if loop that is not working properly, row should be deleted based on several criterias but do not
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
hi,
I have a matrix with 8 columns and I would like to delete those rows where the value of column 8 exceeds 1.1 or is below 0.9 and apply a similar criteria on column 5, here is the respective code:
while i < size(data8, 1)
i = i + 1 ;
if(data8(i,8)>1.1)
data8(i,:) = [];
elseif(data8(i,8)<0.9)
data8(i,:) = [];
elseif(data8(i,5)<5*1/365)
data8(i,:) = [];
elseif(data8(i,5)>120*1/365)
data8(i,:) = [];
end
end
although the code is running, I get in the resulting matrix values in column 8 that exceeds 1.1 and are below 0.9, whan do I need to change?
0 Kommentare
Akzeptierte Antwort
Image Analyst
am 13 Apr. 2013
Try it this way
rowsToDelete = data(:,8) > 1.1 | data(:,8) < 0.9;
data(rowsToDelete, :) = [];
Do the same for column #5. You could combine column 5 into the calculation of rowsToDelete if you want.
2 Kommentare
Image Analyst
am 13 Apr. 2013
No. The code was all there was. The whole point of vectorizing it was to get rid of the while loop - and you put it back in. Don't use it. Try this, which I constructed to use sample data in the rang 0.8 - 1.2.
% Create array of numbers between 0.8 and 1.2.
data = .8 + .4 * rand(20,8)
% Display rows 8 and 5
data(:,8)
data(:,5)
% Find the rows to delete.
rowsToDelete8 = data(:,8) > 1.1 | data(:,8) < 0.9
rowsToDelete5 = data(:,5) > 1.1 | data(:,5) < 0.9
rowsToDelete = rowsToDelete8 & rowsToDelete5
% Go ahead and delete them.
data(rowsToDelete, :) = [];
% Display final data.
data
Weitere Antworten (1)
Locks
am 13 Apr. 2013
3 Kommentare
Image Analyst
am 13 Apr. 2013
Bearbeitet: Image Analyst
am 13 Apr. 2013
That's correct. If we're done, then mark my answer (not yours) as Accepted.
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!