How to find first two maximum number in the matrix
5 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Moe
am 19 Okt. 2015
Bearbeitet: Andrei Bobrov
am 23 Okt. 2015
Hello everyone,
I have the following matrix A (right table):
A = [1248,30,12;1248,20,13;1248,5,14;177,5,12;177,25,13;230,10,14;230,40,15;274,60,12;274,5,14];
I want to find the matrix B (left table) in the way that first, check the column "ID" to find the similar ID, then find the first two max number and finally find the "PE" related to the found first two max number. For example, first it will find that ID = 1248 has three repetitions. Then, from TE, it will find that 30 and 20 are the first two max numbers. And finally, it will find 12 (for max 30) and 13 (for max 20). Can anyone help how to search for that unique id
0 Kommentare
Akzeptierte Antwort
Andrei Bobrov
am 23 Okt. 2015
Bearbeitet: Andrei Bobrov
am 23 Okt. 2015
[a,~,c] = unique(A(:,1),'stable');
a0 = sortrows([c,A(:,2:end)],[1,-(2:3)]);
i1 = bsxfun(@plus,find([1;diff(A(:,1))~=0]),0:1)';
out = [a,reshape(permute(reshape(a0(i1,2:end),2,[],2),[1,3,2]),4,[])'];
or
a = unique(A(:,1),'stable');
n = numel(a);
out = zeros(n,5);
for ii = 1:n
l0 = a(ii) == A(:,1);
b = sortrows(A(l0,2:end),-(1:2));
out(ii,:) = [a(ii),reshape(b(1:2,:),1,[])];
end
0 Kommentare
Weitere Antworten (1)
TastyPastry
am 19 Okt. 2015
Assuming there are at least 2 values for TE/PE for each ID:
uniqueVals = unique(A(:,1),'stable');
output = zeros(numel(uniqueVals),5);
for i = 1:numel(uniqueVals)
mask = A(:,1) == uniqueVals(i);
[sorted,ind] = sort(A(mask,2),'descend');
PE = A(mask,3);
newRow = [uniqueVals(i) sorted(1:2) PE(ind)];
output(i,:) = newRow;
end
2 Kommentare
TastyPastry
am 23 Okt. 2015
newRow = [uniqueVals(i) sorted(1:2)' PE(ind(1:2))'];
This code still only works if each ID has two values of TE and PE associated with it. It will error on the last line ID = 811 as shown above.
Siehe auch
Kategorien
Mehr zu Logical 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!