Group matrix with this condition

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busra gogen
busra gogen 2022-5-2
评论： Star Strider 2022-5-2
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.

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采纳的回答

Star Strider
Star Strider 2022-5-2
Ran in:
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
.

更多回答（2 个）

Image Analyst
Image Analyst 2022-5-2
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);
  2 个评论
busra gogen
busra gogen 2022-5-2
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
Image Analyst 2022-5-2
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

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Torsten
Torsten 2022-5-2
first_column = m3(:,1);
v = unique(first_column);
for i = 1:numel(v)
M{i} = m3(first_column == v(i),2:end)
end

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