[R] Linear Discriminant Analysis

2007-06-06 Thread Soare Marcian-Alin
Hello,

I want to make a linear discriminant analysis for the dataset olive, and I
get always this error:#
Warning message:
variables are collinear in: lda.default(x, grouping, ...)

## Loading Data
library(MASS)
olive - url(
http://www.statistik.tuwien.ac.at/public/filz/students/multi/ss07/olive.R;)
print(load(olive))

y - 1:572
x - sample(y)
y1 - x[1:286]

train - olive[y1,-11]
test  - olive[-y1,-11]

summary(train)
summary(test)

table(train$Region)
table(test$Region)

# Linear Discriminant Analysis
z - lda(Region ~ . , train)
predict(z, train)

z - lda(Region ~ . , test)
predict(z, test)

Thanks in advance!

-- 
Mit freundlichen Grüssen / Best Regards

Soare Marcian-Alin

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Re: [R] Linear Discriminant Analysis

2007-06-06 Thread Soare Marcian-Alin
Thanks for explaining...

Im just sitting at the homework for 6 hours after taking for one week
antibiotica, because i had an amygdalitis...
I just wanted some tipps for solving this homework, but thanks, I will try
to get help on another way :)

I think i solved it, but I still get this Error :(

## Loading Data
library(MASS)
olive - url(
http://www.statistik.tuwien.ac.at/public/filz/students/multi/ss07/olive.R;)
print(load(olive))
dim(olive)
summary(olive)

index - sample(nrow(olive), 286)

train  - olive[index,-11]
test   - olive[-index,-11]

summary(train)
summary(test)

table(train$Region)
table(test$Region)

# Linear Discriminant Analysis
z - lda(Region ~ . , train)
zn - predict(z, newdata=test)$class
mean(zn != test$Region)

2007/6/6, Uwe Ligges [EMAIL PROTECTED]:


 So what about asking your teacher (who seems to be Peter Filzmoser) and
 try to find out your homework yourself?
 You might want to think about some assumptions that must hold for LDA
 and look at the class of your explaining variables ...

 Uwe Ligges



 Soare Marcian-Alin wrote:
  Hello,
 
  I want to make a linear discriminant analysis for the dataset olive, and
 I
  get always this error:#
  Warning message:
  variables are collinear in: lda.default(x, grouping, ...)
 
  ## Loading Data
  library(MASS)
  olive - url(
 
 http://www.statistik.tuwien.ac.at/public/filz/students/multi/ss07/olive.R
 )
  print(load(olive))
 
  y - 1:572
  x - sample(y)
  y1 - x[1:286]
 
  train - olive[y1,-11]
  test  - olive[-y1,-11]
 
  summary(train)
  summary(test)
 
  table(train$Region)
  table(test$Region)
 
  # Linear Discriminant Analysis
  z - lda(Region ~ . , train)
  predict(z, train)
 
  z - lda(Region ~ . , test)
  predict(z, test)
 
  Thanks in advance!
 
 
 
  
 
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-- 
Mit freundlichen Grüssen / Best Regards

Soare Marcian-Alin

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Re: [R] canoncial correlation

2007-06-02 Thread Soare Marcian-Alin
hello,

Does nobody know what the problem could be? :

2007/5/26, Soare Marcian-Alin [EMAIL PROTECTED]:

 Hello,

 I have a problem with the function concar:

 data set: 
 http://www.statistik.tuwien.ac.at/public/filz/students/multi/ss07/world2.R


 source(world2.R)

 world[,8] - log(world[,8])
 world[,9] - log(world[,9])
 x - world[,-c(1,2)]
 x - scale(x)

 a - cancor(x[,-c(6:9)],x[,-c(1:5)])
 attributes(a)
 a

 How do I plot the first two canonial variables of a? And I want to take
 the rownames of world as pch ...

 plot(..., pch=rownames(world), col=as.numeric(world[,1]))

 Thanks in advance!


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[R] confidence band

2007-05-31 Thread Soare Marcian-Alin
Hello,

I made a function, which calculates the confidence interval and the
prediction interval, but if I want to plot it, then it plots only the
regressionline.
Maybe somebody can help me:

conf.band - function(x,y) {
  res.lsfit - lsfit(x,y)
  xi - seq(from=40, to=160, length=200)
  n - length(x)
  MSE - sqrt(sum(res.lsfit$residuals^2)/(n-2))

  # confidence interval
  band - sqrt(2*qf(0.95,2,n-2)) * MSE *
sqrt(1/n+(xi-length(x))^2/(var(x)*(n-1)))
  uiv  - res.lsfit$coef[1] + res.lsfit$coef[2]*xi - band
  oiv  - res.lsfit$coef[1] + res.lsfit$coef[2]*xi + band

  # prediction interval
  band - sqrt(2*qf(0.95,2,n-2)) * MSE * sqrt(1+1/n
+(xi-mean(x))^2/(var(x)*(n-1)))
  uip - res.lsfit$coef[1] + res.lsfit$coef[2]*xi - band
  oip - res.lsfit$coef[1] + res.lsfit$coef[2]*xi + band

  # creating the graphik
  plot(x, y, xlab=colnames(x), ylab=colnames(y), pch=19)
  abline(res.lsfit, col=1)
  matlines(xi, cbind(uiv,oiv), col=3, lty=2, lwd=2)
  matlines(xi, cbind(uiv,oip), col=2, lty=3, lwd=2)
}

-- 
Mit freundlichen Grüssen / Best Regards

Soare Marcian-Alin

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[R] canoncial correlation

2007-05-26 Thread Soare Marcian-Alin
Hello,

I have a problem with the function concar:

data set:
http://www.statistik.tuwien.ac.at/public/filz/students/multi/ss07/world2.R

source(world2.R)

world[,8] - log(world[,8])
world[,9] - log(world[,9])
x - world[,-c(1,2)]
x - scale(x)

a - cancor(x[,-c(6:9)],x[,-c(1:5)])
attributes(a)
a

How do I plot the first two canonial variables of a? And I want to take the
rownames of world as pch ...

plot(..., pch=rownames(world), col=as.numeric(world[,1]))

Thanks in advance!

-- 
Mit freundlichen Grüssen / Best Regards

Soare Marcian-Alin

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Re: [R] Excel data into R

2007-05-14 Thread Soare Marcian-Alin
library(xlsReadWrite)
?read.xls

KR,
Alin Soare

2007/5/12, Ozlem Ipekci [EMAIL PROTECTED]:

 Hello to all,
 How can I make R read the data from an Excel sheet?
 thanks,
 ozlem

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[R] Biplot

2007-05-13 Thread Soare Marcian-Alin
Hello everybody,

How could I extract the components of of a PCA to plot it into a biplot?
I want to make pairwise biplots of the first three main components:

biplot(princomp(x, cor=TRUE))
biplot(princomp(x[,c(1,2)], cor=TRUE))
biplot(princomp(x[,c(1,3)], cor=TRUE))
biplot(princomp(x[,c(2,3)], cor=TRUE))

Is this the only way, how I could plot them?
I thought, that it could be possible to extract them of the full PCA ..

KR,
Alin

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Re: [R] Value at Risk

2007-05-11 Thread Soare Marcian-Alin
Oh sry, I forgot to add the R-help list.

Thanks a lot for your help Guarav!

2007/5/11, [EMAIL PROTECTED] [EMAIL PROTECTED]:


 Hi Saore

 ***Please always mark a copy to the R-help list, it may be helpfull to
 many, you have forgotten twice***
 *** Do Reply to all with history ***
  :-) cheers




 YOU ARE THE BEST :)

 I have some problems understanding R, but R and I will be friends in the
 future hehe :)

 I have another problem with investing in one of this stockfonds.
 Lets say I would invest in the europe stockfonds on 2.5.2006 1000 dollar.
 What would be the 99%VaR/1 day under historical simulation? The same as
 above?
 @@@ So what do you think, -2.86 number means Lets investigate

 suppose you have invested 1000, one way to realize is
 100*ln(Tommorow/Todays) = -2.86 (out return) where Todays = 1000
 hence Tommorow = 1000*exp(-2.86/100) = 971.8051 , therefore Loss = 1000 -
 971.8051 = 28.19489

 I hope this would make your understanding much better and clear about VaR
 :-)
 HTH

 KR,
 Alin

 2007/5/11, [EMAIL PROTECTED] * [EMAIL PROTECTED]*
 [EMAIL PROTECTED] [EMAIL PROTECTED]:

 reply is inline





 Hello Gaurav,

 The function:

  VaR(tstock[,2],alpha=0.01) # gives the same VaR as above with historical
 simulation
 VaR
 -2.86

 but i tried this function for normal distribution:

  VaR.norm (tstock[,2],p=0.99)$VaR
 Error in VaR.norm(tstock[, 2], p = 0.99) :
   Negative value in parameter 'ydat'
 @@@ if you have seen the help manual then you mus have got that you dont
 need to give the retun series.
 R is trying to calucate the logarithm of a negative number which is why it
 is throwing you error.
 try this instead
  XXX-VaR.norm(stock$ESPA.STOCK.EUROPE,p=0.01)
  XXX$VaR
 [1] -3.11079
 

 I dont understand the way with the normal distribution :( Maybe you can
 help me a littble bit.
 Cheers :-)

 KR,
 Alin Soare

 2007/5/11, [EMAIL PROTECTED] [EMAIL PROTECTED] *
 [EMAIL PROTECTED] [EMAIL PROTECTED]:

 reply is inline

 -  Regards,

\\\|///
 \\   --   //
  (  o   o  )
 oOOo-(_)-oOOo
 |
 | Gaurav Yadav
 | Assistant Manager, CCIL, Mumbai (India)
 | Mob: +919821286118 Email: [EMAIL PROTECTED] *[EMAIL PROTECTED]
 | Man is made by his belief, as He believes, so He is.
 |   --- Bhagavad Gita
 |___Oooo
   oooO(  )
   (  )   )   /
\   ((_/
  \_ )


 Hello Mr. Gaurav Yadav,
 Hi Soare,
 1. I want to calculate the 99%VaR/1 day for the stock fonds, after sorting
 the values the 5th or 6th value is it?
 In Historical simulation it is the 5th value.. because it tells you to
 be more cautious that a higher loss 'may' be there, secondly VaR only shows
 the possibility and not the maximum loss which you can incur :-) cheers
 2. How do I calculate it under normal distribution aproximation?
 Well there is also a normal method or variance - covariance method which
 assumes normal distribution :-)
 if you want to incorporate recency effect then you can also see boudhouks
 method

 try this  paper which will give you very good understanding of various
 methods of VaR *

 **http://papers.ssrn.com/sol3/papers.cfm?abstract_id=51420#PaperDownload 
 *http://papers.ssrn.com/sol3/papers.cfm?abstract_id=51420#PaperDownload




 ++
 apply(tstock,2,function(x) VaR(x,alpha=0.01)) # are these the right VaR's
 for the stockfonds?
 @@@ you can yourself see it, you have around 579 observation and 1% of it
 mean 5.79th observation
 Thus if you become risk averse then you take the 5th smallest value and
 otherwise 6th value.
 So just sort the returns in ascending order and then see the 5th and the
 6th values


  sorted_espa_stock_europe-sort(tstock[,2])
  sorted_espa_stock_europe[5]
 [1] -2.86
  sorted_espa_stock_europe[6]
 [1] -2.74
 

 your code gives -2.86 thus you can get the rest :-) cheers

 
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[R] Value at Risk

2007-05-10 Thread Soare Marcian-Alin
Hello,

I have a problem with calculating the VaR of stockfonds.

Here the stockfonds dataset:
http://www.ci.tuwien.ac.at/~weingessel/FStat2006/stockfonds.csv

library(VaR)
library(fPortfolio)
library(e1071)

stock - read.table(stockfonds.csv, header=TRUE, sep=,)
tstock = ts(impute(stock[,2:6]), start=c(2004, 1), end=c(2006, 68),
frequency=256) # imputing the NA's
plot(tstock)
tstock - diff(tstock)

apply(tstock,2,function(x) VaR(x,alpha=0.01)) # are these the right VaR's
for the stockfonds?

apply(tstock,2,function(x) VaR.norm(x,p=0.99)$VaR) # this doesnt work :( I
want to calculate the same VaR as above, but under normal distribution

And I also have a problem with the historical simulation.
If I would invest 1000 dollars in one of these stockfonds on 2.5.2006, what
would be the 99%VaR/1day under historical simulation.
Is there a function for calculating this? I cant find something usefull on
google :(

KR,
Alin

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[R] How do i calculate emp Returns

2007-05-08 Thread Soare Marcian-Alin
Hello,

I want to build empirical Returns of stockfonds, but I cant find a function
in R, which calculate this.

Alin

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[R] Sharpe-Ratio

2007-05-03 Thread Soare Marcian-Alin
Hello,

Is there any possibility how i can calculate the Sharpe Ratio of a ts?

KR,
Alin Soare

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Re: [R] simulate values

2007-04-25 Thread Soare Marcian-Alin
Hello,

Yes I tried arima.sim and everything works fine. Thanks for the help!

Alin Soare

2007/4/25, Leeds, Mark (IED) [EMAIL PROTECTED]:

 he's just being a wise guy bcause I'm sure you meant to have an
 epsilon_t on the end. And he
 Surely knows that also. Did you
 Check out ?arima.sim.


 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] On Behalf Of Uwe Ligges
 Sent: Wednesday, April 25, 2007 3:15 AM
 To: Soare Marcian-Alin
 Cc: R-help@stat.math.ethz.ch
 Subject: Re: [R] simulate values



 Soare Marcian-Alin wrote:
  Hello,
 
  I want to simulate 100 values of the ARMA Process with this function:
 
  x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 *
  x[i-2] +
  0.3 * x[i-3]

 There is no kind of noise in your model, hence no need to do any
 simulation so far ...

 Uwe Ligges


 
  which possibilities do I have?
 
  Alin Soare
 
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[R] Log-Returns

2007-04-24 Thread Soare Marcian-Alin
Hello,

I have a Problem to make Log-Returns of the dataset EuStockMarkets.
Is there any function which could calculate it for me?

data(EuStockMarkets)

Thanks!

Alin Soare

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[R] simulate values

2007-04-24 Thread Soare Marcian-Alin
Hello,

I want to simulate 100 values of the ARMA Process with this function:

x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 * x[i-2] +
0.3 * x[i-3]

which possibilities do I have?

Alin Soare

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[R] robust correlation

2007-04-21 Thread Soare Marcian-Alin
Hello,

How do I calculate a robust correlation in R?
I want to compare it to the pearson method.

library(mvoutlier)
data(chorizon)
cor(log10(chorizon$Al), log10(chorizon$Na), method=c(pearson))

KR,
Alin Soare

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[R] distance-distance plot and multivariate outliers

2007-04-21 Thread Soare Marcian-Alin
Hello,

I want to make a distance-distance plot of the classical covariance and the
robust covariance.
How can I create a distance-distance plot?
And by the distance-distance plot I want to look for multivariate outliers,
how can I do it? With identify?

Kind Regards,
Alin Soare

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Re: [R] Change the mode of a list

2007-04-20 Thread Soare Marcian-Alin
Hello,

I dont know if it works, but try: storage.mode(xxx) = numeric

Alin

2007/4/20, Felix Wave [EMAIL PROTECTED]:

 Hello,
 can anybody tell me a easy way to change the mode of an aggregate list
 to numeric?
 I found a solution but I looks cruel.

 Thank's
 Felix

 PS: In the past you have asked what I am doing. I have to evaluate
 measures of two
 gauges of our university. The aim is to get an answer which one is better.



  mode(MEAN)
 [1] list
  mode(MEASURE)
 [1] numeric


 MEAN- aggregate(INPUT[,3], by=list(INPUT[,2],INPUT[,1]), FUN=mean)

 MODE - matrix(c(MEAN[,2],MEAN[,1],MEAN[,3]), ncol=3, byrow=FALSE )
 x -  MODE[,1] -1
 y - (MODE[,2] -1) / 10

 MEASURE - matrix(c(MODE[,2],MODE[,1],MODE[,3]), ncol=3, byrow=FALSE )

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Re: [R] print command

2007-04-19 Thread Soare Marcian-Alin
Hello Corinna,

NullPointRevertante=plan1[3,2] #0.859
cat(null-pint measurement Revertante:, NullPointRevertante ,\n)

Have Fun!

Kind Regards,
Soare Alin

2007/4/19, Schmitt, Corinna [EMAIL PROTECTED]:

 Dear R-Experts,

 I have the following command lines:

 NullPointRevertante=plan1[3,2] #0.859
 print(null-pint measurement Revertante: )
 print(NullPointRevertante)

 The result looks like:

 [1] null-pint measurement Revertante: 
 [1] 0.859


 It is ok but I achieve the following outlook:

 [1] null-pint measurement Revertante: 0.859

 How can I achieve this?

 Thanks, Corinna

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Re: [R] print command

2007-04-19 Thread Soare Marcian-Alin
Yes :)

\t ... tabulator

Alin

2007/4/19, Schmitt, Corinna [EMAIL PROTECTED]:

   Thanks!

 If I want to add a new line behind I need the command

 cat(null-pint measurement Revertante:, NullPointRevertante ,\n\n)

 Right?

 Corinna






   --

 *Von:* Soare Marcian-Alin [mailto:[EMAIL PROTECTED]
 *Gesendet:* Donnerstag, 19. April 2007 10:46
 *An:* Schmitt, Corinna; R-help@stat.math.ethz.ch
 *Betreff:* Re: [R] print command



 Hello Corinna,

 NullPointRevertante=plan1[3,2] #0.859
 cat(null-pint measurement Revertante:, NullPointRevertante ,\n)

 Have Fun!

 Kind Regards,
 Soare Alin

 2007/4/19, Schmitt, Corinna [EMAIL PROTECTED]:

 Dear R-Experts,

 I have the following command lines:

 NullPointRevertante=plan1[3,2] #0.859
 print(null-pint measurement Revertante: )
 print(NullPointRevertante)

 The result looks like:

 [1] null-pint measurement Revertante: 
 [1] 0.859


 It is ok but I achieve the following outlook:

 [1] null-pint measurement Revertante: 0.859

 How can I achieve this?

 Thanks, Corinna

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[R] Filtering

2007-04-19 Thread Soare Marcian-Alin
Hello Everybody,

How can I filter a dataset?

Im trying to filter the dataset EuStockMarkets monthly and quarter.

data(EuStockMarkets)
ftse = EuStockMarkets[,4]

Alin

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Re: [R] importing excel-file

2007-04-18 Thread Soare Marcian-Alin
Hello Everybody,

Install the package: install.packages(xlsReadWrite)
Load it: library(xlsReadWrite)

testfile = read.xls(TesFile.xls)

Have Fun!

Kind Regards,
Soare Marcian-Alin

PS: If dont works, then install also the package xtable, but it should
work without installing it!

2007/4/18, John C Frain [EMAIL PROTECTED]:

 To avoid complications, save your file as comma separated and use one
 of the instructions for reading delimited files.  If you are using a
 comma as a decimal point you are probably using ; as a separator.  If
 this is so use read.csv2.  Please see the help files for read.table.

 Best Regards

 John

 On 18/04/07, Schmitt, Corinna [EMAIL PROTECTED] wrote:
  Dear R-experts,
 
  It is a quite stupid question but please help me. I am very confuced. I
  am able to import normal txt ant mat-files to R but unable to import
  .xls-file
 
  I do not understand the online help. Can please anyone send me the
  corresponding command lines? The .xls-file is attached. In my file we
  use commas for the decimal format (example: 0,712), changes might be
  needed.
 
  Thanks, Corinna
 
 
  __
  R-help@stat.math.ethz.ch mailing list
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  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 --
 John C Frain
 Trinity College Dublin
 Dublin 2
 Ireland
 www.tcd.ie/Economics/staff/frainj/home.html
 mailto:[EMAIL PROTECTED]
 mailto:[EMAIL PROTECTED]

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Re: [R] How do I print a string without the initial [1]?

2007-04-18 Thread Soare Marcian-Alin
Hello Steve,

You can print strings in R with the method cat()

\n . new line
\t  . tabulator

Try:
name - c(Steve)
age=22
cat(\tHello my name is, name ,and I am, age ,years old.\n)

Have Fun!

Kind Regards,
Soare Marcian-Alin

2007/4/18, steve [EMAIL PROTECTED]:

 If I print a sting I get an initial [1]:

  xx=a
  xx
 [1] a


 How do I get it to print just

 a

 with no [1]?

 I tried looking this up, but I don't know what the initial [1] is called.

 Steve

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Re: [R] Use of argument '...'

2007-04-17 Thread Soare Marcian-Alin
Dear Juan,

You have to define your function like this:

fun - function (x,y=0, ...) {
   x+y
}

The ... means that u can use the other parameters in par!
The variables, which you want to use, these you have to define in your
function!

In this function i put y=0 on a standard value zero,

if u try now fun(2), the result will be: 2
if u try fun(2,4), then the result will be: 6

Kind Regards,
Alin Soare

2007/4/17, Juan Lewinger [EMAIL PROTECTED]:

 Dear R list,

 I've read the function writing sections on both An introduction to R and
 R language Definition manuals but still don't understand why the following
 gives an error message:

 fun - function(x, ...) x + y
 fun(1, y=2)

 I get:

 Error in fun(1, y = 2) : object y not found

 I'd appreciate any help in understanding this.

 R version 2.4.1 (2006-12-18)
 i386-pc-mingw32
 ...

 Juan Pablo Lewinger
 Department of Preventive Medicine
 Keck School of Medicine
 University of Southern California
 1540 Alcazar Street, CHP-220
 Los Angeles, CA 90089-9011, USA

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