No dublicates in for loop.

6 vues (au cours des 30 derniers jours)
Kian Kirchhof
Kian Kirchhof le 29 Déc 2016
Commenté : Kian Kirchhof le 29 Déc 2016
Im doing a project on directed networks. I have a 2000x2 array in total created by a loop, running 9 times (once per week.) I need for every week to have no dublicated rows, but dublicated rows are allowed in the array for all weeks. I dont doubt the answer is simple, but i cant come up with new answers right now.
the loop is like this
for k=1:9
for i=1:length(NumeriskData)
if NumeriskData(i,DateRow)>= Ar(k,1) && NumeriskData(i,DateRow)<= Ar(k,2)
if NumeriskData(i,RowSorter)== 1 || NumeriskData(i,RowSorter2)== 1 || NumeriskData(i,RowSorter3)== 1
if NumeriskData(i,1)~= NumeriskData(i,2)
WeightLink(Count+1,:)= NumeriskData(i,1:2) ;
Count=Count+1;
end
end
end
end
end
Unique(WeightLink) after here would just take uniques of the total array, i need it to take of every k run, and still make the proper total WeightLink
EDIT. Sorry, not sure how to format the code properly.

Réponse acceptée

David J. Mack
David J. Mack le 29 Déc 2016
Modifié(e) : David J. Mack le 29 Déc 2016
Hey Kian,
I am not sure if I really understood your problem, but maybe the following solution helps:
%Assuming NumeriskData is a nx2 matrix.
for k=1:9
IsData = NumeriskData(:,DateRow)>= Ar(k,1) ...
& NumeriskData(:,DateRow)<= Ar(k,2) ...
& NumeriskData(:,1)~= NumeriskData(:,2) ...
& (NumeriskData(:,RowSorter )== 1 ...
| NumeriskData(:,RowSorter2)== 1 ...
| NumeriskData(:,RowSorter3)== 1);
if any(IsData)
NumDatWeek = unique(NumeriskData(IsData,:),'rows');
nNumDataWeek = size(NumDatWeek,1);
WeightLink(Count+1:Count+nNumDataWeek,:) = NumDatWeek;
Count = Count+nNumDataWeek;
end
end
Greetings, David
  1 commentaire
Kian Kirchhof
Kian Kirchhof le 29 Déc 2016
Never done if statements like that. You actually did what i needed, only had to change very small amount. Sweet, thanks!

Connectez-vous pour commenter.

Plus de réponses (0)

Community Treasure Hunt

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

Start Hunting!

Translated by