Group matrix with this condition

I have a matrix with three columns like that:

1 92.1433186490000 23.9154939160001

1 92.9534543080001 23.9122345640001

1 92.6530971670001 23.0872503970001

1 93.0291740250000 21.9333670360001

1 93.2541111270000 21.2386782580001

1 94.2215646970001 19.4112785390000

1 94.1402179610001 18.4163064250001

1 92.1429410350000 18.4263430520001

1 92.1433186490000 23.9154939160001

2 92.9534543080001 23.9122345640001

2 95.7343830470001 23.9198671710001

2 95.4631112670000 23.3802137790000

2 95.3625913070001 22.0275655970000

2 95.4085413600001 21.7528253520001

2 93.1248683680001 21.6382351500001

2 92.6530971670001 23.0872503970001

2 92.9534543080001 23.9122345640001

3 93.1248683680001 21.6382351500001

3 95.4085413600001 21.7528253520001

3 95.8866160070000 19.4626209220000

3 95.6639418050000 18.4131395020000

3 94.1402179610001 18.4163064250001

3 94.2215646970001 19.4112785390000

3 93.2541111270000 21.2386782580001

3 93.1248683680001 21.6382351500001

4 95.7343830470001 23.9198671710001

4 96.9171762360001 23.9192869440001

4 96.8640078520000 23.7995176290000

4 96.5947759490001 22.5219473650001

4 96.4500026700001 21.8299999240001

........

6 96.4500026700001 21.8299999240001

6 96.5947759490001 22.5219473650001

6 97.6237908480001 22.5663499820001

I want to obtain 6 grouped matrix with the condition that if the first column values are same. For example, one of the grouped matrix is should be like that:

a1=[92.1433186490000 23.9154939160001

92.9534543080001 23.9122345640001

92.6530971670001 23.0872503970001

93.0291740250000 21.9333670360001

93.2541111270000 21.2386782580001

94.2215646970001 19.4112785390000

94.1402179610001 18.4163064250001

92.1429410350000 18.4263430520001

92.1433186490000 23.9154939160001]

a1 belongs to values of 1.

Answer 1

[Star Strider]

Using accumarray and mat2cell —

A = [1	92.1433186490000	23.9154939160001
     1	92.9534543080001	23.9122345640001
     1	92.6530971670001	23.0872503970001
     1	93.0291740250000	21.9333670360001
     1	93.2541111270000	21.2386782580001
     1	94.2215646970001	19.4112785390000
     1	94.1402179610001	18.4163064250001
     1	92.1429410350000	18.4263430520001
     1	92.1433186490000	23.9154939160001
     2	92.9534543080001	23.9122345640001
     2	95.7343830470001	23.9198671710001
     2	95.4631112670000	23.3802137790000
     2	95.3625913070001	22.0275655970000
     2	95.4085413600001	21.7528253520001
     2	93.1248683680001	21.6382351500001
     2	92.6530971670001	23.0872503970001
     2	92.9534543080001	23.9122345640001
     3	93.1248683680001	21.6382351500001
     3	95.4085413600001	21.7528253520001
     3	95.8866160070000	19.4626209220000
     3	95.6639418050000	18.4131395020000
     3	94.1402179610001	18.4163064250001
     3	94.2215646970001	19.4112785390000
     3	93.2541111270000	21.2386782580001
     3	93.1248683680001	21.6382351500001
     4	95.7343830470001	23.9198671710001
     4	96.9171762360001	23.9192869440001
     4	96.8640078520000	23.7995176290000
     4	96.5947759490001	22.5219473650001
     4	96.4500026700001	21.8299999240001];

[Au,~,ix] = unique(A(:,1),'stable');                        % May Not Be Necessary If 'Column 1' Values Are Continuous And Stepwise Monotonically Increasing, If So, Substitute 'A(:,1)' For 'ix'
Col1Nrs = accumarray(ix,1);                                 % Tally Of 'Column 1' Occurrences
Uvals = table(Au,Col1Nrs)                                   % Table Of Results (Optional, Requires The 'unique' Call)

Uvals = 4×2 table
    Au    Col1Nrs
    __    _______

    1        9   
    2        8   
    3        8   
    4        5   

Ac = mat2cell(A, Col1Nrs, 3)                                % Partition 'A' As Requested

Ac = 4×1 cell array
    {9×3 double}
    {8×3 double}
    {8×3 double}
    {5×3 double}

Ac{1}                                                       % Results For 'Group 1'

ans = 9×3

1.0000 92.1433 23.9155 1.0000 92.9535 23.9122 1.0000 92.6531 23.0873 1.0000 93.0292 21.9334 1.0000 93.2541 21.2387 1.0000 94.2216 19.4113 1.0000 94.1402 18.4163 1.0000 92.1429 18.4263 1.0000 92.1433 23.9155

.

[busra gogen]

Thank you! This is what I want..

[Star Strider]

My pleasure!

Answer 2

[Image Analyst]

Try using a logical vector to mask your N-by-3 array, m3:

a1 = m3(m3(:, 1) == 1, 2:end); % Extract only rows where column1 of m3 has the value 1.
a2 = m3(m3(:, 1) == 2, 2:end); % Extract only rows where column1 of m3 has the value 2.
a3 = m3(m3(:, 1) == 3, 2:end);
a4 = m3(m3(:, 1) == 4, 2:end);
a5 = m3(m3(:, 1) == 5, 2:end);
a6 = m3(m3(:, 1) == 6, 2:end);

[busra gogen]

Thank you! It works but I need to use for loop instead of writing a1,a2,..a6 because my column values (1,2,3...6) may change depending on my data

[Image Analyst]

OK, so put it in a loop

for k = 1 : whatever
    % First update column 1 of m3 somehow with the new group numbers.
    m3 = whatever
    % Now extract the 6 groups we need for this iteration.
    a1 = m3(m3(:, 1) == 1, 2:end); % Extract only rows where column1 of m3 has the value 1.
    a2 = m3(m3(:, 1) == 2, 2:end); % Extract only rows where column1 of m3 has the value 2.
    a3 = m3(m3(:, 1) == 3, 2:end);
    a4 = m3(m3(:, 1) == 4, 2:end);
    a5 = m3(m3(:, 1) == 5, 2:end);
    a6 = m3(m3(:, 1) == 6, 2:end);
end

Answer 3

[Torsten]

first_column = m3(:,1);
v = unique(first_column);
for i = 1:numel(v)
  M{i} = m3(first_column == v(i),2:end)
end