Matrix problem for same values of column

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Anu Sharma
Anu Sharma el 2 de Sept. de 2019
Respondida: Andrei Bobrov el 2 de Sept. de 2019
A=[29.78 5 8
24.97 8 11
22.98 4 12
21.05 12 13
24.78 1 16
25.53 26 29
21.43 2 32
29.94 11 33
29.57 15 35
28.43 17 36
11.49 23 37
13.69 37 38
26.97 28 39
16.25 25 40
27.36 36 41
4.24 18 42
19.39 39 44
29.93 16 45
25.83 30 46
26.09 40 47
27.58 24 48
28.61 41 49
29.41 48 50]
and i want
output =[22.98 4 12
25.53 26 29
21.43 2 32
29.94 11 33
29.57 15 35
28.43 17 36
13.69 37 38
26.97 28 39
4.24 18 42
29.93 16 45
25.83 30 46
26.09 40 47
28.61 41 49
29.41 48 50 ]
No value in column 2, 3 get repeated and in case of repeated value in any of the column(2,3) the higest value of column 1 is as the output.
For example, in row 1, 2 and 8. column (2,3) have values as
[ 5 8
8 11
11 33]
Among these 3 rows row 8, ie. [29.94 11 33] have the highest value so only this row will be the output. all other row like [29.78 5 8] and [24.97 8 11]will be elimanted.
simillarly,
for row 3 = [22.98 4 12]
And 4 = [21.05 12 13]
row 3= [22.98 4 12]
will be output and row 4 will get eliminated.
  1 comentario
Stephen23
Stephen23 el 2 de Sept. de 2019
Editada: Stephen23 el 2 de Sept. de 2019
Is the row
28.43 17 36
correct in your example output array? Following your explanation, these rows are one group:
28.43 17 36
...
27.36 36 41
...
28.61 41 49
of which the last row has the highest values in the first column (and the last row is in your output array). But why do you keep the first row as well?

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Respuesta aceptada

Stephen23
Stephen23 el 2 de Sept. de 2019
Editada: Stephen23 el 2 de Sept. de 2019
N = size(A,1);
X = ones(N,1); % group numbers
Z = true(N,1); % logical index
V = 1; % group number
for k = 2:N % for each row...
Y = A(k,2)==A(1:k-1,3); % check if any matching rows.
if any(Y)
X(k) = X(Y); % copy group number (assumed scalar).
W = X(k)==X(1:k-1); % logical index of that group.
if all(A(k,1)>A(W,1))
Z(W) = false; % current val > prev vals.
else
Z(k) = false; % prev val > current val.
end
else % no matching rows:
V = V+1; % increment group number.
X(k) = V;
end
end
B = A(Z,:) % output matrix
Giving:
B =
22.98 4 12
25.53 26 29
21.43 2 32
29.94 11 33
29.57 15 35
13.69 37 38
26.97 28 39
4.24 18 42
29.93 16 45
25.83 30 46
26.09 40 47
28.61 41 49
29.41 48 50
  3 comentarios
Stephen23
Stephen23 el 2 de Sept. de 2019
"In which variable ,output matrix is storing. "
B
Anu Sharma
Anu Sharma el 2 de Sept. de 2019
Thankyou so much..

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Más respuestas (1)

Andrei Bobrov
Andrei Bobrov el 2 de Sept. de 2019
[m,n] = size(A);
B = [(1:m)',A(:,2:3)];
k = B(1,2:3);
ii = 1;
C{1} = [];
while ~isempty(B)
i0 = ismember(B(:,2:3),k);
lo = any(i0,2);
if any(lo)
C{ii} = [C{ii};[repmat(ii,nnz(lo),1),B(lo,1)]];
k = B(xor(i0(:,1),i0(:,2)),2:3);
B = B(~lo,:);
else
ii = ii + 1;
k = B(1,2:3);
C{ii} = [];
end
end
iii = cat(1,C{:});
T = array2table(A);
T = T(iii(:,2),:);
T.g = iii(:,1);
T = sortrows(T,{'g','A1'},{'ascend','descend'});
T = rowfun(@(x,y,z)[x(1),y(1),z(1)],T,'GroupingVariables','g');

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