Filter löschen
Filter löschen

Matrix problem for same values of column

1 Ansicht (letzte 30 Tage)
Anu Sharma
Anu Sharma am 2 Sep. 2019
Beantwortet: Andrei Bobrov am 2 Sep. 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 Kommentar
Stephen23
Stephen23 am 2 Sep. 2019
Bearbeitet: Stephen23 am 2 Sep. 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?

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Stephen23
Stephen23 am 2 Sep. 2019
Bearbeitet: Stephen23 am 2 Sep. 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 Kommentare
Stephen23
Stephen23 am 2 Sep. 2019
"In which variable ,output matrix is storing. "
B
Anu Sharma
Anu Sharma am 2 Sep. 2019
Thankyou so much..

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Andrei Bobrov
Andrei Bobrov am 2 Sep. 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');

Kategorien

Mehr zu Resizing and Reshaping Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by