date:20121004

Hello,

If the dataset 'data_ostrya' is kept in a data.frame, then 
length(data_ostrya) is equal to the number of columns, not the number of 
rows. This is because a data.frame is a special type of list and 
length(some_list) is the number of its elements (obvious). if you want 
to sample rows, try sampling row numbers:

row_ostrya - sample(nrow(ostrya),200, replace=F)


Hope this helps,

Rui Barradas
Em 04-10-2012 11:06, Gian Maria Niccolò Benucci escreveu:
 Hi again to everybody,

 I incountered the following error when I try to make a sample inside a
 dataset.
 My code is:

 data_ostrya - sample(ostrya,200, replace=F)
 Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace,  :
cannot take a sample larger than the population when 'replace = FALSE'

 Why it does not work?
 The whole dataset is composed of 536 rows and I just want to sample
 randomly 200 of them...

 Thank you in advance,

 Gian




 On 13 September 2012 14:01, Duncan Murdoch murdoch.dun...@gmail.com wrote:

 On 12-09-13 7:43 AM, Gian Maria NiccolÃ² Benucci wrote:

 Thank you Duncan,

 I got the result of sampling, but it gave me only the row numbers. Is it
 possible to have the entire row with variables and other information?
 Because I need to re-sample inside my matrix the whole rows in reason to
 have 20 samples (i.e., rows) each year.
 Thank you for your invaluable help!

 Use those row numbers to index the dataframe or matrix, e.g.

 a[rows,]

 Duncan Murdoch


 Gian


 On 13 September 2012 13:32, Duncan Murdoch murdoch.dun...@gmail.com
 wrote:

   On 12-09-13 7:18 AM, Gian Maria NiccolÅ^ Benucci wrote:
   Thank you very much for your help,
 I was wondering if is possible to sample randomly specifying to select
 in
 a
 particular group of data inside the matrix, for example only within the
 whole samples collected in 2011 I would randomly choose 20 random
 samples...


 You need two steps:  find the rows that meet your condition, then sample
 from those.  For example,

 rows - which( a$year == 2011 )
 sample(rows, 20)

 There is one thing to watch out for:  if you have a condition that only
 matches one row, you will get unexpected results here, because the sample
 will be taken from 1:rows.  See the examples in ?sample for the
 workaround
 that uses sample.int.

 Duncan Murdoch


   Thanks a again,

 Gian

 On 13 September 2012 12:26, anna freni sterrantino annafr...@yahoo.it

 wrote:

Hello Gian,

 sure sample function
 will do it for your sampling.

 a=as.data.frame(matrix(1:20,4))

 sample(rownames(a),2)

 see ?sample for more details.
 Hope it helps

 Cheers

 Anna


 Anna Freni Sterrantino
 Department of Statistics
 University of Bologna, Italy
 via Belle Arti 41, 40124 BO.
  --
 *Da:* Gian Maria NiccolÅ^ Benucci gian.benu...@gmail.com
 *A:* r-help@r-project.org
 *Inviato:* GiovedÄ 13 Settembre 2012 10:42
 *Oggetto:* [R] random sampling inside a dataset


 Hi all,

 I am wondering if do exist a function in R that allow me to sample or
 choose randomly the rows (i.e., samples) inside a given matrix.
 Thank you very much in advance.
 Cheers,

 --
 Gian

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Re: [R] random sampling inside a dataset

2012-10-04 Thread Duncan Murdoch


On 12-10-04 6:06 AM, Gian Maria Niccolò Benucci wrote:

Hi again to everybody,

I incountered the following error when I try to make a sample inside a
dataset.
My code is:


data_ostrya - sample(ostrya,200, replace=F)

Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace,  :
   cannot take a sample larger than the population when 'replace = FALSE'

Why it does not work?
The whole dataset is composed of 536 rows and I just want to sample
randomly 200 of them...


The sample() works on vectors, not dataframes.  Since dataframes are 
lists containing the columns, it was trying to sample columns, not rows.


This would give you a sample of rows:

ostrya[sample(1:536, 200, replace=FALSE),]

Duncan Murdoch



Thank you in advance,

Gian




On 13 September 2012 14:01, Duncan Murdoch murdoch.dun...@gmail.com wrote:


On 12-09-13 7:43 AM, Gian Maria NiccolÃ² Benucci wrote:


Thank you Duncan,

I got the result of sampling, but it gave me only the row numbers. Is it
possible to have the entire row with variables and other information?
Because I need to re-sample inside my matrix the whole rows in reason to
have 20 samples (i.e., rows) each year.
Thank you for your invaluable help!



Use those row numbers to index the dataframe or matrix, e.g.

a[rows,]

Duncan Murdoch



Gian


On 13 September 2012 13:32, Duncan Murdoch murdoch.dun...@gmail.com
wrote:

  On 12-09-13 7:18 AM, Gian Maria NiccolÅˆ Benucci wrote:


  Thank you very much for your help,


I was wondering if is possible to sample randomly specifying to select
in
a
particular group of data inside the matrix, for example only within the
whole samples collected in 2011 I would randomly choose 20 random
samples...



You need two steps:  find the rows that meet your condition, then sample
from those.  For example,

rows - which( a$year == 2011 )
sample(rows, 20)

There is one thing to watch out for:  if you have a condition that only
matches one row, you will get unexpected results here, because the sample
will be taken from 1:rows.  See the examples in ?sample for the
workaround
that uses sample.int.

Duncan Murdoch


  Thanks a again,



Gian

On 13 September 2012 12:26, anna freni sterrantino annafr...@yahoo.it


wrote:



   Hello Gian,


sure sample function
will do it for your sampling.

a=as.data.frame(matrix(1:20,4))

sample(rownames(a),2)

see ?sample for more details.
Hope it helps

Cheers

Anna


Anna Freni Sterrantino
Department of Statistics
University of Bologna, Italy
via Belle Arti 41, 40124 BO.
 --
*Da:* Gian Maria NiccolÅˆ Benucci gian.benu...@gmail.com
*A:* r-help@r-project.org
*Inviato:* GiovedÄ› 13 Settembre 2012 10:42
*Oggetto:* [R] random sampling inside a dataset


Hi all,

I am wondering if do exist a function in R that allow me to sample or
choose randomly the rows (i.e., samples) inside a given matrix.
Thank you very much in advance.
Cheers,

--
Gian

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Re: [R] random sampling inside a dataset

2012-10-04 Thread Marianna Chimienti

Hi,

I guess it is because when you have the FALSE option the sample function
take off the number from the whole population so your data_frame becomes
different in length. If you try with the option TRUE it works (of course)
but then you have the possibility that you sample the same row twice. Hope
this can help



On 4 October 2012 11:06, Gian Maria NiccolÃ² Benucci
gian.benu...@gmail.comwrote:

 Hi again to everybody,

 I incountered the following error when I try to make a sample inside a
 dataset.
 My code is:

  data_ostrya - sample(ostrya,200, replace=F)
 Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace,  :
   cannot take a sample larger than the population when 'replace = FALSE'

 Why it does not work?
 The whole dataset is composed of 536 rows and I just want to sample
 randomly 200 of them...

 Thank you in advance,

 Gian




 On 13 September 2012 14:01, Duncan Murdoch murdoch.dun...@gmail.com
 wrote:

  On 12-09-13 7:43 AM, Gian Maria NiccolÃ² Benucci wrote:
 
  Thank you Duncan,
 
  I got the result of sampling, but it gave me only the row numbers. Is it
  possible to have the entire row with variables and other information?
  Because I need to re-sample inside my matrix the whole rows in reason to
  have 20 samples (i.e., rows) each year.
  Thank you for your invaluable help!
 
 
  Use those row numbers to index the dataframe or matrix, e.g.
 
  a[rows,]
 
  Duncan Murdoch
 
 
  Gian
 
 
  On 13 September 2012 13:32, Duncan Murdoch murdoch.dun...@gmail.com
  wrote:
 
   On 12-09-13 7:18 AM, Gian Maria NiccolÅ Benucci wrote:
 
   Thank you very much for your help,
 
  I was wondering if is possible to sample randomly specifying to select
  in
  a
  particular group of data inside the matrix, for example only within
 the
  whole samples collected in 2011 I would randomly choose 20 random
  samples...
 
 
  You need two steps:  find the rows that meet your condition, then
 sample
  from those.  For example,
 
  rows - which( a$year == 2011 )
  sample(rows, 20)
 
  There is one thing to watch out for:  if you have a condition that only
  matches one row, you will get unexpected results here, because the
 sample
  will be taken from 1:rows.  See the examples in ?sample for the
  workaround
  that uses sample.int.
 
  Duncan Murdoch
 
 
   Thanks a again,
 
 
  Gian
 
  On 13 September 2012 12:26, anna freni sterrantino 
 annafr...@yahoo.it
 
  wrote:
 
 
Hello Gian,
 
  sure sample function
  will do it for your sampling.
 
  a=as.data.frame(matrix(1:20,4))
 
  sample(rownames(a),2)
 
  see ?sample for more details.
  Hope it helps
 
  Cheers
 
  Anna
 
 
  Anna Freni Sterrantino
  Department of Statistics
  University of Bologna, Italy
  via Belle Arti 41, 40124 BO.
  --
  *Da:* Gian Maria NiccolÅ Benucci gian.benu...@gmail.com
  *A:* r-help@r-project.org
  *Inviato:* GiovedÄ 13 Settembre 2012 10:42
  *Oggetto:* [R] random sampling inside a dataset
 
 
  Hi all,
 
  I am wondering if do exist a function in R that allow me to sample or
  choose randomly the rows (i.e., samples) inside a given matrix.
  Thank you very much in advance.
  Cheers,
 
  --
  Gian
 
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Re: [R] Intermittent connectivity issues for JDBC / Oracle

On Thu, Oct 4, 2012 at 4:29 AM, dh_ss da...@serendipityscience.com wrote:
 I have been experiencing issues with an R script hanging when connecting to
 an Oracle database.  To help debug, I found a function here in the nabble
 forum and made a small test script:

   #!/usr/bin/Rscript --no-restore --no-save --no-init-file
   library(RJDBC)
   drv-JDBC(oracle.jdbc.OracleDriver,/home/oracle/lib/ojdbc6.jar, ')

   dbCheck = function(url, user='', pwd='') {
 j = .jcall(java/sql/DriverManager, Ljava/sql/Connection;,
   getConnection, url, user, pwd, check=FALSE)
 x = .jgetEx()
 .jcheck()
 x
   }
   dbCheck(jdbc:oracle:thin:@192.168.3.10:1521:psrndpt1,myuser,mypass)


 When I run the file - sometimes it works, returning NULL, and sometimes it
 hangs for between 1 and 2.5 minutes before throwing an error.  The database
 is new and stable, on the same network - with no connection issues from any
 other clients.  This machine isn't currently doing anything else except for
 running this script.  Output from running it several times looks like this:

 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 Error in .jcall(java/sql/DriverManager, Ljava/sql/Connection;,
 getConnection,  :
   ignoring SIGPIPE signal
 Calls: dbCheck - .jcall - .External
 Execution halted
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 Error in .jcall(java/sql/DriverManager, Ljava/sql/Connection;,
 getConnection,  :
   ignoring SIGPIPE signal
 Calls: dbCheck - .jcall - .External
 Execution halted
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 NULL
 R@ss-dev:~/scripts$ ./dbCheck.R
 Loading required package: methods
 Loading required package: DBI
 Loading required package: rJava
 Error in .jcall(java/sql/DriverManager, Ljava/sql/Connection;,
 getConnection,  :
   ignoring SIGPIPE signal
 Calls: dbCheck - .jcall - .External
 Execution halted


 The times that it returns NULL, it returns immediately.  When it throws the
 error, it is a varying amount of time between roughly 1 and 2.5 minutes.
 Both machines are not busy at all.

 Any ideas would greatly appreciated.  Thanks.


I'd suggest you take this to the R-SIG-DB (Databases) mailing list
instead. I don't think it's mirrored on the abomination that is
Nabble, so you'll have to subscribe and post there:
https://stat.ethz.ch/mailman/listinfo/r-sig-db

Note that they'll most certainly want your OS, rJava version, JVM
version, R version, and relevant info about the Oracle DB as well.

Cheers,
Michael

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Re: [R] t-test


Hello,

I'm not quite sure I understand, but something like this?

tapply(dat$Score, dat$Name, FUN = mean)
sapply(unique(dat$Name), function(un){
with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})

My doubt is in what you mean by mean for CTK and all. The ?t.test 
gives a confidence interval for the difference in the means, so maybe 
you'll have to look there for what you want.


Hope this helps,

Rui Barradas
Em 04-10-2012 10:34, Nico Met escreveu:

Dear Group,

I want to do a t-test calculation on a large data set.

I am pasting some part of it

structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), .Label = c(CTK100, CTK103, CTK121), class = factor),
 Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
 206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
 249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
 196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
 )), .Names = c(Name, Score), class = data.frame, row.names =
c(NA,
24L))


I want to compare groups with CTK100 and with all the groups and want to
save the p-values and mean for each of that particular group (for example:
mean for CTK and all)
Similarly, for other groups like that CTK121 etc...

Is there any way to automate this process?

Thanks for your advice !

Nico

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Re: [R] integrate / Vectorize question

2012-10-04 Thread Gerrit Draisma


Dag Rui,
Yes, it helped a lot!
Thanks and greetings,
Gerrit.

Op 10/4/2012 12:06 AM, Rui Barradas schreef:

Hello,

Try


Fnum - function(x, beta){
sapply(x, function(x) integrate(f,lower=-x,upper=0,beta=beta)$value)
}


And all results were the same.

Hope this helps,

Rui Barradas
Em 03-10-2012 21:59, Gerrit Draisma escreveu:

Dear R-users,

I want to use the function Fnum below
in another integrate procedure.

How do I write Fnum so that it returns a vector of values?
And will the last integrate return the right answer?

(I tried Vectorize() in several places,
but got all sorts of errors.)

Thanks for helping out.

Gerrit.


This my code
=
beta=c(3,0.3)*1E-2
mu-5

f-function(x,beta){beta[1]+beta[2]*x}

Fnum-function(x,beta){
   integrate(f,lower=-x,upper=0,beta=beta)$value
   }

Fsym-function(x,beta){x*(beta[1]-beta[2]/2*x)}

y-1:4
Fsym(y,beta)
Fnum(y,beta)

g-function(y,mu){exp(-y/mu)/mu}

integrate((function(y,beta,mu){Fsym(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

#This fails
integrate((function(y,beta,mu){Fnum(y,beta)*g(y,mu)}),lower=0,upper=2*mu,
  beta=beta,mu=mu)

t-seq(0,2*mu, by=0.2)
0.2*sum(tapply(t,t,function(y,beta,mu){Fnum(y,beta)*g(y,mu)},
   beta=beta,mu=mu))

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Re: [R] t-test

2012-10-04 Thread Nico Met

Dear Rui,

Many thanks for help.

mean for CTK and all  = comparison between mean of all groups ( which
means second col) vs. each groups like CTK100, CTK121 etc.

Regards

Nico

On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 I'm not quite sure I understand, but something like this?

 tapply(dat$Score, dat$Name, FUN = mean)
 sapply(unique(dat$Name), function(un){
 with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})

 My doubt is in what you mean by mean for CTK and all. The ?t.test gives
 a confidence interval for the difference in the means, so maybe you'll have
 to look there for what you want.

 Hope this helps,

 Rui Barradas
 Em 04-10-2012 10:34, Nico Met escreveu:

 Dear Group,

 I want to do a t-test calculation on a large data set.

 I am pasting some part of it

 structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L), .Label = c(CTK100, CTK103, CTK121), class = factor),
  Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
  206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
  249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
  196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
  )), .Names = c(Name, Score), class = data.frame, row.names =
 c(NA,
 24L))


 I want to compare groups with CTK100 and with all the groups and want to
 save the p-values and mean for each of that particular group (for example:
 mean for CTK and all)
 Similarly, for other groups like that CTK121 etc...

 Is there any way to automate this process?

 Thanks for your advice !

 Nico

 [[alternative HTML version deleted]]

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 posting-guide.html http://www.R-project.org/posting-guide.html
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Re: [R] Ops.factor(point1, point2) : - not meaningful for factors


Hello,

First of all, it's not a good idea to use  'dist' and 'diff' since they 
already are names of R functions.
As for the problem, your column DSi$Si is not a column of 2-dim points, 
it's not even a vector of numbers, it's a categorical variable,  a 
factor, like you can see in the output of str(). See the difference with 
an example, using your first two points.



mydist - function (point1, point2){
dff - point1 - point2
distance - sqrt(sum(dff^2))
distance
}

x - c(5, 20)  # these are vectors representing
y - c(20, 2)  # 2-dim points

dist(matrix(c(x, y), ncol=2))
mydist(x, y)


Also, it's better to post data using ?dput().

Hope this helps,

Rui Barradas
Em 04-10-2012 09:07, nadiah escreveu:

Hi, can anyone help me in this problem :(. I am a total beginner in R
software.  It took me 2 days just to look into this problem.  Due to this
problem. I cant do looping. i want to find the distance between x and DSi



DSi

   i   Si
1 1  (5, 20)
2 2  (20, 2)
3 3 (25, 32)
4 4  (8, 39)
5 5 (10, 17)
6 6 (35, 20)
7 7 (38, 10)

str (DSi)

'data.frame':   7 obs. of  2 variables:
  $ i : int  1 2 3 4 5 6 7
  $ Si: Factor w/ 7 levels (10, 17),(20, 2),..: 6 2 3 7 1 4 5

x - c(15, 45)
y - DSi [,2]
dist - function (point1, point2)

+ {
+ diff - point1-point2
+ distance - sqrt(sum(diff^2))
+ distance
+ }

dist (x, y)

[1] NA
Warning message:
In Ops.factor(point1, point2) : - not meaningful for factors
Regards,
Nadiah



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Re: [R] Class for time series

2012-10-04 Thread Gabor Grothendieck

On Thu, Oct 4, 2012 at 3:07 AM, Hasan Diwan hasan.di...@gmail.com wrote:
 Mr. Emmanuel,

 On 4 October 2012 02:43, Poizot Emmanuel emmanuel.poi...@cnam.fr wrote:

 Dear all,

 I have a time serie dataset such as the following with data acquired every
 15 minutes:

 DateHeure Profondeur Température Salinité Turbidité Chloration
 1 2012-07-06 08:47:22 -0.144  22.4690.011 0.000  0
 2 2012-07-06 09:02:21 -0.147  22.4760.011 0.000  0
 3 2012-07-06 09:17:21 -0.139  22.4980.011 19.323  0
 4 2012-07-06 09:32:21 -0.136  22.5400.011 19.343  0
 5 2012-07-06 09:47:21 -0.141  22.5100.011 19.321  0
 6 2012-07-06 10:02:21 -0.139  22.3720.011 19.280  0

 I wonder what is the best class to use to manage such time series


 Use xts whenever dealing with timeseries, to construct:
 xts(data.in[,-1:2], order.by=as.POSIXct(paste(data.in[,1:2])))


If you are using xts and reading in the data from an external file
then note that xts loads the zoo package and read.zoo can be used to
do the actually reading:

Lines - DateHeure Profondeur Température Salinité Turbidité Chloration
1 2012-07-06 08:47:22 -0.144  22.4690.011 0.000  0
2 2012-07-06 09:02:21 -0.147  22.4760.011 0.000  0
3 2012-07-06 09:17:21 -0.139  22.4980.011 19.323  0
4 2012-07-06 09:32:21 -0.136  22.5400.011 19.343  0
5 2012-07-06 09:47:21 -0.141  22.5100.011 19.321  0
6 2012-07-06 10:02:21 -0.139  22.3720.011 19.280  0

library(xts) # also pulls in zoo

# z - read.zoo(myfile.dat, header = TRUE, index = 1:2, tz = )
z - read.zoo(text = Lines, header = TRUE, index = 1:2, tz = )

x - as.xts(z)

Here index = 1:2 says that the date/time index is in the first two
columns and tz =  says to interpret it as POSIXct with the indicated
time zone.  tz = GMT is another possibility.

For more info on read.zoo see ?read.zoo .  Also there is a document
entirely devoted to read.zoo examples obtained by issuing:
vignette(zoo-read) .

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

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Re: [R] Creating vegetation distance groups from one column

2012-10-04 Thread Jean V Adams

Jean,

Take a look at the cut() function,
?cut

For example ...

mydf - data.frame(nest=1:100, d2veg=runif(100, 0, 60))
mydf$dgroup - cut(mydf$d2veg, breaks=seq(0, 70, 5), include.lowest=TRUE)
head(mydf)

(another) Jean



Jhope jeanwaij...@gmail.com wrote on 10/04/2012 02:27:38 AM:
 
 Hi R listers, 
 
 I am trying to group distances of nests to the vegetation into classes 
that
 are define by (0-5m, 6-10m, 11-15m, 16-20m, 21-25m, 26-30m, 31-35m, 
36-40m,
 41-45m, 46-50m, 51-55m, 56-60m). Each row is a nest and all the 
distances to
 the vegetation is in a column. 
 
 In plyr, I have tried - below script but I think I am going about this 
the
 wrong way and am not successful. 
 
 #Veg index star = resp
 Veg.index - function(values, weights=c(0, 5, 10, 15, 20, 25, 30, 35, 
40,
 45, 50, 55, 60, 65, 70)) {
   star - values*weights
   return(apply(star, 1, sum) / apply(values, 1, sum))
 }
 data.to.analyze$VegIndex - Veg.index(data.to.analyze[,c(0, 5, 10,
 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 
70)])
 write.csv(data.to.analyze, 3turtlehatch.csv, row.names=FALSE)
 
 Please advise, Jean

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[R] Problem with colors in contour plot

2012-10-04 Thread Loukia Spineli

Dear R users,

I have a 51 by 51 matrix of p-values (named as pvalue_MA). I want to
present graphically this matrix in a plot (filled contour plot) where both
axes represent probabilities. I have also added a grid in this plot. I want
to highlight in white the cells of the grid that represent p-values smaller
than the (common) significance threshold, 0.05. The code from this plot is
colored in blue (I had to copy-pasted all this code in order to reach to
the plot). I suspect that the problem might be in the col parameter of
the filled.contour function. Honestly, I cannot understand why the plot
appears to be completely white!! I checked the values of my variable and it
has a great range from 0 to 1.
Any suggestion/ comment would really help me to move on with my project.

All the best,

Loukia
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Re: [R] Creating vegetation distance groups from one column


Hello,

Without sample data it's difficult to give an answer but see ?cut.
To give a data example, the best way is to use ?dput().

dput( head(mydata, 30) )  # Paste the output of this in a post

Hope this helps,

Rui Barradas
Em 04-10-2012 08:27, Jhope escreveu:

Hi R listers,

I am trying to group distances of nests to the vegetation into classes that
are define by (0-5m, 6-10m, 11-15m, 16-20m, 21-25m, 26-30m, 31-35m,  36-40m,
41-45m, 46-50m, 51-55m, 56-60m). Each row is a nest and all the distances to
the vegetation is in a column.

In plyr, I have tried - below script but I think I am going about this the
wrong way and am not successful.

#Veg index star = resp
Veg.index - function(values, weights=c(0, 5, 10, 15, 20, 25, 30, 35, 40,
45, 50, 55, 60, 65, 70)) {
   star - values*weights
   return(apply(star, 1, sum) / apply(values, 1, sum))
}
data.to.analyze$VegIndex - Veg.index(data.to.analyze[,c(0, 5, 10,
15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70)])
write.csv(data.to.analyze, 3turtlehatch.csv, row.names=FALSE)

Please advise, Jean



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Re: [R] Problem with mutli-dimensional array

2012-10-04 Thread Loukia Spineli

Thank you very much!! Apparently, If I had defined my main variable
(results) as a list a priori, I hadn't had consumed so much time writing
and running codes after codes!
Your comments were very useful both for the code and my document! Thank you!

On Wed, Oct 3, 2012 at 4:35 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 I'm glad it helped.
 Now you have what is a data structure problem. The computations are the
 same but in a different output.

 First of all a terminology issue. In R the correct names for the data
 structures you've refered to are 'list', 'matrix' and 'array'. To make it
 short,

 vector --- 1-dim
 matrix --- 2-dim
 array --- n-dim

 and

 list --- no dim attribute, can hold any type of data structure.

 Your first sentence of this last post would read (changes in uppercase)

 Actually I am trying to make an LIST of length 7, where each LIST ELEMENT
  IS AN ARRAY containing a different number of 2 by 2 matrices.

 In what follows, for the 3-dim array case I'll use the naming convention
 [row, column, slice].
 Each of your output list elements is an array with a possibly varying
 number of slices. Right now all of them have 64 slices.

 1. You need a list of 7 elements, each of them indexed by your current 'i';
 2. Each of those elements is a 3-dim array, each slice is a 2x2 matrix and
 each slice is indexed by your current 'l'.
 3. Let's try it.


 Nx - rep(0, length(x))
 Ny - rep(0, length(y))
 n - (x+1)*(y+1)

 results - vector(list, 7)


 for(i in 1:length(x)){
 Nx[i] - length(1:(x[i]+1))
 Ny[i] - length(1:(y[i]+1))
 results[[i]] - array(dim = c(2, 2, n[i]))
 l - 1

 for(j in 1:(Nx[i])){
 for(k in 1:(Ny[i])){
 tmp - c((0:x[i])[j], (0:y[i])[k],-(0:x[i])[j], -(0:y[i])[k])
 results[[i]][,, l] - mat.stat[,,i] + matrix(tmp, nrow=2,
 ncol=2,byrow=T)
 l - l + 1
 }
 }
 }
 results


 I think this is it. Let us now if not.

 Rui Barradas
 Em 03-10-2012 08:48, Loukia Spineli escreveu:

 Thank you very much for your suggestions. They both worked!

 Actually I am trying to make an array of length 7, where each array
 contains a different number of 2 by 2 matrices. The vector n defines the
 different number of matrices across the arrays. Therefore, the parameter
 l has to change in every array i. Of course  I cannot put the vector n
 directly into the dimensions of the array (under the name results).

 First, I wanted to see if the code gives rational results assuming that
 each array has the same number of 2 by 2 matrices. Fortunately, the
 results
 are rational (thanks to your help, as well). Now I am taking the problem
 to
 the next level and I want to construct this array of different number of 2
 by 2 matrices!

 Any suggestions would be really very helpful!

 On Tue, Oct 2, 2012 at 6:23 PM, Rui Barradas ruipbarra...@sapo.pt
 wrote:

  Hello,
 See if this is it.


 Nx - rep(0,length(x))
 Ny - rep(0,length(y))
 n - (x+1)*(y+1)
 results - array(0, dim=c(2,2,64,7))
 # l - 1 # --- This changed place

 for(i in 1:length(x)){
  Nx[i] - length(1:(x[i]+1))
  Ny[i] - length(1:(y[i]+1))
  l - 1 # --- To here

  for(j in 1:(Nx[i])){
  for(k in 1:(Ny[i])){


 Hope this helps,

 Rui Barradas
 Em 02-10-2012 15:20, Loukia Spineli escreveu:

  I want to make a multi-dimensional array. To be specific I want to make
 the
 following array

 results-array(0,dim=c(2,2,64,7))


 This is the code I have created but it gives no result due to the error
 subscript out of bound.

 x-rep(7,7)  # Missingness in intervention
 y-rep(7,7) # Missingness in control

 arraynames-list(Group=c(Success,Failure),Outcome=c(
 Intervention,Control),Trial=c(1:7))
 mat.stat-array(c(9,16,10,15,66,12,44,23,102,88,66,104,**277,**
 60,247,119,23,43,20,41,201,162,122,
 263,14,41,4,41),dim=c(2,2,7),dimnames=arraynames);mat.stat


 Nx-rep(0,length(x))
 Ny-rep(0,length(y))
 n-(x+1)*(y+1)
 results-array(0,dim=c(2,2,64,7))

 l-1
 for(i in 1:length(x)){
 Nx[i] - length(1:(x[i]+1))
 Ny[i] - length(1:(y[i]+1))
 for(j in 1:(Nx[i])){
 for(k in 1:(Ny[i])){
 results[,,l,i]-mat.stat[,,i]+matrix(c(c(0:x[i])[j],c(0:y[**i]**

 )[k],-c(0:x[i])[j],
 -c(0:y[i])[k]),nrow=2,ncol=2,byrow=T)

 l-l+1
 }
 }
 }
 results

 Any suggestion would be really welcome!
 Thank you very much!

 Loukia

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[R] Text to Speech In R

2012-10-04 Thread Alex Zhang

Dear all,

Excuse me for my searching skills: I just couldn't figure out how to do any 
simple text to speech in R console.

What I want to do is, out put simple English words or sentences to speaker as 
voice. No fanciness. For example, I want the R console to speak Hello world 
or Howdy.

Do you know of any related function/package? Thanks,

- Alex
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Re: [R] t-test

HI,
Try this:
sapply(split(dat,dat$Name),function(x) t.test(x[,2],dat[,2])$p.value)
#    CTK100 CTK103 CTK121 
#0.86330310 0.32706859 0.02023357 
A.K.



- Original Message -
From: Nico Met nicome...@gmail.com
To: Rui Barradas ruipbarra...@sapo.pt
Cc: r-help@r-project.org; r-help r-h...@stat.math.ethz.ch
Sent: Thursday, October 4, 2012 6:37 AM
Subject: Re: [R] t-test

Dear Rui,

Many thanks for help.

mean for CTK and all  = comparison between mean of all groups ( which
means second col) vs. each groups like CTK100, CTK121 etc.

Regards

Nico

On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 I'm not quite sure I understand, but something like this?

 tapply(dat$Score, dat$Name, FUN = mean)
 sapply(unique(dat$Name), function(un){
     with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})

 My doubt is in what you mean by mean for CTK and all. The ?t.test gives
 a confidence interval for the difference in the means, so maybe you'll have
 to look there for what you want.

 Hope this helps,

 Rui Barradas
 Em 04-10-2012 10:34, Nico Met escreveu:

 Dear Group,

 I want to do a t-test calculation on a large data set.

 I am pasting some part of it

 structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L), .Label = c(CTK100, CTK103, CTK121), class = factor),
      Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
      206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
      249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
      196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
      )), .Names = c(Name, Score), class = data.frame, row.names =
 c(NA,
 24L))


 I want to compare groups with CTK100 and with all the groups and want to
 save the p-values and mean for each of that particular group (for example:
 mean for CTK and all)
 Similarly, for other groups like that CTK121 etc...

 Is there any way to automate this process?

 Thanks for your advice !

 Nico

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Re: [R] Text to Speech In R

2012-10-04 Thread Jeff Newmiller

Why do you think this is something R should be able to do?
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Alex Zhang alex.zh...@ymail.com wrote:

Dear all,

Excuse me for my searching skills: I just couldn't figure out how to do
any simple text to speech in R�console.

What I want to do is, out put simple English words or sentences to
speaker as voice. No fanciness. For example, I want the R console to
speak Hello world or Howdy.

Do you know of any related function/package? Thanks,

- Alex
   [[alternative HTML version deleted]]





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Re: [R] Class for time series

Hi,
library(xts)

xts(dat1[,-1:2],order.by=as.POSIXct(paste(dat1[,1],dat1[,2],sep= 
),format=%Y-%m-%d %H:%M:%S))
#Error in .subset(x, j) : only 0's may be mixed with negative subscripts


xts(dat1[,3:7],order.by=as.POSIXct(paste(dat1[,1],dat1[,2],sep= 
),format=%Y-%m-%d %H:%M:%S))
#    Profondeur Température Salinité Turbidité Chloration
#2012-07-06 00:00:22 -0.145  22.468    0.011 0.000  0
#2012-07-06 00:02:21 -0.143  22.475    0.011 0.000  0
#2012-07-06 01:17:21 -0.132  22.456    0.011 0.323  0
#2012-07-06 08:47:22 -0.144  22.469    0.011 0.000  0
#2012-07-06 09:02:21 -0.147  22.476    0.011 0.000  0
#2012-07-06 09:17:21 -0.139  22.498    0.011    19.323  0
#2012-07-06 09:32:21 -0.136  22.540    0.011    19.343  0
#2012-07-06 09:47:21 -0.141  22.510    0.011    19.321  0
#2012-07-06 10:02:21 -0.139  22.372    0.011    19.280  0
A.K.







- Original Message -
From: Hasan Diwan hasan.di...@gmail.com
To: Poizot Emmanuel emmanuel.poi...@cnam.fr
Cc: r-help@r-project.org
Sent: Thursday, October 4, 2012 3:07 AM
Subject: Re: [R] Class for time series

Mr. Emmanuel,

On 4 October 2012 02:43, Poizot Emmanuel emmanuel.poi...@cnam.fr wrote:

 Dear all,

 I have a time serie dataset such as the following with data acquired every
 15 minutes:

 Date    Heure Profondeur Température Salinité Turbidité Chloration
 1 2012-07-06 08:47:22     -0.144      22.469    0.011 0.000          0
 2 2012-07-06 09:02:21     -0.147      22.476    0.011 0.000          0
 3 2012-07-06 09:17:21     -0.139      22.498    0.011 19.323          0
 4 2012-07-06 09:32:21     -0.136      22.540    0.011 19.343          0
 5 2012-07-06 09:47:21     -0.141      22.510    0.011 19.321          0
 6 2012-07-06 10:02:21     -0.139      22.372    0.011 19.280          0

 I wonder what is the best class to use to manage such time series


Use xts whenever dealing with timeseries, to construct:
xts(data.in[,-1:2], order.by=as.POSIXct(paste(data.in[,1:2])))

Also, when you post help requests, use dput on your data set. I'm assuming
it's a data.frame, but I'd be able to actually test my code if there were
dput output in your question. -- H
-- 
Sent from my mobile device
Envoyait de mon portable

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Re: [R] Text to Speech In R

On Thu, Oct 4, 2012 at 1:31 PM, Alex Zhang alex.zh...@ymail.com wrote:
 Dear all,

 Excuse me for my searching skills: I just couldn't figure out how to do any 
 simple text to speech in R console.

 What I want to do is, out put simple English words or sentences to speaker as 
 voice. No fanciness. For example, I want the R console to speak Hello world 
 or Howdy.

 Do you know of any related function/package? Thanks

On Mac OS X, you can access the system faculties to do so with
system(say hello) or what have you.

M


 - Alex
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Re: [R] Problem with colors in contour plot

Hello,

Something like this?


myRamp - colorRampPalette(c(lightblue, darkblue))
mypal - myRamp(20)
mypal[1] - rgb(1, 1, 1)

#?filled.contour
## Persian Rug Art:
x - y - seq(-4*pi, 4*pi, len = 27)
r - sqrt(outer(x^2, y^2, +))
filled.contour(cos(r^2)^2, frame.plot = FALSE,
plot.axes = {}, col = mypal)
grid()


Hope this helps,

Rui Barradas
Em 04-10-2012 13:25, Loukia Spineli escreveu:
 Dear R users,

 I have a 51 by 51 matrix of p-values (named as pvalue_MA). I want to
 present graphically this matrix in a plot (filled contour plot) where both
 axes represent probabilities. I have also added a grid in this plot. I want
 to highlight in white the cells of the grid that represent p-values smaller
 than the (common) significance threshold, 0.05. The code from this plot is
 colored in blue (I had to copy-pasted all this code in order to reach to
 the plot). I suspect that the problem might be in the col parameter of
 the filled.contour function. Honestly, I cannot understand why the plot
 appears to be completely white!! I checked the values of my variable and it
 has a great range from 0 to 1.
 Any suggestion/ comment would really help me to move on with my project.

 All the best,

 Loukia


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Re: [R] Text to Speech In R

2012-10-04 Thread Alex Zhang

Jeff - so, do you mean that you are sure it can NOT AND SHOULD NOT? Thanks,

- Alex

 From: Jeff Newmiller jdnew...@dcn.davis.ca.us
To: Alex Zhang alex.zh...@ymail.com; Alex Zhang alex.zh...@ymail.com; 
r-help@r-project.org r-help@r-project.org 
Sent: Thursday, October 4, 2012 8:46 AM
Subject: Re: [R] Text to Speech In R

Why do you think this is something R should be able to do?
---
Jeff NewmillerÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  TheÂ  Â   .Â  Â  Â   
.Â  Go Live...
DCN:jdnew...@dcn.davis.ca.usÂ  Â  Â  Â  Basics: ##.#.Â  Â  Â   ##.#.Â  Live 
Go...
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Live:Â   OO#.. Dead: 
OO#..Â  Playing
Research Engineer (Solar/BatteriesÂ  Â  Â  Â  Â  Â  O.O#.Â  Â  Â   #.O#.Â  with
/Software/Embedded Controllers)Â  Â  Â  Â  Â  Â  Â   .OO#.Â  Â  Â   .OO#.Â  
rocks...1k
---
Sent from my phone. Please excuse my brevity.

Alex Zhang alex.zh...@ymail.com wrote:

Dear all,

Excuse me for my searching skills: I just couldn't figure out how to do
any simple text to speech in Rï¿½console.

What I want to do is, out put simple English words or sentences to
speaker as voice. No fanciness. For example, I want the R console to
speak Hello world or Howdy.

Do you know of any related function/package? Thanks,

- Alex
Â Â Â  [[alternative HTML version deleted]]

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Re: [R] Text to Speech In R

2012-10-04 Thread Alex Zhang

Thanks, Mike! I am on Windows for this project tho...

- Alex

 From: R. Michael Weylandt michael.weyla...@gmail.com
To: Alex Zhang alex.zh...@ymail.com 
Cc: r-help@r-project.org r-help@r-project.org 
Sent: Thursday, October 4, 2012 8:50 AM
Subject: Re: [R] Text to Speech In R

On Thu, Oct 4, 2012 at 1:31 PM, Alex Zhang alex.zh...@ymail.com wrote:
 Dear all,

 Excuse me for my searching skills: I just couldn't figure out how to do any 
 simple text to speech in R console.

 What I want to do is, out put simple English words or sentences to speaker as 
 voice. No fanciness. For example, I want the R console to speak Hello world 
 or Howdy.

 Do you know of any related function/package? Thanks

On Mac OS X, you can access the system faculties to do so with
system(say hello) or what have you.

M

 - Alex
         [[alternative HTML version deleted]]

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Re: [R] t-test

Hi John,
You are right.  With more than two groups, the type 1 error rate should be a 
problem.
A.K.

- Original Message -
From: John Kane jrkrid...@inbox.com
To: arun smartpink...@yahoo.com
Cc: 
Sent: Thursday, October 4, 2012 8:40 AM
Subject: Re: [R] t-test

My stats are lousy but isnt Nico doing some multiple t-tests when an anova with 
some post hoc comparisons complete with a Tukey or Bonferroni correction looks 
more suitable?  

Of course I have no idea of the topic area and maybe he has already done thi.

John Kane
Kingston ON Canada

 -Original Message-
 From: smartpink...@yahoo.com
 Sent: Thu, 4 Oct 2012 05:31:55 -0700 (PDT)
 To: nicome...@gmail.com
 Subject: Re: [R] t-test

 HI,
 Try this:
 sapply(split(dat,dat$Name),function(x) t.test(x[,2],dat[,2])$p.value)
 #    CTK100 CTK103 CTK121
 #0.86330310 0.32706859 0.02023357
 A.K.

 - Original Message -
 From: Nico Met nicome...@gmail.com
 To: Rui Barradas ruipbarra...@sapo.pt
 Cc: r-help@r-project.org; r-help r-h...@stat.math.ethz.ch
 Sent: Thursday, October 4, 2012 6:37 AM
 Subject: Re: [R] t-test

 Dear Rui,

 Many thanks for help.

 mean for CTK and all  = comparison between mean of all groups ( which
 means second col) vs. each groups like CTK100, CTK121 etc.

 Regards

 Nico

 On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas ruipbarra...@sapo.pt
 wrote:

 Hello,

 I'm not quite sure I understand, but something like this?

 tapply(dat$Score, dat$Name, FUN = mean)
 sapply(unique(dat$Name), function(un){
      with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})

 My doubt is in what you mean by mean for CTK and all. The ?t.test
 gives
 a confidence interval for the difference in the means, so maybe you'll
 have
 to look there for what you want.

 Hope this helps,

 Rui Barradas
 Em 04-10-2012 10:34, Nico Met escreveu:

 Dear Group,

 I want to do a t-test calculation on a large data set.

 I am pasting some part of it

 structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L), .Label = c(CTK100, CTK103, CTK121), class = factor),
       Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
       206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
       249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
       196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
       )), .Names = c(Name, Score), class = data.frame, row.names
 =
 c(NA,
 24L))

 I want to compare groups with CTK100 and with all the groups and want
 to
 save the p-values and mean for each of that particular group (for
 example:
 mean for CTK and all)
 Similarly, for other groups like that CTK121 etc...

 Is there any way to automate this process?

 Thanks for your advice !

 Nico

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[R] Is there any package for Vector Auto-regressive with exogenous variable other than fastVAR?

2012-10-04 Thread Paul Gilbert



Package dse also does vector auto-regression with exogenous variables 
(and also vector ARMA and state-space models with exogenous variables).
But I don't understand what you mean by not taking the base in the 
model so I don't know if it will solve your problem.


Paul

On 12-10-04 06:00 AM, r-help-requ...@r-project.org wrote:

Date: Wed, 3 Oct 2012 23:08:54 -0700 (PDT)
From: Sankalpsankalpsing...@gmail.com
To:r-help@r-project.org
Subject: [R] Is there any package for Vector Auto-regressive with
exogenous variable other than fastVAR?
Message-ID:1349330934183-4644964.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii

Is there any package for Vector Auto-regressive with exogenous variable other
than fastVAR?
Because it is not able to solve my problem of not taking the base in the
model.
Please suggest some appropriate solution



--
View this message in 
context:http://r.789695.n4.nabble.com/Is-there-any-package-for-Vector-Auto-regressive-with-exogenous-variable-other-than-fastVAR-tp4644964.html
Sent from the R help mailing list archive at Nabble.com.




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Re: [R] Text to Speech In R

2012-10-04 Thread Jeff Newmiller

Please don't shout.

Speech synthesis is associated in my mind with sound generation, which is 
rather platform-specific. R is a platform agnostic research tool. R can be 
interfaced with many kinds of external libraries, so many things are possible. 
However, they aren't necessarily simple, and hiding operating system 
differences can take a lot of work. Your expectation that this be simple is 
what seems unwarranted to me.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Alex Zhang alex.zh...@ymail.com wrote:

Jeff - so, do you mean that you are sure it can NOT AND SHOULD NOT?
Thanks,

- Alex



 From: Jeff Newmiller jdnew...@dcn.davis.ca.us
To: Alex Zhang alex.zh...@ymail.com; Alex Zhang
alex.zh...@ymail.com; r-help@r-project.org r-help@r-project.org 
Sent: Thursday, October 4, 2012 8:46 AM
Subject: Re: [R] Text to Speech In R
 
Why do you think this is something R should be able to do?
---
Jeff Newmiller                        The     .       .  Go
Live...
DCN:jdnew...@dcn.davis.ca.us        Basics: ##.#.       ##.#.  Live
Go...
                                      Live:   OO#.. Dead: OO#.. 
Playing
Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
/Software/Embedded Controllers)               .OO#.       .OO#. 
rocks...1k
---

Sent from my phone. Please excuse my brevity.

Alex Zhang alex.zh...@ymail.com wrote:

Dear all,

Excuse me for my searching skills: I just couldn't figure out how to
do
any simple text to speech in R�console.

What I want to do is, out put simple English words or sentences to
speaker as voice. No fanciness. For example, I want the R console to
speak Hello world or Howdy.

Do you know of any related function/package? Thanks,

- Alex
    [[alternative HTML version deleted]]





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

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Re: [R] Text to Speech In R

2012-10-04 Thread Bhupendrasinh Thakre

Integrating question and my 2 cents.

Use R2WD to print your output in word file 
Use TTS facility of windows to do the speech 

Never implemented myself but may work

Best Regards,

Bhupendrasinh Thakre
Sent from my iPhone

On Oct 4, 2012, at 7:55 AM, Alex Zhang alex.zh...@ymail.com wrote:

 Thanks, Mike! I am on Windows for this project tho...
 
 - Alex
 
 
 
 From: R. Michael Weylandt michael.weyla...@gmail.com
 To: Alex Zhang alex.zh...@ymail.com 
 Cc: r-help@r-project.org r-help@r-project.org 
 Sent: Thursday, October 4, 2012 8:50 AM
 Subject: Re: [R] Text to Speech In R
 
 On Thu, Oct 4, 2012 at 1:31 PM, Alex Zhang alex.zh...@ymail.com wrote:
 Dear all,
 
 Excuse me for my searching skills: I just couldn't figure out how to do any 
 simple text to speech in R console.
 
 What I want to do is, out put simple English words or sentences to speaker 
 as voice. No fanciness. For example, I want the R console to speak Hello 
 world or Howdy.
 
 Do you know of any related function/package? Thanks
 
 On Mac OS X, you can access the system faculties to do so with
 system(say hello) or what have you.
 
 M
 
 
 - Alex
  [[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.
[[alternative HTML version deleted]]
 
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[R] (minor) R syntax error in help page to the function makeCluster of library(snow)

2012-10-04 Thread Wolfgang Raffelsberger

Dear list,

I just realized that one of the examples given in the help page to the
function makeCluster of the library(snow) has a small syntax error :

## to get started
library(snow)
?makeCluster

 .. will open a halp page containing the command towards the end of the
examples :

cl - makeCluster(c(rep(list(macOptions), 2), rep(list(lnxOptions), 2),
rep(list(winOptions), 2), type = SOCK)

 .. which is missing a closing parenthesis.


For the rest I'd like to congratulate the developpers of this library for
the great job they have done !!

Best regards,
Wolfgang


For completness (according to the posing guide):
 sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-unknown-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=fr_FR.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=fr_FR.UTF-8LC_COLLATE=fr_FR.UTF-8
 [5] LC_MONETARY=fr_FR.UTF-8LC_MESSAGES=fr_FR.UTF-8
 [7] LC_PAPER=C LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=fr_FR.UTF-8 LC_IDENTIFICATION=C

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

loaded via a namespace (and not attached):
[1] snow_0.3-10


. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wolfgang Raffelsberger, PhD
IGBMC,
1 rue Laurent Fries,  67404 Illkirch  Strasbourg,  France
wolfgang.raffelsberger (a t) igbmc.fr

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Re: [R] smoothScatter plot

2012-10-04 Thread John Kane


   Hi,
   Do you mean something like this?
   =
   scatter.smooth(x,y)scatter.smooth(x,y)
   =
   It looks like invoking that dcols - densCols(x,y) is callling in some
   package that is masking the basic::smoothScatter()  and applying some other
   version of smoothScatter, but I am not expert enough to be sure.
   Another way to get the same result as mine with smoothScatter is to use the
   ggplot2 package.  it looks a bit more complicated but it is very good and in
   some ways easier to see exactly what is happening.
   To   try   it   you   would   need  to  install  the  ggplot2  package
   (install.packages(ggplot2)  then with your original x and y data frames
   ===
   library(ggplot2)
   xy  -  cbind(x, y)
   names(xy)  -  c(xx, yy)
   p  -  ggplot(xy , aes(xx, yy )) + geom_point( ) +
geom_smooth( method=loess, se =FALSE)
   p
   
   Thanks for the data set.  However it really is easier to use dput()
   To use dput() simply issue the command dput(myfile) where myfile is the file
   you are working with.  It will give you something like this:
   ==
   1 dput(x)
   structure(c(0.4543462924, 0.2671718761, 0.1641577016, 1.1593356462,
   0.0421177346, 0.3127782861, 0.4515537795, 0.5332559665, 0.0913911528,
   0.1472054054, 0.1340672893, 1.2599304224, 0.3872026125, 0.0368560053,
   0.0371828779, 0.3999714282, 0.0175815783, 0.8871547761, 0.2706762487,
   0.7401904063, 0.0991320236, 0.2565567348, 0.5854167363, 0.7515717421,
   0.7220388222, 1.3528297744, 0.9339971349, 0.0128652431, 0.4102527051
   ), .Dim = c(29L, 1L), .Dimnames = list(NULL, V1))
   1 dput(y)
   structure(list(V1 = c(0.8669898448, 0.6698647266, 0.1641577016,
   0.4779091929, 0.2109900366, 0.2915241414, 0.2363116664, 0.3808731568,
   0.379908928, 0.2565868263, 0.1986675964, 0.7589866876, 0.6496236922,
   0.1327986663, 0.4196107999, 0.3436442638, 0.1910728051, 0.5625817464,
   0.1429791079, 0.6441837334, 0.1477153617, 0.369079266, 0.3839842979,
   0.39044223, 0.4186374286, 0.7611640016, 0.446291999, 0.2943343355,
   0.3019098386)), .Names = V1, class = data.frame, row.names = c(NA,
   -29L))
   1
   ===
   That is your x in dput() form.  You just copy it from the R terminal and
   paste it into your email message.  It is handy if you add the x  -  and y
   -  to the output.
   Your method works just fine but it's a bit more cumbersome with a lot of
   data.
   Also, please reply to the R-help list as well.  It is a source of much more
   expertise than me and it also can reply when a single person is unavailable.
   I hope this helps
   John Kane
   Kingston ON Canada

   -Original Message-
   From: zhyjiang2...@hotmail.com
   Sent: Thu, 4 Oct 2012 05:19:14 +0800
   To: jrkrid...@inbox.com
   Subject: RE: [R] smoothScatter plot

   Hi John,

   Thanks for your reply. But I cannot figure out how to use dput(). I included
   data and code below. Is that possible to make a plot similar to attached
   smoothing effect.

   Zhengyu
   ###

   x-read.table(text=0.4543462924
   0.2671718761
   0.1641577016
   1.1593356462
   0.0421177346
   0.3127782861
   0.4515537795
   0.5332559665
   0.0913911528
   0.1472054054
   0.1340672893
   1.2599304224
   0.3872026125
   0.0368560053
   0.0371828779
   0.3999714282
   0.0175815783
   0.8871547761
   0.2706762487
   0.7401904063
   0.0991320236
   0.2565567348
   0.5854167363
   0.7515717421
   0.7220388222
   1.3528297744
   0.9339971349
   0.0128652431
   0.4102527051,header=FALSE)
   y-read.table(text=0.8669898448
   0.6698647266
   0.1641577016
   0.4779091929
   0.2109900366
   0.2915241414
   0.2363116664
   0.3808731568
   0.379908928
   0.2565868263
   0.1986675964
   0.7589866876
   0.6496236922
   0.1327986663
   0.4196107999
   0.3436442638
   0.1910728051
   0.5625817464
   0.1429791079
   0.6441837334
   0.1477153617
   0.369079266
   0.3839842979
   0.39044223
   0.4186374286
   0.7611640016
   0.446291999
   0.2943343355
   0.3019098386,header=FALSE)
   x-data.matrix(x)
   y-data.matrix(y)
   dcols - densCols(x,y)
   smoothScatter(x,y, col = dcols, pch=20,xlab=A,ylab=B)

Date: Tue, 2 Oct 2012 05:19:27 -0800
From: jrkrid...@inbox.com
Subject: RE: [R] smoothScatter plot
To: zhyjiang2...@hotmail.com; r-help@r-project.org
   
It's hard to know what's wrong with your code since you did not supply it.
   
Please supply a small working example and some data. To supply data use
   the dput() function, see ?dput() for details.
   
John Kane
Kingston ON Canada
   
   
 -Original Message-
 From: zhyjiang2...@hotmail.com
 Sent: Tue, 2 Oct 2012 11:38:31 +0800
 To:

[R] How to convert array dim names to expressions for plotting?

2012-10-04 Thread Hofert Jan Marius

Hi,

I would like to use the dim names of an array as labels in a plot. In the case 
of character labels, this is no problem. However, if I would like to pass 
expressions through array dim names, they only appear as character labels in 
the plot and are not evaluated as expressions (see the minimal example below). 
How can this be done?

Cheers,

Marius


require(grid)

tau - c(0.25, 0.75)
alpha - c(0.95, 0.99, 0.999)
a - array(1:6, dim=c(2,3), dimnames=list(tau=tau, alpha=alpha))

taulabs - as.expression(lapply(tau, function(t) bquote(tau==.(t
dimnames(a)$tau - taulabs # does not create expressions
dimnames(a) # = no expressions
dimnames(a)$tau[[1]] - taulabs[1]
dimnames(a)$tau[[2]] - taulabs[2]
dimnames(a) # = now characters

gl - grid.layout(1, 2, default.units=inch, widths=c(2,2), heights=c(2,2))
## grid.show.layout(gl)
pushViewport(viewport(layout=gl))

pushViewport(viewport(layout.pos.row=1, layout.pos.col=1))
grid.rect(gp=gpar(col=NA, fill=gray80))
grid.text(dimnames(a)[[tau]][1], x=0.5, y=0.5)
upViewport()

pushViewport(viewport(layout.pos.row=1, layout.pos.col=2))
grid.rect(gp=gpar(col=NA, fill=gray50))
grid.text(dimnames(a)[[tau]][2], x=0.5, y=0.5)
upViewport()

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Re: [R] How to convert array dim names to expressions for plotting?

2012-10-04 Thread Bert Gunter

It is not clear to me what you mean by expressions in this context.
Dimension names but be character strings -- see ?dimnames, which says:

The dimnames of a matrix or array can be NULL or a list of the same
length as dim(x). If a list, its components are either NULL or a
character vector with positive length of the appropriate dimension of
x. 

If somehow you expect plotmath type drawn objects or R expression
objects, then you are confused. If I am just misunderstanding what you
want, then I am confused.

-- Bert

On Thu, Oct 4, 2012 at 7:27 AM, Hofert  Jan Marius
marius.hof...@math.ethz.ch wrote:
 Hi,

 I would like to use the dim names of an array as labels in a plot. In the 
 case of character labels, this is no problem. However, if I would like to 
 pass expressions through array dim names, they only appear as character 
 labels in the plot and are not evaluated as expressions (see the minimal 
 example below). How can this be done?

 Cheers,

 Marius


 require(grid)

 tau - c(0.25, 0.75)
 alpha - c(0.95, 0.99, 0.999)
 a - array(1:6, dim=c(2,3), dimnames=list(tau=tau, alpha=alpha))

 taulabs - as.expression(lapply(tau, function(t) bquote(tau==.(t
 dimnames(a)$tau - taulabs # does not create expressions
 dimnames(a) # = no expressions
 dimnames(a)$tau[[1]] - taulabs[1]
 dimnames(a)$tau[[2]] - taulabs[2]
 dimnames(a) # = now characters

 gl - grid.layout(1, 2, default.units=inch, widths=c(2,2), heights=c(2,2))
 ## grid.show.layout(gl)
 pushViewport(viewport(layout=gl))

 pushViewport(viewport(layout.pos.row=1, layout.pos.col=1))
 grid.rect(gp=gpar(col=NA, fill=gray80))
 grid.text(dimnames(a)[[tau]][1], x=0.5, y=0.5)
 upViewport()

 pushViewport(viewport(layout.pos.row=1, layout.pos.col=2))
 grid.rect(gp=gpar(col=NA, fill=gray50))
 grid.text(dimnames(a)[[tau]][2], x=0.5, y=0.5)
 upViewport()

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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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Re: [R] (minor) R syntax error in help page to the function makeCluster of library(snow)

On Thu, Oct 4, 2012 at 2:30 PM, Wolfgang Raffelsberger
wolfgang.raffelsber...@gmail.com wrote:
 Dear list,

 I just realized that one of the examples given in the help page to the
 function makeCluster of the library(snow) has a small syntax error :

 ## to get started
 library(snow)
 ?makeCluster

  .. will open a halp page containing the command towards the end of the
 examples :

 cl - makeCluster(c(rep(list(macOptions), 2), rep(list(lnxOptions), 2),
 rep(list(winOptions), 2), type = SOCK)

  .. which is missing a closing parenthesis.

You should report this to the maintainer of snow (the completely
awesome and altogether remarkable Luke Tierney) who I don't believe
reads R-help regularly. Use the maintainer() function to get his
email.

Cheers,
Michael

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Re: [R] t-test

2012-10-04 Thread Jose Iparraguirre

Hi,

You can also have a look at this paper (no subscription needed):

Herberich, E.; Sikorski, J.; and Hothorn, T. (2010). A Robust Procedure for 
Comparing Multiple Means under Heteroscedasticity in Unbalanced Designs. 
PLoSONE, Vol. 5, Issue 3, e9788. doi:10.1371/journal.pone.0009788

In this paper, Herberich et al have included a piece of R code to run their 
procedure. Their procedure is a modification of Tukey, and as it says in the 
title of the paper, it can be used regardless of whether the samples had 
different sizes or distributions.

José


José Iparraguirre
Chief Economist
Age UK

T 020 303 31482
E jose.iparragui...@ageuk.org.uk
Twitter @jose.iparraguirre@ageuk


Tavis House, 1- 6 Tavistock Square
London, WC1H 9NB
www.ageuk.org.uk | ageukblog.org.uk | @ageukcampaigns 


For a copy of our new Economic Monitor and the full Chief Economist's report, 
visit the Age UK Knowledge Hub 
http://www.ageuk.org.uk/professional-resources-home/knowledge-hub-evidence-statistics/


For evidence and statistics on the older population, visit the Age UK Knowledge 
Hub 
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of arun
Sent: 04 October 2012 13:59
To: John Kane
Cc: R help
Subject: Re: [R] t-test

Hi John,
You are right.  With more than two groups, the type 1 error rate should be a 
problem.
A.K.



- Original Message -
From: John Kane jrkrid...@inbox.com
To: arun smartpink...@yahoo.com
Cc: 
Sent: Thursday, October 4, 2012 8:40 AM
Subject: Re: [R] t-test

My stats are lousy but isnt Nico doing some multiple t-tests when an anova with 
some post hoc comparisons complete with a Tukey or Bonferroni correction looks 
more suitable?  

Of course I have no idea of the topic area and maybe he has already done thi.

John Kane
Kingston ON Canada


 -Original Message-
 From: smartpink...@yahoo.com
 Sent: Thu, 4 Oct 2012 05:31:55 -0700 (PDT)
 To: nicome...@gmail.com
 Subject: Re: [R] t-test
 
 HI,
 Try this:
 sapply(split(dat,dat$Name),function(x) t.test(x[,2],dat[,2])$p.value)
 #    CTK100 CTK103 CTK121
 #0.86330310 0.32706859 0.02023357
 A.K.
 
 
 
 - Original Message -
 From: Nico Met nicome...@gmail.com
 To: Rui Barradas ruipbarra...@sapo.pt
 Cc: r-help@r-project.org; r-help r-h...@stat.math.ethz.ch
 Sent: Thursday, October 4, 2012 6:37 AM
 Subject: Re: [R] t-test
 
 Dear Rui,
 
 Many thanks for help.
 
 mean for CTK and all  = comparison between mean of all groups ( which
 means second col) vs. each groups like CTK100, CTK121 etc.
 
 Regards
 
 Nico
 
 On Thu, Oct 4, 2012 at 12:28 PM, Rui Barradas ruipbarra...@sapo.pt
 wrote:
 
 Hello,
 
 I'm not quite sure I understand, but something like this?
 
 tapply(dat$Score, dat$Name, FUN = mean)
 sapply(unique(dat$Name), function(un){
      with(dat, t.test(Score[Name == un], Score[Name != un])$p.value)})
 
 My doubt is in what you mean by mean for CTK and all. The ?t.test
 gives
 a confidence interval for the difference in the means, so maybe you'll
 have
 to look there for what you want.
 
 Hope this helps,
 
 Rui Barradas
 Em 04-10-2012 10:34, Nico Met escreveu:
 
 Dear Group,
 
 I want to do a t-test calculation on a large data set.
 
 I am pasting some part of it
 
 structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 3L,
 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L), .Label = c(CTK100, CTK103, CTK121), class = factor),
       Score = c(236.9726, 207.0055, 237.3464, 224.4774, 236.5034,
       206.7382, 233.94, 240.31, 240.9, 235.15, 223.36, 248.67,
       249.25, 201.4051, 244.1689, 182.2756, 229.001, 241.3211,
       196.0453, 232.6055, 225.0783, 196.0453, 232.6055, 225.0783
       )), .Names = c(Name, Score), class = data.frame, row.names
 =
 c(NA,
 24L))
 
 
 I want to compare groups with CTK100 and with all the groups and want
 to
 save the p-values and mean for each of that particular group (for
 example:
 mean for CTK and all)
 Similarly, for other groups like that CTK121 etc...
 
 Is there any way to automate this process?
 
 Thanks for your advice !
 
 Nico
 
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 __

[R] convert multi dimensional array to list

2012-10-04 Thread anto.r

I have an 4D named array (the result of recast() on a dataframe) and would
like to convert it to a list with the names retained. 

Example:
my.df-expand.grid(name=c(Alf,Que), month=c(May,Jun),
year=c(2011,2012))
my.df$tmax-sample(15:20,4)
my.df$tmin-sample(10:15,4)
my.df.melt=melt(my.df)
my.df.cast=cast(my.df.melt, year~name~month~variable)

I want to convert it to a list so I can access the elements by name like 
my.df.cast$tmax$May 
to get the corresponding year-name matrix

This assuming, lists provide an easier way to handle this data. Does it?
Also, may be there is a way to get the list I want from my.df without going
through the melt and cast.  Any suggestions are welcome.

Thanks in advance
Anto

PS: I came across threads for converting list to array but not the other way
round. 



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[R] running a regression

2012-10-04 Thread Dereje Bacha

Hi all,

When I run a regression, I got the following response.  your help is 
appreciated.  

Error in constants (lhs, cnames_symb) : 
  The hypothesis
Cost_PKL-QK.X_(Intercept)=0 is not well formed: contains bad
coefficient/variable names.  In addition:
Warning message:
In constants(lhs, cnames_symb) : NAs introduced by coercion

Dereje
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[R] (no subject)

2012-10-04 Thread dick liang

producing a multi-figure plot, i am try to add beta_1, beta_2,.. beta_9 to
ylab using expression or substitution, but cannot work out like

for (i in 1:9){
   plot(seq(1/m, 1-1/m, 1/m), beta.q[,i], type=l, col=1,
   ylim=range(beta.q),
   xlab=quantile, ylab=expresion(beta[i]))
}

any suggestions will be greatly appreciated.

DL

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[R] Rainflow, range pair counting

2012-10-04 Thread Dandy Ehlert

Hello



I got some question about R. Im searching for a function or R-code which
makes me some collectives. In the literature they are called:



 level crossing counting

 rainflow counting

 range pair counting



Cant believe thats not available for R. May I just looking for the wrong
word.


There functions are used to count the cycles (Range and Means) of a time
serie.



Thx for helping :)

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[R] Error in installing package wordcloud in R

2012-10-04 Thread M Bindu/TVM/TCS

Dear Sir,

I faced problem during the installation of wordcloud package in R -2.15 . I 
have mentioned the error below. After selecting the cran mirror it shows the 
error. I will be thankful for your response to this solution.


Error

Selection: 34
Warning: unable to access index for repository 
http://ftp.iitm.ac.in/cran/src/contrib
Error in apply(available[p1, dependencies, drop = FALSE], 1L, function(x) 
paste(x[!is.na(x)],  : 
  subscript out of bounds
In addition: Warning messages:
1: In open.connection(con, r) : unable to resolve 'cran.r-project.org'
2: package âwordcloudâ is not available (for R version 2.15.1) 


Regards 
Bindu M
=-=-=
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information contained in this e-mail message 
and/or attachments to it are strictly prohibited. If 
you have received this communication in error, 
please notify us by reply e-mail or telephone and 
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Re: [R] convert multi dimensional array to list

On Thu, Oct 4, 2012 at 2:54 PM, anto.r anto.ra...@gmail.com wrote:
 I have an 4D named array (the result of recast() on a dataframe) and would
 like to convert it to a list with the names retained.

 Example:
my.df-expand.grid(name=c(Alf,Que), month=c(May,Jun),
 year=c(2011,2012))
my.df$tmax-sample(15:20,4)
my.df$tmin-sample(10:15,4)
my.df.melt=melt(my.df)
my.df.cast=cast(my.df.melt, year~name~month~variable)

 I want to convert it to a list so I can access the elements by name like
my.df.cast$tmax$May
 to get the corresponding year-name matrix

 This assuming, lists provide an easier way to handle this data. Does it?

I might actually stick with an array, but avoid the dollar sign syntax:

my.df.cast[,,May,]

would be one way.

Cheers,
M


 Also, may be there is a way to get the list I want from my.df without going
 through the melt and cast.  Any suggestions are welcome.

 Thanks in advance
 Anto

 PS: I came across threads for converting list to array but not the other way
 round.



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/convert-multi-dimensional-array-to-list-tp4645011.html
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 __
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[R] Subsetting a group of data

2012-10-04 Thread Ian Arvin

Hi,


I am making my way down the learning curve of R, and find it a great
language with so many helpful users!

Below is an example of what I'm trying to do, but can't quite figure out the
right path to go down.

 

Here's what I have:

 

Main is a time series of data with columns Cola and Colb

Cola   Colb

1  1  1

2  1  1

3  1  1

4  -11

5  -1-1

6  -1-1

7  1  -1

8  -1-1

9  -1-1

10   -1-1

 

What I would like is to subset the data into groups -  rows 1-3 into one
group (Cola and Colb both =1), then row 4 by itself, then rows 5-6, then row
7, then rows 8-10.

Group 1

Cola   Colb

1  1  1

2  1  1

3  1  1

 

Group 2

Cola   Colb

1  -11

 

Group 3

Cola   Colb

1  -1-1

2  -1-1

 

Thank you for your help!

 

Ian

 

 


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[R] RPostgreSQL Date format problem

2012-10-04 Thread scottyjock

Hi, I've just started using RPostgreSQL within R to interrogate a PostgreSQL
database. The problem I have is that a column containing dates shows up
differently in R than in the database:

here is the output direct from the psql database in Linux:

db= select pat_dob_dt from patients where pat_dob_dt like '%1980';
 pat_dob_dt 

 04/05/1980
 01/12/1980
 04/05/1980
 02/10/1980
 05/01/1980
 25/09/1980
 14/12/1980
 22/07/1980
 24/01/1980
 01/07/1980
(10 rows)

and here is output from within R:

R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-redhat-linux-gnu (64-bit)

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

  Natural language support but running in an English locale

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

 library(RPostgreSQL)
Loading required package: DBI
  dbhost - Sys.getenv(PGHOST)
  m - dbDriver(PostgreSQL)
  con - dbConnect(m, dbname = db, host = dbhost)
 query - SELECT pat_dob_dt from patients where pat_dob_dt like '%1980';
 pat_data - dbGetQuery(con, statement = query)
 pat_data
   pat_dob_dt
1 4-05-19
2 1-12-19
3 4-05-19
4 2-10-19
5 5-01-19
625-09-19
714-12-19
822-07-19
924-01-19
101-07-19

As you can see it strips off the last two digits in the year, and converts
the / to a -.
The as.Date function does not help as it cannot magically make the 2 digits
appear.is this a setting in RPostgreSQL?
Completely stuckgoogle not helped. 

thanks in advance
Scott






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Re: [R] Error in installing package wordcloud in R

Try using a different repo: something seems off about whichever one
you've tried.

M

On Thu, Oct 4, 2012 at 9:33 AM, M Bindu/TVM/TCS m.bi...@tcs.com wrote:
 Dear Sir,

 I faced problem during the installation of wordcloud package in R -2.15 . I 
 have mentioned the error below. After selecting the cran mirror it shows the 
 error. I will be thankful for your response to this solution.


 Error

 Selection: 34
 Warning: unable to access index for repository 
 http://ftp.iitm.ac.in/cran/src/contrib
 Error in apply(available[p1, dependencies, drop = FALSE], 1L, function(x) 
 paste(x[!is.na(x)],  :
   subscript out of bounds
 In addition: Warning messages:
 1: In open.connection(con, r) : unable to resolve 'cran.r-project.org'
 2: package ‘wordcloud’ is not available (for R version 2.15.1)


 Regards
 Bindu M
 =-=-=
 Notice: The information contained in this e-mail
 message and/or attachments to it may contain
 confidential or privileged information. If you are
 not the intended recipient, any dissemination, use,
 review, distribution, printing or copying of the
 information contained in this e-mail message
 and/or attachments to it are strictly prohibited. If
 you have received this communication in error,
 please notify us by reply e-mail or telephone and
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[R] R combining vectors into a data frame but without a continuous common variable

2012-10-04 Thread lucy88

Hello,

I have two different files which I'd like to combine to make one data frame
but I've no idea how to do it! The first file has two columns; one is the
date, the following is a binary code for debris flow events. Then my other
file has also two columns; the date and then precipitation data.

The thing is, is that the two date columns don't all contain the same dates.
The binary one is every day from April - October from 1900 - 2005, yet the
precipitation file has dates from from say, 1911 to 2004, with some missing
data on certain months and during certain years.

So my question is how to make a data frame which would have the date, the
binary 0 or 1, and then the corresponding precip value from that particular
date. I only want the precip information for the days where I have
information in the binary file; the others can be disregarded.

I have tried using codes which I found in answer to other questions asked
but none of them work with my issue. If I'm honest I don't really know if
this is what I need. I'm hoping to end up doing a logistic regression. I've
uploaded the two files in case I've not been very clear...

I'd be really grateful if anyone could help me and suggest a way to do it!
I'm also really not very technical and am not at all comfortable with R so
if you could be really basic in your advice I'd appreciate it!

Many thanks in advance,
Lucy

Landeck_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Landeck_vec.txt

Kaurnetal_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Kaurnetal_vec.txt

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Re: [R] How to convert array dim names to expressions for plotting?

2012-10-04 Thread Hofert Jan Marius

Dear Bert,

thanks, I was confused (of course). So the right question would have been: How 
to convert strings to expressions (plotmath expressions). But that I could 
easily find, thanks.

Cheers,

Marius

Here is the working example (showing plotmath and just-character labels):

require(grid)

tau - c(0.25, 0.75)
alpha - c(0.95, 0.99, 0.999)
a - array(1:6, dim=c(2,3), dimnames=list(tau=tau, alpha=alpha))

dimnames(a)$tau[1] - tau==0.25 ## dimnames(a)$tau - paste0(tau==, tau)
dimnames(a)
gl - grid.layout(1, 2, default.units=inch, widths=c(2,2), heights=c(2,2))
## grid.show.layout(gl)
pushViewport(viewport(layout=gl))

pushViewport(viewport(layout.pos.row=1, layout.pos.col=1))
grid.rect(gp=gpar(col=NA, fill=gray80))
grid.text(parse(text=dimnames(a)[[tau]][1]), x=0.5, y=0.5)
upViewport()

pushViewport(viewport(layout.pos.row=1, layout.pos.col=2))
grid.rect(gp=gpar(col=NA, fill=gray50))
grid.text(parse(text=dimnames(a)[[tau]][2]), x=0.5, y=0.5)
upViewport()


From: Bert Gunter [gunter.ber...@gene.com]
Sent: Thursday, October 04, 2012 4:42 PM
To: Hofert  Jan Marius
Cc: mailman, r-help
Subject: Re: [R] How to convert array dim names to expressions for plotting?

It is not clear to me what you mean by expressions in this context.
Dimension names but be character strings -- see ?dimnames, which says:

The dimnames of a matrix or array can be NULL or a list of the same
length as dim(x). If a list, its components are either NULL or a
character vector with positive length of the appropriate dimension of
x. 

If somehow you expect plotmath type drawn objects or R expression
objects, then you are confused. If I am just misunderstanding what you
want, then I am confused.

-- Bert

On Thu, Oct 4, 2012 at 7:27 AM, Hofert  Jan Marius
marius.hof...@math.ethz.ch wrote:
 Hi,

 I would like to use the dim names of an array as labels in a plot. In the 
 case of character labels, this is no problem. However, if I would like to 
 pass expressions through array dim names, they only appear as character 
 labels in the plot and are not evaluated as expressions (see the minimal 
 example below). How can this be done?

 Cheers,

 Marius


 require(grid)

 tau - c(0.25, 0.75)
 alpha - c(0.95, 0.99, 0.999)
 a - array(1:6, dim=c(2,3), dimnames=list(tau=tau, alpha=alpha))

 taulabs - as.expression(lapply(tau, function(t) bquote(tau==.(t
 dimnames(a)$tau - taulabs # does not create expressions
 dimnames(a) # = no expressions
 dimnames(a)$tau[[1]] - taulabs[1]
 dimnames(a)$tau[[2]] - taulabs[2]
 dimnames(a) # = now characters

 gl - grid.layout(1, 2, default.units=inch, widths=c(2,2), heights=c(2,2))
 ## grid.show.layout(gl)
 pushViewport(viewport(layout=gl))

 pushViewport(viewport(layout.pos.row=1, layout.pos.col=1))
 grid.rect(gp=gpar(col=NA, fill=gray80))
 grid.text(dimnames(a)[[tau]][1], x=0.5, y=0.5)
 upViewport()

 pushViewport(viewport(layout.pos.row=1, layout.pos.col=2))
 grid.rect(gp=gpar(col=NA, fill=gray50))
 grid.text(dimnames(a)[[tau]][2], x=0.5, y=0.5)
 upViewport()

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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.



--

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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Re: [R] Problem with colors in contour plot

2012-10-04 Thread Loukia Spineli

The contpour plot is very impressive!!! The contour plot I want to display
should have only 2 colors: white a grey. I have created this plot for only
one trial (see, ppts) and now I am attempting to do the same for many
trials.

On Thu, Oct 4, 2012 at 3:51 PM, Rui Barradas ruipbarra...@sapo.pt wrote:

  Hello,

 Something like this?


 myRamp - colorRampPalette(c(lightblue, darkblue))
 mypal - myRamp(20)
 mypal[1] - rgb(1, 1, 1)

 #?filled.contour
 ## Persian Rug Art:
 x - y - seq(-4*pi, 4*pi, len = 27)
 r - sqrt(outer(x^2, y^2, +))
 filled.contour(cos(r^2)^2, frame.plot = FALSE,
plot.axes = {}, col = mypal)
 grid()


 Hope this helps,

 Rui Barradas
 Em 04-10-2012 13:25, Loukia Spineli escreveu:

 Dear R users,

 I have a 51 by 51 matrix of p-values (named as pvalue_MA). I want to
 present graphically this matrix in a plot (filled contour plot) where both
 axes represent probabilities. I have also added a grid in this plot. I want
 to highlight in white the cells of the grid that represent p-values smaller
 than the (common) significance threshold, 0.05. The code from this plot is
 colored in blue (I had to copy-pasted all this code in order to reach to
 the plot). I suspect that the problem might be in the col parameter of
 the filled.contour function. Honestly, I cannot understand why the plot
 appears to be completely white!! I checked the values of my variable and it
 has a great range from 0 to 1.
 Any suggestion/ comment would really help me to move on with my project.

 All the best,

 Loukia



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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Subsetting a group of data

On Thu, Oct 4, 2012 at 3:20 PM, Ian Arvin i...@innovativequant.com wrote:
 Hi,


 I am making my way down the learning curve of R, and find it a great
 language with so many helpful users!

Up it hopefully, no?


 Below is an example of what I'm trying to do, but can't quite figure out the
 right path to go down.



 Here's what I have:



 Main is a time series of data with columns Cola and Colb

 Cola   Colb

 1  1  1

 2  1  1

 3  1  1

 4  -11

 5  -1-1

 6  -1-1

 7  1  -1

 8  -1-1

 9  -1-1

 10   -1-1



 What I would like is to subset the data into groups -  rows 1-3 into one
 group (Cola and Colb both =1), then row 4 by itself, then rows 5-6, then row
 7, then rows 8-10.

 Group 1

 Cola   Colb

 1  1  1

 2  1  1

 3  1  1



 Group 2

 Cola   Colb

 1  -11



 Group 3

 Cola   Colb

 1  -1-1

 2  -1-1




Short answer is to ask why you want to do this? There are many
functions which operate on the split--apply--combine paradigm (see
the JSS article of that name for a short intro) but I find it easier
to do this in conjunction with specifying what my data transformation
is.

To directly answer your question, you'll probably need to combine
split() and interaction() in some way.

Cheers,
Michael

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[R] Explore SPSS function in R

2012-10-04 Thread MARIA RODRIGUEZ

Hi everyone,

Does anybody knows if there is an equivalent R function  that gives the same
outcome as in Explore function in SPSS ?
(Analize-Descriptive Statistics-Explore)

It does a categorical vs quantitative variables analysis. ( But not linear
regression)

I need to compare intragroup (categorical variable with 4 values) means and
confidence intervals of a quantitative variable.
Just like Explore function does.

Thanks a lot




--
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http://r.789695.n4.nabble.com/Explore-SPSS-function-in-R-tp4645013.html
Sent from the R help mailing list archive at Nabble.com.

__
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Re: [R] running a regression

You'll need to supply a reproducible example:

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

Cheers,
Michael

On Thu, Oct 4, 2012 at 2:19 PM, Dereje Bacha d_ba...@yahoo.com wrote:
 Hi all,

 When I run a regression, I got the following response.  your help is 
 appreciated.

 Error in constants (lhs, cnames_symb) :
   The hypothesis
 Cost_PKL-QK.X_(Intercept)=0 is not well formed: contains bad
 coefficient/variable names.  In addition:
 Warning message:
 In constants(lhs, cnames_symb) : NAs introduced by coercion

 Dereje
 [[alternative HTML version deleted]]


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Re: [R] Help with R Fitting an inverse Gamma

On Thu, Oct 4, 2012 at 11:42 AM, kmammasis mammasi...@hotmail.com wrote:
 Dear all,

 I am new in R and would like to ask for someone's help in understanding
 where I go wrong with the following code:

#  rm(list=ls())
#

Please don't include this line into your postings: it's quite rude to
clobber the data of folks trying to help you.


 # Required packages
 library(MCMCpack)

 # Simulated data
 set.seed(1)
 data = rinvgamma(n=250, shape = 5, scale = 2) + 2


# Bad name for a data variable: see
library(fortunes)
fortune(dog)

 hist(data)

 # log-likelihood
 ll = function(par){
 if(par[1]0  par[2]0  par[3]min(data)) return( -sum(log(dinvgamma(data-
 par[3],par[1],par[2]))) )
 else return(Inf)
 }

 # MLE
 mle = optim(c(5,2,2),ll)
 params = mle$par

 # Fit
 hist(data,probability=T,ylim=c(0,2.5))
 points(seq(2,4.5,0.001),dinvgamma(seq(2,4.5,0.001)-params[3],params[1],params[2]),type=l,col=red)

 This code fits an Inverse Gamma distribution to the randomly generated data
 and plots the associated histogram. My problem is that when I try to import
 a dataset saved locally it gives me an error. Here is the modification of
 the code:



 # Required packages
 library(MCMCpack)

 data=my.csv.data$V1
 hist(data)

 # log-likelihood
 ll = function(par){
 if(par[1]0  par[2]0  par[3]min(data)) return( -sum(log(dinvgamma(data-
 par[3],par[1],par[2]))) )
 else return(Inf)
 }

 # MLE
 mle = optim(c(5,2,2),ll)
 params = mle$par

 # Fit
 hist(data,probability=T,ylim=c(0,2.5))
 points(seq(2,4.5,0.001),dinvgamma(seq(2,4.5,0.001)-params[3],params[1],params[2]),type=l,col=red)



 I have attached the file which I import directly from the menu, named:
 excel250.csv. When I run the code it gives me the following error:

No you haven't. The R-help servers scrub most attachments. Use dput()
to make a nice reproducible example:

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

Cheers,
M


 Error in optim(c(5, 2, 2), ll) :
   function cannot be evaluated at initial parameters

Totally random guess: your function isn't being evaluated properly
either because i) you used the name data() which is also a function
and R is having trouble finding your data; ii) your data is such that
dinvgamma() gives bad value at the initial point c(5,2,2)

Cheers,
Michael


 I would appreciate your help.

 Thank you.

 Konstantinos

 nabble_a href=excel250.csvexcel250.csv



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[R] problems with plotting and printing aareg

2012-10-04 Thread Asa Johannesen

Hi all,

I've ventured into the world of nonparametric survival and I would like to use 
the maxtime option for printing and plotting my aareg fit.

However, my fit does not have test.var2 and this stops the print and plot 
when adding a maxtime.

My code is as follows:

Response-Surv(Time,Event)
Model-aareg(Response~Factor1*Factor2)
Model2-aareg(Response~Factor1+Factor2)  #Just did this to see if the 
interaction term had anything to do with it

Model, print(Model), summary(Model), and plot(Model) seem to work fine, but as 
soon as I try summary/print/plot(Model, maxtime=400) it tells me that test.var2 
is not found and when I look at  summary(Model), there is indeed a NULL under 
test.var2.

Anyone know why it doesn't include test.var2? Is this a compatibility problem? 
I'm using R version 2.13 (I know it's quite old, but updating is a pain when 
you don't have admin rights to your computer) and just updated the survival 
package (no warning messages).

Any input would be much appreciated.

Cheers,

Freya

[[alternative HTML version deleted]]

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Re: [R] Explore SPSS function in R

On Thu, Oct 4, 2012 at 3:28 PM, MARIA RODRIGUEZ mrodrigu...@imim.es wrote:
 Hi everyone,

 Does anybody knows if there is an equivalent R function  that gives the same
 outcome as in Explore function in SPSS ?
 (Analize-Descriptive Statistics-Explore)

 It does a categorical vs quantitative variables analysis. ( But not linear
 regression)

I'm afraid I don't know what that means like a t.test()? Or ANOVA?


 I need to compare intragroup (categorical variable with 4 values) means and
 confidence intervals of a quantitative variable.
 Just like Explore function does.

 Thanks a lot




 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Explore-SPSS-function-in-R-tp4645013.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.

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

Re: [R] Rainflow, range pair counting

R is used by many many folks, and most of us don't have the domain
knowledge to make heads or tails of what you're mentioning here.
You'll need to greatly clarify and perhaps ask on the appropriate
R-SIG-* list. (Ecology maybe?) Perhaps rseek.org is also of help.

Cheers,
Michael

On Thu, Oct 4, 2012 at 1:19 PM, Dandy Ehlert dandy.ehl...@gmail.com wrote:
 Hello



 I got some question about R. Im searching for a function or R-code which
 makes me some collectives. In the literature they are called:



  level crossing counting

  rainflow counting

  range pair counting



 Cant believe that’s not available for R. May I just looking for the wrong
 word.


 There functions are used to count the cycles (Range and Means) of a time
 serie.



 Thx for helping :)

 [[alternative HTML version deleted]]


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[R] Failure of sas7bdat package

2012-10-04 Thread Fisher Dennis

R 2.15.1
OS 10.7

Colleagues

I have been an enthusiastic user of the sas7bdat package in R.  However, 
several recent sas7bdat files sent to me from different sources cannot be read 
by the package.  The error message is:
Error in read.sas7bdat(FILENAME) : 
  unknown host W32_7PRO please report bugs to sas7bdatrb...@gmail.com

I examined the file with a text editor and it contains 
9.0301M1W32_7PRO 
in the second line.  An older file that works successfully with the package 
contains
9.0101M3XP_PRO
in the same position.

Is this causing the failure?  Is this something easy to fix?  For the moment, I 
am using StatTransfer to convert the files.  StatTransfer has no problem with 
the newer files suggesting that the problem might be trivial.

I have contacted the package maintainer (matt.shotw...@vanderbilt.edu) twice 
without a response.

Dennis

Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com

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Re: [R] Subsetting a group of data


Hello,

Try the following.


dat - read.table(text=
Cola   Colb
1  1  1
2  1  1
3  1  1
4  -11
5  -1-1
6  -1-1
7  1  -1
8  -1-1
9  -1-1
10   -1-1
, header=TRUE)

idx - dat$Cola != dat$Colb
split(dat, 2*cumsum(idx) - idx)


Hope this helps,

Rui Barradas
Em 04-10-2012 15:20, Ian Arvin escreveu:

Hi,


I am making my way down the learning curve of R, and find it a great
language with so many helpful users!

Below is an example of what I'm trying to do, but can't quite figure out the
right path to go down.

  


Here's what I have:

  


Main is a time series of data with columns Cola and Colb

 Cola   Colb

1  1  1

2  1  1

3  1  1

4  -11

5  -1-1

6  -1-1

7  1  -1

8  -1-1

9  -1-1

10   -1-1

  


What I would like is to subset the data into groups -  rows 1-3 into one
group (Cola and Colb both =1), then row 4 by itself, then rows 5-6, then row
7, then rows 8-10.

Group 1

 Cola   Colb

1  1  1

2  1  1

3  1  1

  


Group 2

 Cola   Colb

1  -11

  


Group 3

 Cola   Colb

1  -1-1

2  -1-1

  


Thank you for your help!

  


Ian

  

  



[[alternative HTML version deleted]]

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Re: [R] Failure of sas7bdat package

2012-10-04 Thread Anthony Damico

I can't help troubleshoot the read.sas7bdat function, but if you have the
fixed-width data and a SAS importation script, the SAScii package might
help you work around this issue.  :)


On Thu, Oct 4, 2012 at 11:55 AM, Fisher Dennis fis...@plessthan.com wrote:

 R 2.15.1
 OS 10.7

 Colleagues

 I have been an enthusiastic user of the sas7bdat package in R.  However,
 several recent sas7bdat files sent to me from different sources cannot be
 read by the package.  The error message is:
 Error in read.sas7bdat(FILENAME) :
   unknown host W32_7PRO please report bugs to sas7bdatrb...@gmail.com

 I examined the file with a text editor and it contains
 9.0301M1W32_7PRO
 in the second line.  An older file that works successfully with the
 package contains
 9.0101M3XP_PRO
 in the same position.

 Is this causing the failure?  Is this something easy to fix?  For the
 moment, I am using StatTransfer to convert the files.  StatTransfer has no
 problem with the newer files suggesting that the problem might be trivial.

 I have contacted the package maintainer (matt.shotw...@vanderbilt.edu)
 twice without a response.

 Dennis

 Dennis Fisher MD
 P  (The P Less Than Company)
 Phone: 1-866-PLessThan (1-866-753-7784)
 Fax: 1-866-PLessThan (1-866-753-7784)
 www.PLessThan.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.


[[alternative HTML version deleted]]

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

Re: [R] Explore SPSS function in R

2012-10-04 Thread William Dunlap

Can you tell us what Explore gives you for the following dataset?

 d-data.frame(Quant=round(log2(1:24), 2), Categ=LETTERS[rep(1:4,c(8,7,6,3))])
 write.table(d, row.names=FALSE, sep=\t)
Quant Categ
0   A
1   A
1.58A
2   A
2.32A
2.58A
2.81A
3   A
3.17B
3.32B
3.46B
3.58B
3.7 B
3.81B
3.91B
4   C
4.09C
4.17C
4.25C
4.32C
4.39C
4.46D
4.52D
4.58D

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of MARIA RODRIGUEZ
 Sent: Thursday, October 04, 2012 7:29 AM
 To: r-help@r-project.org
 Subject: [R] Explore SPSS function in R
 
 Hi everyone,
 
 Does anybody knows if there is an equivalent R function  that gives the same
 outcome as in Explore function in SPSS ?
 (Analize-Descriptive Statistics-Explore)
 
 It does a categorical vs quantitative variables analysis. ( But not linear
 regression)
 
 I need to compare intragroup (categorical variable with 4 values) means and
 confidence intervals of a quantitative variable.
 Just like Explore function does.
 
 Thanks a lot
 
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Explore-SPSS-function-in-
 R-tp4645013.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
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[R] Analysis of Means in R

2012-10-04 Thread Data Analytics Corp.


Hi,

Does anyone know of a package for doing Analysis of Means in R? Or have 
any functions they would like to share?


Thanks,

Walt



Walter R. Paczkowski, Ph.D.
Data Analytics Corp.
44 Hamilton Lane
Plainsboro, NJ 08536

(V) 609-936-8999
(F) 609-936-3733
w...@dataanalyticscorp.com
www.dataanalyticscorp.com

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Re: [R] convert multi dimensional array to list

2012-10-04 Thread anto.r

Hi Michael

thanks! That was the option if I kept it an array. The list format with $
sign since it leaves me feeling that the names are there and can be easily
accessed. Why would you rather not use the $ sign?

I use R-Studio and there names can be selected from a drop-down list, I have
found it easier but that could be my lack of proper training in R. 

Cheers
Anto








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Re: [R] Find the average line from several lines

2012-10-04 Thread David Winsemius


On Oct 3, 2012, at 10:31 PM, Greg Lollback wrote:

 Hi people,
 
 I'm looking at the influence of sample size on a dataset, so I randomly 
 sampled from the data with different sample sizes multiple times using 
 the sample and replicate function. For example:
 
 /replicate(1000, sample(variable name, sample size, replace=T))/.

That does not look like it is valid R code.

 
 I saved the data, and now I have 1000 replicates of data of various 
 sample sizes (2-33 in my case). If you can picture that one (of 1000) 
 iteration is a line of mean values of the variable for each sample size, 
 i.e. the sample size (2-33) is on the x-axis and the average variable 
 value (number of trees, dogs, cats, whatever) is on the y-axis.
 
 This means I have 1000 replicates of a line. *I'm looking to find the 
 average line from several lines *(1000 lines in my case).

If your code to do this had been provided we might now have an idea about what 
the structure of you objects might be. 

 Does anybody 
 know how to do this in R? There was a topic on this using the function 
 approx, but I don't have missing data and my data is on the same scale 
 range.

It sounds as though you are redoing the invention of the bootstrap. You could 
look at package 'boot', although in this case it does sound rather simple. (And 
it sounds like homework and this mailing list frowns on being used as a 
surreptitious help line.)

-- 
David Winsemius, MD
Alameda, CA, USA

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Re: [R] Explore SPSS function in R

2012-10-04 Thread Ista Zahn

Hi Maria,

I suggest starting with the by() function. Here is an example to get
you started:

by(mtcars$mpg,
   INDICES = mtcars[c(cyl, am)],
   FUN = function(x) {
 list(summary = summary(x),
  mean.CI95 = confint(lm(x ~ 1))),

 })

Best,
Ista

On Thu, Oct 4, 2012 at 10:28 AM, MARIA RODRIGUEZ mrodrigu...@imim.es wrote:
 Hi everyone,

 Does anybody knows if there is an equivalent R function  that gives the same
 outcome as in Explore function in SPSS ?
 (Analize-Descriptive Statistics-Explore)

 It does a categorical vs quantitative variables analysis. ( But not linear
 regression)

 I need to compare intragroup (categorical variable with 4 values) means and
 confidence intervals of a quantitative variable.
 Just like Explore function does.

 Thanks a lot




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Re: [R] Failure of sas7bdat package

2012-10-04 Thread Barry Rowlingson

On Thu, Oct 4, 2012 at 4:55 PM, Fisher Dennis fis...@plessthan.com wrote:
 R 2.15.1
 OS 10.7

 Colleagues

 I have been an enthusiastic user of the sas7bdat package in R.  However, 
 several recent sas7bdat files sent to me from different sources cannot be 
 read by the package.  The error message is:
 Error in read.sas7bdat(FILENAME) :
   unknown host W32_7PRO please report bugs to sas7bdatrb...@gmail.com

 I examined the file with a text editor and it contains
 9.0301M1W32_7PRO
 in the second line.  An older file that works successfully with the package 
 contains
 9.0101M3XP_PRO
 in the same position.

 Is this causing the failure?  Is this something easy to fix?  For the moment, 
 I am using StatTransfer to convert the files.  StatTransfer has no problem 
 with the newer files suggesting that the problem might be trivial.

 Just having a look at the source code for Matt's sas7bdat package,
there's a special bug report email:

 sas7bdat:::BUGREPORT
[1] please report bugs to sas7bdatrb...@gmail.com

Further, if you read the vignette(sas7bdat) you'll see a lot of
technical stuff that has been reverse engineered about the sas format.
The source code contains this chunk:

# Host systems known to work
KNOWNHOST - c(WIN_PRO, WIN_NT, WIN_NTSV, WIN_SRV,
   WIN_ASRV, XP_PRO, XP_HOME, NET_ASRV,
   NET_DSRV, NET_SRV, WIN_98, W32_VSPR,
   WIN, WIN_95, X64_VSPR, X64_ESRV)

and it seems your data doesn't have a known host code. I suspect Matt
has never seen a SAS file from a Windows 7 Pro box.

Now, the code currently doesn't do anything dependent on the host,
there's just a cautionary stop() that you have hit. You could edit
read.sas7bdat, remove the stop (or add W32_7PRO to the known hosts),
reinstall, and see if it works.

 Or more hackerly, edit the file and replace W32_7PRO with a valid
KNOWNHOST of exactly the same length (so as not to upset the delicate
binary balance). W32_VSPR would work. Be careful editing binary
files with a text editor - it might try to wrap lines or mung your
file in nasty ways.

 Barry

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Re: [R] Failure of sas7bdat package

2012-10-04 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Barry Rowlingson
 Sent: Thursday, October 04, 2012 9:52 AM
 To: Fisher Dennis
 Cc: r-h...@stat.math.ethz.ch
 Subject: Re: [R] Failure of sas7bdat package

 On Thu, Oct 4, 2012 at 4:55 PM, Fisher Dennis fis...@plessthan.com
 wrote:
  R 2.15.1
  OS 10.7

  Colleagues

  I have been an enthusiastic user of the sas7bdat package in R.
 However, several recent sas7bdat files sent to me from different
 sources cannot be read by the package.  The error message is:
  Error in read.sas7bdat(FILENAME) :
unknown host W32_7PRO please report bugs to sas7bdatrb...@gmail.com

  I examined the file with a text editor and it contains
  9.0301M1W32_7PRO
  in the second line.  An older file that works successfully with the
 package contains
  9.0101M3XP_PRO
  in the same position.

  Is this causing the failure?  Is this something easy to fix?  For the
 moment, I am using StatTransfer to convert the files.  StatTransfer has
 no problem with the newer files suggesting that the problem might be
 trivial.

  Just having a look at the source code for Matt's sas7bdat package,
 there's a special bug report email:

  sas7bdat:::BUGREPORT
 [1] please report bugs to sas7bdatrb...@gmail.com

 Further, if you read the vignette(sas7bdat) you'll see a lot of
 technical stuff that has been reverse engineered about the sas format.
 The source code contains this chunk:

 # Host systems known to work
 KNOWNHOST - c(WIN_PRO, WIN_NT, WIN_NTSV, WIN_SRV,
WIN_ASRV, XP_PRO, XP_HOME, NET_ASRV,
NET_DSRV, NET_SRV, WIN_98, W32_VSPR,
WIN, WIN_95, X64_VSPR, X64_ESRV)

 and it seems your data doesn't have a known host code. I suspect Matt
 has never seen a SAS file from a Windows 7 Pro box.

 Now, the code currently doesn't do anything dependent on the host,
 there's just a cautionary stop() that you have hit. You could edit
 read.sas7bdat, remove the stop (or add W32_7PRO to the known hosts),
 reinstall, and see if it works.

  Or more hackerly, edit the file and replace W32_7PRO with a valid
 KNOWNHOST of exactly the same length (so as not to upset the delicate
 binary balance). W32_VSPR would work. Be careful editing binary
 files with a text editor - it might try to wrap lines or mung your
 file in nasty ways.

  Barry

Dennis,

Looking at the source code on github, Matt has already added W32_7PRO to the 
list of known hosts.  So, you could download the source tarball and build the 
package yourself, or just extract the SAS7bdat.R file from the tarball and 
source it into your R session.  Otherwise, you will probably need to wait for 
the next update to be submitted to CRAN.

Hope this is helpful,

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

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Re: [R] Failure of sas7bdat package

2012-10-04 Thread Barry Rowlingson

On Thu, Oct 4, 2012 at 6:07 PM, Nordlund, Dan (DSHS/RDA)
nord...@dshs.wa.gov wrote:

 Looking at the source code on github, Matt has already added W32_7PRO to the 
 list of known hosts.  So, you could download the source tarball and build the 
 package yourself, or just extract the SAS7bdat.R file from the tarball and 
 source it into your R session.  Otherwise, you will probably need to wait for 
 the next update to be submitted to CRAN.

Easy peasy. No need to wait. Just use devtools:

 install.packages(devtools) # from CRAN
 library(devtools)
 install_github(sas7bdat,BioStatMatt)

Barry

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Re: [R] Legend Truncated Using filled.contour

2012-10-04 Thread David Winsemius


On Oct 3, 2012, at 12:58 PM, Kirsten wrote:

 Hey everyone,
 
 I'm working on a contour plot depicting asymptomatic prevalence at varying
 durations of infectiousness and force of infection.  I've been able to work
 everything out except for this one - my legend title keeps getting cut off. 
 Here's what I have:
 
 filled.contour(x=seq(2,30,length.out=nrow(asym_matrix)),
 y=seq(1,2,length.out=ncol(asym_matrix)),
 asym_matrix, 
 color = function(x)rev(heat.colors(x)),
 plot.title = title(main=Asymptomatic Prevalence in 0-4 Year Olds with\n
 Increasing Duration of Infectiousness,
 xlab = Duration of Infectiousness,
 ylab = Relative Force of Infection),
 key.title = title(main = Asymptomatic\n Prevalence))
 
 My first thought was to make the legend title text smaller using cex = 0.75
 (or similar), but it doesn't change the text size at all.  In fact, none of
 the modifiers that I've tried to add to the key.title line (size, color,
 font, etc) seems to be making a bit of difference.
 
 key.title = title(main = Asymptomatic\n Prevalence, cex = 0.75)

I couldn't see it until now (that I got your csv file) but you should not be 
qusoting the cex parameter. Also it should be `cex.main` rather than just 
`cex`. As explained in the filled.contour help page (which you clearly have 
read) the legend is really a second plot and you should be using the title 
function(as you clearly have figured out.) But the details and examples are in 
?title .

Try:
..., 
key.title = title(main = Asymptomatic\n Prevalence, cex.main=0.70)   )

#   0.75 still got cut off a bit. =, and I think we are both on Macs so this 
should be the same on a default quartz() window.

 
 I assume there's an override earlier in the code, but I have no idea what it
 is.  Any suggestions?

People who choose to go to Nabble , a distinctly smaller group, would have been 
able to find you data, but most people could not. You should have used a .txt 
extension so you mailer would not improperly label it as something other than 
MIME-text. Or you could have simply copied the output of dput() into your email.

-- 

David Winsemius, MD
Alameda, CA, USA

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Re: [R] Rainflow, range pair counting

2012-10-04 Thread Marc Schwartz

A quick Google search suggests that the methods are used in material fatigue 
studies where varying stress loads are placed upon the material, which puts 
this into a materials engineering context.

Using rseek.org does not give any joy. Perhaps someone with experience in that 
domain will chime in.

Regards,

Marc Schwartz

On Oct 4, 2012, at 10:27 AM, R. Michael Weylandt michael.weyla...@gmail.com 
wrote:

 R is used by many many folks, and most of us don't have the domain
 knowledge to make heads or tails of what you're mentioning here.
 You'll need to greatly clarify and perhaps ask on the appropriate
 R-SIG-* list. (Ecology maybe?) Perhaps rseek.org is also of help.
 
 Cheers,
 Michael
 
 On Thu, Oct 4, 2012 at 1:19 PM, Dandy Ehlert dandy.ehl...@gmail.com wrote:
 Hello
 
 
 
 I got some question about R. Im searching for a function or R-code which
 makes me some collectives. In the literature they are called:
 
 
 
 level crossing counting
 
 rainflow counting
 
 range pair counting
 
 
 
 Cant believe that’s not available for R. May I just looking for the wrong
 word.
 
 
 There functions are used to count the cycles (Range and Means) of a time
 serie.
 
 
 
 Thx for helping :)

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Re: [R] R combining vectors into a data frame but without a continuous common variable

Hello,

Try the following.

url1 - http://r.789695.n4.nabble.com/file/n4644986/Landeck_vec.txt;
url2 - http://r.789695.n4.nabble.com/file/n4644986/Kaurnetal_vec.txt;

dat1 - read.table(url1, header = TRUE)
dat2 - read.table(url2, header = TRUE)

str(dat1)
str(dat2) # Precip is a factor, so convert to numeric
dat2$Precip - as.numeric(levels(dat2$Precip)[dat2$Precip])

dat1$Landeck - as.Date(dat1$Landeck, format = %d.%m.%Y)
dat2$Date - as.Date(dat2$Date, format = %d.%m.%Y)

dat3 - merge(dat1, dat2, by.x = Landeck, by.y = Date)
str(dat3)
head(dat3, 20) # See first 20 rows

Hope this helps,

Rui Barradas
Em 04-10-2012 12:18, lucy88 escreveu:

Hello,

Many thanks in advance,
Lucy

Landeck_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Landeck_vec.txt

Kaurnetal_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Kaurnetal_vec.txt

Re: [R] Errors when saving output from WinBUGS to R

2012-10-04 Thread Uwe Ligges


So have you closed WinBUGS regularly (and not via Esc in R) now?

Uwe Ligges


On 03.10.2012 21:29, yiyiwang wrote:

Sorry for any cross messaging. Here are what I've tried and more info. The
error sticks :(

I installed the latest R2.15.1 (64bit) on my machine, per Uwe's comments.
The same error happened. Log file of WinBUGS look perfectly normal:
###
display(log)
check(C:/Jinmodel/test_codes/model_poisson.txt)
model is syntactically correct
data(C:/Users/Yiyi/AppData/Local/Temp/RtmpSAhD25/data.txt)
data loaded
compile(1)
model compiled
inits(1,C:/Users/Yiyi/AppData/Local/Temp/RtmpSAhD25/inits1.txt)
model is initialized
gen.inits()
command #Bugs:gen.inits cannot be executed (is greyed out)
thin.updater(1)
update(0)
set(beta0)
set(beta1)
set(beta2)
set(beta3)
set(deviance)
dic.set()
update(30)
coda(*,C:/Users/Yiyi/AppData/Local/Temp/RtmpSAhD25/coda)
stats(*)

Node statistics
 nodemeansd  MC error   2.5%median  97.5%   start   
sample
beta0   1.652   0.7417  0.3092  1.025   1.381   4.121   1   30
beta1   1.383   0.1343  0.05416 0.8926  1.442   1.491   30
beta2   -0.8984 0.1197  0.0496  -1.008  -0.9378 -0.4977 1   30
beta3   1.745   0.2898  0.1212  0.7972  1.874   1.988   1   30
deviance2461.0  4410.0  1458.0  745.0   880.1   23240.0 1   
30
dic.stats()

DIC
Dbar = post.mean of -2logL; Dhat = -2LogL at post.mean of stochastic nodes
DbarDhatpD  DIC 
Y20 2461.0401205.0701255.9703717.010
total   2461.0401205.0701255.9703717.010
history(*,C:/Users/Yiyi/AppData/Local/Temp/RtmpSAhD25/history.odc)
###



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Re: [R] Error in if (any(ch)) { : missing value where TRUE/FALSE needed

2012-10-04 Thread Uwe Ligges




On 03.10.2012 19:47, liang@us.pwc.com wrote:

Can someone please help with the error message below?


Not really, the message says it all: You cannot improve the AIC since it 
is numerically -Inf already!


Uwe Ligges






thanks!



Start:  AIC=-Inf
value ~ 1 + Core_CPI__ + GDP_change + Unemployment + housing +
 interest + S_P + d1 + d2 + d3

Error in if (any(ch)) { : missing value where TRUE/FALSE needed
In addition: Warning message:
attempting model selection on an essentially perfect fit is nonsense


__
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or entities other than the intended recipient is prohibited, and all liability 
arising therefrom is disclaimed. If you received this in error, please contact 
the sender and delete the material from any computer. PricewaterhouseCoopers 
LLP is a Delaware limited liability partnership.  This communication may come 
from PricewaterhouseCoopers LLP or one of its subsidiaries.

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Re: [R] Rainflow, range pair counting

2012-10-04 Thread Bert Gunter

I'm going to make a wild guess that this might have something to do
with bump hunting =local conditional extrema. If so, the prim
package might be applicable.

Otherwise (...cycles in time series) time series methods might be appropriate.

But I basically agree with others that the OP has not clearly
specified the issue.

Cheers,
Bert

On Thu, Oct 4, 2012 at 10:34 AM, Marc Schwartz marc_schwa...@me.com wrote:
A quick Google search suggests that the methods are used in material fatigue
studies where varying stress loads are placed upon the material, which puts
this into a materials engineering context.

Using rseek.org does not give any joy. Perhaps someone with experience in
that domain will chime in.

Regards,

Marc Schwartz

On Oct 4, 2012, at 10:27 AM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:

R is used by many many folks, and most of us don't have the domain
knowledge to make heads or tails of what you're mentioning here.
You'll need to greatly clarify and perhaps ask on the appropriate
R-SIG-* list. (Ecology maybe?) Perhaps rseek.org is also of help.

Cheers,
Michael

On Thu, Oct 4, 2012 at 1:19 PM, Dandy Ehlert dandy.ehl...@gmail.com wrote:
Hello

I got some question about R. Im searching for a function or R-code which
makes me some collectives. In the literature they are called:

level crossing counting

rainflow counting

range pair counting

Cant believe that’s not available for R. May I just looking for the wrong
word.

There functions are used to count the cycles (Range and Means) of a time
serie.

Thx for helping :)

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] convert multi dimensional array to list

Hi,

You can try this:
list2-lapply(1:dim(my.df.cast)[3],function(i) my.df.cast[,,i,])
 names(list2)-unlist(dimnames(my.df.cast)[3])
 list2$May
#, , variable = tmax
#
#  name
#year   Alf Que
 # 2011  18  16
 # 2012  18  16

#, , variable = tmin
#
 #     name
#year   Alf Que
 # 2011  12  11
 # 2012  12  11

list2$May[,,tmax]
#  name
#year   Alf Que
 # 2011  18  16
 # 2012  18  16


A.K.



- Original Message -
From: anto.r anto.ra...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Thursday, October 4, 2012 11:57 AM
Subject: Re: [R] convert multi dimensional array to list

Hi Michael

thanks! That was the option if I kept it an array. The list format with $
sign since it leaves me feeling that the names are there and can be easily
accessed. Why would you rather not use the $ sign?

I use R-Studio and there names can be selected from a drop-down list, I have
found it easier but that could be my lack of proper training in R. 

Cheers
Anto








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Re: [R] Failure of sas7bdat package

2012-10-04 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: b.rowling...@gmail.com [mailto:b.rowling...@gmail.com] On Behalf
 Of Barry Rowlingson
 Sent: Thursday, October 04, 2012 10:12 AM
 To: Nordlund, Dan (DSHS/RDA)
 Cc: Fisher Dennis; r-h...@stat.math.ethz.ch
 Subject: Re: [R] Failure of sas7bdat package

 On Thu, Oct 4, 2012 at 6:07 PM, Nordlund, Dan (DSHS/RDA)
 nord...@dshs.wa.gov wrote:

  Looking at the source code on github, Matt has already added W32_7PRO
 to the list of known hosts.  So, you could download the source tarball
 and build the package yourself, or just extract the SAS7bdat.R file
 from the tarball and source it into your R session.  Otherwise, you
 will probably need to wait for the next update to be submitted to CRAN.

 Easy peasy. No need to wait. Just use devtools:

  install.packages(devtools) # from CRAN
  library(devtools)
  install_github(sas7bdat,BioStatMatt)

 Barry

Thanks Barry.  

I haven't done this before so this saves me from figuring it out myself. :-)

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

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[R] can stepAIC be customized to exclude coefficients with p-value less than certain values?

2012-10-04 Thread liang . che

For example, if coefficient's p-value is less than 0.1 I want the stepwise 
to automatically drop that variable.  Can the stepAIC be customized to do 
that?  SAS seems to be able to customized stepwise function with p-value 
or cooks'd.

thanks!

__
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the person or entity to which it is addressed and may contain confidential 
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or entities other than the intended recipient is prohibited, and all liability 
arising therefrom is disclaimed. If you received this in error, please contact 
the sender and delete the material from any computer. PricewaterhouseCoopers 
LLP is a Delaware limited liability partnership.  This communication may come 
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Re: [R] Broken Links on http://www.r-project.org

2012-10-04 Thread Uwe Ligges




On 26.09.2012 10:56, Viechtbauer Wolfgang (STAT) wrote:

I was not sure who I should contact about this, so I am posting this here.

There are a few broken links on the R website.

1) http://www.r-project.org/search.html - link to the Nabble R Forum. I belive 
the correct/new URL should be: http://r.789695.n4.nabble.com/

2) http://www.r-project.org/other-docs.html - link to Auswertung ökologischer 
Daten. Not sure if there is a new URL.

3) http://www.r-project.org/other-projects.html - link to Jim Lindsey's R 
page. I believe the correct/new URL should be: http://www.commanster.eu/rcode.html



Thank for letting us know. Changed in the master sources and will go 
online within 24 hours,

Uwe Ligges



Best,
Wolfgang

--
Wolfgang Viechtbauer, Ph.D., Statistician
Department of Psychiatry and Psychology
School for Mental Health and Neuroscience
Faculty of Health, Medicine, and Life Sciences
Maastricht University, P.O. Box 616 (VIJV1)
6200 MD Maastricht, The Netherlands
+31 (43) 388-4170 | http://www.wvbauer.com

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Re: [R] R combining vectors into a data frame but without a continuous common variable

Hi Lucy,

No problem.

Just a correction to my earlier email.

dat1-read.table(Landeck_vec.txt,sep=,header=TRUE,stringsAsFactors=FALSE)
dat2-read.table(Kaurnetal_vec.txt,sep=,header=TRUE,stringsAsFactors=FALSE)
colnames(dat1)[1]-Date

(Rui:
#dat2 Date format is inconsistent.)
dat2$Date-gsub(\\.,\\/,dat2$Date)
dat1$Date-as.POSIXct(dat1$Date,format=%d.%m.%Y)
dat2$Date-as.POSIXct(dat2$Date,format=%d/%m/%Y)

 str(dat1)
#'data.frame':    22623 obs. of  2 variables:
# $ Date : POSIXct, format: 1900-04-01 1900-04-02 ...
# $ Event: int  0 0 0 0 0 0 0 0 0 0 ...
 str(dat2)
#'data.frame':    36598 obs. of  2 variables:
# $ Date  : POSIXct, format: 1900-01-01 1900-01-02 ...
# $ Precip: chr  0 0 0 0 ...

Precip is character, which I convert it to numeric
 #dat2-within(dat2,{Precip-as.numeric(Precip)})
#Warning message:
#In eval(expr, envir, enclos) : NAs introduced by coercion

The reason is that there are datapoints which has some unusual characters.

which(is.na(dat2$Precip))
# [1]  7060  8584  8798 11235 12848 13701 14006 14038 14098 14311 16016 16748
#[13] 18575 19307 19489 19702 19764 21196
dat2[8584,]
#   Date Precip
#8584 1923-09-01 NA

When I looked into the data, I found this:

01/09/1923  L�cke

  count(is.na(dat2$Precip))
#  x  freq
#1 FALSE 36580
#2  TRUE    18
#Removed those rows.
dat3-subset(dat2,!is.na(Precip))
 nrow(dat3)
#[1] 36580

dat4-merge(dat1,dat3,by=Date)
 dat5-subset(dat4,Event!=0)
 nrow(dat5)
#[1] 132
 rownames(dat5)-1:nrow(dat5)
 head(dat5)
#    Date Event Precip
#1 1901-06-02 1    0.0
#2 1905-06-02 1    0.0
#3 1906-08-03 1   15.6
#4 1908-05-08 1    0.0
#5 1911-06-02 1    3.0
#6 1911-09-15 1   23.2

A.K.





- Original Message -
From: lucy88 lucy.fog...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Thursday, October 4, 2012 7:18 AM
Subject: [R] R combining vectors into a data frame but without a continuous 
common variable

Hello,

I have two different files which I'd like to combine to make one data frame
but I've no idea how to do it! The first file has two columns; one is the
date, the following is a binary code for debris flow events. Then my other
file has also two columns; the date and then precipitation data.

The thing is, is that the two date columns don't all contain the same dates.
The binary one is every day from April - October from 1900 - 2005, yet the
precipitation file has dates from from say, 1911 to 2004, with some missing
data on certain months and during certain years.

So my question is how to make a data frame which would have the date, the
binary 0 or 1, and then the corresponding precip value from that particular
date. I only want the precip information for the days where I have
information in the binary file; the others can be disregarded.

I have tried using codes which I found in answer to other questions asked
but none of them work with my issue. If I'm honest I don't really know if
this is what I need. I'm hoping to end up doing a logistic regression. I've
uploaded the two files in case I've not been very clear...

I'd be really grateful if anyone could help me and suggest a way to do it!
I'm also really not very technical and am not at all comfortable with R so
if you could be really basic in your advice I'd appreciate it!

Many thanks in advance,
Lucy

Landeck_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Landeck_vec.txt  

Kaurnetal_vec.txt
http://r.789695.n4.nabble.com/file/n4644986/Kaurnetal_vec.txt  




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Re: [R] Broken Links on http://www.r-project.org

2012-10-04 Thread Hasan Diwan

The R Graphics Gallery has moved to http://gallery.r-enthusiasts.com/ and
there's another R Graphics Manual at http://rgm2.lab.nig.ac.jp/RGM2 -- H

On 26 September 2012 04:56, Viechtbauer Wolfgang (STAT) 
wolfgang.viechtba...@maastrichtuniversity.nl wrote:

 I was not sure who I should contact about this, so I am posting this here.

 There are a few broken links on the R website.

 1) http://www.r-project.org/search.html - link to the Nabble R Forum. I
 belive the correct/new URL should be: http://r.789695.n4.nabble.com/

 2) http://www.r-project.org/other-docs.html - link to Auswertung
 Ã¶kologischer Daten. Not sure if there is a new URL.

 3) http://www.r-project.org/other-projects.html - link to Jim Lindsey's
 R page. I believe the correct/new URL should be:
 http://www.commanster.eu/rcode.html

 Best,
 Wolfgang

 --
 Wolfgang Viechtbauer, Ph.D., Statistician
 Department of Psychiatry and Psychology
 School for Mental Health and Neuroscience
 Faculty of Health, Medicine, and Life Sciences
 Maastricht University, P.O. Box 616 (VIJV1)
 6200 MD Maastricht, The Netherlands
 +31 (43) 388-4170 | http://www.wvbauer.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.




-- 
Sent from my mobile device
Envoyait de mon portable

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[R] problem with the installation of r commander on a mac

2012-10-04 Thread Marco Mello

Dear list members,

Im trying to install R Commander under Mac OSX Mountain Lion (10.8.2). After 
following all the steps described in the installation notes 
(http://socserv.socsci.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html), I 
got this error message:

=
Loading required package: tcltk
Loading Tcl/Tk interface ... done
Loading required package: car
Loading required package: MASS
Loading required package: nnet
Error : .onLoad failed in loadNamespace() for 'Rcmdr', details:
  call: structure(.External(dotTclObjv, objv, PACKAGE = tcltk), class = 
tclObj)
  error: [tcl] invalid command name tk_messageBox.

In addition: Warning message:
In fun(libname, pkgname) : couldn't connect to display :0
Error: package/namespace load failed for Rcmdr
=

The strange thing is that Ive installed Tcl/Tk and XQuark as recommended.

Ive also tried uninstalling R, reinstalling the latest version (2.15.1 signed) 
and all packages, and following the same official steps again. It didnt work 
neither.

Could you please help me? R Commander always worked fine on my iMac, until I 
updated to Mountain Lion. 


Best regards,

Marco



Prof. Dr. Marco A. R. Mello
Universidade Federal de Minas Gerais
ICB - Depto. Biologia Geral
http://marcomello.casadosmorcegos.org
marme...@gmail.com



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Re: [R] R combining vectors into a data frame but without a continuous common variable

2012-10-04 Thread lucy88

Oh my word, you're a genius!! That is absolutely perfect, thank you so much!! 

I've no idea how you've learnt these things but I would never ever have been
able to do that. You've just made my day so much better after the horror of
confusion before. 

Thank you!!





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[R] (no subject)

2012-10-04 Thread Chia-Yi Chiu

Hi

I would like to learn how the R function hclust deals with ties. It is
written in Fortran, so I cannot access the code. Thanks!!

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[R] data structure for plsr

2012-10-04 Thread Emma Jones

I am having a similar problem understanding the data structure of the
yarn dataset described in the [R] data structure for plsr posts. I have
spectroscopic data I'd like to run through a PLSR and have read the
tutorial series, but still do not understand the data format required for
the code to process my data. My current data structure consists of a .csv
file read into R containing 15 columns (a charcoal dilution series going
from 100% to 0%) and 1050 rows of absorbance data from 400 nm to 2500 nm at
 2 nm interval. I think I need to transpose the data such that the specific
wavelengths become my columns and dilutions are defined in rows, but after
that point I am lost. Should I (and how do I) make my absorbance data into
individual matrices that read into a data frame with only two columns
(dilution vs spectra or something of the sort that both the yarn and
gasoline data use)? Thank you for any pointers.

Emma

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[R] Coda, HPDinterval and multiple chains

2012-10-04 Thread NORRIS Paul

Dear all,

I'm not 100% sure if this question is best directed at the r-list, or a mailing 
list concerned with Bayesian analysis, so please accept my apologies if another 
audience may be more appropriate.

I have been using the rjags package to run Jags models with multiple chains and 
store the results in a Coda based mcmc list.  For instance, having created a 
jags model and done initial adapting and updating, I run the following command:

coda_odp_gini_only - coda.samples(odp_gini_only,
   variable.names=c(beta,sigma2.u2, 
deviance),
   n.iter=itercount, thin=thincount)

This create an object with 4 separate chains as requested in my initial call to 
Jags which created the object odp_gini_only.  I then use the  Coda package to 
look at the results stored in coda_odp_gini_only.  Prior to running any 
analysis with Coda, I used the command, 

coda.options(combine.plots=TRUE, combine.stats=TRUE)

to ask for results that combine the four separate chains.  Sure enough, if I 
enter summary(coda_odp_gini_only), I am given a single set of output 
combining the four chains.  However, if I enter 
HPDinterval(coda_odp_gini_only) I receive 4 sets of HPD figures, one for each 
chain.  Is it possible to combine the four chains together to receive a single 
set of HPD estimates? 

In a similar vein, is it possible to use the Coda object to estimate the 
proportion of a given parameters distribution which is above (or below) a given 
value, for instance, the proportion of the distribution of beta[1] greater than 
zero?  Again, in doing so, is it possible to combine the results of the four 
chains into a single estimate?

Kind regards, and many thanks in advance for any advice anyone can offer me.

Paul
Dr Paul Norris
Lecturer in Social Policy
University of Edinburgh
-- 
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.

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[R] P value discrepancy plots

2012-10-04 Thread bilelsan

Hello R-users,
I am trying to plot P value discrepancy plots as describe by Russell
Davidson and James G. MacKinnon in Graphical Methods for Investigating the
Size and Power of Hypothesis Tests (1998).
I know ecdf () to obtain the empirical cumulative distribution frequencies
and knots() to extract its values.
Supposing p.value as a vector obtained by DGP (out of the loop); is there
anyone who already write this short function to obtain the same graph than
presented in the mentioned paper (figures P value discrepancy plots)?
I am sorry in advance because I know it's not correct to ask the list to do
something. But I am pretty sure this could be useful for some of us.
TFH



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[R] R help - Adding a column in a data frame with multiple conditions

2012-10-04 Thread Libby M Gertken

Hi,

I am trying to add a column of numbers to a data frame in R with multiple
conditions.

Here is a simplified example df:

[A]  [B] [C] [D] [E]
[1] 1 X 90 88
[2] 1 Y 72 70
[3] 1 Z 67 41
[4] 2 X 74 49
[5] 2 Y 42 50
[6] 2 Z 81 56
[7] 3 X 92 59
[8] 3 Y 94 80
[9] 3 Z 80 82

I would like column [E] to have a certain value (found either in [C] or
[D]) based on conditions in columns [A] *and* [B].

E.g. :
if [A] = 1 and [B] = X, then [E] = the entry in [C] for that row (i.e., 90)
if [A] = 1 and [B] = Y, then [E] = the entry in [D] for that row (i.e., 70)
if [A] = 1 and [B] = Z, then [E] = the entry in [C] for that row (i.e., 67)

if [A] = 2 and [B] = X, then [E] = the entry in [C] for that row (i.e., 74)
if [A] = 2 and [B] = Y, then [E] = the entry in [D] for that row (i.e., 50)
if [A] = 2 and [B] = Z, then [E] = the entry in [C] for that row (i.e., 81)

and so on.

ATTEMPT TO RESOLVE:

The following code allowed me to add values for column [E] when [A] ==1,
but I can't figure out how to keep the code going in order to get a value
for column [E] based on all of the numbers in column [A] and the secondary
condition for [B] ([A] goes from 1:48).

df$[E] -


ifelse((df$A == 1)  (df$B == X), df$C[df$A == 1],

ifelse((df$A == 1)  (df$B == Y), df$D[df$A == 1],

ifelse((df$A == 1)  (df$B == Z),  df$C[df$A == 1],

NA


Thank you for any advice you can give.


Libby G

libbymg[at]utexas.edu

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Re: [R] (no subject)

2012-10-04 Thread Sarah Goslee

On Thu, Oct 4, 2012 at 1:40 PM, Chia-Yi Chiu
chia-yi.c...@gse.rutgers.edu wrote:
 Hi

 I would like to learn how the R function hclust deals with ties. It is
 written in Fortran, so I cannot access the code. Thanks!!

Sure you can. Just download the source from r-project.org.



-- 
Sarah Goslee
http://www.functionaldiversity.org

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Re: [R] Creating vegetation distance groups from one column

2012-10-04 Thread Jhope

Hi,

Allow me to recap my question. In plyr I am trying to group distances of
nests to the vegetation into classes that are defined by (0-5m, 6-10m,
11-15m, 16-20m, 21-25m, 26-30m, 31-35m). Each row is a nest and all the
distances to the vegetation is in a column. 

As well I would like to create classes from the HTL starting at (-5-0m,
1-5m, 6-10m, 11-15m, 16-20m, 21-25m, 26m-30m). 
 
I have included a sample of the data through dput() - see below. If that is
too tiresome, I've also included head(data.to.analyze) below that. 

Best, Jean


 dput(head(data.to.analyze, 30))
structure(list(QuadratEvent = c(8.1A, 10.1A, 14.1A, 19.1A, 
23.1A, 24.1A, 25.1A, 29.1A, 30.1A, 31.1A, 32.1A, 
33.1A, 38.1A, 39.1A, 41.1A, 47.1A, 48.1A, 50.1A, 
52.1A, 53.1A, 58.1A, 63.1A, 68.1A, 73.1A, 75.1A, 
79.1A, 29BA, 90.1B, 90.2B, 82.1B), ExDate = c(Nov. 26, 2010, 
Nov. 26, 2010, Nov. 18, 2010, Nov. 22, 2010, Nov. 20, 2010, 
Nov. 28, 2010, Nov. 28, 2010, Nov. 27, 2010, Nov. 20, 2010, 
Nov. 20, 2010, Nov. 26, 2010, Nov. 26, 2010, Nov. 20, 2010, 
Nov. 26, 2010, Nov. 27, 2010, Nov. 27, 2010, Nov. 26, 2010, 
Nov. 27, 2010, Nov. 27, 2010, Nov. 27, 2010, Nov. 27, 2010, 
Nov. 26, 2010, Nov. 26, 2010, Nov. 28, 2010, Nov. 28, 2010, 
Nov. 26, 2010, Nov. 28, 2010, Jan. 1, 2011, Jan. 1, 2011, 
Dec. 23, 2010), Sector = c(14, 13, 20, 79, 34, 34, 
34, 36, 67, 66, 67, 67, 67, 66, 70, 70, 68, 
57, 55, 56, 40, 27, 26, 27, 28, 19, 36, 70, 
70, 47), Quadrat = c(8, 10, 14, 19, 23, 24, 25, 
29, 30, 31, 32, 33, 38, 39, 41, 47, 48, 50, 
52, 53, 58, 63, 68, 73, 75, 79, 29B, 90, 
90, 82), Aeventexhumed = c(A, A, A, A, A, A, 
A, A, A, A, A, A, A, A, A, A, A, A, A, 
A, A, A, A, A, A, A, A, B, B, B), ST0 = c(6L, 
4L, 6L, 0L, 135L, 660L, 321L, 608L, 50L, 5L, 88L, 48L, 7L, 60L, 
0L, 69L, 195L, 68L, 162L, 47L, 595L, 144L, 111L, 478L, 636L, 
136L, 515L, 213L, 0L, 175L), ST1 = c(0L, 0L, 0L, 0L, 23L, 0L, 
0L, 0L, 4L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 0L, 39L, 2L, 1L, 4L, 
9L, 1L, 5L, 25L, 4L, 139L, 0L, 0L, 0L), ST2 = c(1L, 0L, 2L, 0L, 
1L, 2L, 2L, 0L, 153L, 0L, 0L, 0L, 117L, 93L, 3L, 2L, 0L, 0L, 
6L, 2L, 0L, 17L, 1L, 11L, 67L, 0L, 31L, 68L, 0L, 0L), ST3 = c(98L, 
109L, 26L, 0L, 6L, 0L, 0L, 0L, 1L, 3L, 11L, 88L, 3L, 342L, 9L, 
3L, 4L, 72L, 7L, 81L, 0L, 33L, 103L, 0L, 0L, 218L, 162L, 0L, 
74L, 0L), ST4 = c(1L, 1L, 11L, 0L, 0L, 0L, 0L, 0L, 27L, 1L, 0L, 
0L, 7L, 0L, 8L, 6L, 0L, 0L, 5L, 4L, 0L, 38L, 150L, 0L, 0L, 19L, 
5L, 0L, 0L, 0L), Shells = c(72L, 74L, 106L, 102L, 125L, 4L, 15L, 
0L, 31L, 111L, 0L, 0L, 86L, 48L, 21L, 35L, 0L, 0L, 12L, 13L, 
0L, 23L, 20L, 3L, 4L, 115L, 5L, 2L, 0L, 0L), Hatchlings = c(3L, 
4L, 1L, 1L, 48L, 0L, 0L, 0L, 45L, 99L, 0L, 0L, 93L, 9L, 0L, 0L, 
0L, 6L, 2L, 6L, 0L, 9L, 13L, 0L, 0L, 9L, 0L, 0L, 0L, 0L), MaxHatch = c(72L, 
74L, 106L, 102L, 125L, 4L, 15L, 0L, 45L, 111L, 0L, 0L, 93L, 48L, 
21L, 35L, 0L, 6L, 12L, 13L, 0L, 23L, 20L, 3L, 4L, 118L, 5L, 2L, 
0L, 0L), DeadHatch = c(2L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 4L, 0L, 
0L, 0L, 1L, 5L, 0L, 0L, 0L, 0L, 0L, 2L, 0L, 0L, 0L, 0L, 0L, 6L, 
0L, 0L, 0L, 0L), Oldeggs = c(0L, 21L, 165L, 0L, 0L, 36L, 0L, 
0L, 96L, 45L, 5L, 142L, 10L, 81L, 103L, 92L, 0L, 0L, 0L, 0L, 
108L, 0L, 0L, 0L, 93L, 0L, 0L, 0L, 0L, 0L), TotalEggs = c(178L, 
188L, 151L, 102L, 290L, 666L, 338L, 608L, 280L, 120L, 99L, 136L, 
227L, 543L, 42L, 117L, 199L, 185L, 194L, 148L, 599L, 264L, 386L, 
497L, 732L, 495L, 857L, 283L, 74L, 175L), QuadratID = structure(c(107L, 
4L, 22L, 38L, 43L, 44L, 46L, 54L, 57L, 58L, 60L, 61L, 67L, 69L, 
72L, 79L, 80L, 81L, 83L, 85L, 91L, 94L, 97L, 103L, 105L, 106L, 
56L, 109L, NA, 108L), .Label = c(, 1.1, 1.2, 10.1, 10.2, 
100.1, 100.2, 108.1, 108.2, 110.1, 110.2, 114.1, 
114.2, 121.1, 121.2, 127.1, 127.2, 128.1, 128.2, 
135.1, 135.2, 14.1, 15.1, 15.2, 151.1, 151.2, 152.1, 
152.2, 153.1, 153.2, 154.1, 154.2, 155.1, 155.2, 
16.1, 16.2, 18.1, 19.1, 21.1, 21.2, 22.1, 22.2, 
23.1, 24.1, 24.2, 25.1, 25.2, 26.1, 26.2, 27.1, 
27.2, 28.1, 28.2, 29.1, 29.2, 29B, 30.1, 31.1, 
31.2, 32.1, 33.1, 33.2, 35.1, 35.2, 37.1, 37.2, 
38.1, 38.2, 39.1, 40.1, 40.2, 41.1, 42.1, 42.2, 
43.1, 43.2, 46.1, 46.2, 47.1, 48.1, 50.1, 50.2, 
52.1, 52.2, 53.1, 53.2, 54.1, 54.2, 56.1, 56.2, 
58.1, 62.1, 62.2, 63.1, 65.1, 65.2, 68.1, 68.2, 
68.3, 68.4, 69.1, 70.1, 73.1, 74.1, 75.1, 79.1, 
8.1, 82.1, 90.1, 94.1, 94.2), class = factor), Sector = c(14L, 
13L, 20L, 79L, 34L, 34L, 34L, 36L, 67L, 66L, 67L, 67L, 67L, 66L, 
70L, 70L, 68L, 57L, 55L, 56L, 40L, 27L, 26L, 27L, 28L, 19L, 36L, 
70L, NA, 47L), Aeventexhumed = structure(c(2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 3L, NA, 3L), .Label = c(, A, B, 
C), class = factor), Veg = c(9.75, 16.34, 3.15, 5.7, 12.2, 
4.82, 3.3, 1.85, 17.75, 27.13, 0.84, 8.45, 16.42, 20.4, 22.4, 
25, 25.4, 15.1, 13.7, 17.1, NA, 17.3, 24.3, 9.35, 5.4, 15.5, 
1.85, 11, NA, 3), Berm = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, 10, NA, 11), HTL = c(4.66, -2.47, 16, NA, 2.35, 22.85, 
24, 15.7, 9.55, 4.2, 9.65,

Re: [R] (no subject)

2012-10-04 Thread Jeff Newmiller

Please use a relevant subject line.

You can easily download source code for all CRAN packages and read it yourself.
---
Jeff NewmillerThe .   .  Go Live...
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--- 
Sent from my phone. Please excuse my brevity.

Chia-Yi Chiu chia-yi.c...@gse.rutgers.edu wrote:

Hi

I would like to learn how the R function hclust deals with ties. It
is
written in Fortran, so I cannot access the code. Thanks!!

   [[alternative HTML version deleted]]

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[R] geoRglm with factor variable as covariable

2012-10-04 Thread Filoche

Dear R users.

I'm trying to fit  a generalised linear spatial mode using the geoRglm
package. To do so, I'm preparing my data (geodata) as follow:

geoData9093 = as.geodata(data9093, coords.col= 17:18, data.col=15,*
covar.col=16*)

where covar.col is a factor variable (years in this case 90-91-92-93)).

Then I run the model as follow:
/
model.5 = list(cov.pars=c(1,1), cov.model='exponential', beta=1,
family=poisson) 
mcmc.5 = mcmc.control(S.scale = 0.25, n.iter = 3, burn.in=5, thin =
100) #trial error
outmcmc.5 = glsm.mcmc(geoData9093, model= model.5, mcmc.input = mcmc.5) 
mcmcobj.5 = prepare.likfit.glsm(outmcmc.5)   
lik.5 = likfit.glsm(mcmcobj.5, ini.phi = 0.3, fix.nugget.rel = F)/

And the summary of lik.5 is:

likfit.glsm: estimated model parameters:
 beta   sigmasq   phi tausq.rel 
 1.2781  0.5193  0.0977  0.0069 

 likfit.glsm : maximised log-likelihood = 43.62

I'm fairly new to geostatistics, but I thought using a factor variable as
covariable would give me 4 intercepts (beta) as I have 4 levels in my covar.
But looking at the summary, we see that I only have 1 beta which is equal to
1.28. I guess I made mistakes in specifying the model description, but I
can't find where. Any advices would be welcome.

With regards,
Phil





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Re: [R] (no subject)