date:20100609




On 09.06.2010 06:15, RICHARD M. HEIBERGER wrote:

Is there a cleaner way of combining two expressions.

This example works and gives what I want

plot(1:10)
aa- expression(alpha==.05)
bb- expression(beta ==.80)
aabb- expression(alpha==.05 ~ ,  ~ beta ==.80)

text(5, 10, aa)
text(5,  9, bb)
text(5,  8, aabb)

text(5,1, parse(text=paste(deparse(aa[[1]]), deparse(bb[[1]]), sep=~)))


text(5, 1, substitute(a ~ b, list(a=aa[[1]], b=bb[[1]])))



text(5,2, parse(text=paste(deparse(aa[[1]]), deparse(bb[[1]]), sep=~',
'~)))


text(5, 1, substitute(a ~'~ b, list(a=aa[[1]], b=bb[[1]])))

Uwe





Is there a cleaner way of combining the expressions aa and bb  to get the
effect of the last two lines?

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[R] equivalent of stata command in R

2010-06-09 Thread mike mick


Dear all,

I need to use R for one estimation, and i have readily available  stata 
command, but i need also the R version of the same command.
the estimation in stata is as following:
   1. Compute mean values of relevant variables



. sum inno lnE lnM



Variable |   ObsMeanStd. Dev.   MinMax

-+

inno |146574.0880374.2833503  0  1

 lnE |146353.92562391.732912  -4.473922   10.51298

 lnM |1462094.2819031.862192  -4.847253   13.71969



2. Estimate model



. xi: xtreg lnLP lnC lnL lnE lnM eco inno eco_inno eco_lnE eco_lnM i.year, fe 
i(stno)

i.year_Iyear_1997-1999(naturally coded; _Iyear_1997 omitted)



Fixed-effects (within) regression   Number of obs  =146167

Group variable (i): stnoNumber of groups   = 48855



R-sq:  within  = 0.9908 Obs per group: min = 1

   between = 0.9122avg =   3.0

   overall = 0.9635max = 3



F(11,97301)= 949024.29

corr(u_i, Xb)  = 0.2166 Prob  F   =0.



--

lnLP |  Coef.   Std. Err.  tP|t| [95% Conf. Interval]

-+

 lnC |   .0304896   .000950932.06   0.000 .0286258.0323533

 lnL |  -.9835998   .0006899 -1425.74   0.000 -.984952   -.9822476

 lnE |   .0652658   .000943969.14   0.000 .0634158.0671159

 lnM |   .6729931   .0012158   553.53   0.000   .67061.6753761

 eco |   .0610348   .0177048 3.45   0.001 .0263336 .095736

inno |   .0173824   .0058224 2.99   0.003 .0059706.0287943

eco_inno |   .0080325   .0110815 0.72   0.469-.0136872.0297522

 eco_lnE |   .0276226.004059 6.81   0.000  .019667.0355781

 eco_lnM |  -.0214237   .0039927-5.37   0.000-.0292494   -.0135981

 _Iyear_1998 |  -.0317684   .0013978   -22.73   0.000 -.034508   -.0290287

 _Iyear_1999 |  -.0647261   .0027674   -23.39   0.000-.0701501   -.0593021

   _cons |   1.802112.009304   193.69   0.000 1.7838761.820348

-+

 sigma_u |  .38142386

 sigma_e |   .2173114

 rho |  .75494455   (fraction of variance due to u_i)

--

F test that all u_i=0: F(48854, 97301) = 3.30Prob  F = 0.



3. Compute marginal effect of eco at sample mean



. nlcom (_b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903)



   _nl_1:  
_b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903



--

lnLP |  Coef.   Std. Err.  tP|t| [95% Conf. Interval]

-+

   _nl_1 |  -.0036011.008167-0.44   0.659-.0196084.0124061

--



in fact i can find the mean of the variables ( step 1) and extimate the model 
(step 2) but i couldnt find the equivalent of step 3 (compute marginal effect 
of eco at sample mean). Can someone help me for this issue?

Cheers!

  
_


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Re: [R] iterating over groups of columns

2010-06-09 Thread Jannis

Seems to get complicated ;-)

As always, i expect more sophisticted solutions  than mine to exist, but I 
would try the following:

If II understood you correct you want to compute the means of all the k.1.x, 
then k.2.x ., correct?

I would do it this way

col.index - sub('[.].*$','',sub('^.?[.]','',colnames(the.data)))

min.values=numeric(length=10)
for (i in 1:10)
   min[i]-min(the.data[,col.index==i])


Perhaps you have to adapt the code slightly as I did not test it. 


HTH
Jannis



col.index - unlist(strsplit(colnames(the.data),'\\.'))[2]

--- 09wkj bill.k.jan...@williams.edu schrieb am Di, 8.6.2010:

 Von: 09wkj bill.k.jan...@williams.edu
 Betreff: Re: [R] iterating over groups of columns
 An: r-help@r-project.org
 CC: Jannis bt_jan...@yahoo.de
 Datum: Dienstag, 8. Juni, 2010 22:32 Uhr
 In the code fragment, I used 'by' to
 actually compute the min value (part of the statement with
 the eval) - and I agree that an apply would work there
 wonderfully. 
 
 However, my hope was to use an apply for the subsetting of
 the data.frame's columns, so that I could then use an apply
 to compute the min across each row of the subsets.
 
 Something that would give me the results of the following,
 but programmatically:
 apply(the.data[,1], 1, min)        #min
 of the first column
 apply(the.data[,2:3], 1, min)    #min of the next
 2 columns
 apply(the.data[,4:6], 1, min)    #min of the next
 3 columns
 apply(the.data[,7:10], 1, min)  #min of the next 4
 columns
 ...
 apply(the.data[,46:55], 1, min)#min of the next 10 columns
 
 
 
 Like, can I make a vector of levels with 'rep(1:10,1:10)',
 and then apply the function across all columns in each
 level? And then how could I cbind them together?
 
 
 Thanks for any help,
 Bill
 
 
 
 
 On Jun 8, 2010, at 5:08 PM, Jannis wrote:
 
  you should have found a solution for that in the help
 page of apply.
  
  just run
  
  min.values = apply(the.data,1,min)
  
  the '1' marks the direction (e.g. whether apply is
 applied to rows or columns), it could be a 2 as well. Check
 that yourself in the apply documentation.
  
  Then run rbind(the.data,min.values) (could be cbind as
 well, I am not sure again ;-) ) and you get what you want.
  
  09wkj schrieb:
  I am mainly a Java/C++ programmer, so my mind is
 used to iterating over data with for loops. After a long
 break, I am trying to get back into the R mindset, but I
 could not find a solution in the documentation for the
 applys, aggregate, or by.
  
  I have a data.frame where each row is an entry
 with 10 groups of measurements. The first measurement spans
 1 column, the second spans 2 columns, third 3, and so on (55
 total columns). What I want to do is add to my data.frame 10
 new columns containing the minimum value of each
 measurement.
  
  dim(the.data)
  [1] 1679  55
  
   
  colnames(the.data)
     
   [1]
 k.1.1   k.2.1   k.2.2   k.3.1   k.3.2   k.3.3   k.4.1 
   [8]
 k.4.2   k.4.3   k.4.4   k.5.1   k.5.2   k.5.3   k.5.4   [15]
 k.5.5   k.6.1   k.6.2   k.6.3   k.6.4   k.6.5   k.6.6   [22]
 k.7.1   k.7.2   k.7.3   k.7.4   k.7.5   k.7.6   k.7.7   [29]
 k.8.1   k.8.2   k.8.3   k.8.4   k.8.5   k.8.6   k.8.7   [36]
 k.8.8   k.9.1   k.9.2   k.9.3   k.9.4   k.9.5   k.9.6   [43]
 k.9.7   k.9.8   k.9.9   k.10.1 
 k.10.2  k.10.3  k.10.4  [50]
 k.10.5  k.10.6  k.10.7  k.10.8 
 k.10.9  k.10.10
  
  I want to add to the.data new columns: min.k.1,
 min.k.2, ..., min.k.10
  
  This is the section of code I would like to
 improve, hopefully getting rid of the eval and the for
 loop:
  
  for(k in 1:10){
     s - subset(the.data,
 select=paste(k, k, 1:k, sep=.))
     eval(parse(text =
 paste(the.data$min.k., k, -as.vector(by(s, 1:nrow(s),
 min)), sep=)))
  }
  
  Thanks for any help,
  Bill
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 mailing list
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  http://www.R-project.org/posting-guide.html
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 reproducible code.
  
   
  
 
 



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Re: [R] how to ignore rows missing arguments of a function when creating a function?

It's difficult to help you if we don't know what the data looks like.
Two more tips :
- look at ?with instead of attach(). The latter causes a lot of
trouble further on.
- use require() instead of library(). See also the help files on this.

I also wonder what you're doing with the na.rm=TRUE. I don't see how
it affects the function, as it is used nowhere. Does it really remove
NA?

This said: estfun takes the estimation function from your model
object. Your model is essentially fit on all observations that are not
NA, whereas I presume your cluster contains all observations,
including NA. Now I don't know what kind of model fm represents, but
normally there is information in the model object about the removed
observations. I illustrate this using lm :

 x - rnorm(100)
 y - c(rnorm(90),NA,rnorm(9))
 test - lm(x~y)
 str(test)
List of 13
 ... (tons of information)
 $ model:'data.frame':  99 obs. of  2 variables:
  ... (more tons of information)
  ..- attr(*, na.action)=Class 'omit'  Named int 91
  .. .. ..- attr(*, names)= chr 91
 - attr(*, class)= chr lm

Now we know that we can do :
 not -attr(test$model,na.action)
 y[-not]

So try this
### NOT TESTED
cl - function(dat, na.rm = TRUE, fm, cluster){
require(sandwich)
require(lmtest)

not - attr(fm$model,na.action)
cluster - cluster[-not]

with( dat ,{
M - length(unique(cluster))
N - length(cluster)
K - fm$rank

dfc - (M/(M-1))*((N-1)/(N-K))

uj - data.frame(apply(estfun(fm),2, function(x) data.frame(tapply(x,
cluster, sum)) ) );

vcovCL - dfc*sandwich(fm, meat=crossprod(uj)/N)

coeftest(fm, vcovCL)
} )
}

Cheers
Joris

On Wed, Jun 9, 2010 at 12:06 AM, edmund jones edmund.j.jo...@gmail.com wrote:
 Hi,

 I am relatively new to R; when creating functions, I run into problems with
 missing values. I would like my functions to ignore rows with missing values
 for arguments of my function) in the analysis (as for example is the case in
 STATA). Note that I don't want my function to drop rows if there are missing
 arguments elsewhere in a row, ie for variables that are not arguments of my
 function.

 As an example: here is a clustering function I wrote:


 cl - function(dat, na.rm = TRUE, fm, cluster){

 attach( dat , warn.conflicts = F)

 library(sandwich)

 library(lmtest)

 M - length(unique(cluster))

 N - length(cluster)

 K - fm$rank

 dfc - (M/(M-1))*((N-1)/(N-K))

 uj - data.frame(apply(estfun(fm),2, function(x) data.frame(tapply(x,
 cluster, sum)) ) );

 vcovCL - dfc*sandwich(fm, meat=crossprod(uj)/N)

 coeftest(fm, vcovCL)

 }


 When I run my function, I get the message:


 Error in tapply(x, cluster, sum) : arguments must have same length


 If I specify instead attach(na.omit(dat), warn.conflicts = F)  and don't
 have the na.rm=TRUE argument, then my function runs; but only for the rows
 where there are no missing values AT ALL; however, I don't care if there are
 missing values for variables on which I am not applying my function.


 For example, I have information on children's size; if I want regress scores
 on age and parents' education, clustering on class, I would like missing
 values in size not to interfere (ie if I have scores, age, parents'
 education, and class, but not size, I don't want to drop this observation).


 I tried to look at the code of lm to see how the na.action part works, but
 I couldn't figure it out... This is exactly how I would like to deal with
 missing values.


 I tried to write

 cl - function(dat, fm, cluster, na.action){

 attach( dat , warn.conflicts = F)

 library(sandwich)

 library(lmtest)

  M - length(unique(cluster))

  N - length(cluster)

  K - fm$rank

  dfc - (M/(M-1))*((N-1)/(N-K))

 uj - data.frame(apply(estfun(fm),2, function(x) data.frame(tapply(x,
 cluster, sum)) ) );

 vcovCL - dfc*sandwich(fm, meat=crossprod(uj)/N)

  coeftest(fm, vcovCL)

 }

  attr(cl,na.action) - na.exclude


 but it still didn't work...


 Any ideas of how to deal with this issue?

 Thank you for your answers!

 Edmund

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-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
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[R] Odp: Extracting Elements By Date

Hi

You have got several suggestions. When I try your code I get only errors, 
not list or any other objects.

 x-unique(Returns.names$date_)
Error in unique(Returns.names$date_) : object 'Returns.names' not found
 n-1
 while(n=19) {
+ Returns.period-lapply(((n-1)*125+1):((n+1)*125), function (i)
+  which(Returns.filter$date_==x[i]))
+ Returns.period-unlist(Returns.period)
+ 
+ 
Returns.period1[n,1:(length(Returns.period))]-Returns.filter[Returns.period,]
+ n-n+1
+ }
Error in which(Returns.filter$date_ == x[i]) : 
  object 'Returns.filter' not found


How do you expect to get reasonable help if you can not present what you 
did and how result differ from what you want on some fake data which we 
can copy to R.

r-help-boun...@r-project.org napsal dne 09.06.2010 03:49:37:

 
 Dear R Gurus,
 
 Thanks for any help in advance!
 
 Date.frame: Returns.names 
 
X   id ticker  date_ adjClose   totret RankStk
 258060 258060 13645T10 CP 2001-06-29   18.125 1877.758 
 
 My data frame is in the above format. I would like to filter by period, 
per

what is filter??
Help page says

Description
Applies linear filtering to a univariate time series or to each series 
separately of a multivariate time series.

 id (every 125 days) each consisting of 250 days, I.e. 1-250, 126-375, 
etc.

snip

 Thus, I tried to adjust by dynamically assigning each period to a 
column.
 However, I get the error: object 'Returns.period1' not found. And R is 
not
 like Java where you can simply declare a variable by typing say
 Returns.period1. 

Why do you expect that one language will behave the same way as another 
language. If R was like Java what would be the reason to have 2 languages?

Try instead to comply posting guide and make a fake data which can be 
directly used e.g. by dput, construct a result which is to be achieved and 
present a code which can be copied to R and which gives other then 
expected result.

After that we can quit guessing and start providing relevant help.

Regards
Petr

 
 I do not know how to create a flexible empty matrix that would allow 
this
 code to work.
 
 If anyone knows how to do this without a loop, that would be even 
better!
 
 ##Filtering by Period
 n-1
 while(n=19) {
Returns.period-lapply(((n-1)*125+1):((n+1)*125), function (i)
 which(Returns.filter$date_==x[i]))
Returns.period-unlist(Returns.period)
 
 
Returns.period1[n,1:(length(Returns.period))]-Returns.filter[Returns.period,]
n-n+1
}
 
 -- 
 View this message in context: http://r.789695.n4.nabble.com/Extracting-
 Elements-By-Date-tp2248227p2248227.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] equivalent of stata command in R

It helps if you translate the Stata commands. Not everybody is fluent
in those. It would even help more if you would enlight us about the
function you used to fit the model. Getting the marginal effects is
not that hard at all, but how depends a bit on the function you used
to estimate the model.

You can try
predict(your_model,type=terms,terms=the_term_you're_interested_in)

For exact information, look at the respective predict function, eg if
you use lme, do ?predict.lme
Be aware of the fact that R normally choses the correct predict
function without you having to specify it. predict() works for most
model objects. Yet, depending on the model eacht predict function can
have different options or different functionality. That information is
in the help files of the specific function.

Cheers
Joris

On Wed, Jun 9, 2010 at 11:28 AM, mike mick saint-fi...@hotmail.com wrote:

 Dear all,

 I need to use R for one estimation, and i have readily available  stata 
 command, but i need also the R version of the same command.
 the estimation in stata is as following:
   1. Compute mean values of relevant variables



 . sum inno lnE lnM



    Variable |       Obs        Mean    Std. Dev.       Min        Max

 -+

        inno |    146574    .0880374    .2833503          0          1

         lnE |    146353    .9256239    1.732912  -4.473922   10.51298

         lnM |    146209    4.281903    1.862192  -4.847253   13.71969



        2. Estimate model



 . xi: xtreg lnLP lnC lnL lnE lnM eco inno eco_inno eco_lnE eco_lnM i.year, fe 
 i(stno)

 i.year            _Iyear_1997-1999    (naturally coded; _Iyear_1997 omitted)



 Fixed-effects (within) regression               Number of obs      =    146167

 Group variable (i): stno                        Number of groups   =     48855



 R-sq:  within  = 0.9908                         Obs per group: min =         1

       between = 0.9122                                        avg =       3.0

       overall = 0.9635                                        max =         3



                                                F(11,97301)        = 949024.29

 corr(u_i, Xb)  = 0.2166                         Prob  F           =    0.



 --

        lnLP |      Coef.   Std. Err.      t    P|t|     [95% Conf. Interval]

 -+

         lnC |   .0304896   .0009509    32.06   0.000     .0286258    .0323533

         lnL |  -.9835998   .0006899 -1425.74   0.000     -.984952   -.9822476

         lnE |   .0652658   .0009439    69.14   0.000     .0634158    .0671159

         lnM |   .6729931   .0012158   553.53   0.000       .67061    .6753761

         eco |   .0610348   .0177048     3.45   0.001     .0263336     .095736

        inno |   .0173824   .0058224     2.99   0.003     .0059706    .0287943

    eco_inno |   .0080325   .0110815     0.72   0.469    -.0136872    .0297522

     eco_lnE |   .0276226    .004059     6.81   0.000      .019667    .0355781

     eco_lnM |  -.0214237   .0039927    -5.37   0.000    -.0292494   -.0135981

  _Iyear_1998 |  -.0317684   .0013978   -22.73   0.000     -.034508   -.0290287

  _Iyear_1999 |  -.0647261   .0027674   -23.39   0.000    -.0701501   -.0593021

       _cons |   1.802112    .009304   193.69   0.000     1.783876    1.820348

 -+

     sigma_u |  .38142386

     sigma_e |   .2173114

         rho |  .75494455   (fraction of variance due to u_i)

 --

 F test that all u_i=0:     F(48854, 97301) =     3.30        Prob  F = 0.



        3. Compute marginal effect of eco at sample mean



 . nlcom (_b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903)



       _nl_1:  
 _b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903



 --

        lnLP |      Coef.   Std. Err.      t    P|t|     [95% Conf. Interval]

 -+

       _nl_1 |  -.0036011    .008167    -0.44   0.659    -.0196084    .0124061

 --



 in fact i can find the mean of the variables ( step 1) and extimate the model 
 (step 2) but i couldnt find the equivalent of step 3 (compute marginal effect 
 of eco at sample mean). Can someone help me for this issue?

 Cheers!


 _


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Re: [R] equivalent of stata command in R‏

2010-06-09 Thread mike mick

From: saint-fi...@hotmail.com
To: saint-fi...@hotmail.com
Subject: RE:
Date: Wed, 9 Jun 2010 09:53:20 +

OK! sorry thats my fault,

here the translations of the stata commands
1st step is to get the mean values of the variables, well that doesnt need 
explanation i guess,

2nd step is to estimate the model on panel data estimation method
which is:
mdl-plm(lnLP~lnC+lnL+lnM+lnE+Eco+Inno+Eco*Inno+Eco*lnM+Eco*lnE+year,data=newdata,model=within)
and basically i need to get the marginal effect of variable Eco at the sample 
mean (step 3) but i am not that good in R so any additional help is wlcome!
Thanks
From: saint-fi...@hotmail.com
To: r-help@r-project.org
Subject: 
Date: Wed, 9 Jun 2010 09:45:16 +

It helps if you translate the Stata commands. Not everybody is fluent
in those. It would even help more if you would enlight us about the
function you used to fit the model. Getting the marginal effects is
not that hard at all, but how depends a bit on the function you used
to estimate the model.

You can try
predict(your_model,type=terms,terms=the_term_you're_interested_in)

For exact information, look at the respective predict function, eg if
you use lme, do ?predict.lme
Be aware of the fact that R normally choses the correct predict
function without you having to specify it. predict() works for most
model objects. Yet, depending on the model eacht predict function can
have different options or different functionality. That information is
in the help files of the specific function.

Cheers
Joris

Dear all,

I need to use R for one estimation, and i have readily available  stata 
command, but i need also the R version of the same command.
the estimation in stata is as following:
   1. Compute mean values of relevant variables

. sum inno lnE lnM

Variable |   ObsMeanStd. Dev.   MinMax

-+

inno |146574.0880374.2833503  0  1

 lnE |146353.92562391.732912  -4.473922   10.51298

 lnM |1462094.2819031.862192  -4.847253   13.71969

2. Estimate model

. xi: xtreg lnLP lnC lnL lnE lnM eco inno eco_inno eco_lnE eco_lnM i.year, fe 
i(stno)

i.year_Iyear_1997-1999(naturally coded; _Iyear_1997 omitted)

Fixed-effects (within) regression   Number of obs  =146167

Group variable (i): stnoNumber of groups   = 48855

R-sq:  within  = 0.9908 Obs per group: min = 1

   between = 0.9122avg =   3.0

   overall = 0.9635max = 3

F(11,97301)= 949024.29

corr(u_i, Xb)  = 0.2166 Prob  F   =0.

--

lnLP |  Coef.   Std. Err.  tP|t| [95% Conf. Interval]

-+

 lnC |   .0304896   .000950932.06   0.000 .0286258.0323533

 lnL |  -.9835998   .0006899 -1425.74   0.000 -.984952   -.9822476

 lnE |   .0652658   .000943969.14   0.000 .0634158.0671159

 lnM |   .6729931   .0012158   553.53   0.000   .67061.6753761

 eco |   .0610348   .0177048 3.45   0.001 .0263336 .095736

inno |   .0173824   .0058224 2.99   0.003 .0059706.0287943

eco_inno |   .0080325   .0110815 0.72   0.469-.0136872.0297522

 eco_lnE |   .0276226.004059 6.81   0.000  .019667.0355781

 eco_lnM |  -.0214237   .0039927-5.37   0.000-.0292494   -.0135981

 _Iyear_1998 |  -.0317684   .0013978   -22.73   0.000 -.034508   -.0290287

 _Iyear_1999 |  -.0647261   .0027674   -23.39   0.000-.0701501   -.0593021

   _cons |   1.802112.009304   193.69   0.000 1.7838761.820348

-+

 sigma_u |  .38142386

 sigma_e |   .2173114

 rho |  .75494455   (fraction of variance due to u_i)

--

F test that all u_i=0: F(48854, 97301) = 3.30Prob  F = 0.

3. Compute marginal effect of eco at sample mean

. nlcom (_b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903)

   _nl_1:  
_b[eco]+_b[inno]*.0880374+_b[eco_lnE]*.9256239+_b[eco_lnM]*4.281903

--

lnLP |  Coef.   Std. Err.  tP|t| [95% Conf. Interval]

[R] Problem with library(SSPA)

2010-06-09 Thread Samuel Okoye


Hello,

I have the fellowing problem and I am thankful for any advice!

Regards,
Samuel


   source(http://bioconductor.org/biocLite.R;)
BioC_mirror = http://www.bioconductor.org
Change using chooseBioCmirror().
 biocLite(SSPA)
Using R version 2.11.0, biocinstall version 2.6.7.
Installing Bioconductor version 2.6 packages:
[1] SSPA
Please wait...

trying URL 
'http://www.bioconductor.org/packages/2.6/bioc/bin/windows/contrib/2.11/SSPA_1.4.0.zip'
Content type 'application/zip' length 305310 bytes (298 Kb)
opened URL
downloaded 298 Kb

package 'SSPA' successfully unpacked and MD5 sums checked

 library(SSPA)
Loading required package: qvalue
Loading
 required package: tcltk
Loading Tcl/Tk interface ... done
Error : .onAttach failed in attachNamespace() for 'SSPA', details:
  call: fun(...)
  error: could not find function addVigs2WinMenu
Error: package/namespace load failed for 'SSPA'
 sessionInfo()
R version 2.11.0 (2010-04-22)
i386-pc-mingw32

locale:
[1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United 
Kingdom.1252    LC_MONETARY=English_United Kingdom.1252
[4] LC_NUMERIC=C    LC_TIME=English_United 
Kingdom.1252   

attached base packages:
[1] tcltk stats graphics  grDevices utils datasets  methods   
base    

other attached
 packages:
[1] qvalue_1.22.0

loaded via a namespace (and not attached):
[1] limma_3.4.3  SSPA_1.4.0   tools_2.11.0

##




  
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[R] Read in dataset without saving it

2010-06-09 Thread Paul Chatfield


A simple question - I have a small dataset to read in and want to copy and
paste part from Excel and paste it into an R script file without creating
more files saving it as a .txt/.csv and then reading that in.  I want to
read in 3 columns e.g.
1 2.5 3.4
1 2.3 3.1
1 2.6 3.9
2 2.9 2.8
2 2.6 2.9
2 2.7 2.9
3 2.3 3.3
3 2.4 3.0
3 2.7 3.2

I thought I could use scan() but don't know how to extend it to multiple
columns?  I thought about using colwise (plyr) but think I am making this
more complicated than it probably should be!  Is there an easy way to do
this?

Any ideas gratefully received,

Paul
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View this message in context: 
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Re: [R] Please help me

Dear Xionqing,

It's still difficult to say without the data, but I guess the error message
reads System computation is singular, meaning you just have too many
variables or some variables are near-aliases of eachother. I did not ask to
read the posting guide to be difficult. Without some reproducible code and
detailed information on the data, it's impossible to say what is the problem
exactly.

Cheers
Joris

On Wed, Jun 9, 2010 at 2:47 AM, Xiongqing Zhang xqzhan...@yahoo.com wrote:

 Dear Joris,

 Thank for your  reply.


 Firstly, I introduced one dummy variable (eg: u=0, the first group; u=1,
 the second group) to the model, and I got the results (parameter estimators
 and AIC value).

 #zero inflated negative binomial model

 library(pscl)

 dat=read.csv(data.csv,header=T)

 zinb - zeroinfl(y ~x1+x2+x3+x4+x5+u | x1+x2+x3+x4+x5+u, data = dat, link=
 logit, dist = negbin)

 summary(zinb)



 But the results are not satisfied to me. So I decided to introduce two
 dummy variables (u1,u2) to the model, and then I could not get the results.



 dat=read.csv(data.csv,header=T)

 zinb - zeroinfl(y ~x1+x2+x3+x4+x5+u1+u2 | x1+x2+x3+x4+x5+u1+u2, data =
 dat, link= logit, dist = negbin)

 summary(zinb)

 And the error message is

 error in solve.default(as.matrix(fit$hessian)) :

   System computation is unsigular

 moreover: Warning message:

 In glm.fit(Z, as.integer(Y0), weights = weights, family = binomial(link =
 linkstr),  :

   algorithm did not converge



 Again, many thanks for you.



 Sincerely,

 Xiongqing















 --- On *Tue, 6/8/10, Joris Meys jorism...@gmail.com* wrote:


 From: Joris Meys jorism...@gmail.com
 Subject: Re: [R] Please help me
 To: Xiongqing Zhang xqzhan...@yahoo.com
 Cc: r-help@r-project.org
 Date: Tuesday, June 8, 2010, 11:22 AM


 First, read the posting guides.
 Then, supply us with a bit more information, like the package you
 used, example code that reproduces the error, information about the
 data, the complete error message, the traceback (use the function
 traceback() right after you got the error).

 Otherwise we ain't going to be able to help you.

 Cheers
 Joris

 On Tue, Jun 8, 2010 at 9:42 AM, Xiongqing Zhang 
 xqzhan...@yahoo.comhttp://cn.mc453.mail.yahoo.com/mc/compose?to=xqzhan...@yahoo.com
 wrote:
 
  Dear Mr. or Ms.,
 
  I used the R-software to run the zero-inflatoin negative binomial model
 (zeroinfl()) .
 
  Firstly, I introduced one dummy variable to the model as an independent
 variable, and I got the estimators of parameters. But the results are not
 satisfied to me. So I introduced three dummy variables to the model. but I
 could not get the results. And the error message is
 solve.default(as.matrix(fit$gaussian)) . I do not know the reasons fully.
 And I want to know what method was used to estimate the parameters when I
 called the function zeroinfl() .
 
  I will be very appreciate if you can help me. Thank you.
 
  Best regards,
 
  Sincerely,
  Xiongqing Zhang
 
 
 
 [[alternative HTML version deleted]]
 
 
  __
  R-help@r-project.orghttp://cn.mc453.mail.yahoo.com/mc/compose?to=r-h...@r-project.orgmailing
   list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 



 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.behttp://cn.mc453.mail.yahoo.com/mc/compose?to=joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php





-- 
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Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] ordisurf (pkg vegan) gives implausible result

Ordisurf uses a gam model to fit the surface (and more exactly, in
your case a thin plate regression spline). Standard these ts splines
try to reduce the dimensionality of your model as much as possible.
Hence in your case they fit indeed a linear plane through the space,
which is supposed to be the best fit. If you want an exact fit of your
data, you shouldn't use a model.

To see what goes on, you can try following right after the ordisurf command :
lK - cut2(K,g=5,onlycuts=T)
gen - (K-min(K)+0.1)/(max(K)+0.2-min(K))
leg - (lK-min(K)+0.1)/(max(K)+0.2-min(K))
colgen - rgb(0,gen,0)
points(sample.mds$points, pch=19,col=colgen)

colleg - rgb(0,leg,0)
legend(topleft,legend=lK,col=colleg,horiz=T,pch=19)

This plots the values in color coding, so you can try to judge the
fit. Also look at the object from ordisurf itself:
ordi.object - ordisurf(sample.mds, edaphic$K, xlab=PC 1, ylab=PC
2, main=K)
str(ordi.object)

This should contain the information on the gam fit.

Cheers
Joris

On Wed, Jun 9, 2010 at 5:33 AM, Matt Bakker matt.g.bak...@gmail.com wrote:
 I'm having trouble with the ordisurf function in the vegan package.

 I have created an ordination plot (cmdscale) of 60 samples based on
 Bray-Curtis dissimilarities, and would like to overlay various soil edaphic
 characteristics as possible clues to the clustering I observe in my plot.
 However, I find that ordisurf creates a surface on the plot that is a
 perfect, even gradient - and doesn't match my data. An example of this
 output is attached (as .ps file), while these are the data for the
 environmental variable of interest:

 edaphic$K
  [1] 28 61 48 29 28 26 45 28 34 33 55 62 44 51 60 68 51 31 54 32 58 50 37 35
 35 34 52 29 53 24 37 50 62 51
 [35] 28 39 47 41 49 83 33 50 55 71 59 50 57 47 46 49 43 30 56 23 49 35 31 27
 38 52

 The contour lines suggest that there are only two values below 32, when in
 fact there are 11.
 Any ideas? Thanks!


 bc - read.table(braycurtis_noheader.dist,header=F)
 sample.mds - cmdscale(bc)
 edaphic - read.table(edaphic.txt, header=T)
 library(vegan)
 This is vegan 1.17-2
 ordisurf(sample.mds, edaphic$K, xlab=PC 1, ylab=PC 2, main=K)
 Loading required package: mgcv
 This is mgcv 1.6-1. For overview type `help(mgcv-package)'.
 Family: gaussian
 Link function: identity
 Formula:
 y ~ s(x1, x2, k = knots)
 Estimated degrees of freedom:
 2  total = 3
 GCV score: 140.9051

 --
 Matthew Bakker
 Ph.D. Candidate
 Department of Plant Pathology
 University of Minnesota
 495 Borlaug Hall
 1991 Upper Buford Circle
 Saint Paul, MN  55108 USA

 612-624-2253
 matt.g.bak...@gmail.com
 http://plpa.cfans.umn.edu/Matt_Bakker.html

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-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
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[R] minor tick marks

2010-06-09 Thread Stéphane Adamowicz

Hi !

I need a plot for data extending over several orders of magnitude on the y axis.
The following command generates a nice looking semi-log plot for my data:
plot(x,y,log=y,type=l,lty=3, ylim=c(0.01,2),yaxp=c(0.01,1,1),las=1)

 I would appreciate having also minor tick marks in-between the 3 major ticks 
obtained with the above command. The minor.tick function in library Hmisc 
gives an error when applied to log axes.

Any solution ?

Stéphane


_
Stéphane Adamowicz
INRA, unité PSH
domaine St Paul, site agroparc
84914 Avignon, cedex 9
France

stephane.adamow...@avignon.inra.fr
tel.  +33 (0)4 32 72 24 35
fax. +33 (0)4 32 72 24 32
do not dial 0 when out of France
web PSH  : http://www.avignon.inra.fr/psh
web INRA : http://www.inra.fr/
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[R] Problem using Rmpi

2010-06-09 Thread Paco Pastor


Hi

Thanks to the help from Uwe Ligges I could update Rmpi package. Now R 
can load the package but it still does not work. Now the problem comes 
when trying to use the first Rmpi command in a basic tutorial:


 library(Rmpi)
 mpi.spawn.Rslaves()
Error en mpi.spawn.Rslaves() : You cannot use MPI_Comm_spawn API

Seaching in the list I have found references to this problem but not the 
solution. More info about R and MPICH installation in my system (Ubuntu 
Hardy Heron in a dual AMD Athlon PC) can be found in 
http://ubuntuone.com/p/6Wr/


Thanks in advance

--
---
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Meteorology department
Fundación CEAM
p...@ceam.es
http://www.ceam.es/ceamet - http://www.ceam.es
Parque Tecnologico, C/ Charles R. Darwin, 14
46980 PATERNA (Valencia), Spain
Tlf. 96 131 82 27 - Fax. 96 131 81 90
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Re: [R] minor tick marks

2010-06-09 Thread Romain Francois



Hello,

Perhaps something like:

 axis( 2, 0:10/10, las = 1, tcl = -.2 )

Romain

Le 09/06/10 11:08, Stéphane Adamowicz a écrit :

Hi !

I need a plot for data extending over several orders of magnitude on the y axis.
The following command generates a nice looking semi-log plot for my data:
plot(x,y,log=y,type=l,lty=3, ylim=c(0.01,2),yaxp=c(0.01,1,1),las=1)

  I would appreciate having also minor tick marks in-between the 3 major ticks obtained 
with the above command. The minor.tick function in library Hmisc gives an 
error when applied to log axes.

Any solution ?

Stéphane


_
Stéphane Adamowicz
INRA, unité PSH
domaine St Paul, site agroparc
84914 Avignon, cedex 9
France

stephane.adamow...@avignon.inra.fr
tel.  +33 (0)4 32 72 24 35
fax. +33 (0)4 32 72 24 32
do not dial 0 when out of France
web PSH  : http://www.avignon.inra.fr/psh
web INRA : http://www.inra.fr/



--
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Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://bit.ly/98Uf7u : Rcpp 0.8.1
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and provide commented, minimal, self-contained, reproducible code.

[R] dealing with heteroscedasticity in lmer: problem with the method weights

2010-06-09 Thread doris gomez

Dear lmer users,

The experiment includes 15 groups of (3 males and 1 female). The female is 
characterized by its quality Q1 and Q2. Each male of a group is characterized 
by the number of MatingAttempts (with Poisson distribution). I want to examine 
if male mating attempts depend on female quality.  I can see from graphic 
exploration  that the within-group heterogeneity of male attempts increases 
with female quality Q1.

When including the method weights in the function lmer,  I get the message that 
variables' length varies and the model does not run.
lmer(MatingAttempts~Q1+Q2+(1|Group),data=file,family=poisson,na.action=na.omit,
 REML=FALSE, weights=varExp(form=~Q1))

If I run the same model (fixed effects and random effects) with lme, it 
works properly, which shows that there is no problem with data structure.
lme(MatingAttempts~Q1+Q2,random=~1|Group,data=file,na.action=na.omit, 
method=ML, weights=varExp(form=~Q1))

I saw on the forum that lmer had problems in taking into account variance 
heterogeneity. Yet, the messages were old and there are maybe new solutions.
How can I correct the analyses for this problem of heteroscedasticity? 
Should I normalise the within group variance before implementing the model? And 
deal with the variance (as a new variable to explain) in another model?
Is there another way to solve this problem?

Thank you in advance for your help
Doris Gomez


  
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Re: [R] Read in dataset without saving it

See ?textConnection, eg
test -1 2.5 3.4
1 2.3 3.1
1 2.6 3.9
2 2.9 2.8
2 2.6 2.9
2 2.7 2.9
3 2.3 3.3
3 2.4 3.0
3 2.7 3.2
x - textConnection(test)
read.table(x)
close(x)

On Wed, Jun 9, 2010 at 10:38 AM, Paul Chatfield
p.s.chatfi...@reading.ac.uk wrote:

 A simple question - I have a small dataset to read in and want to copy and
 paste part from Excel and paste it into an R script file without creating
 more files saving it as a .txt/.csv and then reading that in.  I want to
 read in 3 columns e.g.
 1 2.5 3.4
 1 2.3 3.1
 1 2.6 3.9
 2 2.9 2.8
 2 2.6 2.9
 2 2.7 2.9
 3 2.3 3.3
 3 2.4 3.0
 3 2.7 3.2

 I thought I could use scan() but don't know how to extend it to multiple
 columns?  I thought about using colwise (plyr) but think I am making this
 more complicated than it probably should be!  Is there an easy way to do
 this?

 Any ideas gratefully received,

 Paul
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Read-in-dataset-without-saving-it-tp2248495p2248495.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
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and provide commented, minimal, self-contained, reproducible code.

[R] Sorting Association Rules on 2 criteria

2010-06-09 Thread markwebb

I'm using library(arules)
Once rules have been created one can inspect  sort them.
Here is an example to sort on support.
I want to sort on support AND THEN confidence - Is this possible ?
inspect (SORT(rules, by = support))

Regards

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Re: [R] Extracting Elements By Date

2010-06-09 Thread Gabor Grothendieck

Try this and next time provide reproducible code and data as per
posting guide (see last line of every message).

First we create a sample data frame, DF.   Then we use the zoo time
series package to read it in and convert it from long form, DF, to
wide form, z, with one column per id.

Then we use split to split it into two day groups noting that %/%
signifies integer division. Since we have 4 successive days here this
gives two groups of two days each.

See ?read.zoo and the three zoo vignettes (pdf documents) that come
with the zoo package.  Also see ?split

 # sample data
 DF - data.frame(Date = as.Date(2000-1-1) + 0:3, id = A, value = 1:4)
 DF - rbind(DF, transform(DF, id = B))
 DF
Date id value
1 2000-01-01  A 1
2 2000-01-02  A 2
3 2000-01-03  A 3
4 2000-01-04  A 4
5 2000-01-01  B 1
6 2000-01-02  B 2
7 2000-01-03  B 3
8 2000-01-04  B 4

 library(zoo)
 # need devel version of read.zoo
 source(http://r-forge.r-project.org/scm/viewvc.php/*checkout*/pkg/zoo/R/read.zoo.R?revision=719root=zoo;)
 z - read.zoo(DF, split = id)
 z
   A B
2000-01-01 1 1
2000-01-02 2 2
2000-01-03 3 3
2000-01-04 4 4
 split(z, as.numeric(time(z) - start(z)) %/% 2)
$`0`
   A B
2000-01-01 1 1
2000-01-02 2 2

$`1`
   A B
2000-01-03 3 3
2000-01-04 4 4


On Tue, Jun 8, 2010 at 9:49 PM, Jeff08 jefferyd...@gmail.com wrote:

 Dear R Gurus,

 Thanks for any help in advance!

 Date.frame: Returns.names

       X       id ticker      date_ adjClose   totret RankStk
 258060 258060 13645T10     CP 2001-06-29   18.125 1877.758

 My data frame is in the above format. I would like to filter by period, per
 id (every 125 days) each consisting of 250 days, I.e. 1-250, 126-375, etc.

 One important thing to note is that not all ID's have the same number of
 dates.

 x-unique(Returns.names$date_) gives me the ordered list of all the dates,
 so for each period i want it would be x[((n-1)*125+1):((n+1)*125)]

 when i tried to filter for just 1 period, it worked fine, but alas, I do not
 know how to work around the problem that you cannot assign dynamic variables
 in loops (say I named the variable for the extracted dates Returns.period,
 it would get overwritten every iteration of the loop, and I cant name
 something Returns.n, where n is the index for the loop)

 Thus, I tried to adjust by dynamically assigning each period to a column.
 However, I get the error: object 'Returns.period1' not found. And R is not
 like Java where you can simply declare a variable by typing say
 Returns.period1.

 I do not know how to create a flexible empty matrix that would allow this
 code to work.

 If anyone knows how to do this without a loop, that would be even better!

 ##Filtering by Period
 n-1
 while(n=19) {
        Returns.period-lapply(((n-1)*125+1):((n+1)*125), function (i)
 which(Returns.filter$date_==x[i]))
        Returns.period-unlist(Returns.period)

 Returns.period1[n,1:(length(Returns.period))]-Returns.filter[Returns.period,]
        n-n+1
        }

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Extracting-Elements-By-Date-tp2248227p2248227.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Read in dataset without saving it

Hi

r-help-boun...@r-project.org napsal dne 09.06.2010 12:44:33:

 See ?textConnection, eg
 test -1 2.5 3.4
 1 2.3 3.1
 1 2.6 3.9
 2 2.9 2.8
 2 2.6 2.9
 2 2.7 2.9
 3 2.3 3.3
 3 2.4 3.0
 3 2.7 3.2
 x - textConnection(test)
 read.table(x)
 close(x)
 
 On Wed, Jun 9, 2010 at 10:38 AM, Paul Chatfield
 p.s.chatfi...@reading.ac.uk wrote:
 
  A simple question - I have a small dataset to read in and want to copy 
and
  paste part from Excel and paste it into an R script file without 
creating
  more files saving it as a .txt/.csv and then reading that in.  I want 
to

In Excel select what you want to copy
press Ctrl C
In R use
read.delim(clipboard)
or
read.table(clipboard)

depending whether you selected or did not selected some header.

Regards
Petr


  read in 3 columns e.g.
  1 2.5 3.4
  1 2.3 3.1
  1 2.6 3.9
  2 2.9 2.8
  2 2.6 2.9
  2 2.7 2.9
  3 2.3 3.3
  3 2.4 3.0
  3 2.7 3.2
 
  I thought I could use scan() but don't know how to extend it to 
multiple
  columns?  I thought about using colwise (plyr) but think I am making 
this
  more complicated than it probably should be!  Is there an easy way to 
do
  this?
 
  Any ideas gratefully received,
 
  Paul
  --
  View this message in context: 
http://r.789695.n4.nabble.com/Read-in-dataset-
 without-saving-it-tp2248495p2248495.html
  Sent from the R help mailing list archive at Nabble.com.
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 
 
 
 -- 
 Joris Meys
 Statistical consultant
 
 Ghent University
 Faculty of Bioscience Engineering
 Department of Applied mathematics, biometrics and process control
 
 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
 
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Re: [R] Read in dataset without saving it

2010-06-09 Thread Gabor Grothendieck

See:

http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windowss=excel

On Wed, Jun 9, 2010 at 4:38 AM, Paul Chatfield
p.s.chatfi...@reading.ac.uk wrote:

 A simple question - I have a small dataset to read in and want to copy and
 paste part from Excel and paste it into an R script file without creating
 more files saving it as a .txt/.csv and then reading that in.  I want to
 read in 3 columns e.g.
 1 2.5 3.4
 1 2.3 3.1
 1 2.6 3.9
 2 2.9 2.8
 2 2.6 2.9
 2 2.7 2.9
 3 2.3 3.3
 3 2.4 3.0
 3 2.7 3.2

 I thought I could use scan() but don't know how to extend it to multiple
 columns?  I thought about using colwise (plyr) but think I am making this
 more complicated than it probably should be!  Is there an easy way to do
 this?

 Any ideas gratefully received,

 Paul
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Read-in-dataset-without-saving-it-tp2248495p2248495.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Efficiency question

On Tue, Jun 8, 2010 at 11:53 PM, Worik R wor...@gmail.com wrote:
 Given the following snippet
  m.nf.xts - xts(rep(0, length(index(m.xts))), order.by=index(m.xts))

 Does R know to cache the index(m.xts) or is it more efficient to say...

 m.i - index(m.xts)
  m.nf.xts - xts(rep(0, length(m.i)), order.by=index(m.i))
this is one call less to a function, and hence more efficient. I
suppose it has to be order.by=m.i

Btw, How should R know what to cache? If it would cache all
calculations, you'd be out of memory in no time.
If you have more questions like this, you can test it by yourself, eg :

 x - sample(rnorm(100))

 id - index(x) # is in this case equivalent to 1:100

 system.time(replicate(10^5,{rep(0,length(index(x)))[index(x)]}))
   user  system elapsed
   7.610.097.76

 system.time(replicate(10^5,{rep(0,length(id))[id]}))
   user  system elapsed
   1.490.061.58

Cheers
Joris



 ?

 cheers
 Worik

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-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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[R] Rglpk

2010-06-09 Thread Kaveh Vakili


Hi list,


in the Rglpk_solve_LP function (::Rglpk), 
on line 26, the function calls a function 
as.glp_bounds() that i cannot access.

i'm trying to alter the Rglpk_solve_LP function
to add a line to retrieve column/row dual values.
everytime i change the slightest line of code inside
Rglpk_solve_LP() [to even add a print]
i get a 
': could not find function as.glp_bounds'

What's the catch here ?

Best,

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[R] bootpred for multinomial

2010-06-09 Thread azam jaafari

I applied bootpred for multinomial logistic reg. (with nnet package). I used 
same as theta.fit and theta.predict of R for my data. but give me error. Can I 
do this with
response vriable;7 levels
predictor variables:5 (1 classifier, 4 continuous)?
 
Thanks alot
Azam
 


  
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[R] strange issue with which on seq

2010-06-09 Thread james.foadi

Dear R community,
I am puzzled by the following lines:

 v - seq(-0.5,0.5,by=0.1)
 v
  [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5
 which(v == -0.4)
  [1] 2
 which(v == 0)
  [1] 6
 which(v == 0.1)
integer(0)
 which(v == 0.2)
integer(0)
 which(v == 0.3)
integer(0)
 which(v == 0.4)
  [1] 10
 which(v == 0.5)
  [1] 11

Why which can only match some of the values in v? Are the numbers generated 
by seq not exact fractional numbers?
Please, help me to understand this.

J

Dr James Foadi PhD
Membrane Protein Laboratory (MPL)
Diamond Light Source Ltd
Diamond House
Harewell Science and Innovation Campus
Chilton, Didcot
Oxfordshire OX11 0DE

Email:  james.fo...@diamond.ac.uk
Alt Email:  j.fo...@imperial.ac.uk

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[R] Odp: strange issue with which on seq

Hi

r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:

 Dear R community,
 I am puzzled by the following lines:
 
  v - seq(-0.5,0.5,by=0.1)
  v
   [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5
  which(v == -0.4)
   [1] 2
  which(v == 0)
   [1] 6
  which(v == 0.1)
 integer(0)
  which(v == 0.2)
 integer(0)
  which(v == 0.3)
 integer(0)
  which(v == 0.4)
   [1] 10
  which(v == 0.5)
   [1] 11
 
 Why which can only match some of the values in v? Are the numbers 
 generated by seq not exact fractional numbers?
 Please, help me to understand this.

Well FAQ 7.31 was not here some time. Computing in binary results in 
finite precision of fractional numbers.

v - seq(-0.5,0.5,by=0.1)
v[7]-0.1
[1] 8.326673e-17

Regards
Petr



 
 J
 
 Dr James Foadi PhD
 Membrane Protein Laboratory (MPL)
 Diamond Light Source Ltd
 Diamond House
 Harewell Science and Innovation Campus
 Chilton, Didcot
 Oxfordshire OX11 0DE
 
 Email:  james.fo...@diamond.ac.uk
 Alt Email:  j.fo...@imperial.ac.uk
 
 -- 
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Re: [R] specifying plot symbol sizes in qplot or ggplot2

2010-06-09 Thread ONKELINX, Thierry

Have a look at ?scale_size_manual()

HTH,

Thierry



ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie  Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics  Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
  

 -Oorspronkelijk bericht-
 Van: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] Namens cfriedl
 Verzonden: woensdag 9 juni 2010 3:22
 Aan: r-help@r-project.org
 Onderwerp: [R] specifying plot symbol sizes in qplot or ggplot2
 
 
 Hi.
 
 first things first ... thanks for ggplot2.
 
 Now my question. I'm using qplot to generate a plot as 
 follows where X,Y,Z, A are columns in a dataframe.
 
 qplot(X, Y, data=XYDATA, color=Z, geom=c(point), size=A)
 
 This works as expected. Factor A has three levels so there 
 are three sizes of the point plot symbol. I understand that 
 the factor levels are mapped to symbol sizes. However the 
 sizes are too small for my liking. Is there any way I can 
 specify the range of plot symbol sizes that factor A levels 
 are mapped to?
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/specifying-plot-symbol-sizes-in-
 qplot-or-ggplot2-tp2248217p2248217.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.

Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer 
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
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and any annex are purely those of the writer and may not be regarded as stating 
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[R] Help needed on switch function

2010-06-09 Thread Megh Dal

Hi all, Here I am trying to implement the switch() function to choose value of 
a variable depending on the value of an input variable :
 
temp1 - 1
  temp1.name - switch(temp1,
  1 == aa,
  2 == bb,
  3 == cc,
  4 == dd,
  5 == ee)
 
Goal is if temp1 equals to 1, then value of temp1.name would be aa. However 
I am getting following answer :
 
 temp1 - 1
   temp1.name - switch(temp1,
+   1 == aa,
+   2 == bb,
+   3 == cc,
+   4 == dd,
+   5 == ee)
 temp1.name
[1] FALSE

 
Can anyone please point me where I am doing wrong?
 
Thanks,


  
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] equivalent of stata command in R‏‏

2010-06-09 Thread mike mick

Thanx for your response,
yeah, i know i didnst specified the indexes
when i wrote the 2nd mail, in fact in the 1st mail i wrote already that
i dont have problem with the estimation of the model... thats the
reason why i didnt write in fact since the issue is not to estimate the
model but to get the marginal effect,
anyway, i figured out that predict(), doesnt work for panel data...
and
well, my problem is that contrary to your guess, i couldnt figure out
the rest of the calculations... since i am not that experienced in R.
one last help of yours would be quite helpful to get rid of this silly problem!
Thanx again...

Date: Wed, 9 Jun 2010 12:40:42 +0200
Subject: Re: [R] equivalent of stata command in Râ
From: jorism...@gmail.com
To: saint-fi...@hotmail.com
CC: r-help@r-project.org

plm does not have a predict function, so forget my former mail. To get
to the coefficients, you just :
coef(mdl)

The rest of the calculations you can figure out I guess.

I'm also not sure if you're doing what you think you're doing. You
never specified the index stno in your pml call. Read the help files
again. And while you're at it, read the posting guide for the list as
well:
http://www.R-project.org/posting-guide.html

Cheers
Joris

On Wed, Jun 9, 2010 at 11:54 AM, mike mick saint-fi...@hotmail.com wrote:

From: saint-fi...@hotmail.com
To: saint-fi...@hotmail.com
Subject: RE:
Date: Wed, 9 Jun 2010 09:53:20 +

OK! sorry thats my fault,

here the translations of the stata commands
1st step is to get the mean values of the variables, well that doesnt need
explanation i guess,

2nd step is to estimate the model on panel data estimation method
which is:
mdl-plm(lnLP~lnC+lnL+lnM+lnE+Eco+Inno+Eco*Inno+Eco*lnM+Eco*lnE+year,data=newdata,model=within)

and basically i need to get the marginal effect of variable Eco
at the sample mean (step 3) but i am not that good in R so any
additional help is wlcome!
Thanks
From: saint-fi...@hotmail.com
To: r-help@r-project.org
Subject:
Date: Wed, 9 Jun 2010 09:45:16 +

It helps if you translate the Stata commands. Not everybody is fluent
in those. It would even help more if you would enlight us about the
function you used to fit the model. Getting the marginal effects is
not that hard at all, but how depends a bit on the function you used
to estimate the model.

You can try
predict(your_model,type=terms,terms=the_term_you're_interested_in)

For exact information, look at the respective predict function, eg if
you use lme, do ?predict.lme
Be aware of the fact that R normally choses the correct predict
function without you having to specify it. predict() works for most
model objects. Yet, depending on the model eacht predict function can
have different options or different functionality. That information is
in the help files of the specific function.

Cheers
Joris

Dear all,

I need to use R for one estimation, and i have readily available
stata command, but i need also the R version of the same command.
the estimation in stata is as following:
1. Compute mean values of relevant variables

. sum inno lnE lnM

Variable | ObsMeanStd. Dev. MinMax

inno |146574.0880374.2833503 0 1

lnE |146353.92562391.732912 -4.473922 10.51298

lnM |1462094.2819031.862192 -4.847253 13.71969

2. Estimate model

. xi: xtreg lnLP lnC lnL lnE lnM eco inno eco_inno eco_lnE eco_lnM i.year,
fe i(stno)

i.year_Iyear_1997-1999(naturally coded; _Iyear_1997 omitted)

Fixed-effects (within) regression Number of obs =
146167

Group variable (i): stnoNumber of groups =
48855

R-sq: within = 0.9908 Obs per group: min =
1

between = 0.9122avg =
3.0

overall = 0.9635max =
3

F(11,97301)=
949024.29

corr(u_i, Xb) = 0.2166 Prob F =
0.

lnLP | Coef. Std. Err. tP|t| [95% Conf.
Interval]

lnC | .0304896 .000950932.06 0.000 .0286258
.0323533

lnL | -.9835998 .0006899 -1425.74 0.000 -.984952
-.9822476

lnE | .0652658 .000943969.14 0.000 .0634158
.0671159

lnM | .6729931 .0012158

[R] Testing for differences in amplitude and phase

2010-06-09 Thread Karin L




Dear
R-helpers,

 

I have time
series data from 16 subjects and 2 treatment groups. The seasonal variation can
be best described by two harmonics, I called the frequencies
omega and omega2.

 

I now want
to test whether (1) the seasonal pattern differs between the treatments (some
kind of overall test). If this is the case, (2) I want to conduct tests to find
out whether the amplitude of the fist harmonic, the amplitude of the second 
harmonic,
the phase of the first harmonic and the phase of the second harmonic differs
between the treatment groups.

 

My attempt
was to create a lme containing the two oscillations and the treatment group:

 

omegaDay=omega*Day # there is probably a more
elegant solution

omega2Day=omega2*Day

 

m=lme(y~(cos(omegaDay)+sin(omegaDay)+cos(omega2Day)+sin(omega2Day))*treatment,
random=~1|subject) 

summary(m)

 

 

The summary
table tells me that the respective elements are significant (except for  
cos(omegaDay):treatmentTRUE).
I know how to calculate the amplitudes and phases for both treatment groups
from the model estimates, but how do I know if they are significantly
different?

 

 

Thanks to
anyone who takes the time to help me with this problem!

Karin


 

  
_
Hotmail: Vertrauenswürdige E-Mails dank leistungsstarkem SPAM-Schutz.
https://signup.live.com/signup.aspx?id=60969
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] equivalent of stata command in R‏

plm does not have a predict function, so forget my former mail. To get
to the coefficients, you just  :
coef(mdl)

The rest of the calculations you can figure out I guess.

I'm also not sure if you're doing what you think you're doing. You
never specified the index stno in your pml call. Read the help files
again. And while you're at it, read the posting guide for the list as
well:
http://www.R-project.org/posting-guide.html

Cheers
Joris


On Wed, Jun 9, 2010 at 11:54 AM, mike mick saint-fi...@hotmail.com wrote:








 From: saint-fi...@hotmail.com
 To: saint-fi...@hotmail.com
 Subject: RE:
 Date: Wed, 9 Jun 2010 09:53:20 +









 OK! sorry thats my fault,

 here the translations of the stata commands
 1st step is to get the mean values of the variables, well that doesnt need 
 explanation i guess,

 2nd step is to estimate the model on panel data estimation method
 which is:
 mdl-plm(lnLP~lnC+lnL+lnM+lnE+Eco+Inno+Eco*Inno+Eco*lnM+Eco*lnE+year,data=newdata,model=within)
 and basically i need to get the marginal effect of variable Eco at the 
 sample mean (step 3) but i am not that good in R so any additional help is 
 wlcome!
 Thanks
 From: saint-fi...@hotmail.com
 To: r-help@r-project.org
 Subject:
 Date: Wed, 9 Jun 2010 09:45:16 +







 It helps if you translate the Stata commands. Not everybody is fluent
 in those. It would even help more if you would enlight us about the
 function you used to fit the model. Getting the marginal effects is
 not that hard at all, but how depends a bit on the function you used
 to estimate the model.

 You can try
 predict(your_model,type=terms,terms=the_term_you're_interested_in)

 For exact information, look at the respective predict function, eg if
 you use lme, do ?predict.lme
 Be aware of the fact that R normally choses the correct predict
 function without you having to specify it. predict() works for most
 model objects. Yet, depending on the model eacht predict function can
 have different options or different functionality. That information is
 in the help files of the specific function.

 Cheers
 Joris

 Dear all,

 I need to use R for one estimation, and i have readily available  stata 
 command, but i need also the R version of the same command.
 the estimation in stata is as following:
   1. Compute mean values of relevant variables



 . sum inno lnE lnM



    Variable |       Obs        Mean    Std. Dev.       Min        Max

 -+

        inno |    146574    .0880374    .2833503          0          1

         lnE |    146353    .9256239    1.732912  -4.473922   10.51298

         lnM |    146209    4.281903    1.862192  -4.847253   13.71969



        2. Estimate model



 . xi: xtreg lnLP lnC lnL lnE lnM eco inno eco_inno eco_lnE eco_lnM i.year, fe 
 i(stno)

 i.year            _Iyear_1997-1999    (naturally coded; _Iyear_1997 omitted)



 Fixed-effects (within) regression               Number of obs      =    146167

 Group variable (i): stno                        Number of groups   =     48855



 R-sq:  within  = 0.9908                         Obs per group: min =         1

       between = 0.9122                                        avg =       3.0

       overall = 0.9635                                        max =         3



                                                F(11,97301)        = 949024.29

 corr(u_i, Xb)  = 0.2166                         Prob  F           =    0.



 --

        lnLP |      Coef.   Std. Err.      t    P|t|     [95% Conf. Interval]

 -+

         lnC |   .0304896   .0009509    32.06   0.000     .0286258    .0323533

         lnL |  -.9835998   .0006899 -1425.74   0.000     -.984952   -.9822476

         lnE |   .0652658   .0009439    69.14   0.000     .0634158    .0671159

         lnM |   .6729931   .0012158   553.53   0.000       .67061    .6753761

         eco |   .0610348   .0177048     3.45   0.001     .0263336     .095736

        inno |   .0173824   .0058224     2.99   0.003     .0059706    .0287943

    eco_inno |   .0080325   .0110815     0.72   0.469    -.0136872    .0297522

     eco_lnE |   .0276226    .004059     6.81   0.000      .019667    .0355781

     eco_lnM |  -.0214237   .0039927    -5.37   0.000    -.0292494   -.0135981

  _Iyear_1998 |  -.0317684   .0013978   -22.73   0.000     -.034508   -.0290287

  _Iyear_1999 |  -.0647261   .0027674   -23.39   0.000    -.0701501   -.0593021

       _cons |   1.802112    .009304   193.69   0.000     1.783876    1.820348

 -+

     sigma_u |  .38142386

     sigma_e |   .2173114

         rho |  .75494455   (fraction of variance due to u_i)

 --

 F test that all

[R] RMySQL package on 64bit R for Windows

2010-06-09 Thread mswierniak

Hi,
I have installed 64bit R version 2.11.1 on 64bit Vista system and I have a 
problem with RMySQL package.
I have installed 64bit verion of MySQL (5.1.47) and set MYSQL_HOME. 
Installation of package looks fine :
install.packages(RMySQL_0.7-4.zip)
inferring 'repos = NULL' from the file name
package 'RMySQL' successfully unpacked and MD5 sums checked

but whet I try to load it I've got following error:
 library(RMySQL)
Error: package 'RMySQL' is not installed for 'arch=x64'
In addition: Warning message:
package 'RMySQL' was built under R version 2.12.0

I have tried also installing package from source using RTools but it's also 
failed.

If somebody succesfully install RMySQL package on 64bit R for Windows?
I'll be thankful for any help.
 

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Taylor Diagram

2010-06-09 Thread Mustafa Tufan Turp

Dear R users,

I need to learn plotting Taylor diagram. I dowloaded the plotrix that
includes taylor.diagram  but I dont know how to study with real datasets.
Anyone help me?

Tufan

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Re: [R] Help needed on switch function

R is confused about the type in the switch, reads it as numeric
instead of a character. Try :
  temp1.name - switch(temp1,
  1 = aa,
  2 = bb,
  3 = cc,
  4 = dd,
  5 = ee)
 temp1.name
[1] aa

cheers
Joris

On Wed, Jun 9, 2010 at 12:36 PM, Megh Dal megh700...@yahoo.com wrote:
 Hi all, Here I am trying to implement the switch() function to choose value 
 of a variable depending on the value of an input variable :

 temp1 - 1
   temp1.name - switch(temp1,
   1 == aa,
   2 == bb,
   3 == cc,
   4 == dd,
   5 == ee)

 Goal is if temp1 equals to 1, then value of temp1.name would be aa. 
 However I am getting following answer :

 temp1 - 1
   temp1.name - switch(temp1,
 +   1 == aa,
 +   2 == bb,
 +   3 == cc,
 +   4 == dd,
 +   5 == ee)
 temp1.name
 [1] FALSE


 Can anyone please point me where I am doing wrong?

 Thanks,



        [[alternative HTML version deleted]]


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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.





-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help needed on switch function

PS : use a single = instead of a double. It's an assignment you do,
not a comparison.
Cheers
Joris

On Wed, Jun 9, 2010 at 2:14 PM, Joris Meys jorism...@gmail.com wrote:
 R is confused about the type in the switch, reads it as numeric
 instead of a character. Try :
              temp1.name - switch(temp1,
                                  1 = aa,
                                  2 = bb,
                                  3 = cc,
                                  4 = dd,
                                  5 = ee)
 temp1.name
 [1] aa

 cheers
 Joris

 On Wed, Jun 9, 2010 at 12:36 PM, Megh Dal megh700...@yahoo.com wrote:
 Hi all, Here I am trying to implement the switch() function to choose value 
 of a variable depending on the value of an input variable :

 temp1 - 1
   temp1.name - switch(temp1,
   1 == aa,
   2 == bb,
   3 == cc,
   4 == dd,
   5 == ee)

 Goal is if temp1 equals to 1, then value of temp1.name would be aa. 
 However I am getting following answer :

 temp1 - 1
   temp1.name - switch(temp1,
 +   1 == aa,
 +   2 == bb,
 +   3 == cc,
 +   4 == dd,
 +   5 == ee)
 temp1.name
 [1] FALSE


 Can anyone please point me where I am doing wrong?

 Thanks,



        [[alternative HTML version deleted]]


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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.





 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Odp: strange issue with which on seq

2010-06-09 Thread Nikhil Kaza


which(abs(v - .1) = .Machine$double.eps)

seems to me too cumbersome to write. Any other easier way? all.equal  
does not quite work


Nikhil

On Jun 9, 2010, at 7:54 AM, Petr PIKAL wrote:


Hi

r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:


Dear R community,
I am puzzled by the following lines:


v - seq(-0.5,0.5,by=0.1)
v

 [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5

which(v == -0.4)

 [1] 2

which(v == 0)

 [1] 6

which(v == 0.1)

integer(0)

which(v == 0.2)

integer(0)

which(v == 0.3)

integer(0)

which(v == 0.4)

 [1] 10

which(v == 0.5)

 [1] 11

Why which can only match some of the values in v? Are the numbers
generated by seq not exact fractional numbers?
Please, help me to understand this.


Well FAQ 7.31 was not here some time. Computing in binary results in
finite precision of fractional numbers.

v - seq(-0.5,0.5,by=0.1)
v[7]-0.1
[1] 8.326673e-17

Regards
Petr





J

Dr James Foadi PhD
Membrane Protein Laboratory (MPL)
Diamond Light Source Ltd
Diamond House
Harewell Science and Innovation Campus
Chilton, Didcot
Oxfordshire OX11 0DE

Email:  james.fo...@diamond.ac.uk
Alt Email:  j.fo...@imperial.ac.uk

--
This e-mail and any attachments may contain confidential... 
{{dropped:8}}


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and provide commented, minimal, self-contained, reproducible code.


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[R] Odp: Help needed on switch function

Hi

if you do not insist on switch you can use factor to get desired result

set.seed(666)
x-sample(1:5,20, replace=T)
x
 [1] 4 1 5 2 2 4 5 3 1 2 4 1 1 1 2 5 1 5 3 3
factor(x, labels=letters[1:5])
 [1] d a e b b d e c a b d a a a b e a e c c
Levels: a b c d e


r-help-boun...@r-project.org napsal dne 09.06.2010 12:36:26:

 Hi all, Here I am trying to implement the switch() function to choose 
value of
 a variable depending on the value of an input variable :
  
 temp1 - 1
   temp1.name - switch(temp1,
   1 == aa,
   2 == bb,
   3 == cc,
   4 == dd,
   5 == ee)
  
 Goal is if temp1 equals to 1, then value of temp1.name would be aa. 
 However I am getting following answer :
  
  temp1 - 1
temp1.name - switch(temp1,
 +   1 == aa,
 +   2 == bb,
 +   3 == cc,
 +   4 == dd,
 +   5 == ee)
  temp1.name
 [1] FALSE
 
  
 Can anyone please point me where I am doing wrong?

Have you looked at help ?switch

Arguments
EXPR
an expression evaluating to a number or a character string.
  ^^ 



...
the list of alternatives. If it is intended that EXPR has a 
character-string value these will be named, perhaps except for one 
alternative to be used as a ‘default’ value.


So in your case

temp1 - 1
   temp1.name - switch(temp1,
   1 = aa,
   2 = bb,
   3 = cc,
   4 = dd,
   5 = ee)


Basically switch compares your temp1 with all LHS alternatives, if it 
match it returns RHS. If it does not it will return NULL and in your case 
it returned probably value 1==aa which is FALSE.

Regards
Petr

  
 Thanks,
 
 
 
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Re: [R] correcting a few data in an unreshaped data frame

Use indices.

Eg:
# a sample dataframe
lupepn1 - data.frame(
  bushno=rep(1:2,4),
  bout=rep(1:4,each=2),
  survival=rep(0,8),
  wwg=rep(1,8)
)
lupepn1
# select the wrong cases
wrong - lupepn1$bushno==1  lupepn1$bout %in% c(1,2)

# make a replacement frame
replacement - rbind(c(4,5),c(9,10))

# do the replacement
lupepn1[wrong,c(survival,wwg)] - replacement
lupepn1

Cheers
Joris

On Wed, Jun 9, 2010 at 3:36 AM, Mr. Natural drstr...@ucdavis.edu wrote:

 Thanks for the excellent help on my recent question on this topic in which
 the data frame had been reshaped by cast.
 Now, I would like to access and change erroneous data in a data frame that
 has not been reshaped.

 The file is lupepn1, with identifier variables bushno  bout and
 dependent variables survival, and wwG
 I know the bushno and bout of the erroneous dependent survival and wwG data.
 I could correct these in the csv file before read.data, but I would like to
 learn some more R

 head(lupepn1)
  bushno bout survival wwG
 2      1    2        0   0
 3      1    3        0   0
 4      1    4        0   0
 5      1    5        0   0
 6      1    6        0   2
 7      1    7        0   0

 str(lupepn1)
 'data.frame':   5023 obs. of  4 variables:
  $ bushno  : int  1 1 1 1 1 1 1 2 2 2 ...
  $ bout    : int  2 3 4 5 6 7 8 1 2 3 ...
  $ survival: int  0 0 0 0 0 0 0 1 1 1 ...
  $ wwG     : int  0 0 0 0 2 0 0 5 1 0 ...
  - attr(*, na.action)=Class 'omit'  Named int [1:81] 1 49 65 177 201 257
 337 417 449 505 ...
  .. ..- attr(*, names)= chr [1:81] 1 49 65 177 ...

 Your kind advice is very much appreciated.
 Mr. Natural.




 --
 View this message in context: 
 http://r.789695.n4.nabble.com/correcting-a-few-data-in-an-unreshaped-data-frame-tp2248219p2248219.html
 Sent from the R help mailing list archive at Nabble.com.

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-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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[R] OOP and passing by value

2010-06-09 Thread michael meyer

Greetings,

I love the R system and am sincerely grateful for the great effort the
product and contributors
are delivering.

My question is as follows:

I am trying to use S4 style classes but cannot write functions that modify
an object
because paramter passing is by value.
For example I want to do this:

setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

setClass(
  test,
  representation(f=numeric, t=numeric)
)
setMethod(setData,test,
  function(this,fcn,k){
t...@t - as.numeric(seq(-k,k))/(2*k+1)
t...@f - sapply(t,FUN=fcn)
  }
)

#---
tst - new(test)
fcn - function(u){ sin(2*pi*u) }
setData(tst,fcn,100)
t...@t   # it's still empty because of pass by value


How can this be handled?


Many thanks,

Michael

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Re: [R] Odp: strange issue with which on seq

Hi

Well, then please explain what you really want to do. I use such 
fractional sequences only for evaluation of some models and in that case 
this finite precision issue is not important.

Regards
Petr


r-help-boun...@r-project.org napsal dne 09.06.2010 14:20:16:

 which(abs(v - .1) = .Machine$double.eps)
 
 seems to me too cumbersome to write. Any other easier way? all.equal 
 does not quite work
 
 Nikhil
 
 On Jun 9, 2010, at 7:54 AM, Petr PIKAL wrote:
 
  Hi
 
  r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:
 
  Dear R community,
  I am puzzled by the following lines:
 
  v - seq(-0.5,0.5,by=0.1)
  v
   [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5
  which(v == -0.4)
   [1] 2
  which(v == 0)
   [1] 6
  which(v == 0.1)
  integer(0)
  which(v == 0.2)
  integer(0)
  which(v == 0.3)
  integer(0)
  which(v == 0.4)
   [1] 10
  which(v == 0.5)
   [1] 11
 
  Why which can only match some of the values in v? Are the numbers
  generated by seq not exact fractional numbers?
  Please, help me to understand this.
 
  Well FAQ 7.31 was not here some time. Computing in binary results in
  finite precision of fractional numbers.
 
  v - seq(-0.5,0.5,by=0.1)
  v[7]-0.1
  [1] 8.326673e-17
 
  Regards
  Petr
 
 
 
 
  J
 
  Dr James Foadi PhD
  Membrane Protein Laboratory (MPL)
  Diamond Light Source Ltd
  Diamond House
  Harewell Science and Innovation Campus
  Chilton, Didcot
  Oxfordshire OX11 0DE
 
  Email:  james.fo...@diamond.ac.uk
  Alt Email:  j.fo...@imperial.ac.uk
 
  -- 
  This e-mail and any attachments may contain confidential... 
  {{dropped:8}}
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
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Re: [R] Odp: strange issue with which on seq

use ?round is the easier way
v - round(seq(-0.5,0.5,by=0.1),1)


On Wed, Jun 9, 2010 at 2:20 PM, Nikhil Kaza nikhil.l...@gmail.com wrote:
 which(abs(v - .1) = .Machine$double.eps)

 seems to me too cumbersome to write. Any other easier way? all.equal does
 not quite work

 Nikhil

 On Jun 9, 2010, at 7:54 AM, Petr PIKAL wrote:

 Hi

 r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:

 Dear R community,
 I am puzzled by the following lines:

 v - seq(-0.5,0.5,by=0.1)
 v

  [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5

 which(v == -0.4)

  [1] 2

 which(v == 0)

  [1] 6

 which(v == 0.1)

 integer(0)

 which(v == 0.2)

 integer(0)

 which(v == 0.3)

 integer(0)

 which(v == 0.4)

  [1] 10

 which(v == 0.5)

  [1] 11

 Why which can only match some of the values in v? Are the numbers
 generated by seq not exact fractional numbers?
 Please, help me to understand this.

 Well FAQ 7.31 was not here some time. Computing in binary results in
 finite precision of fractional numbers.

 v - seq(-0.5,0.5,by=0.1)
 v[7]-0.1
 [1] 8.326673e-17

 Regards
 Petr




 J

 Dr James Foadi PhD
 Membrane Protein Laboratory (MPL)
 Diamond Light Source Ltd
 Diamond House
 Harewell Science and Innovation Campus
 Chilton, Didcot
 Oxfordshire OX11 0DE

 Email    :  james.fo...@diamond.ac.uk
 Alt Email:  j.fo...@imperial.ac.uk

 --
 This e-mail and any attachments may contain confidential...{{dropped:8}}

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide

 http://www.R-project.org/posting-guide.html

 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Odp: strange issue with which on seq

2010-06-09 Thread Henrik Bengtsson

On Wed, Jun 9, 2010 at 2:20 PM, Nikhil Kaza nikhil.l...@gmail.com wrote:
 which(abs(v - .1) = .Machine$double.eps)

 seems to me too cumbersome to write. Any other easier way? all.equal does
 not quite work

See isZero() in R.utils, e.g.

isZero(abs(v - 0.1));

You can adjust the precision with argument 'neps' (a scale factor of
'eps' specifying how close to close means) by:

isZero(abs(v - 0.1), neps=5);

The default is eps=.Machine$double.eps and neps=1.

You can also do things as:

isZero(abs(v - 0.1), eps=1e-4);

It returns a logical vector.  You have to apply which() to get the TRUE indices.

/Henrik


 Nikhil

 On Jun 9, 2010, at 7:54 AM, Petr PIKAL wrote:

 Hi

 r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:

 Dear R community,
 I am puzzled by the following lines:

 v - seq(-0.5,0.5,by=0.1)
 v

  [1] -0.5 -0.4 -0.3 -0.2 -0.1  0.0  0.1  0.2  0.3  0.4  0.5

 which(v == -0.4)

  [1] 2

 which(v == 0)

  [1] 6

 which(v == 0.1)

 integer(0)

 which(v == 0.2)

 integer(0)

 which(v == 0.3)

 integer(0)

 which(v == 0.4)

  [1] 10

 which(v == 0.5)

  [1] 11

 Why which can only match some of the values in v? Are the numbers
 generated by seq not exact fractional numbers?
 Please, help me to understand this.

 Well FAQ 7.31 was not here some time. Computing in binary results in
 finite precision of fractional numbers.

 v - seq(-0.5,0.5,by=0.1)
 v[7]-0.1
 [1] 8.326673e-17

 Regards
 Petr




 J

 Dr James Foadi PhD
 Membrane Protein Laboratory (MPL)
 Diamond Light Source Ltd
 Diamond House
 Harewell Science and Innovation Campus
 Chilton, Didcot
 Oxfordshire OX11 0DE

 Email    :  james.fo...@diamond.ac.uk
 Alt Email:  j.fo...@imperial.ac.uk

 --
 This e-mail and any attachments may contain confidential...{{dropped:8}}

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide

 http://www.R-project.org/posting-guide.html

 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
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 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] OOP and passing by value

by returning the object ?

setMethod(setData,test,
 function(this,fcn,k){
   t...@t - as.numeric(seq(-k,k))/(2*k+1)
   t...@f - sapply(t...@t,FUN=fcn)   #changed!
   return(this)  #changed!
 }
)

#---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 New - setData(tst,fcn,5)
 n...@t
[1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


On Wed, Jun 9, 2010 at 2:28 PM, michael meyer mjhme...@googlemail.com wrote:
 Greetings,

 I love the R system and am sincerely grateful for the great effort the
 product and contributors
 are delivering.

 My question is as follows:

 I am trying to use S4 style classes but cannot write functions that modify
 an object
 because paramter passing is by value.
 For example I want to do this:

 setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

 setClass(
  test,
  representation(f=numeric, t=numeric)
 )
 setMethod(setData,test,
  function(this,fcn,k){
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t,FUN=fcn)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,100)
 t...@t   # it's still empty because of pass by value


 How can this be handled?


 Many thanks,

 Michael

        [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] bootpred for multinomial




On 09.06.2010 13:26, azam jaafari wrote:

I applied bootpred for multinomial logistic reg. (with nnet package). I used 
same as theta.fit and theta.predict of R for my data. but give me error. Can I 
do this with
response vriable;7 levels
predictor variables:5 (1 classifier, 4 continuous)?

Thanks alot
Azam




[[alternative HTML version deleted]]




__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.



Yes, PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html and provide commented, 
minimal, self-contained, reproducible code.


otherwise it is hard to help without your code and without any error 
message.


Uwe Ligges

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with library(SSPA)

Same for me, but since this is a BioC package, why do you ask here? 
First you may report to the package maintainer or if that fails to the 
BioC mailing list.


Best,
Uwe Ligges


On 09.06.2010 12:01, Samuel Okoye wrote:


Hello,

I have the fellowing problem and I am thankful for any advice!

Regards,
Samuel



source(http://bioconductor.org/biocLite.R;)

BioC_mirror = http://www.bioconductor.org
Change using chooseBioCmirror().

  biocLite(SSPA)

Using R version 2.11.0, biocinstall version 2.6.7.
Installing Bioconductor version 2.6 packages:
[1] SSPA
Please wait...

trying URL 
'http://www.bioconductor.org/packages/2.6/bioc/bin/windows/contrib/2.11/SSPA_1.4.0.zip'
Content type 'application/zip' length 305310 bytes (298 Kb)
opened URL
downloaded 298 Kb

package 'SSPA' successfully unpacked and MD5 sums checked


library(SSPA)

Loading required package: qvalue
Loading
  required package: tcltk
Loading Tcl/Tk interface ... done
Error : .onAttach failed in attachNamespace() for 'SSPA', details:
   call: fun(...)
   error: could not find function addVigs2WinMenu
Error: package/namespace load failed for 'SSPA'

sessionInfo()

R version 2.11.0 (2010-04-22)
i386-pc-mingw32

locale:
[1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United 
Kingdom.1252LC_MONETARY=English_United Kingdom.1252
[4] LC_NUMERIC=CLC_TIME=English_United Kingdom.1252

attached base packages:
[1] tcltk stats graphics  grDevices utils datasets  methods   base

other attached
  packages:
[1] qvalue_1.22.0

loaded via a namespace (and not attached):
[1] limma_3.4.3  SSPA_1.4.0   tools_2.11.0

##





[[alternative HTML version deleted]]




__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] OOP and passing by value

In case you want the function setData to change the object itself
(which is often a dangerous idea!), you can use instead :

setMethod(setData,test,
 function(this,fcn,k){
Name - deparse(substitute(this))
t...@t - as.numeric(seq(-k,k))/(2*k+1)
t...@f - sapply(t...@t,FUN=fcn)
assign(Name,this,.GlobalEnv)
 }
)

#---
tst - new(test)
fcn - function(u){ sin(2*pi*u) }
setData(tst,fcn,5)
t...@t
[1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


On Wed, Jun 9, 2010 at 2:36 PM, Joris Meys jorism...@gmail.com wrote:
 by returning the object ?

 setMethod(setData,test,
  function(this,fcn,k){
   t...@t - as.numeric(seq(-k,k))/(2*k+1)
   t...@f - sapply(t...@t,FUN=fcn)   #changed!
   return(this)                                          #changed!
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 New - setData(tst,fcn,5)
 n...@t
 [1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
 0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


 On Wed, Jun 9, 2010 at 2:28 PM, michael meyer mjhme...@googlemail.com wrote:
 Greetings,

 I love the R system and am sincerely grateful for the great effort the
 product and contributors
 are delivering.

 My question is as follows:

 I am trying to use S4 style classes but cannot write functions that modify
 an object
 because paramter passing is by value.
 For example I want to do this:

 setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

 setClass(
  test,
  representation(f=numeric, t=numeric)
 )
 setMethod(setData,test,
  function(this,fcn,k){
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t,FUN=fcn)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,100)
 t...@t   # it's still empty because of pass by value


 How can this be handled?


 Many thanks,

 Michael

        [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] OOP and passing by value

2010-06-09 Thread Henrik Bengtsson

Pass an object of, or containing an, environment.  Then whenever you
modify any object inside the environment, the changes will remain
also when exiting from the function(s).  This has been used by many
for quite some time and is the standard way to do it, if you need this
feature.  See packages such as R.oo and proto for complete solutions.

Using substitute(this) is not safe when doing nested calling, and
assigning to a global environment is not something you really want to
do.

/Henrik

On Wed, Jun 9, 2010 at 3:05 PM, Joris Meys jorism...@gmail.com wrote:
 In case you want the function setData to change the object itself
 (which is often a dangerous idea!), you can use instead :

 setMethod(setData,test,
  function(this,fcn,k){
    Name - deparse(substitute(this))
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t...@t,FUN=fcn)
    assign(Name,this,.GlobalEnv)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,5)
 t...@t
 [1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
 0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


 On Wed, Jun 9, 2010 at 2:36 PM, Joris Meys jorism...@gmail.com wrote:
 by returning the object ?

 setMethod(setData,test,
  function(this,fcn,k){
   t...@t - as.numeric(seq(-k,k))/(2*k+1)
   t...@f - sapply(t...@t,FUN=fcn)   #changed!
   return(this)                                          #changed!
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 New - setData(tst,fcn,5)
 n...@t
 [1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
 0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


 On Wed, Jun 9, 2010 at 2:28 PM, michael meyer mjhme...@googlemail.com 
 wrote:
 Greetings,

 I love the R system and am sincerely grateful for the great effort the
 product and contributors
 are delivering.

 My question is as follows:

 I am trying to use S4 style classes but cannot write functions that modify
 an object
 because paramter passing is by value.
 For example I want to do this:

 setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

 setClass(
  test,
  representation(f=numeric, t=numeric)
 )
 setMethod(setData,test,
  function(this,fcn,k){
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t,FUN=fcn)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,100)
 t...@t   # it's still empty because of pass by value


 How can this be handled?


 Many thanks,

 Michael

        [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php




 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] bootpred for multinomial

2010-06-09 Thread azam jaafari

Thank you very much
 
I used bootpred for multinomial logistic reg. similar below:
theta.fit - function(x,y){lsfit(x,y)}
   theta.predict - function(fit,x){
   cbind(1,x)%*%fit$coef 
   }    
   sq.err - function(y,yhat) { (y-yhat)^2}   
   results - bootpred(x,y,20,theta.fit,theta.predict,
 err.meas=sq.err)  

x=data
mlr=multinom(y~x1+x2+x3+x4+x5, data)
 
after run results:
 results - bootpred(data, mlr,50,theta.fit,theta.predict, 
 sq.err)   
Error in as.vector(data) : 
  no method for coercing this S4 class to a vector
 
Is it incorrect?
 
Azam

 
 
 
 


--- On Wed, 6/9/10, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:


From: Uwe Ligges lig...@statistik.tu-dortmund.de
Subject: Re: [R] bootpred for multinomial
To: azam jaafari azamjaaf...@yahoo.com
Cc: R-help r-help@r-project.org
Date: Wednesday, June 9, 2010, 5:46 AM




On 09.06.2010 13:26, azam jaafari wrote:
 I applied bootpred for multinomial logistic reg. (with nnet package). I used 
 same as theta.fit and theta.predict of R for my data. but give me error. Can 
 I do this with
 response vriable;7 levels
 predictor variables:5 (1 classifier, 4 continuous)?

 Thanks alot
 Azam




     [[alternative HTML version deleted]]




 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.


Yes, PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html and provide commented, 
minimal, self-contained, reproducible code.

otherwise it is hard to help without your code and without any error 
message.

Uwe Ligges




  
[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] OOP and passing by value

2010-06-09 Thread Philippe Grosjean

I think you should be interested by section 5.4 (non-local assignments; 
closures) in Chamber's Software for Data Analysis book published in 
2008 by Springer 
(http://www.springer.com/statistics/computanional+statistics/book/978-0-387-75935-7).


Besides the solution proposed in the book, which uses assignment outside 
functions, there is another possibility: environment objects. They are 
passed by reference and there are tricks to use them in this situation, 
but you have to be extremely careful because you break the rules for 
functional programming.


Don't forget generics like Data- to assign in R. See for instance 
help(names-). This is the most conventional and easier way to do this.


Best,

Philippe Grosjean

On 09/06/10 14:28, michael meyer wrote:

Greetings,

I love the R system and am sincerely grateful for the great effort the
product and contributors
are delivering.

My question is as follows:

I am trying to use S4 style classes but cannot write functions that modify
an object
because paramter passing is by value.
For example I want to do this:

setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

setClass(
   test,
   representation(f=numeric, t=numeric)
)
setMethod(setData,test,
   function(this,fcn,k){
 t...@t- as.numeric(seq(-k,k))/(2*k+1)
 t...@f- sapply(t,FUN=fcn)
   }
)

#---
tst- new(test)
fcn- function(u){ sin(2*pi*u) }
setData(tst,fcn,100)
t...@t   # it's still empty because of pass by value


How can this be handled?


Many thanks,

Michael

[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.




__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with library(SSPA)

2010-06-09 Thread Samuel Okoye

I can't find his email and I have asked the same question to 

bioconduc...@stat.math.ethz.ch

Regards,
Samuel

--- On Wed, 9/6/10, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:

From: Uwe Ligges lig...@statistik.tu-dortmund.de
Subject: Re: [R] Problem with library(SSPA)
To: Samuel Okoye samu...@yahoo.com
Cc: r-help@r-project.org
Date: Wednesday, 9 June, 2010, 5:43

Same for me, but since this is a BioC package, why do you ask here? 
First you may report to the package maintainer or if that fails to the 
BioC mailing list.

Best,
Uwe Ligges


On 09.06.2010 12:01, Samuel Okoye wrote:

 Hello,

 I have the fellowing problem and I am thankful for any advice!

 Regards,
 Samuel

 
     source(http://bioconductor.org/biocLite.R;)
 BioC_mirror = http://www.bioconductor.org
 Change using chooseBioCmirror().
       biocLite(SSPA)
 Using R version 2.11.0, biocinstall version 2.6.7.
 Installing Bioconductor version 2.6 packages:
 [1] SSPA
 Please wait...

 trying URL 
 'http://www.bioconductor.org/packages/2.6/bioc/bin/windows/contrib/2.11/SSPA_1.4.0.zip'
 Content type 'application/zip' length 305310 bytes (298 Kb)
 opened URL
 downloaded 298 Kb

 package 'SSPA' successfully unpacked and MD5 sums checked

 library(SSPA)
 Loading required package: qvalue
 Loading
   required package: tcltk
 Loading Tcl/Tk interface ... done
 Error : .onAttach failed in attachNamespace() for 'SSPA', details:
    call: fun(...)
    error: could not find function addVigs2WinMenu
 Error: package/namespace load failed for 'SSPA'
 sessionInfo()
 R version 2.11.0 (2010-04-22)
 i386-pc-mingw32

 locale:
 [1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United 
 Kingdom.1252    LC_MONETARY=English_United Kingdom.1252
 [4] LC_NUMERIC=C                            LC_TIME=English_United 
 Kingdom.1252

 attached base packages:
 [1] tcltk     stats     graphics  grDevices utils     datasets  methods   base

 other attached
   packages:
 [1] qvalue_1.22.0

 loaded via a namespace (and not attached):
 [1] limma_3.4.3  SSPA_1.4.0   tools_2.11.0

 ##





     [[alternative HTML version deleted]]




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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.



  
[[alternative HTML version deleted]]

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with library(SSPA)




On 09.06.2010 15:19, Samuel Okoye wrote:

I can't find his email


 packageDescription(SSPA)




and I have asked the same question to

bioconduc...@stat.math.ethz.ch

Regards,
Samuel

--- On Wed, 9/6/10, Uwe Liggeslig...@statistik.tu-dortmund.de  wrote:

From: Uwe Liggeslig...@statistik.tu-dortmund.de
Subject: Re: [R] Problem with library(SSPA)
To: Samuel Okoyesamu...@yahoo.com
Cc: r-help@r-project.org
Date: Wednesday, 9 June, 2010, 5:43

Same for me, but since this is a BioC package, why do you ask here?
First you may report to the package maintainer or if that fails to the
BioC mailing list.

Best,
Uwe Ligges


On 09.06.2010 12:01, Samuel Okoye wrote:


Hello,

I have the fellowing problem and I am thankful for any advice!

Regards,
Samuel



  source(http://bioconductor.org/biocLite.R;)

BioC_mirror = http://www.bioconductor.org
Change using chooseBioCmirror().

biocLite(SSPA)

Using R version 2.11.0, biocinstall version 2.6.7.
Installing Bioconductor version 2.6 packages:
[1] SSPA
Please wait...

trying URL 
'http://www.bioconductor.org/packages/2.6/bioc/bin/windows/contrib/2.11/SSPA_1.4.0.zip'
Content type 'application/zip' length 305310 bytes (298 Kb)
opened URL
downloaded 298 Kb

package 'SSPA' successfully unpacked and MD5 sums checked


library(SSPA)

Loading required package: qvalue
Loading
required package: tcltk
Loading Tcl/Tk interface ... done
Error : .onAttach failed in attachNamespace() for 'SSPA', details:
 call: fun(...)
 error: could not find function addVigs2WinMenu
Error: package/namespace load failed for 'SSPA'

sessionInfo()

R version 2.11.0 (2010-04-22)
i386-pc-mingw32

locale:
[1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United 
Kingdom.1252LC_MONETARY=English_United Kingdom.1252
[4] LC_NUMERIC=CLC_TIME=English_United Kingdom.1252

attached base packages:
[1] tcltk stats graphics  grDevices utils datasets  methods   base

other attached
packages:
[1] qvalue_1.22.0

loaded via a namespace (and not attached):
[1] limma_3.4.3  SSPA_1.4.0   tools_2.11.0

##





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[R] RGoogleDocs not working for me with wise service

2010-06-09 Thread Harlan Harris

Hello,

I'm trying to figure out how to use the RGoogleDocs package from OmegaHat,
and am having a bit of trouble. I emailed Duncan Temple Lang directly, but
didn't receive a response, so I thought I'd try here to see if anyone else
can help.

I'm using 32-bit R 2.10.1 (Mac OS X), I built RGoogleDocs 0.4-0 from source,
and I'm using XML 3.1-0 and RCurl 1.2-0.

If I use service=wise in getGoogleDocsConnection, I then get this error:

 getDocs(con)
Error in getDocs(con) : problems connecting to get the list of documents

tracing through I get:

Browse[1] status

 WWW-Authenticate
GoogleLogin 
realm=\http://www.google.com/accounts/ClientLogin\http://www.google.com/accounts/ClientLogin%5C,
service=\writely\

Content-Type
 text/html;
charset=UTF-8

Date
Fri, 04 Jun 2010
16:22:19 GMT

Expires
Fri, 04 Jun 2010
16:22:19 GMT

Cache-Control
   private,
max-age=0

X-Content-Type-Options

nosniff

X-Frame-Options

SAMEORIGIN

X-XSS-Protection
1;
mode=block

Server

GSE

Transfer-Encoding

chunked

status

401

statusMessage

Token invalid

If I don't use service=wise, getDocs(con) properly returns a
GoogleDocList. (But, of course, I can't manipulate spreadsheet content...)

Any suggestions? Is it a package version issue, perhaps? Has Google updated
their API in a way that hasn't been reflected in updates to this package?

Thanks a lot!

 -Harlan

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[R] changing the number of elements in a list

2010-06-09 Thread Luis Borda de Agua

I would like to have a list where each element is a matrix, for example:

my.list - list(matrix(0, ncol=3, nrow=3),
   matrix(0, ncol=3, nrow=3),
   matrix(0, ncol=3, nrow=3))

The problem is, I would like to be able to change automatically the number of 
elements in the list (not only three as in the above example).

That is, the instruction creating the list would be a part of a function that 
has an inputing parameter stating how many elements (matrices) the list has. 

I have tried several things, but none worked. 
However, this seems to be a rather simple problem.
All help is welcome!

LBA

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with library(SSPA)

2010-06-09 Thread Samuel Okoye

Dear Maarten,

I have problem to use your package and I would be very thankful if you could 
help me to solve this.

Regards,
Samuel

--- On Wed, 9/6/10, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:

From: Uwe Ligges lig...@statistik.tu-dortmund.de
Subject: Re: [R] Problem with library(SSPA)
To: Samuel Okoye samu...@yahoo.com
Cc: r-help@r-project.org
Date: Wednesday, 9 June, 2010, 6:43



On 09.06.2010 15:19, Samuel Okoye wrote:
 I can't find his email

  packageDescription(SSPA)



and I have asked the same question to

 bioconduc...@stat.math.ethz.ch

 Regards,
 Samuel

 --- On Wed, 9/6/10, Uwe Liggeslig...@statistik.tu-dortmund.de  wrote:

 From: Uwe Liggeslig...@statistik.tu-dortmund.de
 Subject: Re: [R] Problem with library(SSPA)
 To: Samuel Okoyesamu...@yahoo.com
 Cc: r-help@r-project.org
 Date: Wednesday, 9 June, 2010, 5:43

 Same for me, but since this is a BioC package, why do you ask here?
 First you may report to the package maintainer or if that fails to the
 BioC mailing list.

 Best,
 Uwe Ligges


 On 09.06.2010 12:01, Samuel Okoye wrote:

 Hello,

 I have the fellowing problem and I am thankful for any advice!

 Regards,
 Samuel

 
       source(http://bioconductor.org/biocLite.R;)
 BioC_mirror = http://www.bioconductor.org
 Change using chooseBioCmirror().
         biocLite(SSPA)
 Using R version 2.11.0, biocinstall version 2.6.7.
 Installing Bioconductor version 2.6 packages:
 [1] SSPA
 Please wait...

 trying URL 
 'http://www.bioconductor.org/packages/2.6/bioc/bin/windows/contrib/2.11/SSPA_1.4.0.zip'
 Content type 'application/zip' length 305310 bytes (298 Kb)
 opened URL
 downloaded 298 Kb

 package 'SSPA' successfully unpacked and MD5 sums checked

 library(SSPA)
 Loading required package: qvalue
 Loading
     required package: tcltk
 Loading Tcl/Tk interface ... done
 Error : .onAttach failed in attachNamespace() for 'SSPA', details:
      call: fun(...)
      error: could not find function addVigs2WinMenu
 Error: package/namespace load failed for 'SSPA'
 sessionInfo()
 R version 2.11.0 (2010-04-22)
 i386-pc-mingw32

 locale:
 [1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United 
 Kingdom.1252    LC_MONETARY=English_United Kingdom.1252
 [4] LC_NUMERIC=C                            LC_TIME=English_United 
 Kingdom.1252

 attached base packages:
 [1] tcltk     stats     graphics  grDevices utils     datasets  
 methods   base

 other attached
     packages:
 [1] qvalue_1.22.0

 loaded via a namespace (and not attached):
 [1] limma_3.4.3  SSPA_1.4.0   tools_2.11.0

 ##





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Re: [R] changing the number of elements in a list

2010-06-09 Thread Ivan Calandra


Hi,

What about:
my.list - vector(mode=list, length=number_of_matrices)
?

HTH,
Ivan

Le 6/9/2010 15:46, Luis Borda de Agua a écrit :

I would like to have a list where each element is a matrix, for example:

my.list- list(matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3))

The problem is, I would like to be able to change automatically the number of 
elements in the list (not only three as in the above example).

That is, the instruction creating the list would be a part of a function that 
has an inputing parameter stating how many elements (matrices) the list has.

I have tried several things, but none worked.
However, this seems to be a rather simple problem.
All help is welcome!

LBA

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

   


--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] changing the number of elements in a list

2010-06-09 Thread 09wkj

I might give this a try as well:

my.list - replicate(3, matrix(0, ncol=3, nrow=3), simplify=FALSE)


args(replicate)
function (n, expr, simplify = TRUE) 

 replicate(3, matrix(0, ncol=3, nrow=3), simplify=FALSE)
[[1]]
 [,1] [,2] [,3]
[1,]000
[2,]000
[3,]000

[[2]]
 [,1] [,2] [,3]
[1,]000
[2,]000
[3,]000

[[3]]
 [,1] [,2] [,3]
[1,]000
[2,]000
[3,]000

 is.list(replicate(3, matrix(0, ncol=3, nrow=3), simplify=FALSE))
[1] TRUE


Bill

On Jun 9, 2010, at 9:54 AM, Ivan Calandra wrote:

 Hi,
 
 What about:
 my.list - vector(mode=list, length=number_of_matrices)
 ?
 
 HTH,
 Ivan
 
 Le 6/9/2010 15:46, Luis Borda de Agua a écrit :
 I would like to have a list where each element is a matrix, for example:
 
 my.list- list(matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3))
 
 The problem is, I would like to be able to change automatically the number 
 of elements in the list (not only three as in the above example).
 
 That is, the instruction creating the list would be a part of a function 
 that has an inputing parameter stating how many elements (matrices) the list 
 has.
 
 I have tried several things, but none worked.
 However, this seems to be a rather simple problem.
 All help is welcome!
 
 LBA
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
   
 
 -- 
 Ivan CALANDRA
 PhD Student
 University of Hamburg
 Biozentrum Grindel und Zoologisches Museum
 Abt. Säugetiere
 Martin-Luther-King-Platz 3
 D-20146 Hamburg, GERMANY
 +49(0)40 42838 6231
 ivan.calan...@uni-hamburg.de
 
 **
 http://www.for771.uni-bonn.de
 http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] OOP and passing by value

Thanks for the correction. Personally, I never really use that kind of
assignments, and the deparse(substitute()) construct I only use in
self-defined plotting functions in simple scripts to be run on series
of equivalent datasets by me, myself and I.

When writing a package with nested functions and the likes, there are
indeed far better solutions than this one. I am -obviously- not an
expert on those. And honestly, I personally don't like the idea of a
function changing my object. So in conclusion, I shouldn't even have
sent my last mail I guess...

Cheers
Joris


On Wed, Jun 9, 2010 at 3:12 PM, Henrik Bengtsson h...@stat.berkeley.edu wrote:
 Pass an object of, or containing an, environment.  Then whenever you
 modify any object inside the environment, the changes will remain
 also when exiting from the function(s).  This has been used by many
 for quite some time and is the standard way to do it, if you need this
 feature.  See packages such as R.oo and proto for complete solutions.

 Using substitute(this) is not safe when doing nested calling, and
 assigning to a global environment is not something you really want to
 do.

 /Henrik

 On Wed, Jun 9, 2010 at 3:05 PM, Joris Meys jorism...@gmail.com wrote:
 In case you want the function setData to change the object itself
 (which is often a dangerous idea!), you can use instead :

 setMethod(setData,test,
  function(this,fcn,k){
    Name - deparse(substitute(this))
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t...@t,FUN=fcn)
    assign(Name,this,.GlobalEnv)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,5)
 t...@t
 [1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
 0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


 On Wed, Jun 9, 2010 at 2:36 PM, Joris Meys jorism...@gmail.com wrote:
 by returning the object ?

 setMethod(setData,test,
  function(this,fcn,k){
   t...@t - as.numeric(seq(-k,k))/(2*k+1)
   t...@f - sapply(t...@t,FUN=fcn)   #changed!
   return(this)                                          #changed!
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 New - setData(tst,fcn,5)
 n...@t
 [1] -0.4545455 -0.3636364 -0.2727273 -0.1818182 -0.0909091  0.000
 0.0909091  0.1818182  0.2727273  0.3636364  0.4545455


 On Wed, Jun 9, 2010 at 2:28 PM, michael meyer mjhme...@googlemail.com 
 wrote:
 Greetings,

 I love the R system and am sincerely grateful for the great effort the
 product and contributors
 are delivering.

 My question is as follows:

 I am trying to use S4 style classes but cannot write functions that modify
 an object
 because paramter passing is by value.
 For example I want to do this:

 setGeneric(setData, function(this,fcn,k){ standardGeneric(setData) })

 setClass(
  test,
  representation(f=numeric, t=numeric)
 )
 setMethod(setData,test,
  function(this,fcn,k){
    t...@t - as.numeric(seq(-k,k))/(2*k+1)
    t...@f - sapply(t,FUN=fcn)
  }
 )

 #---
 tst - new(test)
 fcn - function(u){ sin(2*pi*u) }
 setData(tst,fcn,100)
 t...@t   # it's still empty because of pass by value


 How can this be handled?


 Many thanks,

 Michael

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 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php




 --
 Joris Meys
 Statistical consultant

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

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.





-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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[R] Odp: changing the number of elements in a list

Hi

one option is to create an empty list with

my.list - vector(list, n)

where n is number of elements in list. Then you can populate each element 
by

my.list[[x]] - something

where x can be number or

my.list[x] -something

where x is numeric vector

Regards
Petr


r-help-boun...@r-project.org napsal dne 09.06.2010 15:46:09:

 I would like to have a list where each element is a matrix, for example:
 
 my.list - list(matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3),
matrix(0, ncol=3, nrow=3))
 
 The problem is, I would like to be able to change automatically the 
number of 
 elements in the list (not only three as in the above example).
 
 That is, the instruction creating the list would be a part of a function 
that 
 has an inputing parameter stating how many elements (matrices) the list 
has. 
 
 I have tried several things, but none worked. 
 However, this seems to be a rather simple problem.
 All help is welcome!
 
 LBA
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
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 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] bootpred for multinomial




On 09.06.2010 15:15, azam jaafari wrote:

Thank you very much

I used bootpred for multinomial logistic reg. similar below:
theta.fit- function(x,y){lsfit(x,y)}
theta.predict- function(fit,x){
cbind(1,x)%*%fit$coef
}
sq.err- function(y,yhat) { (y-yhat)^2}
results- bootpred(x,y,20,theta.fit,theta.predict,
  err.meas=sq.err)

x=data
mlr=multinom(y~x1+x2+x3+x4+x5, data)

after run results:

results- bootpred(data, mlr,50,theta.fit,theta.predict, sq.err)

Error in as.vector(data) :
   no method for coercing this S4 class to a vector

Is it incorrect?



First guess: bootpred() is from the bootstrap packages (you just told 
about nnet).



From that help page (?bootpred):
x 	a matrix containing the predictor (regressor) values. Each row 
corresponds to an observation.


And your data is some S4 class object, obviously, but bootpred expects a 
matrix according to the docs 


Uwe Ligges



Azam







--- On Wed, 6/9/10, Uwe Liggeslig...@statistik.tu-dortmund.de  wrote:


From: Uwe Liggeslig...@statistik.tu-dortmund.de
Subject: Re: [R] bootpred for multinomial
To: azam jaafariazamjaaf...@yahoo.com
Cc: R-helpr-help@r-project.org
Date: Wednesday, June 9, 2010, 5:46 AM




On 09.06.2010 13:26, azam jaafari wrote:

I applied bootpred for multinomial logistic reg. (with nnet package). I used 
same as theta.fit and theta.predict of R for my data. but give me error. Can I 
do this with
response vriable;7 levels
predictor variables:5 (1 classifier, 4 continuous)?

Thanks alot
Azam




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and provide commented, minimal, self-contained, reproducible code.



Yes, PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.

otherwise it is hard to help without your code and without any error
message.

Uwe Ligges





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and provide commented, minimal, self-contained, reproducible code.

[R] How to add a new plot in the same graph using add=T at the command plot?

Hi, there!

I'm having kind this same problem

https://stat.ethz.ch/pipermail/r-help/2008-October/178221.html

but I want to display another plot of my data, which is a point with two
arrows indicating confidence interval, in the same graph that I've just
plotted another, but the add=T is not functioning, I'm getting the same
error

Warning messages:
1: In plot.window(...) : add não é um parâmetro gráfico
2: In plot.xy(xy, type, ...) : add não é um parâmetro gráfico
3: In axis(side = side, at = at, labels = labels, ...) :
  add não é um parâmetro gráfico
4: In axis(side = side, at = at, labels = labels, ...) :
  add não é um parâmetro gráfico
5: In box(...) : add não é um parâmetro gráfico
6: In title(...) : add não é um parâmetro gráfico

In all tutorials I search, I see that the form of doing that is using the
add=T... Is there another way?

Thanks in advance!

-- 
Larissa

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Re: [R] How to add a new plot in the same graph using add=T at the command plot?

I forgot to show my code... it is like this

 jpeg()
 plot(x,y,main=str, xlab=str,ylab=str, axes=FALSE)
 axis(1,at=seq(0,4.5,by=0.5),pos=0)
 axis(2,at=seq(0,100,by=1.0),pos=0)
 arrows(x, y, x, ic_esq, length=0.1, angle=90, code=3)
 arrows(x, y, x, ic_dir, length=0.1, angle=90, code=3)
 plot(x1,y1, add=T)
Warning messages:
1: In plot.window(...) : add não é um parâmetro gráfico
2: In plot.xy(xy, type, ...) : add não é um parâmetro gráfico
3: In axis(side = side, at = at, labels = labels, ...) :
  add não é um parâmetro gráfico
4: In axis(side = side, at = at, labels = labels, ...) :
  add não é um parâmetro gráfico
5: In box(...) : add não é um parâmetro gráfico
6: In title(...) : add não é um parâmetro gráfico

Thanks


On Wed, Jun 9, 2010 at 11:29, Larissa Lucena larissaluc...@gmail.comwrote:

 Hi, there!

 I'm having kind this same problem

 https://stat.ethz.ch/pipermail/r-help/2008-October/178221.html

 but I want to display another plot of my data, which is a point with two
 arrows indicating confidence interval, in the same graph that I've just
 plotted another, but the add=T is not functioning, I'm getting the same
 error

 Warning messages:
 1: In plot.window(...) : add não é um parâmetro gráfico
 2: In plot.xy(xy, type, ...) : add não é um parâmetro gráfico
 3: In axis(side = side, at = at, labels = labels, ...) :
   add não é um parâmetro gráfico
 4: In axis(side = side, at = at, labels = labels, ...) :
   add não é um parâmetro gráfico
 5: In box(...) : add não é um parâmetro gráfico
 6: In title(...) : add não é um parâmetro gráfico

 In all tutorials I search, I see that the form of doing that is using the
 add=T... Is there another way?

 Thanks in advance!

 --
 Larissa




-- 
Larissa

[[alternative HTML version deleted]]

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Re: [R] How to add a new plot in the same graph using add=T at the command plot?

Hi

where did you find parameter add=T.

You can use

par(new=T)
before using new plot command

or use

points, lines 

Regards
Petr

r-help-boun...@r-project.org napsal dne 09.06.2010 16:42:25:

 I forgot to show my code... it is like this
 
  jpeg()
  plot(x,y,main=str, xlab=str,ylab=str, axes=FALSE)
  axis(1,at=seq(0,4.5,by=0.5),pos=0)
  axis(2,at=seq(0,100,by=1.0),pos=0)
  arrows(x, y, x, ic_esq, length=0.1, angle=90, code=3)
  arrows(x, y, x, ic_dir, length=0.1, angle=90, code=3)
  plot(x1,y1, add=T)
 Warning messages:
 1: In plot.window(...) : add năo é um parâmetro gráfico
 2: In plot.xy(xy, type, ...) : add năo é um parâmetro gráfico
 3: In axis(side = side, at = at, labels = labels, ...) :
   add năo é um parâmetro gráfico
 4: In axis(side = side, at = at, labels = labels, ...) :
   add năo é um parâmetro gráfico
 5: In box(...) : add năo é um parâmetro gráfico
 6: In title(...) : add năo é um parâmetro gráfico
 
 Thanks
 
 
 On Wed, Jun 9, 2010 at 11:29, Larissa Lucena 
larissaluc...@gmail.comwrote:
 
  Hi, there!
 
  I'm having kind this same problem
 
  https://stat.ethz.ch/pipermail/r-help/2008-October/178221.html
 
  but I want to display another plot of my data, which is a point with 
two
  arrows indicating confidence interval, in the same graph that I've 
just
  plotted another, but the add=T is not functioning, I'm getting the 
same
  error
 
  Warning messages:
  1: In plot.window(...) : add năo é um parâmetro gráfico
  2: In plot.xy(xy, type, ...) : add năo é um parâmetro gráfico
  3: In axis(side = side, at = at, labels = labels, ...) :
add năo é um parâmetro gráfico
  4: In axis(side = side, at = at, labels = labels, ...) :
add năo é um parâmetro gráfico
  5: In box(...) : add năo é um parâmetro gráfico
  6: In title(...) : add năo é um parâmetro gráfico
 
  In all tutorials I search, I see that the form of doing that is using 
the
  add=T... Is there another way?
 
  Thanks in advance!
 
  --
  Larissa
 
 
 
 
 -- 
 Larissa
 
[[alternative HTML version deleted]]
 
 __
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Re: [R] How to add a new plot in the same graph using add=T at the command plot?

Thanks so much!!! I'm using R for the first time, and so, I have many stupid
doubts! Sorry and thanks again!

 Regards!

2010/6/9 Petr PIKAL petr.pi...@precheza.cz

 Hi

 where did you find parameter add=T.

 You can use

 par(new=T)
 before using new plot command

 or use

 points, lines

 Regards
 Petr

 r-help-boun...@r-project.org napsal dne 09.06.2010 16:42:25:

  I forgot to show my code... it is like this
 
   jpeg()
   plot(x,y,main=str, xlab=str,ylab=str, axes=FALSE)
   axis(1,at=seq(0,4.5,by=0.5),pos=0)
   axis(2,at=seq(0,100,by=1.0),pos=0)
   arrows(x, y, x, ic_esq, length=0.1, angle=90, code=3)
   arrows(x, y, x, ic_dir, length=0.1, angle=90, code=3)
   plot(x1,y1, add=T)
  Warning messages:
  1: In plot.window(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
  2: In plot.xy(xy, type, ...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
  3: In axis(side = side, at = at, labels = labels, ...) :
add nÄo Ã© um parÃ¢metro grÃ¡fico
  4: In axis(side = side, at = at, labels = labels, ...) :
add nÄo Ã© um parÃ¢metro grÃ¡fico
  5: In box(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
  6: In title(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
 
  Thanks
 
 
  On Wed, Jun 9, 2010 at 11:29, Larissa Lucena
 larissaluc...@gmail.comwrote:
 
   Hi, there!
  
   I'm having kind this same problem
  
   https://stat.ethz.ch/pipermail/r-help/2008-October/178221.html
  
   but I want to display another plot of my data, which is a point with
 two
   arrows indicating confidence interval, in the same graph that I've
 just
   plotted another, but the add=T is not functioning, I'm getting the
 same
   error
  
   Warning messages:
   1: In plot.window(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
   2: In plot.xy(xy, type, ...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
   3: In axis(side = side, at = at, labels = labels, ...) :
 add nÄo Ã© um parÃ¢metro grÃ¡fico
   4: In axis(side = side, at = at, labels = labels, ...) :
 add nÄo Ã© um parÃ¢metro grÃ¡fico
   5: In box(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
   6: In title(...) : add nÄo Ã© um parÃ¢metro grÃ¡fico
  
   In all tutorials I search, I see that the form of doing that is using
 the
   add=T... Is there another way?
  
   Thanks in advance!
  
   --
   Larissa
  
 
 
 
  --
  Larissa
 
 [[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.




-- 
Larissa

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[R] comparing two regression models with different dependent variable

2010-06-09 Thread Or Duek

Hi,
I would like to compare to regression models - each model has a different
dependent variable.
The first model uses a number that represents the learning curve for reward.
The second model uses a number that represents the learning curve from
punishment stimuli.
The first model is significant and the second isn't.
I want to compare those two models and show that they are significantly
different.
How can I do that?
Thank you.

[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] equivalent of stata command in R‏‏

If you don't know how to proceed, you should look for a good book on
statistics. I checked the .nlcom, and what it does is give estimates
and tests on a (nonlinear) combination of estimated parameters. That's
doable, but a bit tedious to program. It is basically using the rules
on the adding and multiplication of random variables and standard
errors. but the interpretation is also not that straightforward.

In the generalized mixed model world, the appropriate way of
estimating the marginal effects you're looking for (I guess...) is
centering your data around the mean before fitting the model. That
way, the parameter of the main effect represent exactly the marginal
effect at the sample mean for the other variables, simply because the
mean is 0 for all of them and the equation you use simplifies to
_b[eco]. Yet, the test statistic only gives you an idea about whether
or not this coefficient differs from zero, assuming it is normally
distributed with se as calculated. That is not the same as testing
whether there is a significant marginal effect.

Marginal effects are in my opinion better tested using likelihood
ratio methods. These are not provided in plm, as that one is based on
generalized least squares and hence does not return a likelihood
value. To use LR tests, you'll have to go to nlme or lme4.

Following is an obligatory read if you're going to use plm methods :
http://cran.r-project.org/web/packages/plm/vignettes/plm.pdf

Maybe you better contact the maintainer of the package
yves.croiss...@let.ish-lyon.cnrs.fr directly to ask for the correct
testing procedure for your hypothesis, because I'm still not sure that
you're doing the fitting correctly in R. Just like you specify fe
i(stno) in Stata, you should specify index=stno in the R command.

Cheers
Joris

On Wed, Jun 9, 2010 at 1:27 PM, mike mick saint-fi...@hotmail.com wrote:


 Thanx for your response,
 yeah, i know i didnst specified the indexes
 when i wrote the 2nd mail, in fact in the 1st mail i wrote already that
 i dont have problem with the estimation of the model... thats the
 reason why i didnt write in fact since the issue is not to estimate the
 model but to get the marginal effect,
 anyway, i figured out that predict(), doesnt work for panel data...
 and
 well, my problem is that contrary to your guess, i couldnt figure out
 the rest of the calculations... since i am not that experienced in R.
 one last help of yours would be quite helpful to get rid of this silly 
 problem!
 Thanx again...




-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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and provide commented, minimal, self-contained, reproducible code.

[R] Plotting Question

2010-06-09 Thread Steve_Friedman


Hello,

I would like to produce a series of graphs comparing the probability
distributions for 8 factors against a continuous metric.

The kind of graph I'm hoping to produce would look like the density
comparison graphs (library sm) using the function sm.density.compare.
However, instead of calculating the density distributions for comparisons,
I'd like this comparison to be based on probabilities.

An example of the graph I'm seeking can be seen here:
http://www.statmethods.net/graphs/density.html   (scroll to the bottom and
examine the Comparing Groups via Kernel Density).

A graph (attached) produced using my data:
 attach(SoilVegHydro)

 vegtype.f - factor(Physiogomy)

  sm.density.compare(meanAnnualDepthAve, vegtype.f, xlab=Mean Annual
Depth)

  # create color fill patterns
  colfill - c(2:(2+length(levels(vegtype.f
   legend(locator(1), levels(vegtype.f), fill=colfill)

(See attached file: MeanAnnualDepth_Density.png)


Thanks
Steve



Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147attachment: MeanAnnualDepth_Density.png__
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Re: [R] comparing two regression models with different dependentvariable

2010-06-09 Thread Bert Gunter

1. This is not an R question.

2. What you have requested is nonsense.

You need to consult your local statistician. 


Bert Gunter
Genentech Nonclinical Statistics

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Or Duek
Sent: Wednesday, June 09, 2010 8:20 AM
To: R-help@r-project.org
Subject: [R] comparing two regression models with different
dependentvariable

Hi,
I would like to compare to regression models - each model has a different
dependent variable.
The first model uses a number that represents the learning curve for reward.
The second model uses a number that represents the learning curve from
punishment stimuli.
The first model is significant and the second isn't.
I want to compare those two models and show that they are significantly
different.
How can I do that?
Thank you.

[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

[R] generate list of variable names

2010-06-09 Thread Jon Erik Ween

Hi!

Would anyone know how to generate a list of variable names from a data frame by 
the class of the variable? 

I have large tables with different numbers of columns and am trying to script 
some rote analyses. There are several categorizing variables (factors) and many 
response variables (integers and numeric). I want to extract a list of 
classifier column names in one list and response variable names in another 
list, then run for-loops to calculate various statistics on the response 
variables in terms of the classifier variables. I thought something like this 
might work (but didn't):


tmp-dataframe
for (i in 1:length(tmp)){
if(class(tmp[[i]]==factor) else tmp[[i]]-NULL
}

factorlist-names(tmp)

...

One problem is that the scope of the counter changes as I drop columns, but 
there seems to be other problems as well

Any suggestions would be welcome. Thanks!

Jon

Soli Deo Gloria

Jon Erik Ween, MD, MS
Scientist, Kunin-Lunenfeld Applied Research Unit 
Director, Stroke Clinic, Brain Health Clinic, Baycrest Centre
Assistant Professor, Dept. of Medicine, Div. of Neurology
University of Toronto Faculty of Medicine

Kimel Family Building, 6th Floor, Room 644 
Baycrest Centre
3560 Bathurst Street 
Toronto, Ontario M6A 2E1
Canada 

Phone: 416-785-2500 x3648
Fax: 416-785-2484
Email: jw...@klaru-baycrest.on.ca


Confidential: This communication and any attachment(s) may contain confidential 
or privileged information and is intended solely for the address(es) or the 
entity representing the recipient(s). If you have received this information in 
error, you are hereby advised to destroy the document and any attachment(s), 
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unauthorized use or disclosure of this information is strictly prohibited.




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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Draw text with a box surround in plot.

2010-06-09 Thread guox

Rectangle R centered at (x,y) with width 2w and height 2h is given by

x1=x-w
y1=y-h
x2=x+w
y2=y-h
x3=x+w
y3=y+h
x4=x-w
y4=y+h
polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4))

Rotating a point (u,v) at (0,0) by theta degree is given by matrix
[cos(theta),-sin(theta)
 sin(theta),cos(theta)]
so we have a new point (u*cos(theta)-v*sin(theta),u*sin(theta)+v*cos(theta)).

Hence rotated R by theta at (x,y) is given by

x.rotated = c(x + (x1-x)*cos(theta)-(y1-y)*sin(theta),
  x + (x2-x)*cos(theta)-(y2-y)*sin(theta),
  x + (x3-x)*cos(theta)-(y3-y)*sin(theta),
  x + (x4-x)*cos(theta)-(y4-y)*sin(theta))
y.rotated = c(y + (x1-x)*sin(theta)+(y1-y)*cos(theta),
  y + (x2-x)*sin(theta)+(y2-y)*cos(theta),
  y + (x3-x)*sin(theta)+(y3-y)*cos(theta),
  y + (x4-x)*sin(theta)+(y4-y)*cos(theta))

polygon(x.rotated,y.rotated)

But it turns out to be a parallelogram with angles not equal to 90,
not a rectangle. See R code below.

Any way to improve this so that the rotated rectangle looks like a
rectangle? Thanks,

-james


plot(1:10,1:10,xlim=c(1,20),ylim=c(1,40),type=n, main = Rotated
rectangle looks like a   )
## a rect at (10,20) with w = 3 and h = 2
x = 10
y = 20
w = 3
h = 2
x1=x-w
y1=y-h
x2=x+w
y2=y-h
x3=x+w
y3=y+h
x4=x-w
y4=y+h
polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4),border=blue)

##Rotate it at (10,10) by 45 degree
theta = 45/180*pi
x.rotated = c(10 + (x1-10)*cos(theta)-(y1-20)*sin(theta),
  10 + (x2-10)*cos(theta)-(y2-20)*sin(theta),
  10 + (x3-10)*cos(theta)-(y3-20)*sin(theta),
  10 + (x4-10)*cos(theta)-(y4-20)*sin(theta))
y.rotated = c(20 + (x1-10)*sin(theta)+(y1-20)*cos(theta),
  20 + (x2-10)*sin(theta)+(y2-20)*cos(theta),
  20 + (x3-10)*sin(theta)+(y3-20)*cos(theta),
  20 + (x4-10)*sin(theta)+(y4-20)*cos(theta))

polygon(x.rotated,y.rotated,border=red)


 On 06/04/2010 01:21 AM, g...@ucalgary.ca wrote:
 boxed.labels draw text with box well.
 But, the box cannot be shadowed and srt = 45 seems not to work:
 text is rotated but the box does not.
 polygon.shadow can rotate and shadow but have to calculate its
 dimensions,
 based on the text length and size.
 Do you have any other way to draw text with rotated and shadowed box?

 The srt argument was intended to allow the user to rotate the text in 90
 degree increments, and the box just changes shape to fit whatever is in
 it. The underlying function that draws the box (rect) doesn't have a
 rotation argument. It would be possible to write a special function
 using polygon, just do the calculations for box size and then rotate the
 text with srt= and the polygon by transforming the coordinates of the
 vertices, as long as the default justification (center) is used. I can't
 do this right at the moment, but if you are really stuck I might be able
 to do it in the near future.

 Jim




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[R] help for generating data from ar1 like model

2010-06-09 Thread li li

Hi all,
   I wrote the following function to generate data following a mixture-ar1
model.
   The model is described as below:

  theta is a m-vector with each entries identically and
independent Bernoulli trials with
success probability pi1.

   x is a m-vector with entries follow a ar1 model x_{i+1}=rho * x_{i}
+ e_{i} + mu1 * theta[i]''.
In other words, the dependence is like ar1 model, however, the mean is mu1
or 0 according to
whether theta is 1 or 0.


I am not too familar with the r functions generating time series data. Can
some
one have a look at the function I wrote? Thank you very much!



rdata.mix.ar1-function(pi1,m, rho, mu1)
{
## generate data from the following mixture-ar1 model:
## {theta[i]} is m-vector of iid bernoulli trials with
## success prob pi1.
## generate observations from a ar1 model:
## x_{i}=rho * x_{i-1}+epslon_{i}+mu1*theta[i]
## in other word; if theta_{i} is 1, mean is mu1;
## otherwise mean is 0; but the
## dependence structure is like a ar1 model.

##pi1 - 0.5
##m - 10
##rho - 0.6
##mu1 - 1
set.seed(66)
## generate states
theta -rbinom(m, 1, pi1)
## generate data
x - arima.sim(list(order=c(1,0,0), ar=rho), n=m)+mu1*theta
data-list(s=theta, o=x, p=p)
return(data)
}

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Re: [R] generate list of variable names

2010-06-09 Thread Erik Iverson




Jon Erik Ween wrote:

Hi!

Would anyone know how to generate a list of variable names from a
data frame by the class of the variable?


a start...

df - data.frame(f1 = factor(1:10),
 f2 = factor(1:10),
 n1 = 1:10,
 n2 = 1:10)


sapply(df, class)



I have large tables with different numbers of columns and am trying
to script some rote analyses. There are several categorizing
variables (factors) and many response variables (integers and
numeric). I want to extract a list of classifier column names in one
list and response variable names in another list, then run for-loops
to calculate various statistics on the response variables in terms of
the classifier variables. I thought something like this might work
(but didn't):


Reproducible example needed.  All this can surely be done more elegantly 
with lapply/mapply instead of for-loops.


__
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Re: [R] comparing two regression models with different dependent variable

On Wed, Jun 9, 2010 at 5:19 PM, Or Duek ord...@gmail.com wrote:
 Hi,
 I would like to compare to regression models - each model has a different
 dependent variable.
 The first model uses a number that represents the learning curve for reward.
 The second model uses a number that represents the learning curve from
 punishment stimuli.
 The first model is significant and the second isn't.
 I want to compare those two models and show that they are significantly
 different.

They are different, forget about significance.

 How can I do that?

You don't. Your problem translates to whether reward or punishment
gives a better learning curve, so you build a model using a factor
that represents whether the learning curve comes from reward or
punishment, and test the significance of the difference using the
correct approach in the framework of your choice. And since curves are
functions, you actually need a method for functional data analysis.

Either you start eg here : http://www.psych.mcgill.ca/misc/fda/ or you
take the advice of Bert. If I was you, I'd go for the latter.

 Thank you.

You're welcome.
Cheers
Joris

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Joris Meys
Statistical consultant

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

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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[R] Calibration and validation for svycoxph

2010-06-09 Thread R user


Hello,

This post is for Dr. Thomas Lumley or anybody familiar with the survey
package. I am estimating a proportional hazards model with weights using
svycoxph. Are there functions already built in the survey package that allow
me to do validation and calibration of the model?

Thanks
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Calibration-and-validation-for-svycoxph-tp2249004p2249004.html
Sent from the R help mailing list archive at Nabble.com.

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[R] marginal structural models

2010-06-09 Thread moleps

Dear listers,

Does anyone have any experience running marginal structural models in r or can 
point me in the direction of any good tutorials on this?

Regards,

//M

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[R] for loop incremented by 0.01

2010-06-09 Thread suman dhara

Sir,
I want to use a for loop which will be incremented by 0.01 in 0 to 1.
Please help.

Regards,
Suman Dhara

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[R] passing local parameters to nls?

2010-06-09 Thread ivo welch

dear R wizards:  environment-related programming question:

myformula = y ~ a*x^2+b*x+c
d= data.frame( x=rnorm(20), y=rnorm(20) )

cat(\nunconstrained works: \n);
nls( myformula, data=d, start=list(a=1, b=1, c=1), trace= TRUE)

cat(\nconstrained works: \n);
b=1; nls( myformula, data=d, start=list(a=1, c=1), trace= TRUE); rm(b)

cat(\nbut how do I pass a local parameter into my formula? \n);
f = function(c) nls( myformula, data=d, start=list(a=1, b=1), trace= TRUE)

f(c=1);

advice, as always, appreciated.  /iaw


Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)

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[R] creating a new variable, conditional on the value of an existing variable, selected conditionally

2010-06-09 Thread Malcolm Fairbrother

Dear all,

I have a data frame f, with four variables:

f - data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
f
  A B C D
1 0 0 1 3
2 0 1 1 1
3 1 0 0 2
4 1 1 1 3

I want to create a new variable (f$E), such that each of its elements is drawn 
from either f$A, f$B, or f$C, according to the value (for each row) of f$D 
(values of which range from 1 to 3).

In the first row, D is 3, so I want the value from the third variable (C), 
which for the first row is 1. In the second row, D is 1, so I want the value 
from the first variable (A), which for the second row is 0. And so forth, such 
that in the end my new data frame looks like:

  A B C D E
1 0 0 1 3 1
2 0 1 1 1 0
3 1 0 0 2 0
4 1 1 1 3 1

My question is: How do I do this for a much larger dataset, where my index 
variable (f$D in this example) actually indexes a much larger number of 
variables (not just three)?

I know that in principle I could do this with a long series of nested ifelse 
statements (as below), but I assume there is some less cumbersome option, and 
I'd like to know what it is. Any help would be much appreciated. Apologies if 
I'm missing something obvious.

f$E - ifelse(f$D==3, f$C, ifelse(f$D==2, f$B, f$A))

Thanks,
Malcolm

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Re: [R] Issue with assigning text to matrix

2010-06-09 Thread Jessica Queree

Thanks for all your help. I found adding the 'stringsAsFactors' condition
solved the problem.

On 1 June 2010 17:09, Joris Meys jorism...@gmail.com wrote:

 Hi Jessica,

 this tells me that your text is saved as a factor.
 Try :
 names - read.csv(file=Names.csv,stringsAsFactors=F)


 Cheers
 Joris

   On Tue, Jun 1, 2010 at 11:04 AM, Jessica Queree 
 j.j.que...@googlemail.com wrote:

  My issue relates to adding text to a matrix and finding that the text is
 converted to a number.



 This is the section of code I'm having trouble with:



 # First, I load in a list of names from a .csv file to 'names'

 names - read.csv(file(Names.csv))



 # Then I define a matrix which will be populated with various test
 statistics, with several rows for each entry in names



 testOutput -matrix(nrow = 200, ncol = 5)

 for (i in 1:nrow(names)){



testOutput[i,1] - names[i,1]

testOutput[i,2] - names[i,2]



# test statistics code here



 }





 If I look at names[,1], I get the following:



 names[,1]

  [1] EQ_Level_UK   EQ_Level_EUR  EQ_Level_US   EQ_Level_Far
 East

  [5] IR_PC 1_UKIR_PC 2_UKIR_PC 3_UKSwap_PC 1_UK

  [9] Swap_PC 2_UK  Swap_PC 3_UK  FX_Level_EUR  FX_Level_US

 [13] FX_Level_Far East Infl_PC 1_UK  Infl_PC 2_UK  Infl_PC 3_UK

 [17] Prop_Level_UK CreditAAA_PC 1_UK CreditAAA_PC 2_UK CreditAAA_PC
 3_UK

 [21] CreditAA_PC 1_UK  CreditAA_PC 2_UK  CreditAA_PC 3_UK  CreditA_PC 1_UK

 [25] CreditA_PC 2_UK   CreditA_PC 3_UK   CreditBBB_PC 1_UK CreditBBB_PC
 2_UK

 [29] CreditBBB_PC 3_UK

 29 Levels: CreditA_PC 1_UK CreditA_PC 2_UK CreditA_PC 3_UK ... Swap_PC
 3_UK



 But if I look at testOutput[,1], I get:



 testOutput[,1]

  [1] 15 13 16 14 23 24 25 27 28 29 17 19 18 20
 21

  [16] 22 26 7  8  9  4  5  6  1  2  3  10 11
 12
 17

  [31] NA   NA   19 18 NA   NA   NA   20 NA   NA   21 NA   NA
 22
 NA

  [46] NA   26 NA   NA   7  NA   NA   8  NA   NA   9  NA   NA   4
 NA

  [61] NA   5  NA   NA   6  NA   NA   1  NA   NA   2  NA   NA   3
 NA

  [76] NA   10 NA   NA   11 NA   NA   12 NA   NA   NA   NA   NA   NA
 NA

  [91] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [106] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [121] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [136] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [151] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [166] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [181] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
 NA

 [196] NA   NA   NA   NA   NA



 That is, the names are now converted to numbers. I think this might have
 something to do with the way I've defined the testOutput matrix, but
 haven't
 been able to find any information about how to fix it. Can anyone help?



 Many thanks.

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 Joris Meys
 Statistical Consultant

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

 Coupure Links 653
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 tel : +32 9 264 59 87
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[R] Dr. Hadley Wickham - Data Visualisation in R: Harnessing the power of ggplot2. London - November 2010

2010-06-09 Thread Sarah Lewis

Dr. Hadley Wickham - Data Visualisation in R: Harnessing the power of ggplot2 
to produce elegant data graphics
London: 1st - 2nd November 2010

Mango Solutions is delighted to offer a one-off 2 day training course with Dr. 
Hadley Wickham, R Project Data Visualisation Guru and creator of ggplot 2. The 
course is a must for any R user looking to improve their
data visualisation skills. Suitable for any level of R experience, it will 
especially benefit those who know how to get data in and out of R and who can 
do some basic modelling.

Hadley Wickham is the Dobelman Family Junior Chair in Statistics at Rice 
University. He is an active member of the R community, has written and 
contributed to over 20 R packages and won the John Chambers Award for 
Statistical Computing for his work developing tools for data reshaping and 
visualisation. His research focuses on how to make data analysis better, faster 
and easier with a particular emphasis on the use of visualisation to better 
understand data and models.

Venue to be confirmed but will be a central London location.

Places will be limited on what will be a very popular course.

For more information please contact: train...@mango-solutions.com or visit 
www.mango-solutions.com 

Sarah Lewis

T: +44 (0)1249 767700 Ext: 200
F: +44 (0)1249 767707
M: +44 (0)7746 224226
www.mango-solutions.com
Unit 2 Greenways Business Park 
Bellinger Close
Chippenham
Wilts
SN15 1BN
UK 


LEGAL NOTICE
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[R] calibration and validation for svycoxph

2010-06-09 Thread R user


Hello,

This post is for Dr. Thomas Lumley or anybody familiar with the survey
package. I am estimating a proportional hazards model with weights using
svycoxph. Are there functions already built in the survey package that allow
me to do validation and calibration of the model?

Thanks 
-- 
View this message in context: 
http://r.789695.n4.nabble.com/calibration-and-validation-for-svycoxph-tp2249118p2249118.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] for loop incremented by 0.01

2010-06-09 Thread Erik Iverson


not tested


for(i in seq(0, 1, by = 0.01)) {
  print(i)
}

Read FAQ 7.31 though.

suman dhara wrote:

Sir,
I want to use a for loop which will be incremented by 0.01 in 0 to 1.
Please help.

Regards,
Suman Dhara

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Re: [R] creating a new variable, conditional on the value of an existing variable, selected conditionally

2010-06-09 Thread Erik Iverson


Can your data.frame be properly coerced to a matrix like your example?

If so,

apply(f, 1, function(x) x[eval(x)[D]])

Malcolm Fairbrother wrote:

Dear all,

I have a data frame f, with four variables:

f - data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
f
  A B C D
1 0 0 1 3
2 0 1 1 1
3 1 0 0 2
4 1 1 1 3

I want to create a new variable (f$E), such that each of its elements is drawn 
from either f$A, f$B, or f$C, according to the value (for each row) of f$D 
(values of which range from 1 to 3).

In the first row, D is 3, so I want the value from the third variable (C), 
which for the first row is 1. In the second row, D is 1, so I want the value 
from the first variable (A), which for the second row is 0. And so forth, such 
that in the end my new data frame looks like:

  A B C D E
1 0 0 1 3 1
2 0 1 1 1 0
3 1 0 0 2 0
4 1 1 1 3 1

My question is: How do I do this for a much larger dataset, where my index 
variable (f$D in this example) actually indexes a much larger number of variables 
(not just three)?

I know that in principle I could do this with a long series of nested ifelse 
statements (as below), but I assume there is some less cumbersome option, and 
I'd like to know what it is. Any help would be much appreciated. Apologies if 
I'm missing something obvious.

f$E - ifelse(f$D==3, f$C, ifelse(f$D==2, f$B, f$A))

Thanks,
Malcolm

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Re: [R] OOP and passing by value

2010-06-09 Thread michael meyer

Thanks to all replies.
I was able to get it running with the R.oo package.

I hope this reply makes it to the proper thread.

Michael

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Re: [R] generate list of variable names

2010-06-09 Thread Jon Erik Ween

Thanks Erik

I can't figure out how to use the various x_apply functions in this setting, 
nor post datasets to reproduce. But anyhow: the table structure is something 
like this:

id (integer), handedness(R,L,A), gender(M,F), cat1(patient, control). 
cat2(stroke, MS, dement, control), accuracy(integer), reaction time(numeric)

so, I want to extract the factor levels from cat1, cat2, etc and run, say, 
ANOVAs or ROCs on each of the response variables (accuracy, reaction_time, etc) 
extracting F-values, AUCs, etc, sticking the results in a table of results. 
Here is an example script I wrote for ROCR:


###
library(ROCR) # Load stats package to use if not standard
varslist-scan(/Users/jween/Desktop/INCAS/INCASvars.txt,list) # Read 
variable list
results-as.data.frame(array(,c(3,length(varslist # Initialize results 
array, one type of stat at a time for now

for (i in 1:length(varslist)){ # Loop through the variables you want to 
process. Determined by varslist
j-noquote(varslist[i])
vars-c(varslist[i],Issue_class) # Variables to be analyzed
temp-na.omit(MSsmv[vars]) # Have to subset to get rid of NA values 
causing ROCR to choke
n-nrow(temp) # Record how many cases the analysis ios based on. Need 
to figure out how to calc cases/controls
#.table-table(temp$SubjClass)  # Maybe for later figure out 
cases/controls
results[1,i]-j # Name particular results column
results[2,i]-n # Number of subjects in analysis
test-try(aucval(i,j),silent=TRUE) # Error handling in case procedure 
craps oust so loop can continue. Supress annoying error messages
if(class(test)==try-error) next else # Run procedure only if OK, 
otherwise skip
pred-prediction(MSsmv[[j]], MSsmv$Issue_cat); # Procedure
perf-performance(pred,auc);
results[3,i]-as.numeric(p...@y.values) # Enter result into appropriate 
row

}
write.table(results,/Users/jween/Desktop/IncasRres_MSsmv.csv,sep=,,col.names=FALSE,row.names=FALSE)
 # Write results to table
rm(aucval,i,n,temp,vars,results,pred,perf,j,varslist) # Clean up test,

aucval-function(i,j){ # Function to trap errors. Should be the same as real 
procedure above
pred-prediction(MSsmv[[j]], MSsmv$Issue_cat); # Don't put any real 
results here, they don't seem to be passed back
perf-performance(pred,auc);
}
###

Cheers


Jon

Soli Deo Gloria

Jon Erik Ween, MD, MS
Scientist, Kunin-Lunenfeld Applied Research Unit 
Director, Stroke Clinic, Brain Health Clinic, Baycrest Centre
Assistant Professor, Dept. of Medicine, Div. of Neurology
University of Toronto Faculty of Medicine

Kimel Family Building, 6th Floor, Room 644 
Baycrest Centre
3560 Bathurst Street 
Toronto, Ontario M6A 2E1
Canada 

Phone: 416-785-2500 x3648
Fax: 416-785-2484
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unauthorized use or disclosure of this information is strictly prohibited.



On 2010-06-09, at 12:20 PM, Erik Iverson wrote:

 
 
 Jon Erik Ween wrote:
 Hi!
 Would anyone know how to generate a list of variable names from a
 data frame by the class of the variable?
 
 a start...
 
 df - data.frame(f1 = factor(1:10),
 f2 = factor(1:10),
 n1 = 1:10,
 n2 = 1:10)
 
 
 sapply(df, class)
 
 I have large tables with different numbers of columns and am trying
 to script some rote analyses. There are several categorizing
 variables (factors) and many response variables (integers and
 numeric). I want to extract a list of classifier column names in one
 list and response variable names in another list, then run for-loops
 to calculate various statistics on the response variables in terms of
 the classifier variables. I thought something like this might work
 (but didn't):
 
 Reproducible example needed.  All this can surely be done more elegantly with 
 lapply/mapply instead of for-loops.

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Re: [R] How to add a new plot in the same graph using add=T at the command plot?

2010-06-09 Thread Peter Ehlers


Soapbox:
Well, if you're just starting out with R it would be
a VERY good idea to learn right away that T is not TRUE
and F is not FALSE, at least not always. Sooner or
later you WILL have problems. So do yourself a favour
and get into the habit of using TRUE/FALSE instead of T/F.

(I know that Petr knows better.)

 -Peter Ehlers

On 2010-06-09 9:08, Larissa Lucena wrote:

Thanks so much!!! I'm using R for the first time, and so, I have many stupid
doubts! Sorry and thanks again!

  Regards!

2010/6/9 Petr PIKALpetr.pi...@precheza.cz


Hi

where did you find parameter add=T.

You can use

par(new=T)
before using new plot command

or use

points, lines

Regards
Petr



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Re: [R] dealing with heteroscedasticity in lmer: problem with the method weights

2010-06-09 Thread Ben Bolker

doris gomez dodogomez at yahoo.fr writes:

 
 Dear lmer users,

  You should probably redirect this sort of question to
r-sig-mixed-mod...@r-project.org , rather than the generic
R help list.

 
 The experiment includes 15 groups of (3 males and 1 female). 
 The female is characterized by its quality Q1
 and Q2. Each male of a group is characterized by the number of 
 MatingAttempts (with Poisson
 distribution). I want to examine if male mating attempts 
 depend on female quality.  I can see from graphic
 exploration  that the within-group heterogeneity of 
 male attempts increases with female quality Q1.

  Is this true above and beyond the expected increase in
variance due to the Poisson distribution of the data?
 
 When including the method weights in the function lmer, 
  I get the message that variables' length varies and
 the model does not run.
 lmer(MatingAttempts~Q1+Q2+(1|Group),data=file,
 family=poisson,na.action=na.omit,
 REML=FALSE, weights=varExp(form=~Q1))

  The REML and weights arguments do not make sense in the context
of a GLMM fitted with [g]lmer.

 If I run the same model (fixed effects and random effects) with lme, it 
 works properly, which shows that there is no problem with data structure.
 lme(MatingAttempts~Q1+Q2,random=~1|Group,
 data=file,na.action=na.omit, method=ML, weights=varExp(form=~Q1))

  As you presumably know, this model does not incorporate the Poisson
distribution of the data.

 
 I saw on the forum that lmer had problems in taking into account 
 variance heterogeneity. Yet, the messages
 were old and there are maybe new solutions.
 How can I correct the analyses for this problem of heteroscedasticity? 
 Should I normalise the within group variance before implementing the model? 
 And deal with the variance (as
 a new variable to explain) in another model?
 Is there another way to solve this problem?

  (1) lmer still does not allow for models of heteroscedasticity
as implemented in the weights() argument of lme, and may not for some
time ...
  (2) You _could_ use glmmPQL() from the MASS package, along the
lines of 

MASS::glmmPQL(MatingAttempts~Q1+Q2,random=~1|Group,
  family=poisson,
  data=file,na.action=na.omit, weights=varExp(form=~Q1))

 **HOWEVER**: it's not entirely clear what statistical model
this represents, or whether it makes sense.
 (3) I would recommend fitting the GLMM, with lmer, without the
heteroscedasticity, and then see whether the heteroscedasticity
remains in the Pearson residuals or not ... (don't forget to
check for overdispersion too).

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Re: [R] how to draw the probability ellipse circle figure?

2010-06-09 Thread Ben Bolker

Jie TANG totangjie at gmail.com writes:

 
 hi ,R user folks .
 
 Nowadays I read a paper which draw a probability ellipse circle  figure
 shown in the appendix.
  I wonder how to draw this figure by R ?
 the x-axis and y-axis both express the error but in different direction .

  Your question is quite vague, but perhaps you might start with the
ellipse package.  Also try

RSiteSearch(ellipse)

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[R] cut edge from surface using persp()

2010-06-09 Thread marlene marchena

Hi R users!

Is it possible to cut the edges from a surface graph without to cut the axes
using persp function?

I am using the following code:

 op - par(bg = white)

persp(phi1, phi2, z,main=Bullwhip generated with AR(2) demand when L=1,
  xlab =phi1 , ylab =phi2, zlab =Bullwhip, theta = 30, phi = 20,
  r=50,d=0.1,expand=0.5,ltheta=90,border = NA, lphi=180, shade=0.75,
ticktype=detailed,nticks=5)

when I put box=FALSE the axes just disappear.

Any help will be appreciated.

Marlene.

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Re: [R] passing local parameters to nls?

2010-06-09 Thread Gabor Grothendieck

It looks in the environment of your formula (which in this case is the
global environment) and then in data.  Thus here are two solutions:

f2 - function(c) {
environment(fo) - environment()
nls(fo, data=d, start=list(a=1, b=1))
}

f3 - function(c) nls(fo, data=c(d, c = c), start=list(a=1, b=1))

fo - y ~ a + x^2 + b * x + c
f2(1)
f3(1)

On Wed, Jun 9, 2010 at 9:39 AM, ivo welch ivo...@gmail.com wrote:
 dear R wizards:  environment-related programming question:

 myformula = y ~ a*x^2+b*x+c
 d= data.frame( x=rnorm(20), y=rnorm(20) )

 cat(\nunconstrained works: \n);
 nls( myformula, data=d, start=list(a=1, b=1, c=1), trace= TRUE)

 cat(\nconstrained works: \n);
 b=1; nls( myformula, data=d, start=list(a=1, c=1), trace= TRUE); rm(b)

 cat(\nbut how do I pass a local parameter into my formula? \n);
 f = function(c) nls( myformula, data=d, start=list(a=1, b=1), trace= TRUE)

 f(c=1);

 advice, as always, appreciated.  /iaw

 
 Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)

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Re: [R] combining expressions in mathplot

2010-06-09 Thread RICHARD M. HEIBERGER

Uwe,

Very nice, thank you.
Rich

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Re: [R] Rglpk

2010-06-09 Thread Peter Ehlers


On 2010-06-09 5:11, Kaveh Vakili wrote:


Hi list,


in the Rglpk_solve_LP function (::Rglpk),
on line 26, the function calls a function
as.glp_bounds() that i cannot access.

i'm trying to alter the Rglpk_solve_LP function
to add a line to retrieve column/row dual values.
everytime i change the slightest line of code inside
Rglpk_solve_LP() [to even add a print]
i get a
': could not find function as.glp_bounds'

What's the catch here ?



No catch: as.glp_bounds() is buried in namespace:Rglpk.
Your modified function is presumably in the global
environment. This should work:

Let's say your modified function is myfun().
After you source myfun, set its environment with

 environment(myfun) - environment(Rglpk_solve_LP)

-Peter Ehlers

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Re: [R] for loop incremented by 0.01

2010-06-09 Thread John Kane

Why?

See  ?seq

seq(0,1,.01)


--- On Wed, 6/9/10, suman dhara suman.dhar...@gmail.com wrote:

 From: suman dhara suman.dhar...@gmail.com
 Subject: [R] for loop incremented by 0.01
 To: r-h...@stat.math.ethz.ch
 Received: Wednesday, June 9, 2010, 12:30 PM
 Sir,
 I want to use a for loop which will be incremented by 0.01
 in 0 to 1.
 Please help.
 
 Regards,
 Suman Dhara
 
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Re: [R] How to add a new plot in the same graph using add=T at the command plot?