if loop that is not working properly, row should be deleted based on several criterias but do not

1 vue (au cours des 30 derniers jours)
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?

Réponse acceptée

Image Analyst
Image Analyst le 13 Avr 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 commentaires
Locks
Locks le 13 Avr 2013
I tried this:
i = 0 ;
while i < size(data8, 1)
i = i + 1 ;
rowsToDelete = data8(:,8) > 1.1 | data8(:,8) < 0.9 | data8(i,5)<5*1/365 | data8(i,5)>120*1/365;
data8(rowsToDelete, :) = [];
%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
but this gives me an empthy data8 matrix, which cannot be true, is there anything else which is incorrect?
What was the problem with the first code?
Image Analyst
Image Analyst le 13 Avr 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

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Plus de réponses (1)

Locks
Locks le 13 Avr 2013
awesome, this is working. Could you shortly explain me what rowsToDelete is doing? I get just zeros and from what I know zero means false, but I don't understand the logic behind it.
What is the | operator doing? is that the same as &...?
Thanks vermy much for the help!
  3 commentaires
Locks
Locks le 13 Avr 2013
that means that this command:
rowsToDelete8 = data8(:,8) > 1.1 | data8(:,8) < 0.9;
data8(rowsToDelete8, :) = [];
each times delete the respective row then the the 8th colum is either above 1.1 or below 0.9?
Image Analyst
Image Analyst le 13 Avr 2013
Modifié(e) : Image Analyst le 13 Avr 2013
That's correct. If we're done, then mark my answer (not yours) as Accepted.

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