
        7       14    17    20    12.166    4.1231    13.038
        8        2     6    19        13    18.868     10.05
        5        5    12    17    7.6158    13.601    5.6569
        1       13     6    15    17.029     11.18     15.62
        3       13     6    15    17.029     11.18     15.62
        6       16     9    13    17.205    7.2801    16.553
       10        3     6    10    13.038    18.028    10.198

Iteration 1 , success!

But for the 2nd iteration (new values are derived from 'unrelated'), it seems that it wants to add new and separate columns at the end of table A that is also named column 'dist1', 'dist2', 'dist3' and as a result, it outputs this:

dist_T =

  10×3 table

    dist1     dist2     dist3 
    ______    ______    ______

    6.5192    14.866    10.912
     15.89    2.8284    13.103
    3.8079    15.133    8.9376
    12.349    3.6056    10.615
    6.5192    16.125    5.7689
    2.5495     11.18    3.3287
    12.349     8.544    6.9771
    12.349     8.544    6.9771
    13.946         5    9.2022
    6.5192    15.264    4.9679

Error using CCP4 (line 27)
Duplicate table variable name: 'dist1'.    

My end-goal is that no matter how many column 'dist#' there is, the code can still run with no problem.

That said, is there a way to just replace the values for columns 'dist1', 'dist2', 'dist3' for every following iteration so that they remain intact as column #5 ,6, 7 for table A?

I've been trying other methods but they all just add new columns at the end of table A named column 'dist1_1', 'dist2_1', 'dist3_1', etc ....

Any help or tips are greatly appreciated. Thank you!

Answer 1

[Peter Perkins]

You are horzcat'ing tables, and table variable names must be unique. But your two tables both have variables named dist1, dist2, dist3. So you have two choices:

1) modify the var names in the new table to dist4, dist5, dist6, then dist7, dist8, dist9, and so on

2) rather than horzcat'ing, create nested tables, like this:

>> t1 = array2table(rand(5,2),'VariableNames',["d1" "d2"]);
>> t2 = array2table(rand(5,2),'VariableNames',["d1" "d2"]);
>> t = table([1;2;3;4;5],t1,t2,'VariableNames',["X" "T1" "T2"])
t =
  5×3 table
    X            T1                     T2        
           d1          d2         d1         d2   
    _    ___________________    __________________
    1    0.97235     0.70763    0.45469    0.79374
    2    0.29313     0.52507    0.42832    0.64548
    3     0.2155     0.27734    0.81563    0.93903
    4    0.46152    0.017199    0.33228    0.17186
    5    0.65587     0.75829    0.76782    0.51153

Or are you trying to overwrite the old dist# variables with the new ones?

A(:,["dist1" "dist2" "dist3"]) = dist_T

[Christoper Angelo Malayan]

Got some ideas on how to implement it excatly how I want thanks to this.

Thank you, Peter!