gmane.comp.lang.r.general

Trouble with SASxport in R.3.0. under 32-bit and 64-bit Windows (both 7 and Vista)

Santosh — 2013-06-20T00:16:27

Dear Rxperts,..

I am using R 3.0 in both 32- and 64-bit Windows 7 environment and have
trouble using SASxport. They seem to work fine under R.2.15.3.

 ....The error/warning messages are provided below...

Following are the messages obtained when trying to read s SAS transport
file (.xpt)

library(SASxport)
Warning message:
package SASxport was built under R version 3.0.1

read.xport(fname)
Error in `label<-.default`(`*tmp*`, value = c("", "", "", "", "", "",  :
  value must be character vector of length 1

Any suggestions/assistance is highly welcome!

Thanks so much..
Santosh

[[alternative HTML version deleted]]

Returning name of dataframe?

Thomas Parr — 2013-06-20T00:16:02

I am writing a function and I would like to return the name of a data frame
in a paste call, but I can't figure out how to just get the name. The names
of the data frames used, won't be the same each time.  I have to be
overlooking the obvious.

#For example:

a<-replicate(5, rnorm(20))
b<-replicate(5, rnorm(20))

xyz<-function(x,y){
  z<-x+y
  print(paste("the sum of", x, "and", y,"is", sep=" "))
  z 
}

xyz(a,b)

#That function should return:


Not surprisingly, it returns all the data in the data frame.

Thanks,
Thomas

Re: how to get growth rate of a (time series) data?

Rolf Turner — 2013-06-19T22:16:24

     <SNIP>

No, "clearly" ( :-) ) it doesn't.  It gives

(Y[2] - Y[1])/Y[2], (Y[3]-Y[2])/Y[3], .... when what is wanted is

(Y[2] - Y[1])/Y[1], (Y[3]-Y[2])/Y[2], ....

The OP needs to do:

     with(test,diff(Y)/Y[-length(Y)])

This is yet another illustration of the rule that if there is a 50-50 
chance of
getting things the wrong way around, then there is actually a probability
of one of getting things the wrong way around.

     cheers,

         Rolf

Re: GLMM predictor variables

Bert Gunter — 2013-06-19T21:49:10

Wrong list -- Your question is not about R.

Post on a statistics list like stats.stackexchange.com

Re: knitr: side-by-side figures in R markdown

Yihui Xie — 2013-06-19T21:31:41

You need to remove out.extra='style="display:block; margin: auto"'. In
CSS, display:block; means this element stands in its own line, and no
other elements can sit by its side. This is applied to individual
images, so the two images will not be arranged side by side.

But you will lose the center alignment without display:block. The
difficulty comes from the fact that Markdown images are not put into
div containers as in HTML.

Regards,
Yihui
--
Yihui Xie <xieyihui< at >gmail.com>
Phone: 206-667-4385 Web: http://yihui.name
Fred Hutchinson Cancer Research Center, Seattle


On Wed, Jun 19, 2013 at 6:23 AM, Michael Friendly <friendly< at >yorku.ca> wrote:

Re: help with text patterns in strings

arun — 2013-06-19T20:46:27

HI Burnette,
As this is continuation of the earlier thread, you could post it on the same thread by cc: to rhelp.


Try this:
res1<-sapply(vec3,function(x) length(vec2New[grep(x,vec2New)]) )
dat1<-data.frame(res1,Name=names(vec3))

 dat1$Name<-factor(dat1$Name,levels=c("early","mid","late","wknd"))
 with(dat1,tapply(res1,list(Name),FUN=sum))
#early   mid  late  wknd 
 #   0     1     4     6 

#or
 sapply(split(res1,names(vec3)),sum)
#early  late   mid  wknd 
 #   0     4     1     6 
A.K.

----- Original Message -----
From: "Crombie, Burnette N" <bcrombie< at >utk.edu>
To: arun <smartpink111< at >yahoo.com>
Cc: 
Sent: Wednesday, June 19, 2013 3:55 PM
Subject: RE: [R] help with text patterns in strings

Arun, let me know if I should post this email separately, but it involves the script from our previous conversation.  I've been messing around as I think of potential scenarios with my data and am unclear how I can recount vec3 after assigning range names to the different days of

Re: alternative to ifelse

arun — 2013-06-19T19:14:57

Hi,
May bet this also helps:
set.seed(12345)
x <- c(1:30)
x1 <- sample(x, 15, replace = TRUE)
x2 <- sample(x, 15, replace = TRUE)
x3 <- sample(x, 15, replace = TRUE)
x4 <- sample(x, 15, replace = TRUE)

indx<-1+2*(x1==1)+4*(x2==1)+8*(x3==1)+16*(x4==1)
 as.numeric(indx!=1)
#[1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


#Speed comparisons 
set.seed(1235)
x <- c(1:1e7)
x1 <- sample(x, 1e7, replace = TRUE)
x2 <- sample(x, 1e7, replace = TRUE)
x3 <- sample(x, 1e7, replace = TRUE)
x4 <- sample(x, 1e7, replace = TRUE)

system.time(res1 <-ifelse(x1 == 1 | x2 == 1 | x3 == 1 | x4 == 1, 1, 0))
 # user  system elapsed 
 # 3.740   0.256   4.003 

system.time(res2<-as.numeric(x1 == 1 | x2 == 1 | x3 == 1 | x4 == 1))
# user  system elapsed 
#  0.996   0.076   1.077 

system.time({indx<-1+2*(x1==1)+4*(x2==1)+8*(x3==1)+16*(x4==1)
 res3<-as.numeric(indx!=1)})
# user  system elapsed 
#  0.596   0.240   0.840 

system.time(res4<-rowSums(cbind(x1, x2, x3, x4) == 1))
# user  system elapsed 
#  0.732   0.192   0.92

Re: alternative to ifelse

Marc Schwartz — 2013-06-19T18:47:38

On Jun 19, 2013, at 1:23 PM, Marc Schwartz <marc_schwartz< at >me.com> wrote:



Actually, correction. That works in the case where only one of the values in the row is a 1, which happened to fit the example data. If there is a chance that more than one value in a row may be a 1, it returns a value equal to the number of 1's found.

Thus, a more generic approach, I believe, would be:

[1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


Regards,

Marc

Re: alternative to ifelse

Marc Schwartz — 2013-06-19T18:47:07

On Jun 19, 2013, at 1:23 PM, Marc Schwartz <marc_schwartz< at >me.com> wrote:



Actually, correction. That works in the case where only one of the values in the row is a 1, which happened to fit the example data. If there is a chance that more than one value in a row may be a 1, it returns a value equal to the number of 1's found.

Thus, a more generic approach, I believe, would be:

[1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


Regards,

Marc

Re: alternative to ifelse

Marc Schwartz — 2013-06-19T18:23:30

On Jun 19, 2013, at 8:24 AM, Brian Perron <beperron< at >umich.edu> wrote:




If each of the vectors will be of the same length, create a matrix that contains each one as a column:

set.seed(12345)
x <- c(1:30)
x1 <- sample(x, 15, replace = TRUE)
x2 <- sample(x, 15, replace = TRUE)
x3 <- sample(x, 15, replace = TRUE)
x4 <- sample(x, 15, replace = TRUE)

     x1 x2 x3 x4
[1,] 22 14 24 10
[2,] 27 12  1  2
[3,] 23 13  6  2
[4,] 27  6 21  2
[5,] 14 29 12 19
[6,]  5 14 11 29
[7,] 10 10 27 25
[8,] 16 29 28 10
[9,] 22 22 19  7
[10,] 30 20  5 22
[11,]  2 12 24 15
[12,]  5 21 13 22
[13,] 23 17 28  3
[14,]  1  7 24 14
[15,] 12 15  8  8


Then you can use:

[1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


which gets you the same result as:

[1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


Regards,

Marc Schwartz

Re: alternative to ifelse

Marc Schwartz — 2013-06-19T18:21:53

On Jun 19, 2013, at 8:24 AM, Brian Perron <beperron< at >umich.edu> wrote:




If each of the vectors will be of the same length, create a matrix that contains each one as a column:

set.seed(12345)
x <- c(1:30)
x1 <- sample(x, 15, replace = TRUE)
x2 <- sample(x, 15, replace = TRUE)
x3 <- sample(x, 15, replace = TRUE)
x4 <- sample(x, 15, replace = TRUE)

      x1 x2 x3 x4
 [1,] 22 14 24 10
 [2,] 27 12  1  2
 [3,] 23 13  6  2
 [4,] 27  6 21  2
 [5,] 14 29 12 19
 [6,]  5 14 11 29
 [7,] 10 10 27 25
 [8,] 16 29 28 10
 [9,] 22 22 19  7
[10,] 30 20  5 22
[11,]  2 12 24 15
[12,]  5 21 13 22
[13,] 23 17 28  3
[14,]  1  7 24 14
[15,] 12 15  8  8


Then you can use:

 [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


which gets you the same result as:

 [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0


Regards,

Marc Schwartz

Re: alternative to ifelse

Enrico Schumann — 2013-06-19T18:20:40


Hi Brian,

put all your x into a matrix and use apply:

  X <- cbind(x1, x2, x3, x4)
  apply(X, 1, function(x) if (any(x == 1L)) 1 else 0)

  ## [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0

or, since TRUE and FALSE evaluate to 1 and 0 when coerced to numeric:

  as.integer(apply(X, 1, function(x) any(x == 1L)))

  ## [1] 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0



Regards,
        Enrico

Re: nls singular gradient ..as always..

Adams, Jean — 2013-06-19T18:17:29

It's hard to say without seeing the data.  It could be the data, it could
be the starting values, it could be the model choice.

Jean


On Wed, Jun 19, 2013 at 10:45 AM, pakoun <pkount< at >bgc-jena.mpg.de> wrote:


[[alternative HTML version deleted]]

Re: alternative to ifelse

Richard M. Heiberger — 2013-06-19T18:12:51

as.numeric(x1 == 1 | x2 == 1 | x3 == 1 | x4 == 1)

On Wed, Jun 19, 2013 at 9:24 AM, Brian Perron <beperron< at >umich.edu> wrote:


[[alternative HTML version deleted]]

Re: knitr without R studio

Yihui Xie — 2013-06-19T18:07:52

knitr is not tied to RStudio, and I believe this StackOverflow post
can answer your question: http://stackoverflow.com/q/10646665/559676

Regards,
Yihui
--
Yihui Xie <xieyihui< at >gmail.com>
Phone: 206-667-4385 Web: http://yihui.name
Fred Hutchinson Cancer Research Center, Seattle


On Wed, Jun 19, 2013 at 4:42 AM, Ista Zahn <istazahn< at >gmail.com> wrote:

Re: nls singular gradient ..as always..

pakoun — 2013-06-19T15:45:29

Yes it should look like that... what i am doing is a variogram fit . But the
data of course are spread almost all over.. I would guess might be problem
with the data only? 



--
View this message in context: http://r.789695.n4.nabble.com/nls-singular-gradient-as-always-tp4669859p4669898.html
Sent from the R help mailing list archive at Nabble.com.

GLMM predictor variables

Bruce Rawlings — 2013-06-19T15:44:32

Hi all,

I have a quick question regarding predictor variables in a GLMM analysis, a
subject I am new to.

I am running a study investigating multi-modal communication in primates.
specifically, primate gestures that accompany vocalizations.

I have measured the call rate, call duration and peak frequency (response
variables)  for 200 calls (100 accompanied by gestures, 100 without
gestures).

I would like to use the following predictor variables: 1) Call type, 2)
Presence of gesture, 3) Gesture type

Is it possible to use  GESTURE TYPE here as a predictor variable? I am not
sure if it's appropriate since this will only be for 50% of the overall
 the data (where calls are accompanied by gestures).

Any advice would be greatly appreciated.

Bruce Rawlings

[[alternative HTML version deleted]]

alternative to ifelse

Brian Perron — 2013-06-19T13:24:50

Greetings:

I am looking for a way to avoid using the ifelse function for
constructing a new variable.  More specifically, assume I have a set
of variables with scores ranging from 1 to 30.

set.seed(12345)
x <- c(1:30)
x1 <- sample(x, 15, replace = TRUE)
x2 <- sample(x, 15, replace = TRUE)
x3 <- sample(x, 15, replace = TRUE)
x4 <- sample(x, 15, replace = TRUE)

I want to construct a dichotomous variable that tests whether any of
the variables contains the value 1.

newVar <-ifelse(x1 == 1 | x2 == 1 | x3 == 1 | x4 == 1, 1, 0)

I want to avoid the ifelse function because I have a number of large
variable lists that will require new variables to be created.  I'm
sure there is a simple way to do this, but I haven't had any luck with
my search!

Thanks in advance.

Brian

sum epsilon-regression only works with certain epsilon values

Joao Quariguasi — 2013-06-19T13:17:32

Hello,

I am trying to run the following piece of command:

svm.linear<-svm(price~., subset = generation >0, data = trainset, cost = C, epsilon = 1, type = 'eps-regression')

If epsilon = 1, the program runs fine. For larger values, e.g. 10, I get the following error:

Error in predict.svm(ret, xhold, decision.values = TRUE) : 
  Model is empty!
Calls: RegressionAnalyses ... svm.formula -> svm.default -> na.action -> predict -> predict.svm
Execution halted

My impression is that sum.linear is returning empty. I initially believed that it could be coursed by an infeasible solution, but I think now that this may not be the case as we allow 
for slacks in both constraints, i.e. xi. Can someone help me here? 

I am a little stuck. 

(btw, C=300)

Re: Vectorize a series of Matrix Multiplications

David Winsemius — 2013-06-19T17:57:49

On Jun 17, 2013, at 8:48 AM, G Vishwanath wrote:


I'm pretty sure you can drop the as.matrix() calls since ToyArray[,,1,i] and ba[,,2,i] will be matrices. Look at the documentation for  ?"[" and pay attention to the "drop" argument (or look at ?drop since it is an R function). That should provide some increase in speed.

     [,1] [,2]
[1,]    1    1
[2,]    1    1



I don't know. I would guess "not" since none of the help pages for array or ?matmult or the ones linking from it suggested that "%*%" should operate on arrays. (I would have tested my half-baked ideas if you had provided a reproducible example of 'ba'.)



This question is unclear to me. Sorry. It's lacking in details and example. (And I apologize if there is a duplicate of this that explains why smarter people than I have not already addressed it.)

Re: Ryacas loads but yacas has an error

Gabor Grothendieck — 2013-06-19T16:57:26

I am running Ryacas using R 3.0.1 on Windows 8 so its unlikely it does
not run on Windows 7.  Read the troubleshooting section on the home
page.  http://ryacas.googlecode.com


--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com