[R] I need help in analyzing

[wojak121]

I'm sory for my weak english. I need to analyze this subject :

x1	x2	x3	x4	x5	x6	x7	x8	x9	x10	y
0	0	1	0	0	1	0	0	1	0	czarne
1	1	0	0	0	0	1	0	0	0	rude
0	0	1	0	0	1	1	0	0	0	braz
0	0	1	0	1	0	1	0	0	0	blond
1	0	0	0	0	1	0	0	0	1	rude
1	1	0	0	0	0	0	0	0	1	blond
0	0	1	1	0	0	0	0	1	0	czarne
1	0	0	1	0	0	1	0	0	0	blond
0	0	1	0	0	1	1	0	0	0	blond
1	0	0	0	0	1	1	0	0	0	czarne
0	0	1	0	0	1	0	0	0	1	czarne
1	0	1	0	0	0	1	0	0	0	czarne
0	0	1	1	0	0	0	0	0	1	braz
0	1	0	1	0	0	0	0	0	1	braz
1	0	1	0	0	0	0	0	1	0	braz
0	0	0	1	1	0	0	0	0	1	blond
1	0	1	0	0	0	0	0	1	0	czarne
0	1	0	0	0	1	0	0	0	1	braz
1	0	0	1	0	0	0	0	0	1	braz
0	0	1	0	0	1	0	0	0	1	braz
0	0	0	1	0	1	0	0	0	1	blond
0	0	1	1	0	0	0	0	0	1	czarne
0	0	1	0	0	1	0	0	0	1	rude
0	0	1	0	0	1	0	0	0	1	braz
0	0	1	0	0	1	1	0	0	0	braz
0	0	1	0	0	1	1	0	0	0	rude
0	0	1	1	0	0	1	0	0	0	braz
1	0	1	0	0	0	0	0	1	0	rude
0	0	0	1	0	1	1	0	0	0	czarne
0	0	1	0	0	1	0	0	1	0	blond
1	0	0	1	0	0	0	0	1	0	blond
0	0	1	0	0	1	1	0	0	0	rude
1	0	0	0	0	1	0	0	0	1	braz
0	0	0	0	1	1	1	0	0	0	blond
0	0	1	1	0	0	0	0	0	1	blond
0	0	1	0	0	1	1	0	0	0	blond
1	0	0	1	0	0	0	0	0	1	blond
1	0	1	0	0	0	1	0	0	0	rude
0	1	0	0	0	1	0	0	1	0	braz
0	1	1	0	0	0	0	0	0	1	czarne
0	0	1	0	0	1	0	1	0	0	blond
0	1	1	0	0	1	0	1	0	0	rude
1	0	0	0	0	1	0	0	0	1	czarne
0	1	1	0	0	0	0	0	1	0	blond
0	0	1	0	0	1	1	0	0	0	rude
0	0	1	0	0	1	0	1	0	0	blond
0	0	1	0	1	0	1	0	0	0	blond
0	1	1	0	0	0	0	1	0	0	braz
0	0	1	1	0	0	0	0	1	0	braz
1	0	0	1	0	0	0	0	0	1	blond
1	1	0	0	0	0	0	0	0	1	czarne
0	1	1	0	0	0	1	0	0	0	rude
1	0	1	0	0	0	0	0	0	1	braz
1	1	0	0	0	0	1	0	0	0	braz
0	0	1	1	0	0	0	0	0	1	czarne
1	1	0	0	0	0	0	0	0	1	blond
1	0	0	1	0	0	1	0	0	0	blond
0	0	1	1	0	0	0	0	0	1	braz
0	0	1	1	0	0	1	0	0	0	czarne
0	0	1	1	0	0	1	0	0	0	czarne

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
> model.lda
Call:
lda(y ~ ., data = dane)

Prior probabilities of groups:
    blond      braz    czarne      rude 
0.3166667 0.2833333 0.2333333 0.1666667 

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

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

Proportion of trace:
   LD1    LD2    LD3 
0.6918 0.2574 0.0508 

> 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

model=rpart(y~.,dane,method="class",control=rpart.control(xval=3,cp=0))
> plot(model)
> text(model)
model$cptable
          CP nsplit rel error   xerror       xstd
1 0.05691057      0 1.0000000 1.097561 0.08180737
2 0.02439024      3 0.8292683 1.219512 0.07040857
3 0.00000000      4 0.8048780 1.195122 0.07310295
npt=which.min(model$table[,4])
> npt
integer(0)



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.

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