apparently the command rmoutliers does not do the job correctly

Hello,

It seems to me that the command "rmoutliers" has some problems. To make things clear I explain using an example as bellow:

>> a=[1 2 3 4 1000 6;2 5000 3 4 0 1];a

[b1,idx1] = rmoutliers(a(1,:));[b2,idx2] = rmoutliers(a(2,:));[b,idx] = rmoutliers(a,2);

idx1

idx2

idx

a =

1 2 3 4 1000 6

2 5000 3 4 0 1

idx1 =

0 0 0 0 1 0

idx2 =

0 1 0 0 0 0

idx =

0 0 0 0 0 0

This does not make sense to me. so, if I am right then this command needs to be corrected. The second problem with this command is that it apparently can no longer have 3 outputs (unlike what is written in the corresponding matlab page). So, if you try the following then you get error message:

[b,idx,u] = rmoutliers(a,2);

Any comment?

thsnks in advance!

Babak

[Stephan]

I give you 2 numbers:

0.001 10000

please tell me which one ist the outlier - can you do this? Or wil you need at least a third number to decide which one is the outlier?

It is only possible to decide this with at least 3 numbers:

0.001 10000 0.006

or

0.001 10000 10029

[Mohammad Shojaei Arani]

So, if I understood correctly the command [b,idx] = rmoutliers(a,2); should find outliers in each row and then remove the corresponding columns? If my understanding is correct then this command does not do the job correctly

[Stephen23]

"So, if I understood correctly the command [b,idx] = rmoutliers(a,2); should find outliers in each row..."

No. Nowhere in the RMOUTLIERS documentation is it stated that RMOUTLIERS checks anything other than columns: the documentation states for a matrix "...then rmoutliers detects outliers in each column of A separately..."

"...and then remove the corresponding columns?"

Yes. The DIM argument is specifically described as "specifies the dimension of A for which to remove entries when an outlier is detected using any of the previous syntaxes. For example, rmoutliers(A,2) removes columns instead of rows for a matrix A" (bold emphasis added). Note that the DIM description does not state that it changes which dimension the MEDIAN is calculated over, all this option changes is whether rows/columns are removed.

Simple solution: transpose the input matrix.

Answer 1

[Steven Lord]

Rather than going directly to rmoutliers I recommend using isoutlier to detect the outliers then process the resulting logical array.

a=[1 2 3 4 1000 6;2 5000 3 4 0 1];
[b1,idx1] = rmoutliers(a(1,:));
[b2,idx2] = rmoutliers(a(2,:));
bRow = isoutlier(a, 2)

bRow = 2×6 logical array
   0   0   0   0   1   0
   0   1   0   0   0   0

columnsWithOutliers = any(bRow, 1)

columnsWithOutliers = 1×6 logical array
   0   1   0   0   1   0

originalData = a % Make a copy so you can compare the original and processed data

originalData = 2×6

1 2 3 4 1000 6 2 5000 3 4 0 1

a(:, columnsWithOutliers) = []

a = 2×4

1 3 4 6 2 3 4 1

[Mohammad Shojaei Arani]

Hi Steven,

Thanks a lot. Indeed, 'isoutlier' is a very useful command and does the job very good.

I do not understand why the commands 'isoutlier' and 'rmoutliers' are not consistant with each other. The following backs up my claim:

>> a=[1 2 3 4 1000 6;2 5000 3 4 0 1];

isoutlier(a, 2)

[c,d]=rmoutliers(a,2)

0 0 0 0 1 0

0 1 0 0 0 0

c =

1 2 3 4 1000 6

2 5000 3 4 0 1

d =

0 0 0 0 0 0

Anyway, it is like this !!!