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

on 13 Apr 2013

### Image Analyst (view profile)

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?

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### Image Analyst (view profile)

on 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.

Locks

### Locks (view profile)

on 13 Apr 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 (view profile)

on 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
```

### Locks (view profile)

on 13 Apr 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!

Image Analyst

### Image Analyst (view profile)

on 13 Apr 2013

This should have been a comment to my answer. | means "or" while & means "and". It tells you which rows should be removed from the original array. You might have to have

```rowsToDelete = rowsToDelete8 | rowsToDelete5
```

```rowsToDelete = rowsToDelete8 & rowsToDelete5
```

depending on exactly what criteria you're after.

Locks

### Locks (view profile)

on 13 Apr 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 (view profile)

on 13 Apr 2013

That's correct. If we're done, then mark my answer (not yours) as Accepted.

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