[R-SIG-Finance] Antwort: loop for multiple regressions

[Matthias.Koberstein at hsbctrinkaus.de]

Hi,
Basically the problem is that the regression returns a LIST not a single
value.
You can access the single values of the list with $valuename.
Please look into help(lm) for the values contained in the list.
If you select for example
Example[i]<-lm(y~x)$coefficient[1]

it should save the intercept to the matrix Example at place i.

Cheers

Matthias S. Koberstein
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HSBC Trinkaus
Structured Solutions Group
Königsalle 21/23, 40212 Düsseldorf

Phone: +49 211 910 4412
e-mail: matthias.koberstein at hsbctrinkaus.de



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             "Denise Xifara"                                               
             <denisexifaras at go                                             
             oglemail.com>                                              An 
             Gesendet von:               r-sig-finance at stat.math.ethz.ch   
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                                         [R-SIG-Finance] loop for multiple 
             Fax-Deckblatt:              regressions                       
             HSBCTuB                                                       
             22.07.2008 17:05                                              
                                                                           
                                                                           
                                                                           
                                                                           




   Dear all,
I have the following data in excel:

    day1   y 1 2 3 2 3 x1 0.2 0.3 0.4 0.3 0.2 x2 7 3.4 2 8 6  day2   y 2 4
3
2 2 x1 0.4 0.5 0.3 0.3 0.2 x2 7 8 9.1 6 5

I have the following problems:
first of all, when I ask R to read the file (with the package xlsReadWrite
and the command read.xls) it has a problem with the fact that the left most
corner is labelled the same way, so in order for it to work I need to do:



 day1



 y1
 1
 2
 3
 2
 3
 x11
 0.2
 0.3
 0.4
 0.3
 0.2
 x21
 7
 3.4
 2
 8
 6


 day2



 y2
 2
 4
 3
 2
 2
 x12
 0.4
 0.5
 0.3
 0.3
 0.2
 x22
 7
 8
 9.1
 6
 5

Next I need to do the regression y~ x1+x2, for the different days.
I have converted the above data in the form of a matrix and labelled it
mat.
The following loop works:

results<-c(0,0,0,0,0)
for (i in c(1,5)){
results[i]<-lm(mat[i,]~mat[i+1,]+mat[i+2,])}
results

And the results are:

Warning messages:
1: In results[i] <- lm(mat[i, ] ~ mat[i + 1, ] + mat[i + 2, ]) :
  number of items to replace is not a multiple of replacement length
2: In results[i] <- lm(mat[i, ] ~ mat[i + 1, ] + mat[i + 2, ]) :
  number of items to replace is not a multiple of replacement length
> results
[[1]]
 (Intercept) mat[i + 1, ] mat[i + 2, ]
   2.4918415    1.5151515   -0.1356220
[[2]]
[1] 0
[[3]]
[1] 0
[[4]]
[1] 0
[[5]]
 (Intercept) mat[i + 1, ] mat[i + 2, ]
  -0.3741935    3.4193548    0.2580645

The method does work but it is not ideal.  The problem is that the real
data
that I will be using will involve regressions of 20 or so variables over
3000 days. I also want to avoid finding the exact values that "i" should
take inside the loop.
Is there a better way of carrying out these regressions so that I do not
need to relabel everything, the results are faster and it is clear and easy
to isolate the results of the linear models for each day?

Thank you very much everyone in advance,
Regards,
Denise

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