Matrix problem for same values of column
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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
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?
Akzeptierte Antwort
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
Weitere Antworten (1)
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');
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