[R] Y in Kohonen xyf function

[chalabi.elahe at yahoo.de]

On Friday, June 17, 2016 12:21 AM, K. Elo <maillists at pp.inet.fi> wrote:
Hi again!

According to '?xyf', the function is expecting following parameters:

(1) data = a matrix, with each row representing an object.

So, please ensure that your data is a matrix

(2) Y = property that is to be modelled. In case of classification, Y is 
a matrix of zeros, with exactly one '1' in each row indicating the 
class. For prediction of continuous properties, Y is a vector. A 
combination is possible, too, but one then should take care of 
appropriate scaling.

Once again, no data frame here, but a scaled vector or a matrix.

Your could try following steps (I assume 'df' to be you data frame):

--- snip ---
set.seed(7)
training <- sample(nrow(df), 120)
Xtraining <- scale(df[training,])
Xtest <- scale(df[-training,],
                center = attr(Xtraining, "scaled:center"),
                scale = attr(Xtraining, "scaled:scale"))

xyf.df <- xyf(Xtraining,
               factor(df.classes[training]),
               grid = somgrid(5, 5, "hexagonal"))

--- snip ---

Let us know - with output, please - what happens. The point is, if this 
works, then you could try in experimenting the parameter 
'factor(df.classes[training]'. It seems to, that also here you need a 
matrix or a list as a base, not a data frame.

This might also be of interest for your: 
https://www.jstatsoft.org/article/view/v021i05/v21i05.pdf

HTH,
Kimmo


16.06.2016, 17:30, chalabi.elahe at yahoo.de wrote:
> Hi Kimmo,
>
> Thanks for your reply, Here is a part of my df:
>
>
>      'data.frame':  562 obs. of 128 variables
>      $ TE         :int 37 37 35 34 37 37 35 33 32 ...
>      $ TR         :int 11 11 8 13 11 8 15 12 8 .....
>      $ BW         :int 150 191 128 145 200 191 ........
>      $speed       :int 4 4 3 3 2 1 4 1 2 3 ..........
> and I want to cluster my data based on speed, to see the coming costumer's protocols fall into which speed group and I think I need to bring this speed column in Y element of xyf
>
>
> On Thursday, June 16, 2016 2:29 PM, K. Elo <maillists at pp.inet.fi> wrote:
> Hi!
>
> Some sample data could help us to help you...
>
> But have you read '?xyf' in order to ensure that your 'Y' is what 'xyf'
> expects it to be?
>
> What kind of error messages do you get?
>
> Regards,
> Kimmo
>
> 16.06.2016, 15:13, ch.elahe via R-help wrote:
>> Is there any answer?
>>
>>
>> Hi all, I have a df and I want to use supervised Self Organizing Map
>> to do classification. I should use Kohonen library and xyf function
>> from it. As you know the xyf function looks like this and I have
>> problem defining my Y:
>>
>> xyf(data,Y,grid=somgrid(),rlen=100,alpha=c(0.05,0.01)) I want to do
>> classification based on a column which shows the speed that a
>> protocols is run, and this column is the following:
>>
>> $speed   :num 4 4 3 3 3 1 1 1 2 1 4 4 3 numbers from 1 to 4 show the
>> speed from very fast to very slow protocols. so the property I want
>> to be modeled is df$speed, but I don't know how should I bring it in
>> xyf function. Does anyone know how to do that? I also added my train
>> set ans test set:
>>
>> dt=sort(sample(nrow(df),nrow(df)*.7)) train=df[dt,]
>> Xtraining=scale(trian) Xtest=scale(-trian)
>> center=attr(Xtrianing,"scaled:center")
>> scale=attr(Xtraining,"scaled:scale")
>> xyf(Xtraining,........,grid=somgrid(10,10,"hexagonal"))
>>
>>
>> Thanks for any Help, Elahe
>>
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>>
>> ______________________________________________ R-help at r-project.org
>> mailing list -- To UNSUBSCRIBE and more, see
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>> posting guide http://www.R-project.org/posting-guide.html and provide
>> commented, minimal, self-contained, reproducible code.
>>
>
> ______________________________________________
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