Re: [R] I need help in analyzing

From: Joris Meys <jorismeys_at_gmail.com>
Date: Sun, 06 Jun 2010 13:31:54 +0200

This help list is not intended for solving your assignments. But you're honest about it, so I'll describe for you shortly what the output is. How to interprete that output is up to you.

see inline comments.

On Sun, Jun 6, 2010 at 12:38 PM, wojak121 <rotworm121_at_op.pl> wrote:
>
> I'm sory for my weak english. I need to analyze this subject :
> last column is my Y.  When i entered this to R i've get
> model.lda=lda(y~.,dane)
> Warning message:
> In lda.default(x, grouping, ...) : variables are collinear

Obvious warning message, and the indication that there are other methods maybe more suited for your data.

>> model.lda
> Call:
> lda(y ~ ., data = dane)
what you put in
>
> Prior probabilities of groups:
>    blond      braz    czarne      rude
> 0.3166667 0.2833333 0.2333333 0.1666667
The proportions of each hair type in the original dataset
>
> Group means:
>              x1        x2        x3        x4        x5        x6        x7
> blond  0.3684211 0.1578947 0.4736842 0.4210526 0.2105263 0.3684211 0.3684211
> braz   0.2941176 0.2941176 0.6470588 0.3529412 0.0000000 0.4117647 0.2352941
> czarne 0.3571429 0.1428571 0.7142857 0.4285714 0.0000000 0.3571429 0.3571429
> rude   0.4000000 0.3000000 0.8000000 0.0000000 0.0000000 0.6000000 0.6000000
>               x8        x9       x10
> blond  0.10526316 0.1578947 0.3684211
> braz   0.05882353 0.1764706 0.5294118
> czarne 0.00000000 0.2142857 0.4285714
> rude   0.10000000 0.1000000 0.2000000

Proportion of "yes" answers for every hair type, and this for each x variable.
>
> Coefficients of linear discriminants:
>           LD1        LD2        LD3
> x1   5.1043768  4.0739211 -2.3626627
> x2   5.1972181  2.9748157 -0.3920615
> x3   5.9721912  3.0080526 -2.1908394
> x4   3.9526576  2.7992826 -2.4115814
> x5   2.0778084  5.5095145 -1.6788562
> x6   4.9891371  3.5497498 -1.4580874
> x7   0.6484504  0.5349203 -0.4412781
> x8  -2.2934686  0.8713075  1.4076988
> x9  -0.3536417 -0.2746371 -0.4208209
> x10  0.2013050 -0.5773421  0.3025799

How the x variables combine into the linear discriminant functions.
>
> Proportion of trace:
>   LD1    LD2    LD3
> 0.6918 0.2574 0.0508
Can be interpreted as the relative importance of every discriminant function (sum up to 1)
>
>> w=sample(1:60,20)
>> test=dane[w,]
>> ucz=dane[-w,]
>> m=lda(y~.,ucz)
>> test.x=test[,-11]
>> klasyfikacja=predict(m,test.x)
>> table(klasyfikacja$class,test$y)
>
>         blond braz czarne rude
>  blond      2    1      1    0
>  braz       2    3      2    1
>  czarne     0    2      2    0
>  rude       1    1      1    1
misclassification table
>
> model=rpart(y~.,dane,method="class",control=rpart.control(xval=3,cp=0))
>> plot(model)
>> text(model)
> model$cptable
>          CP nsplit rel error   xerror       xstd
> 1 0.05691057      0 1.0000000 1.097561 0.08180737
> 2 0.02439024      3 0.8292683 1.219512 0.07040857
> 3 0.00000000      4 0.8048780 1.195122 0.07310295
the table of optimal prunings
> npt=which.min(model$table[,4])
>> npt
> integer(0)
nothing. there's no "table" in the model object. try npt = which.min(model$cptable[,4])
This gives you the rownumber of the pruning with minimal standard deviation.

>
>
>
> I need to describe this subject, but i don't know what R is saying to me.
> This subject is about what women hairs mens like. x1 to x10 are answers to
> questions 1 is yes,0 is no, but there was 2 groups of questions; from x1 to
> x6 it must be choisen 2 answers on yes and from x7 to x 10 only 1 on yes.
>  Help me please, i need this to pass this subject.
>
> --
> View this message in context: http://r.789695.n4.nabble.com/I-need-help-in-analyzing-tp2244886p2244886.html
> Sent from the R help mailing list archive at Nabble.com.
>
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>

-- 
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
Joris.Meys_at_Ugent.be
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Received on Sun 06 Jun 2010 - 11:34:03 GMT

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