date:20110405

On Tue, Apr 5, 2011 at 6:12 AM, David Scott d.sc...@auckland.ac.nz wrote:

[...]

 I am not sure where I read it and I can't find it again, but my
 understanding is that expressions using bquote with lattice need to be
 enclosed in as.expression() to work. That is in contrast to what happens in
 base graphics.

 Here is a simple example.

 a - 2
 plot(1:10, a*(1:10), main = bquote(alpha == .(a)))
 require(lattice)
 xyplot(a*(1:10)~ 1:10, main = bquote(alpha == .(a)))
 xyplot(a*(1:10)~ 1:10, main = as.expression(bquote(alpha == .(a

 Which produces:

 a - 2
 plot(1:10, a*(1:10), main = bquote(alpha == .(a)))
 require(lattice)
 Loading required package: lattice
 xyplot(a*(1:10)~ 1:10, main = bquote(alpha == .(a)))
 Error in trellis.skeleton(formula = a * (1:10) ~ 1:10, cond = list(c(1L,  :
  object 'alpha' not found
 xyplot(a*(1:10)~ 1:10, main = as.expression(bquote(alpha == .(a

 Using expression() rather than as.expression() doesn't produce the desired
 affect. Try it yourself.

 As to why this is the case .

Let's see: ?xyplot says

  'main': Typically a character string or expression describing
  the main title to be placed on top of each page. [...]

So, lattice is fairly explicit, by R standards, in requiring 'main' to
be character or expression. On the other hand, ?title says

 The labels passed to 'title' can be character strings or language
 objects (names, calls or expressions), or [...]

so it additionally accepts names and calls.

Now, we have

 a - 2
 foo - bquote(alpha == .(a))

 foo # Looks OK
alpha == 2
 mode(foo) # But
[1] call
 is.expression(foo) # not an expression
[1] FALSE

 is.expression(expression(foo)) ## YES, but
[1] TRUE
 expression(foo) ## not what we want
expression(foo)

 is.expression(as.expression(foo))
[1] TRUE
 as.expression(foo) ## This IS what we want
expression(alpha == 2)

So I submit that lattice is behaving exactly as suggested by its documentation.

Now you would naturally argue that this is hiding behind
technicalities, and if call objects work for plot(), it should work
for lattice as well. But watch this:

 plot(1:10, main = foo) # works perfectly

 arglist - list(1:10, main = foo)
 arglist # Looks like what we want
[[1]]
 [1]  1  2  3  4  5  6  7  8  9 10

$main
alpha == 2

 do.call(plot, arglist)
Error in as.graphicsAnnot(main) : object 'alpha' not found

...which I would say is unexpected behaviour, if not a bug.

The moral of the story is that unevaluated calls are dangerous objects
(try this one out for fun:

foo - bquote(q(.(x)), list(x = no))
do.call(plot, list(1:10, main = foo))

), and carrying them around is not a good idea.

Lattice does use the do.call paradigm quite a bit, and I think it
might be quite difficult to fix it up to handle non-expression
language objects (which will still not fix the type of problem shown
above).

-Deepayan

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Re: [R] system() command in R

2011-04-05 Thread Mike Marchywka













 Date: Tue, 5 Apr 2011 13:37:12 +0530
 From: nandan.a...@gmail.com
 To: rasanpreet.k...@gmail.com
 CC: r-help@r-project.org
 Subject: Re: [R] system() command in R

 On 4 April 2011 16:54, rasanpreet kaur suri wrote:

  Hi all,
  I have a local server insalled on my system and have to start that from
  within my R function.
 
  here is how I start it:
 
  cmd-sh start-server.sh
 
  system(cmd, wait=FALSE)
 
  My function has to start the server and proceed with further steps. The
  server starts but the further steps of the program are not executed.The
  cursor keeps waiting after the server is started.
 
  How r u executing further steps after starting server, meant for server
 from R ??


 i tried removing the wait=FALSE, but it still keeps waiting.
 
  I also tried putting the start-server in a separate function and my further
  script in a separate function and then run them together, but it still
  waits. The transition from the start of server to next step is not
  happening.
 
  Please help. I have been stuck on this for quite some time now.
 
  --

I hadn't done this in R but expect to do so soon.
I just got done with some java code to do something similar
and you can expect in any implementation these things will be 
system dependent. It often helps to have simple test cases
to isolate the problem. Here I made a tst script called foo
that takes a minute or so to exevute and generates some output.

If I type
system(./foo,wait=F) 

the prompt comes back right away but stdout seems to still
go to my console and maybe stdin is not redicrected either
and it could eat your input ( no idea, but this is probably not what you
want).

I did try this that could fix your problem, on debian anyway
it seems to work,

system(nohup ./foo )

you can man nohup for details. 














  Rasanpreet Kaur
 
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 --
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 http://aknandan.co.nr

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[R] how to label customized y axis when using lattice parallel parameter common.scale=TRUE

2011-04-05 Thread Pengcheng Yang


Dear all,

When I use parallel function in lattice package, I want to label the 
y-axis with customized numbers. Like this:


parallel(~iris[1:4] | Species, iris,horiz=FALSE,common.scale=TRUE,
scales=list(y=list(at=c(0,2,3

But only Min label in the y-axis, nothing happened. Could anyone help me?

Thanks.
Regards,

Pengcheng Yang

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[R] Value between which elements of a vector?

2011-04-05 Thread beatleb

Dear R-useRs,

I am looking for a why to perform the following:

specialweeks-c(0,2,5,12,18,19,20)
weeks-c(1:30)

Now I would like that for every week it is even between which elements of
vector special weeks it is.
For weeks after 20, the value NA or 20, or even 20-30is fine.

Thus for 
week 1: 0-2
week 2: 2-5
week 3: 2-5
week 4: 2-5
week 5: 5-12 
ect 

It is not relevant if those intervals are captured in a matrix, in a vector
or whatever. 

I hope that you can help me!

With best regards,

Brenda Grondman



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[R] Time series example in Koop

2011-04-05 Thread Ravi Kulkarni

I am trying to reproduce the output of a time series example in Koop's book
Analysis of Financial Data. Koop does the example in Excel and I used the
ts function followed by the lm function. 
I am unable to get the exact coefficients that Koop gives - my coefficients
are slightly different.
After loading the data file and attaching the frame, my code reads:

 y = ts(m.cap)
 x = ts(oil.price)
 d = ts.union(y,x,x1=lag(x,-1),x2=lag(x,-2),x3=lag(x,-3),x4=lag(x,-4))
 mod1 = lm(y~x+x1+x2+x3+x4, data=d)
 summary(mod1)

Koop gives an intercept of 92001.51, while the code above gives 91173.32.
The other coefficients are also  slightly off.

This is the example in Table 8.3 of Koop. I also attach a plain text version
of the tab separated file badnews.txt.
http://r.789695.n4.nabble.com/file/n3427897/badnews.txt badnews.txt 

Any light on why I do not get Koop's coefficients is most welcome...

Ravi

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[R] Confidence interval for the difference between proportions - method used in prop.test()

2011-04-05 Thread Stefanie Von Felten

Hello,

Does anyone know which method from Newcombe (1998)* is implemented in prop.test 
for comparing two proportions?
I would guess it is the method based on the Wilson score (for single 
proportion), with and without continuity correction  for prop.test(..., 
correct=FALSE) and prop.test(..., correct=TRUE). These methods would correspond 
to no. 10 and 11 tested in Newcombe, respectively. Can someone confirm this? If 
not, which other methods are implemented by prop.test?

* Newcombe R.G. (1998) Two-Sided Confidence Intervals for the Single 
Proportion: Comparison of Seven Methods.  Statistics in Medicine  *17*, 857-872.

There is also the function ci.pd() from the R-package Epi, which should 
implement method no. 10 from Newcombe. However, prop.test(..., correct=FALSE) 
and ci.pd do not give the same result if I do the following:

successes - c(21, 41)
total - c(345, 345)
prop.test(successes, total, correct=FALSE)
library(Epi)
ci.pd(matrix(c(successes, total-successes),ncol=2, byrow=TRUE))

Can someone explain why?

Best wishes
Stefanie von Felten


Stefanie von Felten, PhD
Statistician
Clinical Trial Unit, CTU
University Hospital Basel
Schanzenstrasse 55
CH-4031 Basel

Phone: ++41(0)61 556 54 98

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Re: [R] lattice xscale.components: different ticks on top/bottom axis

On Sat, Apr 2, 2011 at 1:29 AM,  boris.vasil...@forces.gc.ca wrote:

 On Fri, Mar 11, 2011 at 12:28 AM,
 boris.vasil...@forces.gc.ca wrote:
  Good afternoon,
 
  I am trying to create a plot where the bottom and top axes have the
  same scale but different tick marks.  I tried user-defined
  xscale.component function but it does not produce desired results.
  Can anybody suggest where my use of xscale.component
  function is incorrect?
 
  For example, the code below tries to create a plot where horizontal
  axes limits are c(0,10), top axis has ticks at odd integers, and
  bottom axis has ticks at even integers.
 
  library(lattice)
 
  df - data.frame(x=1:10,y=1:10)
 
  xscale.components.A - function(...,user.value=NULL) {
   # get default axes definition list; print user.value
   ans - xscale.components.default(...)
   print(user.value)
 
   # start with the same definition of bottom and top axes
   ans$top - ans$bottom
 
   # - bottom labels
   ans$bottom$labels$at - seq(0,10,by=2)
   ans$bottom$labels$labels - paste(B,seq(0,10,by=2),sep=-)
 
   # - top labels
   ans$top$labels$at - seq(1,9,by=2)
   ans$top$labels$labels - paste(T,seq(1,9,by=2),sep=-)
 
   # return axes definition list
   return(ans)
  }
 
  oltc - xyplot(y~x,data=df,
 
  scales=list(x=list(limits=c(0,10),at=0:10,alternating=3)),
                xscale.components=xscale.components.A,
                user.value=1)
  print(oltc)
 
  The code generates a figure with incorrectly placed bottom and top
  labels.  Bottom labels B-0, B-2, ... are at 0, 1, ... and top
  labels T-1, T-3, ... are at 0, 1, ...  When axis-function runs out
  of labels, it replaces labels with NA.
 
  It appears that lattice uses top$ticks$at to place labels and
  top$labels$labels for labels.  Is there a way to override this
  behaviour (other than to expand the labels$labels vector to be as
  long as ticks$at vector and set necessary elements to )?

 Well, top|bottom$ticks$at is used to place the ticks, and
 $labels$at is used to place the labels. They should typically
 be the same, but you have changed one and not the other.
 Everything seems to work if you set $ticks$at to the same
 values as $labels$at:


     ##  - bottom labels
 +   ans$bottom$ticks$at - seq(0,10,by=2)
     ans$bottom$labels$at - seq(0,10,by=2)
     ans$bottom$labels$labels - paste(B,seq(0,10,by=2),sep=-)

     ##  - top labels
 +   ans$top$ticks$at - seq(1,9,by=2)
     ans$top$labels$at - seq(1,9,by=2)
     ans$top$labels$labels - paste(T,seq(1,9,by=2),sep=-)


  Also, can user-parameter be passed into xscale.components()
  function? (For example, locations and labels of ticks on the top
  axis).  In the  code above, print(user.value) returns NULL even
  though in the xyplot() call user.value is 1.

 No. Unrecognized arguments are passed to the panel function
 only, not to any other function. However, you can always
 define an inline
 function:

 oltc - xyplot(y~x,data=df,
                scales=list(x=list(limits=c(0,10), at = 0:10,
                            alternating=3)),
                xscale.components = function(...)
                            xscale.components.A(..., user.value=1))

 Hope that helps (and sorry for the late reply).

 -Deepayan


 Deepyan,

 Thank you very much for your reply.  It makes things a bit clearer.

 It other words in the list prepared by xscale.components(), vectors
 top|bottom$ticks$at and top|bottom$labels$at must be the same.
 If only every second tick is to be labelled then every second label
 should be set explicitly to empty strings:

Now when you put it that way, the current behaviour does seem wrong (I
didn't read your original post carefully enough). I guess this was one
of the not-yet-implemented things mentioned in the Details section of
?xscale.components.default.

I have added support for different ticks$at and labels$at in the SVN
sources in r-forge. You can test it from there (your original code
works as expected). I won't make a new release on CRAN until after R
2.13 is released (we are almost in code freeze now).

-Deepayan

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Re: [R] Value between which elements of a vector?

2011-04-05 Thread andrija djurovic

Hi:

try this and have a look at ?cut just to change the lables:

data.frame(
weeks=sprintf('%s %d','week',1:30),
specialweeks=cut(1:30,c(0,2,5,12,18,19,20),right = FALSE))

Andrija

On Tue, Apr 5, 2011 at 1:20 PM, beatleb rhelpfo...@gmail.com wrote:

 Dear R-useRs,

 I am looking for a why to perform the following:

 specialweeks-c(0,2,5,12,18,19,20)
 weeks-c(1:30)

 Now I would like that for every week it is even between which elements of
 vector special weeks it is.
 For weeks after 20, the value NA or 20, or even 20-30is fine.

 Thus for
 week 1: 0-2
 week 2: 2-5
 week 3: 2-5
 week 4: 2-5
 week 5: 5-12
 ect

 It is not relevant if those intervals are captured in a matrix, in a vector
 or whatever.

 I hope that you can help me!

 With best regards,

 Brenda Grondman



 --
 View this message in context:
 http://r.789695.n4.nabble.com/Value-between-which-elements-of-a-vector-tp3427751p3427751.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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Re: [R] how to label customized y axis when using lattice parallel parameter common.scale=TRUE

On Tue, Apr 5, 2011 at 4:39 PM, Pengcheng Yang peng...@gmail.com wrote:
 Dear all,

 When I use parallel function in lattice package, I want to label the y-axis
 with customized numbers. Like this:

 parallel(~iris[1:4] | Species, iris,horiz=FALSE,common.scale=TRUE,
    scales=list(y=list(at=c(0,2,3

Parallel does not directly support that, and will insist on scaling
the data. However, you can control the scaling (using 'lower' and
'upper'), and override a couple of other arguments to get what you
want:

parallel(~iris[1:4] | Species, iris,
 xlim = extendrange(range(iris[1:4])),
 scales = list(x = list(at = NULL, labels = NULL)),
 lower = 0, upper = 1)

This is for horizontal.axis = TRUE, adjust accordingly for FALSE.

-Deepayan


 But only Min label in the y-axis, nothing happened. Could anyone help me?

 Thanks.
 Regards,

 Pengcheng Yang

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Re: [R] Time series example in Koop

2011-04-05 Thread Gabor Grothendieck

On Tue, Apr 5, 2011 at 8:35 AM, Ravi Kulkarni ravi.k...@gmail.com wrote:
 I am trying to reproduce the output of a time series example in Koop's book
 Analysis of Financial Data. Koop does the example in Excel and I used the
 ts function followed by the lm function.
 I am unable to get the exact coefficients that Koop gives - my coefficients
 are slightly different.
 After loading the data file and attaching the frame, my code reads:

 y = ts(m.cap)
 x = ts(oil.price)
 d = ts.union(y,x,x1=lag(x,-1),x2=lag(x,-2),x3=lag(x,-3),x4=lag(x,-4))
 mod1 = lm(y~x+x1+x2+x3+x4, data=d)
 summary(mod1)

 Koop gives an intercept of 92001.51, while the code above gives 91173.32.
 The other coefficients are also  slightly off.

 This is the example in Table 8.3 of Koop. I also attach a plain text version
 of the tab separated file badnews.txt.
 http://r.789695.n4.nabble.com/file/n3427897/badnews.txt badnews.txt

 Any light on why I do not get Koop's coefficients is most welcome...


It looks like he erroneously left out the first point.

 URL - http://r.789695.n4.nabble.com/file/n3427897/badnews.txt;
 BAD - read.table(URL, header = TRUE)
 library(dyn)
 dyn$lm(m.cap ~ lag(oil.price, -(0:4)), as.zoo(BAD))

Call:
lm(formula = dyn(m.cap ~ lag(oil.price, -(0:4))), data = as.zoo(BAD))

Coefficients:
(Intercept)  lag(oil.price, -(0:4))1  lag(oil.price, -(0:4))2
   91173.32  -131.99  -449.86
lag(oil.price, -(0:4))3  lag(oil.price, -(0:4))4  lag(oil.price, -(0:4))5
-422.52  -187.10   -27.77


 # without first point
 dyn$lm(m.cap ~ lag(oil.price, -(0:4)), tail(as.zoo(BAD), -1))

Call:
lm(formula = dyn(m.cap ~ lag(oil.price, -(0:4))), data = tail(as.zoo(BAD),
-1))

Coefficients:
(Intercept)  lag(oil.price, -(0:4))1  lag(oil.price, -(0:4))2
92001.5   -145.0   -462.1
lag(oil.price, -(0:4))3  lag(oil.price, -(0:4))4  lag(oil.price, -(0:4))5
 -424.5   -199.5-36.9




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

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Re: [R] Confidence interval for the difference between proportions - method used in prop.test()

Hi Stefanie,

Just to be clear, we are talking about differences in the third or
lower decimal place (at least with R version 2.13.0 alpha (2011-03-17
r54849), Epi_1.1.20).  This strikes me as small enough that both
functions may be implementing the same method, but maybe slightly
different ways of going about it?

If you are really concerned and need to know *exactly*, look at the
source code for both functions.  In case you did not know, if you type
the function name at the console with parentheses or any arguments,
just like:

 prop.test
 ci.pd

it will show the actual function code.  It looks to me like both of
them are implemented purely in R, and without even calling any other
complex functions (at least based on a quick glance through).  This
means if you have the Newscomb text, you should be able to sit down
and go through the code step by step comparing it.

Cheers,

Josh

FYI, you can use a matrix with prop.test, and then its transpose for ci.pd.
##
mymat - cbind(Successes = c(21, 41), Failures = c(345, 345) - c(21, 41))
require(Epi)
results - list(prop.test(mymat, correct=FALSE), ci.pd(t(mymat)))
results[[1]][[conf.int]] - results[[2]][6:7]

On Tue, Apr 5, 2011 at 3:38 AM, Stefanie Von Felten svonfel...@uhbs.ch wrote:
 Hello,

 Does anyone know which method from Newcombe (1998)* is implemented in 
 prop.test for comparing two proportions?
 I would guess it is the method based on the Wilson score (for single 
 proportion), with and without continuity correction  for prop.test(..., 
 correct=FALSE) and prop.test(..., correct=TRUE). These methods would 
 correspond to no. 10 and 11 tested in Newcombe, respectively. Can someone 
 confirm this? If not, which other methods are implemented by prop.test?

 * Newcombe R.G. (1998) Two-Sided Confidence Intervals for the Single 
 Proportion: Comparison of Seven Methods.  Statistics in Medicine  *17*, 
 857-872.

 There is also the function ci.pd() from the R-package Epi, which should 
 implement method no. 10 from Newcombe. However, prop.test(..., correct=FALSE) 
 and ci.pd do not give the same result if I do the following:

 successes - c(21, 41)
 total - c(345, 345)
 prop.test(successes, total, correct=FALSE)
 library(Epi)
 ci.pd(matrix(c(successes, total-successes),ncol=2, byrow=TRUE))

 Can someone explain why?

 Best wishes
 Stefanie von Felten


 Stefanie von Felten, PhD
 Statistician
 Clinical Trial Unit, CTU
 University Hospital Basel
 Schanzenstrasse 55
 CH-4031 Basel

 Phone: ++41(0)61 556 54 98

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




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] RODBC excel - need to preserve (or extract) numeric column names



On Apr 5, 2011, at 4:53 AM, Enrico Schumann wrote:

At least for Excel 2003 on my computer (Win XP) I can persuade  
Excel to

treat cells like text by prepending a ' to the entry (eg, '1000). Then
sqlFetch/RODBC should import these cells as character. [But a number  
would

not be valid column name for a data.frame, and you may run into other
trouble. See ?make.names]



If the problem is that Excel is failing to treat cells as text, then  
you could also try selecting the cells, then using the Format/Cells  
panel to specify Text rather than General. But as I suggested  
before, I suspect the problem is you effort to defeat the usual  
checking for valid R names.


--
David.



regards,
enrico




-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Im Auftrag von Folkes, Michael
Gesendet: Dienstag, 5. April 2011 00:05
An: r-help@r-project.org
Betreff: [R] RODBC excel - need to preserve (or extract)
numeric column names

I'm using RODBC to read an excel file (not mine!).  But I'm
struggling to find a way to preserve the column names that
have a numeric value.  sqlFetch() drops the value and calls
them f1, f2, f3,... (ie field number).  this is a different
approach from read.csv, which will append V prior to the
numeric column name.  sqlFetch isn't so helpful.

Is there a way to get the first line of data from the excel
file and place it in a vector?  Perhaps I can use that method
and rename the dataframe column names later?

thanks!
Michael

___
Michael Folkes
Salmon Stock Assessment
Canadian Dept. of Fisheries  Oceans
Pacific Biological Station
3190 Hammond Bay Rd.
Nanaimo, B.C., Canada
V9T-6N7
Ph (250) 756-7264 Fax (250) 756-7053
michael.fol...@dfo-mpo.gc.ca mailto:michael.fol...@dfo-mpo.gc.ca

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David Winsemius, MD
West Hartford, CT

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Re: [R] system() command in R

2011-04-05 Thread Jeff Newmiller

The ampersand is a good idea, but nohup is best avoided in scripts. Rather the 
server itself should handle the transition to daemon status to avoid ending up 
with many duplicate server processes running.
---
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.

Mike Marchywka marchy...@hotmail.com wrote:

_
 Date: Tue, 5 Apr 2011 13:37:12 +0530  From: nandan.a...@gmail.com  To: 
 rasanpreet.k...@gmail.com  CC: r-help@r-project.org  Subject: Re: [R] 
 system() command in R   On 4 April 2011 16:54, rasanpreet kaur suri wrote: 
Hi all,   I have a local server insalled on my system and have to 
 start that from   within my R function. here is how I start it:   
   cmd-sh start-server.sh system(cmd, wait=FALSE) My 
 function has to start the server and proceed with further steps. The   
 server starts but the further steps of the program are not executed.The   
 cursor keeps waiting after the server is started. How r u executing 
 further steps after starting server, meant for server  from R ??i 
 tried removing the wait=FALSE, but it still keeps waiting. I also 
 tried putting the start-server in a separate function and my further   
 script in a separate function and then run them together, but it still   
 waits. The transitio
 n from
the start of server to next step is not   happening. Please help. I 
have been stuck on this for quite some time now. -- I hadn't done this 
in R but expect to do so soon. I just got done with some java code to do 
something similar and you can expect in any implementation these things will be 
system dependent. It often helps to have simple test cases to isolate the 
problem. Here I made a tst script called foo that takes a minute or so to 
exevute and generates some output. If I type system(./foo,wait=F) the prompt 
comes back right away but stdout seems to still go to my console and maybe 
stdin is not redicrected either and it could eat your input ( no idea, but this 
is probably not what you want). I did try this that could fix your problem, on 
debian anyway it seems to work, system(nohup ./foo ) you can man nohup for 
details.   Rasanpreet Kaur [[alternative HTML version deleted]]
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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.   --  
  Amar Kumar Nandan  Karnataka, India, 560100  http://aknandan.co.nr   
  [[alternative HTML version deleted]]  
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Re: [R] Structural equation modeling in R(lavaan,sem)

2011-04-05 Thread yrosseel


On 04/04/2011 07:14 PM, jouba wrote:



Thanks you for your response
For lavaan package can i have more information about this example you have 
applied in the section 7
the meanings  of The variables (c1,c2,c3,c4, i ,s ,x1,x2)
I think i have need more information to learn  more about how  able to apply 
growth model  in my data (longitudianl data)


In the example, c1-c4 are time-varying covariates, i and s are the 
random intercept and slope respectively, and x1 and x2 are two exogenous 
covariates influencing the intercept and slope.


Please note: the lavaanIntroduction document is hardly useful to _learn_ 
about growth models (or any SEM model for that matter). It only explains 
how to fit them using the lavaan package. To learn about growth models, 
you may want to read any one of the books below:


Latent Curve Models: A Structural Equation Perspective (Wiley Series in 
Probability and Statistics) by Kenneth A. Bollen and Patrick J. Curran 
(Hardcover - Dec 23, 2005)


Latent Growth Curve Modeling (Quantitative Applications in the Social 
Sciences) by Dr. Kristopher J. Preacher, Aaron Lee Wichman, Robert 
Charles MacCallum and Dr. Nancy E. Briggs (Paperback - Jun 27, 2008)


An Introduction to Latent Variable Growth Curve Modeling: Concepts, 
Issues, and Applications (Quantitative Methodology) (Quantitative 
Methodology Series) by Terry E. Duncan, Susan C. Duncan and Lisa A. 
Strycker (Paperback - May 23, 2006)


Yves.

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Re: [R] Confidence interval for the difference between proportions - method used in prop.test()

Dear Steffi,

On Tue, Apr 5, 2011 at 7:26 AM, Stefanie Von Felten svonfel...@uhbs.ch wrote:
 Dear Josh,

 Thanks for your help!

 Does your answer mean, that you agree the two methods should do the same,
 and what I was guessing, despite the small differences?

That would be my guess, but I have not actually read the reference in
discussion.  Still, the documentation for prop.test uses the same
Newcombe reference as ci.pd, so if method 10 is the clear winner, it
seems reasonable that prop.test is an implementation of it.


 What I prefer about ci.pd is, that the help clearly says which method is
 implemented, which is not the case for prop.test. But I do not know who has
 programmed the function.

Then for reporting, use ci.pd, and say it is method 10 from Newcombe.
You can always check the results with prop.test() which is part of R
core so you can be pretty confident whatever it does, it does it
correctly (and will be updated if necessary with future releases of
R).

Sincerely,

Josh

 Best wishes
 Steffi


 Stefanie von Felten, PhD
 Statistician
 Clinical Trial Unit, CTU
 University Hospital Basel
 Schanzenstrasse 55
 CH-4031 Basel

 Phone: ++41(0)61 556 54 98
 Joshua Wiley jwiley.ps...@gmail.com 05.04.11 15.59 Uhr 
 Hi Stefanie,

 Just to be clear, we are talking about differences in the third or
 lower decimal place (at least with R version 2.13.0 alpha (2011-03-17
 r54849), Epi_1.1.20). This strikes me as small enough that both
 functions may be implementing the same method, but maybe slightly
 different ways of going about it?

 If you are really concerned and need to know *exactly*, look at the
 source code for both functions. In case you did not know, if you type
 the function name at the console with parentheses or any arguments,
 just like:

 prop.test
 ci.pd

 it will show the actual function code. It looks to me like both of
 them are implemented purely in R, and without even calling any other
 complex functions (at least based on a quick glance through). This
 means if you have the Newscomb text, you should be able to sit down
 and go through the code step by step comparing it.

 Cheers,

 Josh

 FYI, you can use a matrix with prop.test, and then its transpose for ci.pd.
 ##
 mymat - cbind(Successes = c(21, 41), Failures = c(345, 345) - c(21, 41))
 require(Epi)
 results - list(prop.test(mymat, correct=FALSE), ci.pd(t(mymat)))
 results[[1]][[conf.int]] - results[[2]][6:7]

 On Tue, Apr 5, 2011 at 3:38 AM, Stefanie Von Felten svonfel...@uhbs.ch
 wrote:
 Hello,

 Does anyone know which method from Newcombe (1998)* is implemented in
 prop.test for comparing two proportions?
 I would guess it is the method based on the Wilson score (for single
 proportion), with and without continuity correction for prop.test(...,
 correct=FALSE) and prop.test(..., correct=TRUE). These methods would
 correspond to no. 10 and 11 tested in Newcombe, respectively. Can someone
 confirm this? If not, which other methods are implemented by prop.test?

 * Newcombe R.G. (1998) Two-Sided Confidence Intervals for the Single
 Proportion: Comparison of Seven Methods. Statistics in Medicine *17*,
 857-872.

 There is also the function ci.pd() from the R-package Epi, which should
 implement method no. 10 from Newcombe. However, prop.test(...,
 correct=FALSE) and ci.pd do not give the same result if I do the following:

 successes - c(21, 41)
 total - c(345, 345)
 prop.test(successes, total, correct=FALSE)
 library(Epi)
 ci.pd(matrix(c(successes, total-successes),ncol=2, byrow=TRUE))

 Can someone explain why?

 Best wishes
 Stefanie von Felten


 Stefanie von Felten, PhD
 Statistician
 Clinical Trial Unit, CTU
 University Hospital Basel
 Schanzenstrasse 55
 CH-4031 Basel

 Phone: ++41(0)61 556 54 98

 [[alternative HTML version deleted]]

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




 --
 Joshua Wiley
 Ph.D. Student, Health Psychology
 University of California, Los Angeles
 http://www.joshuawiley.com/




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] Time series example in Koop

2011-04-05 Thread Mike Marchywka

 Date: Tue, 5 Apr 2011 07:35:04 -0500
 From: ravi.k...@gmail.com
 To: r-help@r-project.org
 Subject: [R] Time series example in Koop

 I am trying to reproduce the output of a time series example in Koop's book
 Analysis of Financial Data. Koop does the example in Excel and I used the
 ts function followed by the lm function.
 I am unable to get the exact coefficients that Koop gives - my coefficients
 are slightly different.

 After loading the data file and attaching the frame, my code reads:

  y = ts(m.cap)
  x = ts(oil.price)
  d = ts.union(y,x,x1=lag(x,-1),x2=lag(x,-2),x3=lag(x,-3),x4=lag(x,-4))
  mod1 = lm(y~x+x1+x2+x3+x4, data=d)
  summary(mod1)

 Koop gives an intercept of 92001.51, while the code above gives 91173.32.
 The other coefficients are also slightly off.

The differences here seem to be of order 1 percent. You could suspect
a number of things, including the published data file being published
to less precision than that used in the book numbers(also look at number
of points and see if any were added or dropped etc ). However, you may want
to judge these based on what they do to your error which they
presumably are both supposed to minimize but the calculation of which could
be subject to various roundoff errors etc. Unless minimization is done
analytically, it is of course subject to limitations of convergence or iteration
count. Plotting both fits over the data and looking at residuals may help too.
Depending on what you are really trying to do, you may want to change
your error calculation etc. 

Details of numerical results often depend on details of implementation.
This is why stats packages that are not open
source have limitations in applicability. With real models of course
things get even more confusing. 
( take a look at credit rating agencies results for example LOL).

 This is the example in Table 8.3 of Koop. I also attach a plain text version
 of the tab separated file badnews.txt.
 http://r.789695.n4.nabble.com/file/n3427897/badnews.txt badnews.txt

 Any light on why I do not get Koop's coefficients is most welcome...

 Ravi

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[R] Help to check data before putting it in a database

2011-04-05 Thread Ulisses.Camargo

The example scene:

I have a database with stats about each goal made by my soccer team. This
database (a data frame in R) is organized in lines (goals) with a set of
columns containing data about these goals (player name, tactic position,
etc). For now, this database will be called data.frame1.

What I need is to feed this data.frame1 with new information about my team
goals. I will call this new information data.frame2. This set of new goals
is organized in the same way as in data.frame1 (equal numbers of cols).

Where help is needed:

I need help in finding a way to check the player-name column in
data.frame2 before feeding data.frame1 with it. What I need is a way to
verify the name of the player on each line of data.frame2 with the names
of players that already exist on a col in data.frame1. Moreover, I need R
to make two main things:

First, the lines of “data.frame2” with player names that already exists in
“data.frame1” must be added to “data.frame1”.

Second: lines of “data.frame2” with player names that does not exist on
“data.frame1” must be listed in an output to be manually checked and
corrected.
After this verification, corrected lines and new-player-names lines must be
incorporated in data.frame1.

What I want is to guarantee that will be no lines with wrong player names in
my database.
At the same time, my script must permit new information to be added (new
player names).

Is there somebody who could help me with this?

Thanks for your attention

Best wishes
Ulisses

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Re: [R] Support Counting

2011-04-05 Thread psombe

well im using the arules package and i'm trying to use the support command.
my data is read form a file using the read.transactions command and a line
of data looks something like this. there are aboutt 88000 rows and 16000
different items
 inspect(dset[3])
  items
1 {33, 
34, 
35} 
 inspect(dset[1])
  items
1 {0, 1,  10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2,  20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 3, 4,5, 6, 7,  8,  9}  

So in order to use support i have to make an object of class itemsets and
im kind of struggling with the new command.
I made an object of class itemsets by first creating a presence/absence
matrix and with something like 16000 items this is really sort of tedious. I
wonder if there is a better way.

//Currently im doing this

avec = array(dim=400) //dim is till the max number of the item im concerned
with
avec[1:400] = 0
avec[27] = 1
avec[63] = 1 //and do on for all the items i want

amat = matrix(data = avec,ncol = 400)
aset = as(amat,transactions) //coercing the matrix as a transactions class

then say my data is dat i can use

support(aset,dat)
[1] 0.001406470


There has to be a better way
Thanks once again

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http://r.789695.n4.nabble.com/Support-Counting-tp3424730p3428062.html
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Re: [R] how to label customized y axis when using lattice parallel parameter common.scale=TRUE

2011-04-05 Thread Pengcheng Yang


Thanks Deepayan, It works!

On 2011-4-5 21:42, Deepayan Sarkar wrote:

On Tue, Apr 5, 2011 at 4:39 PM, Pengcheng Yangpeng...@gmail.com  wrote:

Dear all,

When I use parallel function in lattice package, I want to label the y-axis
with customized numbers. Like this:

parallel(~iris[1:4] | Species, iris,horiz=FALSE,common.scale=TRUE,
scales=list(y=list(at=c(0,2,3

Parallel does not directly support that, and will insist on scaling
the data. However, you can control the scaling (using 'lower' and
'upper'), and override a couple of other arguments to get what you
want:

parallel(~iris[1:4] | Species, iris,
  xlim = extendrange(range(iris[1:4])),
  scales = list(x = list(at = NULL, labels = NULL)),
  lower = 0, upper = 1)

This is for horizontal.axis = TRUE, adjust accordingly for FALSE.

-Deepayan


But only Min label in the y-axis, nothing happened. Could anyone help me?

Thanks.
Regards,

Pengcheng Yang

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[R] Antw: Re: Confidence interval for the difference between proportions - method used in prop.test()

2011-04-05 Thread Stefanie Von Felten

Dear Josh,

Thanks for your help!

Does your answer mean, that you agree the two methods should do the same, and 
what I was guessing, despite the small differences?

What I prefer about ci.pd is, that the help clearly says which method is 
implemented, which is not the case for prop.test. But I do not know who has 
programmed the function.

Best wishes
Steffi

Stefanie von Felten, PhD
Statistician
Clinical Trial Unit, CTU
University Hospital Basel
Schanzenstrasse 55
CH-4031 Basel

Phone: ++41(0)61 556 54 98
 Joshua Wiley  05.04.11 15.59 Uhr 
Hi Stefanie,

Just to be clear, we are talking about differences in the third or
lower decimal place (at least with R version 2.13.0 alpha (2011-03-17
r54849), Epi_1.1.20).  This strikes me as small enough that both
functions may be implementing the same method, but maybe slightly
different ways of going about it?

If you are really concerned and need to know *exactly*, look at the
source code for both functions.  In case you did not know, if you type
the function name at the console with parentheses or any arguments,
just like:

 prop.test
 ci.pd

it will show the actual function code.  It looks to me like both of
them are implemented purely in R, and without even calling any other
complex functions (at least based on a quick glance through).  This
means if you have the Newscomb text, you should be able to sit down
and go through the code step by step comparing it.

Cheers,

Josh

FYI, you can use a matrix with prop.test, and then its transpose for ci.pd.
##
mymat - cbind(Successes = c(21, 41), Failures = c(345, 345) - c(21, 41))
require(Epi)
results - list(prop.test(mymat, correct=FALSE), ci.pd(t(mymat)))
results[[1]][[conf.int]] - results[[2]][6:7]

On Tue, Apr 5, 2011 at 3:38 AM, Stefanie Von Felten  wrote:
 Hello,

 Does anyone know which method from Newcombe (1998)* is implemented in 
 prop.test for comparing two proportions?
 I would guess it is the method based on the Wilson score (for single 
 proportion), with and without continuity correction  for prop.test(..., 
 correct=FALSE) and prop.test(..., correct=TRUE). These methods would 
 correspond to no. 10 and 11 tested in Newcombe, respectively. Can someone 
 confirm this? If not, which other methods are implemented by prop.test?

 * Newcombe R.G. (1998) Two-Sided Confidence Intervals for the Single 
 Proportion: Comparison of Seven Methods.  Statistics in Medicine  *17*, 
 857-872.

 There is also the function ci.pd() from the R-package Epi, which should 
 implement method no. 10 from Newcombe. However, prop.test(..., correct=FALSE) 
 and ci.pd do not give the same result if I do the following:

 successes - c(21, 41)
 total - c(345, 345)
 prop.test(successes, total, correct=FALSE)
 library(Epi)
 ci.pd(matrix(c(successes, total-successes),ncol=2, byrow=TRUE))

 Can someone explain why?

 Best wishes
 Stefanie von Felten


 Stefanie von Felten, PhD
 Statistician
 Clinical Trial Unit, CTU
 University Hospital Basel
 Schanzenstrasse 55
 CH-4031 Basel

 Phone: ++41(0)61 556 54 98

[[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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/


[[alternative HTML version deleted]]

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

Re: [R] Help to check data before putting it in a database

Hi Ulisses,

Look at the functions ?match and ?rbind

If you do not want to do it by hand, you can make a little function as below.

HTH,

Josh

d1 - data.frame(goals = 4:1, players = LETTERS[1:4])
d2 - data.frame(goals = c(1, 3, 2, 5), players = LETTERS[3:6])

f - function(old, new, check) {
  index - new[, check] %in% old[, check]
  dat - rbind(old, new[index, ])
  tocheck - new[!index, ]
  list(merged = dat, tocheck = tocheck)
}

dmerged - f(d1, d2, players)
## check tocheck and once it is correct
dfinal - do.call(rbind, dmerged)

On Tue, Apr 5, 2011 at 8:06 AM, Ulisses.Camargo
moliterno.cama...@gmail.com wrote:
 The example scene:

 I have a database with stats about each goal made by my soccer team. This
 database (a data frame in R) is organized in lines (goals) with a set of
 columns containing data about these goals (player name, tactic position,
 etc). For now, this database will be called data.frame1.

 What I need is to feed this data.frame1 with new information about my team
 goals. I will call this new information data.frame2. This set of new goals
 is organized in the same way as in data.frame1 (equal numbers of cols).

 Where help is needed:

 I need help in finding a way to check the player-name column in
 data.frame2 before feeding data.frame1 with it. What I need is a way to
 verify the name of the player on each line of data.frame2 with the names
 of players that already exist on a col in data.frame1. Moreover, I need R
 to make two main things:

 First, the lines of “data.frame2” with player names that already exists in
 “data.frame1” must be added to “data.frame1”.

 Second: lines of “data.frame2” with player names that does not exist on
 “data.frame1” must be listed in an output to be manually checked and
 corrected.
 After this verification, corrected lines and new-player-names lines must be
 incorporated in data.frame1.

 What I want is to guarantee that will be no lines with wrong player names in
 my database.
 At the same time, my script must permit new information to be added (new
 player names).

 Is there somebody who could help me with this?

 Thanks for your attention

 Best wishes
 Ulisses

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-to-check-data-before-putting-it-in-a-database-tp3428318p3428318.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.




-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
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Re: [R] Help to check data before putting it in a database

2011-04-05 Thread Jeff Newmiller

I would recommend using R to check your input and identify bad input and to 
only load data that passes validation. Then go back to some other tool for 
editing the data and save/reload/reverify the edited data. The merge command 
with the all.x argument and is.na() can be used, or the ! and %in% logical 
operators can be used, to find non-matching values.

If you are determined to modify the data in R, then you probably need the tk 
library, the use of which is not really a topic for this forum.
---
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/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
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Sent from my phone. Please excuse my brevity.

Ulisses.Camargo moliterno.cama...@gmail.com wrote:

The example scene: I have a database with stats about each goal made by my 
soccer team. This database (a data frame in R) is organized in lines (goals) 
with a set of columns containing data about these goals (player name, tactic 
position, etc). For now, this database will be called data.frame1. What I 
need is to feed this data.frame1 with new information about my team goals. I 
will call this new information data.frame2. This set of new goals is 
organized in the same way as in data.frame1 (equal numbers of cols). Where 
help is needed: I need help in finding a way to check the player-name column in 
data.frame2 before feeding data.frame1 with it. What I need is a way to 
verify the name of the player on each line of data.frame2 with the names of 
players that already exist on a col in data.frame1. Moreover, I need R to 
make two main things: First, the lines of âdata.frame2â with player names 
that already exists in âdata.frame1â must be added to âdata.fram
 e1â.
Second: lines of âdata.frame2â with player names that does not exist on 
âdata.frame1â must be listed in an output to be manually checked and 
corrected. After this verification, corrected lines and new-player-names lines 
must be incorporated in data.frame1. What I want is to guarantee that will be 
no lines with wrong player names in my database. At the same time, my script 
must permit new information to be added (new player names). Is there somebody 
who could help me with this? Thanks for your attention Best wishes Ulisses -- 
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Re: [R] Help to check data before putting it in a database

2011-04-05 Thread Ulisses.Camargo

Hello Josh,
It works. Thanks a lot!!
I have now the basic tools to advance!
Best!
Ulisses

2011/4/5 Joshua Wiley-2 [via R]
ml-node+3428356-844576439-227...@n4.nabble.com:
 Hi Ulisses,

 Look at the functions ?match and ?rbind

 If you do not want to do it by hand, you can make a little function as
 below.

 HTH,

 Josh

 d1 - data.frame(goals = 4:1, players = LETTERS[1:4])
 d2 - data.frame(goals = c(1, 3, 2, 5), players = LETTERS[3:6])

 f - function(old, new, check) {
 Â  index - new[, check] %in% old[, check]
 Â  dat - rbind(old, new[index, ])
 Â  tocheck - new[!index, ]
 Â  list(merged = dat, tocheck = tocheck)
 }

 dmerged - f(d1, d2, players)
 ## check tocheck and once it is correct
 dfinal - do.call(rbind, dmerged)

 On Tue, Apr 5, 2011 at 8:06 AM, Ulisses.Camargo
 [hidden email] wrote:
 The example scene:

 I have a database with stats about each goal made by my soccer team. This
 database (a data frame in R) is organized in lines (goals) with a set of
 columns containing data about these goals (player name, tactic position,
 etc). For now, this database will be called data.frame1.

 What I need is to feed this data.frame1 with new information about my
 team
 goals. I will call this new information data.frame2. This set of new
 goals
 is organized in the same way as in data.frame1 (equal numbers of cols).

 Where help is needed:

 I need help in finding a way to check the player-name column in
 data.frame2 before feeding data.frame1 with it. What I need is a way
 to
 verify the name of the player on each line of data.frame2 with the names
 of players that already exist on a col in data.frame1. Moreover, I need
 R
 to make two main things:

 First, the lines of âdata.frame2â with player names that already exists 
 in
 âdata.frame1â must be added to âdata.frame1â.

 Second: lines of âdata.frame2â with player names that does not exist on
 âdata.frame1â must be listed in an output to be manually checked and
 corrected.
 After this verification, corrected lines and new-player-names lines must
 be
 incorporated in data.frame1.

 What I want is to guarantee that will be no lines with wrong player names
 in
 my database.
 At the same time, my script must permit new information to be added (new
 player names).

 Is there somebody who could help me with this?

 Thanks for your attention

 Best wishes
 Ulisses

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 --
 Joshua Wiley
 Ph.D. Student, Health Psychology
 University of California, Los Angeles
 http://www.joshuawiley.com/

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-- 
Ulisses Moliterno de Camargo
Instituto Nacional de Pesquisas da AmazÃ´nia


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Re: [R] how to label customized y axis when using lattice parallel parameter common.scale=TRUE

2011-04-05 Thread Pengcheng Yang

I have readjust the script as follows to retain the complete information 
of original graph, the background vertical bar.



parallel(~iris[1:4] | Species, iris,horizon=FALSE,
 ylim = extendrange(range(iris[1:4])),
 scales = list(y = list(at = NULL, labels = NULL),x=list(rot=45)),
 lower = 0, upper = 1,
 panel=function(x,y,z,...){
   panel.abline(v=1:4,col=gray90)
   panel.parallel(x,y,z,...)
 })

On 2011-4-5 21:42, Deepayan Sarkar wrote:

On Tue, Apr 5, 2011 at 4:39 PM, Pengcheng Yangpeng...@gmail.com  wrote:

Dear all,

When I use parallel function in lattice package, I want to label the y-axis
with customized numbers. Like this:

parallel(~iris[1:4] | Species, iris,horiz=FALSE,common.scale=TRUE,
scales=list(y=list(at=c(0,2,3

Parallel does not directly support that, and will insist on scaling
the data. However, you can control the scaling (using 'lower' and
'upper'), and override a couple of other arguments to get what you
want:

parallel(~iris[1:4] | Species, iris,
  xlim = extendrange(range(iris[1:4])),
  scales = list(x = list(at = NULL, labels = NULL)),
  lower = 0, upper = 1)

This is for horizontal.axis = TRUE, adjust accordingly for FALSE.

-Deepayan


But only Min label in the y-axis, nothing happened. Could anyone help me?

Thanks.
Regards,

Pengcheng Yang

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[R] loop question

2011-04-05 Thread Thomas


Dear all,

I am trying to set up a list with 1:c objects each meant to capture the 
coefficients for one coefficient and 100 replications. I receive the 
following error message:


Error in betaboot[[p]] : subscript out of bounds.

My code is below. Where is my mistake?

Many thanks,

Thomas

_
betaboot-list(NULL)

for (i in 1:c) {
betaboot[[i]]-cbind()
}


num - 100 # this is the number of bootstraps

for (i in 1:num) {

[BOOTSTRAP]

  coef.temp - coef(model.temp, data=newdata)

  for (p in 1:c){
  betaboot[[p]] - cbind(betaboot[[p]], coef.temp[,p])
  }

  }

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Re: [R] loop question

Dear Thomas,

On Tue, Apr 5, 2011 at 8:33 AM, Thomas thomas.tri...@cantab.net wrote:
 Dear all,

 I am trying to set up a list with 1:c objects each meant to capture the
 coefficients for one coefficient and 100 replications. I receive the
 following error message:

 Error in betaboot[[p]] : subscript out of bounds.

 My code is below. Where is my mistake?

 Many thanks,

 Thomas

 _
 betaboot-list(NULL)

if you know the number of bootstraps (which you seem to later on), a
preferred way to instatiate the list would be:

betaboot - vector(mode = list, length = yourlength)


 for (i in 1:c) {

because c() is such an important function, I would strongly
encourage you not to use it also as a variable.

 betaboot[[i]]-cbind()

Don't use this to build an empty list.

 }


 num - 100 # this is the number of bootstraps

 for (i in 1:num) {

    [BOOTSTRAP]

  coef.temp - coef(model.temp, data=newdata)

  for (p in 1:c){
  betaboot[[p]] - cbind(betaboot[[p]], coef.temp[,p])

This should work assuming betaboot is instatiated properly.  That
said, it looks like you have a nested for loop and then just keep
cbind()ing each element of betaboot bigger and bigger.  You may get a
performance increase if you also instantiate each matrix/dataframe
inside betaboot.  Then the call would become something like:

betaboot[[i]][,p] - coef.temp[,p]

that is, you can use a chained series of extraction operators to get
to the appropriate column in the matrix/dataframe inside the
appropriate list element.  Then rather than constantly using cbind(),
you just place coef.temp[,p] where you want it.  The only requirement
is that you know the sizes of the matrices/dataframes going in so you
can create empty ones from the get go.

Cheers,

Josh

  }

  }

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

-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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[R] Gibbs sampling

2011-04-05 Thread Liliana Pacheco

HI R users, perhaps you can help me with this:

I am planning on using the Gibbs sampler for the correlation coefficient of
a bivariate normal. I have a posterior distribution for rho, besides that,
the conditional distribution for all the parameters of this posterior
distribution. The thing is that, since I have to get a sample, size 1
for rho, obtain a 95% confidence interval, and repeat this procedure 1000
times; and repeat this procedure for 50 scenaries, I'n thinking this is
going to take forever.

Is there a library for making this work faster? I've heard of gibbs.met, but
I don't know if it's going to work or even more, I didn't understand the
examples.

Thanks!

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[R] Precision of summary() when summarizing variables in a data frame

2011-04-05 Thread Daniel Malter

Hi, 

I summary() a variable with 409908 numeric observations. The variable is
part of a data.frame. The problem is that the min and max returned by
summary() do not equal the ones returned by min() and max(). Does anybody
know why that is? 

 min(data$vc)
[1] 15452
 max(data$vc)
[1] 316148
 summary(data$vc)
   Min. 1st Qu.  MedianMean 3rd Qu.Max. 
  15450   21670   40980   55500   63880  316100 


sessionInfo()
R version 2.11.1 (2010-05-31) 
x86_64-apple-darwin9.8.0 

locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8

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

other attached packages:
[1] sqldf_0.3-5   chron_2.3-39  gsubfn_0.5-5 
[4] proto_0.3-8   RSQLite.extfuns_0.0.1 RSQLite_0.9-4
[7] DBI_0.2-5 

Thanks much,
Daniel  

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[R] lorena

2011-04-05 Thread lorena delgadillo

 Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

personally work the series as a SARIMA

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.

M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))

where

D1 = diff (series)

Many thanks in advance
Dear
I would like to know how to use the Croston method in R, consaltarte if I
download a package?

Personally work the series as a SARIMA, this correct?

In the present instalment have many values zeros proposed the following
model, but I have many doubts with his predictions.
M3 = arima (d1, order = c (2,1,4), n. ahead = 4, seasonal = list (order = c
(2,1,4), period = 4))
where
D1 = diff (series)

Many thanks in advance
*
saludos cordiales,

LORENA DELGADILLO AGUIRRE*
Licenciada en Estadística PUCV
*www.lorenadelgadilloaguirre.blogspot.com*
*lod...@gmail.com*   *09-668.48.60*
ID  Serie
1.000.00
2.000.00
3.000.00
4.00152.10
5.00179.58
6.00195.474000
7.00197.937200
8.00216.097160
9.00223.071448
10.00   245.966874
11.00   252.880644
12.00   0.00
13.00   273.109027
14.00   281.577342
15.00   287.372776
16.00   280.555955
17.00   286.182042
18.00   293.301229
19.00   303.959187
20.00   303.197473
21.00   312.053897
22.00   0.00
23.00   322.755681
24.00   0.00
25.00   321.304233
26.00   329.709470
27.00   335.897999
28.00   0.00
29.00   329.701277
30.00   325.349956
31.00   338.850093
32.00   343.215070
33.00   0.00
34.00   0.00
35.00   334.615434
36.00   343.551063
37.00   337.402394
38.00   336.177021
39.00   340.881856
40.00   348.523187
41.00   349.106735
42.00   351.637707
43.00   353.120839
44.00   344.560442
45.00   0.00
46.00   0.00
47.00   0.00
48.00   0.00
49.00   345.231065
50.00   351.480541
51.00   342.707539
52.00   348.914086
53.00   354.102022
54.00   356.273026
55.00   345.428623
56.00

Re: [R] Precision of summary() when summarizing variables in a data frame

2011-04-05 Thread jim holtman

They are probably the same.  It isjust that summary is printing out 4
significant digits.  Try:

options(digits = 20)



On Tue, Apr 5, 2011 at 12:38 PM, Daniel Malter dan...@umd.edu wrote:
 Hi,

 I summary() a variable with 409908 numeric observations. The variable is
 part of a data.frame. The problem is that the min and max returned by
 summary() do not equal the ones returned by min() and max(). Does anybody
 know why that is?

 min(data$vc)
 [1] 15452
 max(data$vc)
 [1] 316148
 summary(data$vc)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  15450   21670   40980   55500   63880  316100


 sessionInfo()
 R version 2.11.1 (2010-05-31)
 x86_64-apple-darwin9.8.0

 locale:
 [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8

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

 other attached packages:
 [1] sqldf_0.3-5           chron_2.3-39          gsubfn_0.5-5
 [4] proto_0.3-8           RSQLite.extfuns_0.0.1 RSQLite_0.9-4
 [7] DBI_0.2-5

 Thanks much,
 Daniel

 --
 View this message in context: 
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 Sent from the R help mailing list archive at Nabble.com.

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-- 
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Data Munger Guru

What is the problem that you are trying to solve?

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Re: [R] Creating multiple vector/list names-novice

2011-04-05 Thread michalseneca

Thanks 

I already found out solution :) 

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[R] Arrangement of Lattice Histograms - Top to bottom and then left to right?

2011-04-05 Thread Les

Hi List,
Using Lattice, I have created a plot of histograms showing Fork Length by
Year. The plot shows the histograms in 3 columns and 5 rows. Using the
as.table=T function I can get the years to start on top. However, what I
would like to do is have the first year start in the top left (column 1, row
1; as it is now) and add the subsequent histograms to the plot going down
the column and then over by row (example and current code is shown below).
Is this possible? I have spent a great deal of time searching, and have not
found any clues. Any help/ideas would be greatly appreciated.

For example, what is being done now:

1974 1975 1976 
1977 1978 1979 
1980 1981 1982

and what I am hoping for:

1974 1977 1980
1975 1978 1981
1976 1979 1982

Thank you kindly,
Les

data=read.csv(AllData.csv,sep=,,header=T)   
data$year=as.factor(data$year)  
ferg=data[data$wtrbody==Ferguson,c(1:12)]
library(lattice)

strip.background=trellis.par.get(strip.background)
trellis.par.set(strip.background = list(col = grey(7:1/8))) 

histogram(~fl|year, data=ferg, as.table=T, type=count,
col='dark grey',layout=c(3,5),lwd=2, lty=1, 
xlab=list(Fork Length (mm), cex=1.4,font=2),  
breaks=seq(from=300,to=900,by=25),  
ylab=list(Frequency,cex=1.4,font=2),  
scales=list(font=2,cex=1.1, tck=c(1,0), alternating=1,  
y=list(relation=free,tick.number=3)), 

par.strip.text=list(cex=1.2),font=2,
aspect=0.6)

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Re: [R] Precision of summary() when summarizing variables in a data frame

2011-04-05 Thread Erik Iverson




jim holtman wrote:

They are probably the same.  It isjust that summary is printing out 4
significant digits.  Try:

options(digits = 20)


FYI, the default summary method also has its own digits argument.





On Tue, Apr 5, 2011 at 12:38 PM, Daniel Malter dan...@umd.edu wrote:

Hi,

I summary() a variable with 409908 numeric observations. The variable is
part of a data.frame. The problem is that the min and max returned by
summary() do not equal the ones returned by min() and max(). Does anybody
know why that is?


min(data$vc)

[1] 15452

max(data$vc)

[1] 316148

summary(data$vc)

  Min. 1st Qu.  MedianMean 3rd Qu.Max.
 15450   21670   40980   55500   63880  316100


sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0

locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8

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

other attached packages:
[1] sqldf_0.3-5   chron_2.3-39  gsubfn_0.5-5
[4] proto_0.3-8   RSQLite.extfuns_0.0.1 RSQLite_0.9-4
[7] DBI_0.2-5

Thanks much,
Daniel

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Re: [R] Arrangement of Lattice Histograms - Top to bottom and then left to right?

2011-04-05 Thread Bert Gunter

Look for the index.cond argument at the bottom of the Help page for xyplot.
-- Bert

On Tue, Apr 5, 2011 at 11:37 AM, Les lnhar...@zoology.ubc.ca wrote:
 Hi List,
 Using Lattice, I have created a plot of histograms showing Fork Length by
 Year. The plot shows the histograms in 3 columns and 5 rows. Using the
 as.table=T function I can get the years to start on top. However, what I
 would like to do is have the first year start in the top left (column 1, row
 1; as it is now) and add the subsequent histograms to the plot going down
 the column and then over by row (example and current code is shown below).
 Is this possible? I have spent a great deal of time searching, and have not
 found any clues. Any help/ideas would be greatly appreciated.

 For example, what is being done now:

 1974 1975 1976
 1977 1978 1979
 1980 1981 1982

 and what I am hoping for:

 1974 1977 1980
 1975 1978 1981
 1976 1979 1982

 Thank you kindly,
 Les

 data=read.csv(AllData.csv,sep=,,header=T)
 data$year=as.factor(data$year)
 ferg=data[data$wtrbody==Ferguson,c(1:12)]
 library(lattice)

 strip.background=trellis.par.get(strip.background)
 trellis.par.set(strip.background = list(col = grey(7:1/8)))

 histogram(~fl|year, data=ferg, as.table=T, type=count,
        col='dark grey',layout=c(3,5),lwd=2, lty=1,
        xlab=list(Fork Length (mm), cex=1.4,font=2),
        breaks=seq(from=300,to=900,by=25),
        ylab=list(Frequency,cex=1.4,font=2),
        scales=list(font=2,cex=1.1, tck=c(1,0), alternating=1,
        y=list(relation=free,tick.number=3)),
        par.strip.text=list(cex=1.2),font=2,
        aspect=0.6)

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-- 
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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Re: [R] Changing parameter in local fdr R code

2011-04-05 Thread Jim Silverton

I am using Efron's local fdr procedure. But, I want to change the null from
N(0,1) to N(0, 0.002). I can access the function but I have no idea what to
change. In other words, I want nulltype to be N(0,0.002) instead of N(0,1)
in his function. Anyone has any ideas. This is his code for the local fdr:

function (zz, bre = 120, df = 7, pct = 0, pct0 = 1/4, nulltype = 1,
type = 0, plot = 1, mult, mlests, main =  , sw = 0)
{
call = match.call()
if (length(bre)  1) {
lo - min(bre)
up - max(bre)
bre - length(bre)
}
else {
if (length(pct)  1) {
lo - pct[1]
up - pct[2]
}
else {
if (pct == 0) {
lo - min(zz)
up - max(zz)
}
if (pct  0) {
med = median(zz)
ra = med + (1 - pct) * (range(zz) - med)
lo = ra[1]
up = ra[2]
}
if (pct  0) {
v - quantile(zz, c(pct, 1 - pct))
lo - v[1]
up - v[2]
}
}
}
zzz - pmax(pmin(zz, up), lo)
breaks - seq(lo, up, length = bre)
zh - hist(zzz, breaks = breaks, plot = F)
x - (breaks[-1] + breaks[-length(breaks)])/2
yall - y - zh$counts
K - length(y)
N - length(zz)
if (pct  0) {
y[1] - min(y[1], 1)
y[K] - min(y[K], 1)
}
if (type == 0) {
X - cbind(1, ns(x, df = df))
f - glm(y ~ ns(x, df = df), poisson)$fit
}
else {
X - cbind(1, poly(x, df = df))
f - glm(y ~ poly(x, df = df), poisson)$fit
}
l - log(f)
Fl - cumsum(f)
Fr - cumsum(rev(f))
D - (y - f)/(f + 1)^0.5
D - sum(D[2:(K - 1)]^2)/(K - 2 - df)
if (D  1.5)
warning(paste(f(z) misfit = , round(D, 1), .  Rerun with
increased df,
sep = ))
if (nulltype == 3) {
fp0 = matrix(NA, 6, 4)
colnames(fp0) = c(delta, sigleft, p0, sigright)
}
else {
fp0 = matrix(NA, 6, 3)
colnames(fp0) = c(delta, sigma, p0)
}
rownames(fp0) = c(thest, theSD, mlest, mleSD, cmest,
cmeSD)
fp0[thest, 1:2] = c(0, 1)
fp0[theSD, 1:2] = 0
imax - seq(l)[l == max(l)][1]
xmax - x[imax]
if (length(pct0) == 1) {
pctup - 1 - pct0
pctlo - pct0
}
else {
pctlo - pct0[1]
pctup - pct0[2]
}
lo0 - quantile(zz, pctlo)
hi0 - quantile(zz, pctup)
nx - length(x)
i0 - (1:nx)[x  lo0  x  hi0]
x0 - x[i0]
y0 - l[i0]
if (nulltype == 3) {
X00 - cbind((x0 - xmax)^2, pmax(x0 - xmax, 0)^2)
}
else {
X00 - cbind(x0 - xmax, (x0 - xmax)^2)
}
lr - lm(y0 ~ X00)
co - lr$coef
if (nulltype == 3) {
cmerror = I(is.na(co[3]) | is.na(co[2]))
if (!cmerror)
cmerror = I(co[2] = 0 | co[2] + co[3] = 0)
}
else {
cmerror = is.na(co[3])
if (!cmerror)
cmerror = I(co[3] = 0)
}
if (cmerror) {
if (nulltype == 3)
stop(CM estimation failed.  Rerun with nulltype = 1 or 2.)
else if (nulltype == 2)
stop(CM estimation failed.  Rerun with nulltype = 1.)
else {
X0 - cbind(1, x - xmax, (x - xmax)^2)
warning(CM estimation failed, middle of histogram non-normal)
}
}
else {
if (nulltype == 3) {
X0 - cbind(1, (x - xmax)^2, pmax(x - xmax, 0)^2)
sigs - 1/sqrt(-2 * (c(co[2], co[2] + co[3])))
fp0[cmest, c(1, 2, 4)] - c(xmax, sigs)
}
else {
X0 - cbind(1, x - xmax, (x - xmax)^2)
xmaxx - -co[2]/(2 * co[3]) + xmax
sighat - 1/sqrt(-2 * co[3])
fp0[cmest, 1:2] - c(xmaxx, sighat)
}
l0 - as.vector(X0 %*% co)
f0 - exp(l0)
p0 - sum(f0)/sum(f)
f0 - f0/p0
fp0[cmest, 3] - p0
}
b = 4.3 * exp(-0.26 * log(N, 10))
if (missing(mlests)) {
med = median(zz)
sc = diff(quantile(zz)[c(2, 4)])/(2 * qnorm(0.75))
mlests = locmle(zz, xlim = c(med, b * sc))
if (N  5e+05) {
warning(length(zz)  500,000: For ML estimation, a wider
interval than optimal was used.  To use the optimal interval, rerun with
mlests = c(,
mlests[1], , , b * mlests[2], ).\n, sep = )
mlests = locmle(zz, xlim = c(med, sc))
}
}
if (!is.na(mlests[1])) {
if (N  5e+05)
b = 1
if (nulltype == 1) {
Cov.in = list(x = x, X = X, f = f, sw = sw)
ml.out = locmle(zz, xlim = c(mlests[1], b * mlests[2]),
d = mlests[1], s = mlests[2], Cov.in = Cov.in)
mlests = ml.out$mle
}
else mlests = locmle(zz, xlim = c(mlests[1], b * mlests[2]),
d = mlests[1], s = mlests[2])
fp0[mlest, 1:3] = mlests[1:3]
fp0[mleSD, 1:3] = mlests[4:6]
}
if (sum(is.na(fp0[c(3,

Re: [R] detect filetype (as in unix 'file')

2011-04-05 Thread Jeroen Ooms

 No, but what is wrong with using system()?

The application is running in a very sandboxed environment and might
not have permission to execute 'file'.

 'file' is large and complex because it tries to be comprehensive (but it
 still does not know about some common systems, e.g. 64-bit Windows
 binaries).  There simply is no point in replicating that in R: which is why
 we chose rather to port 'file' to Windows and provide in in Rools.

Alright that makes sense, thanks.

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Re: [R] Precision of summary() when summarizing variables in a data frame

2011-04-05 Thread Daniel Malter

Thanks all. No I wasn't aware of the fact that summary is rounding in this
case. 

Da.

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[R] Hazard ratio calculation and KM plot p-value:

2011-04-05 Thread Angel Russo

I have two questions:

1) Can anyone provide me with a reference regarding calculation of Hazard
ratio for two groups of data. How is it being manually calculated with an
example. Unlike median time ratio which is the ratio of median times in two
groups, at what time is the hazard ratio calculation done?

2) In kaplan-meier statistics of stratifying into two groups, a p-value is
often calculated (e.g. log-rank p-value). p-value is the test is how
significant the separation is between two groups compared to random. What is
random in kaplan-meier statistics. How is p-value calculated?

Thanks for everyone's time to read and hopefully respond as well.

[[alternative HTML version deleted]]

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Re: [R] Arrangement of Lattice Histograms - Top to bottom and then left to right?

2011-04-05 Thread Les

Thank you.
The help is much appreciated.
Les

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Re: [R] Structural equation modeling in R(lavaan,sem)

2011-04-05 Thread jouba


Thanks a lot 

Antra EL MOUSSELLY 


 


Date: Tue, 5 Apr 2011 09:49:08 -0500
From: ml-node+3428265-1156197921-225...@n4.nabble.com
To: antr...@hotmail.com
Subject: Re: Structural equation modeling in R(lavaan,sem)

On 04/04/2011 07:14 PM, jouba wrote: 
 
 
 Thanks you for your response 
 For lavaan package can i have more information about this example you have 
 applied in the section 7 
 the meanings  of The variables (c1,c2,c3,c4, i ,s ,x1,x2) 
 I think i have need more information to learn  more about how  able to apply 
 growth model  in my data (longitudianl data) 

In the example, c1-c4 are time-varying covariates, i and s are the 
random intercept and slope respectively, and x1 and x2 are two exogenous 
covariates influencing the intercept and slope. 

Please note: the lavaanIntroduction document is hardly useful to _learn_ 
about growth models (or any SEM model for that matter). It only explains 
how to fit them using the lavaan package. To learn about growth models, 
you may want to read any one of the books below: 

Latent Curve Models: A Structural Equation Perspective (Wiley Series in 
Probability and Statistics) by Kenneth A. Bollen and Patrick J. Curran 
(Hardcover - Dec 23, 2005) 

Latent Growth Curve Modeling (Quantitative Applications in the Social 
Sciences) by Dr. Kristopher J. Preacher, Aaron Lee Wichman, Robert 
Charles MacCallum and Dr. Nancy E. Briggs (Paperback - Jun 27, 2008) 

An Introduction to Latent Variable Growth Curve Modeling: Concepts, 
Issues, and Applications (Quantitative Methodology) (Quantitative 
Methodology Series) by Terry E. Duncan, Susan C. Duncan and Lisa A. 
Strycker (Paperback - May 23, 2006) 

Yves. 

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[R] IFELSE function XXXX

2011-04-05 Thread Dan Abner

Hello everyone,

This IFELSE function call is not working properly. I do not receive an error
message, but the actions are not executed conditional as I was hoping. Any
assistance is appreciated.

set.seed(12345)
res1-rbinom(1,1,.1)
rdata3-transform(data.frame(res1),input1=rnorm(1,50,10))
data3
#inducing correlation between res1  input1
ifelse(data3$res1==1,data3$input1-data3$input1+10,data3$input1-data3$input1)
data3

Thank you,

Dan

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[R] simple save question

2011-04-05 Thread xueke


Hi,

When I run the survfit function, I want to get the restricted mean  
value and the standard error also. I found out using the print  
function to do so, as shown below,


 print(km.fit,print.rmean=TRUE)
Call: survfit(formula = Surv(diff, status) ~ 1, type = kaplan-meier)

   records  n.maxn.start events *rmean *se(rmean) median
   200.000200.000200.000129.000  0.145  0.237  1.158
   0.95LCL0.95UCL
 0.450  1.730
* restricted mean with upper limit =  2.97

The questions is, is there any way to extract these values from the  
print command?


Thanks a lot.

Xueke

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Re: [R] IFELSE function XXXX

2011-04-05 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Dan Abner
 Sent: Tuesday, April 05, 2011 12:58 PM
 To: r-help@r-project.org
 Subject: [R] IFELSE function 

 Hello everyone,

 This IFELSE function call is not working properly. I do not 
 receive an error
 message, but the actions are not executed conditional as I 
 was hoping. Any
 assistance is appreciated.

 set.seed(12345)
 res1-rbinom(1,1,.1)
 rdata3-transform(data.frame(res1),input1=rnorm(1,50,10))
 data3
 #inducing correlation between res1  input1
 ifelse(data3$res1==1,data3$input1-data3$input1+10,data3$input
1-data3$input1)

Avoid assignments in arguments to function calls, especially
multiple assignments to the same object, except when you know
what you are doing and want to write obscure code.

Change the above line to
  data3$input1 - ifelse(data3$res1==1, data3$input1+10, data3input1)

There is nothing too special about ifelse here, you would get
similar results if you put assignment statements in other function
calls.  The assignment gets evaluated just as things like log(12)
or data3$input1+10 get evaulated when given as arguments.  E.g.,
   x - 1
   cat(x - 2:1, x - 6:3, x - 10:4, \n)
  2 1 6 5 4 3 10 9 8 7 6 5 4
   x
  [1] 10  9  8  7  6  5  4
Which version of x you end up with depends on which argument
the function evaluates last.
   f - function(arg1, arg2) arg2 + arg1
   f( x - 7, x - 101 )
  [1] 108
   x
  [1] 7

An example of obscure code involving multiple assignments
to one object is
  if (is.null(tms - x$terms)  is.null(tms - attr(x, terms))) {
 stop(Cannot find a list component or attribute called terms)
  }
  tms # x$terms if it exists, otherwise attr(x, terms)
Since the order of evaluation of the 2 arguments to  is well
defined (left then right, and the right won't be evaluated unless
the left if TRUE), this produces a trustworthy answer.  Most functions
don't promise any particular order of evaluation.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 data3

 Thank you,

 Dan

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Re: [R] grImport/ghostscript problems

2011-04-05 Thread Paul Murrell


Hi

On 5/04/2011 9:30 p.m., guillaume Le Ray wrote:

Hi Al,

I'm facing exactly the same problem as you are, have you manage to
fix it? If yes I eager to know the trick.


Al's problem turned out to be a bug in 'grImport', so one thing you can 
try is to install the latest version of 'grImport'.


If that still fails, you might be able to get more information about the 
problem by looking at the end of the XML file that is created by 
PostScriptTrace().  If ghostscript has hit trouble it's error messages 
will hopefully be at the end of that XML file.


Paul



Regards,

Guillaume

2011/3/27 Al Roarkhrbuil...@hotmail.com


Paul Murrellp.murrellat  auckland.ac.nz  writes:



Hi

On 28/03/2011 8:13 a.m., Al Roark wrote:


Hi All: I've been struggling for a while trying to get grImport
up and running.  I'm on a Windows 7 (home premium 64 bit)
machine running R-2.12.2 along with GPL Ghostscript 9.01. I've
set my Windows PATH variable to point to the Ghostscript \bin
and \lib directories, and I've created the R_GSCMD environment
variable pointing to gswin32c.exe. I don't have any experience
with Ghostscript, but with the setup described above I can view
the postscript file with the following command to the Windows
command prompt: gswin32c.exe D:\Sndbx\vasarely.ps However, I
can't get the PostScriptTrace() function to work on the same
file.  Submitting PostScriptTrace(D:/Sndbx/vasarely.ps) gives
me the error: Error in PostScriptTrace(D:/Sndbx/vasarely.ps)
:   status 127 in running command 'gswin32c.exe -q -dBATCH
-dNOPAUSE -sDEVICE=pswrite
-sOutputFile=C:\Users\Al\AppData\Local\Temp\RtmppPjDAf\file5db99cb



-sstdout=vasarely.ps.xml capturevasarely.ps' Your suggestions are

much appreciated. Cheers, Al [[alternative HTML version
deleted]]


You could try running the ghostscript command that is printed in
the error message at the Windows command prompt to see more info
about the problem (might need to remove the '-q' so that
ghostscript prints messages to the screen).

Paul



Thanks for your reply.

Perhaps this is a Ghostscript problem. When I run the Ghostscript
command, I'm met with the rather unhelpful error: 'GPL Ghostscript
9.01: Unrecoverable error, exit code 1 (occurs whether or not I
remove the -q)'.

Interestingly, if I remove the final argument (in this case,
capturevasarely.ps) the Ghostscript command executes, placing a
file (appears to be xml) in the temporary directory. However, I'm
not sure what to do with this result.

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--
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Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
p...@stat.auckland.ac.nz
http://www.stat.auckland.ac.nz/~paul/

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Re: [R] lattice: how to center a subtitle?

2011-04-05 Thread David Scott


On 6/04/2011 12:47 a.m., Deepayan Sarkar wrote:

On Tue, Apr 5, 2011 at 6:12 AM, David Scottd.sc...@auckland.ac.nz  wrote:

[...]


I am not sure where I read it and I can't find it again, but my
understanding is that expressions using bquote with lattice need to be
enclosed in as.expression() to work. That is in contrast to what happens in
base graphics.

Here is a simple example.

a- 2
plot(1:10, a*(1:10), main = bquote(alpha == .(a)))
require(lattice)
xyplot(a*(1:10)~ 1:10, main = bquote(alpha == .(a)))
xyplot(a*(1:10)~ 1:10, main = as.expression(bquote(alpha == .(a

Which produces:


a- 2
plot(1:10, a*(1:10), main = bquote(alpha == .(a)))
require(lattice)

Loading required package: lattice

xyplot(a*(1:10)~ 1:10, main = bquote(alpha == .(a)))

Error in trellis.skeleton(formula = a * (1:10) ~ 1:10, cond = list(c(1L,  :
  object 'alpha' not found

xyplot(a*(1:10)~ 1:10, main = as.expression(bquote(alpha == .(a


Using expression() rather than as.expression() doesn't produce the desired
affect. Try it yourself.

As to why this is the case .


Let's see: ?xyplot says

   'main': Typically a character string or expression describing
   the main title to be placed on top of each page. [...]

So, lattice is fairly explicit, by R standards, in requiring 'main' to
be character or expression. On the other hand, ?title says

  The labels passed to 'title' can be character strings or language
  objects (names, calls or expressions), or [...]

so it additionally accepts names and calls.

Now, we have


a- 2
foo- bquote(alpha == .(a))



foo # Looks OK

alpha == 2

mode(foo) # But

[1] call

is.expression(foo) # not an expression

[1] FALSE


is.expression(expression(foo)) ## YES, but

[1] TRUE

expression(foo) ## not what we want

expression(foo)


is.expression(as.expression(foo))

[1] TRUE

as.expression(foo) ## This IS what we want

expression(alpha == 2)

So I submit that lattice is behaving exactly as suggested by its documentation.

Now you would naturally argue that this is hiding behind
technicalities, and if call objects work for plot(), it should work
for lattice as well. But watch this:


plot(1:10, main = foo) # works perfectly



arglist- list(1:10, main = foo)
arglist # Looks like what we want

[[1]]
  [1]  1  2  3  4  5  6  7  8  9 10

$main
alpha == 2


do.call(plot, arglist)

Error in as.graphicsAnnot(main) : object 'alpha' not found

...which I would say is unexpected behaviour, if not a bug.

The moral of the story is that unevaluated calls are dangerous objects
(try this one out for fun:

foo- bquote(q(.(x)), list(x = no))
do.call(plot, list(1:10, main = foo))

), and carrying them around is not a good idea.

Lattice does use the do.call paradigm quite a bit, and I think it
might be quite difficult to fix it up to handle non-expression
language objects (which will still not fix the type of problem shown
above).

-Deepayan


Thanks very much for this explanation Deepayan. Part of my intention in 
contributing to this thread was to have something explicit in the 
archives for future reference, and your reply is excellent in that regard.


And many thanks for your work on lattice.

David Scott

--
_
David Scott Department of Statistics
The University of Auckland, PB 92019
Auckland 1142,NEW ZEALAND
Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055
Email:  d.sc...@auckland.ac.nz,  Fax: +64 9 373 7018

Director of Consulting, Department of Statistics

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[R] solveRsocp in fPortfolio

2011-04-05 Thread Luis Felipe Parra

Hello I am trying to use solveRsocp to optimize a Portfolio maximizing
return. I was checking the code since I would like to solve it for a short
Portfolio and I found this:

# C - Cone Constraints:
C1 - rep(0, nAssets) # xCx
C2 - eqsumW[2, -1]   # sum(x)
C3 - rbind(diag(nAssets), -diag(nAssets) )   # x[i]0

# d - Cone Constraints:
d1 - targetRisk  # xCx = risk
d2 - eqsumW[2, 1]# sum(x) = 1
d3 - c(rep(0, nAssets), rep(-1, nAssets))# x[i]  0

I was wondering if the conditions C3 and d3 aren't forcing the portfolio to
be always a LongOnly portfolio? I tried modifying the code to take out
these conditions and got the following error:

Error en .socp.phase1(f, A, b, N, control) : Phase 1 failed, alpha=0

Does anybody knkow what might be going on? Is there any problem with this
solver for negative weights?

Thank you

Felipe Parra

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[R] Search arrays based on similar values

2011-04-05 Thread mjdubya

Hey folks,
I have two arrays: A (1X100) with non-ordered values ranging 1-14  
  B (2X14)  containing 14 decimal values.  
I would like to create a new array (1X100) that contains only the decimal
values from array B by associating the integers from A and B.  In other
words, for each value of A find the same integer value of B, select the
associated decimal value of B.  

A  B newarray
1   1  0.1 0.1
1   2  0.3 0.1
1   3  0.14   0.1
2   4  0.2 0.3
3   5  0.82   0.14
3   6  0.21   0.14
4.   .  0.2
7.   .   .
14  .   ..
4  14 0.03.
3. 
5. 
. .
. . 

Any suggestions are greatly appreciated!  

--
View this message in context: 
http://r.789695.n4.nabble.com/Search-arrays-based-on-similar-values-tp3429381p3429381.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] IFELSE function XXXX

2011-04-05 Thread Ben Bolker

William Dunlap wdunlap at tibco.com writes:

 Avoid assignments in arguments to function calls, especially
 multiple assignments to the same object, except when you know
 what you are doing and want to write obscure code.
 
 Change the above line to
   data3$input1 - ifelse(data3$res1==1, data3$input1+10, data3input1)
 

  data3 - transform(data3,input1=ifelse(res==1,input1+10,input1))

 might be even clearer

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Re: [R] Search arrays based on similar values



On Apr 5, 2011, at 6:34 PM, mjdubya wrote:


Hey folks,
I have two arrays: A (1X100) with non-ordered values ranging 1-14
 B (2X14)  containing 14 decimal values.
I would like to create a new array (1X100) that contains only the  
decimal
values from array B by associating the integers from A and B.  In  
other
words, for each value of A find the same integer value of B, select  
the

associated decimal value of B.


Sounds like a job for merge(). Tested solutions offered when  
reproducible examples are provided.




A  B newarray
1   1  0.1 0.1
1   2  0.3 0.1
1   3  0.14   0.1
2   4  0.2 0.3
3   5  0.82   0.14
3   6  0.21   0.14
4.   .  0.2
7.   .   .
14  .   ..
4  14 0.03.
3.
5.
. .
. .


David Winsemius, MD
West Hartford, CT

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Re: [R] R gui on windows how to force to always show the last line of output

2011-04-05 Thread stan zimine

thank you, David, for your answer, which has a therapeutic effect.

Otherwise i was launching my prod code  from Emacs ESS  jas as Janice
suggested.


On Sat, Apr 2, 2011 at 2:27 PM, David Winsemius dwinsem...@comcast.net wrote:

 On Apr 2, 2011, at 4:21 AM, stan zimine wrote:

 Hi.
 Googled but did not found the answer for the following little issue.

 how to force R gui  on windows (maybe a specific setting)  to always
 show the last line of output in the window console.


 My program in R makes measurements every 5 mins in indefinite loop and
 prints  results in the console.

 The problem:  last messages are not visible,  The scrolling bar of the
 gui console  gets shorter. I.e.  you have to scroll for the last
 messages.

 Thanks if anybody knows the sol to this prob.

 You may want to add flush.console() to the code.

 --

 David Winsemius, MD
 West Hartford, CT



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Re: [R] merging 2 frames while keeping all the entries from the reference frame

2011-04-05 Thread Dimitri Liakhovitski

Thanks a lot - these solutions are much more elegant than my own:

new.data-merge(mydata[mydata$group %in%
levels(mydata$group)[1],],reference,by=mydate,all.x=T,all.y=T)
new.data[[group]][is.na(new.data[[group]])]-levels(mydata$group)[1]
new.data[[values]][is.na(new.data[[values]])]-0

# Continue Merging - starting with Group2:
for(i in 2:nlevels(mydata$group)){  #i-2
temp-merge(mydata[mydata$group %in%
levels(mydata$group)[i],],reference,by=mydate,all.x=T,all.y=T)
temp[[group]][is.na(temp[[group]])]-levels(mydata$group)[i]
temp[[values]][is.na(temp[[values]])]-0
new.data-rbind(new.data,temp)
}

Dimitri

On Mon, Apr 4, 2011 at 3:07 PM, Henrique Dallazuanna www...@gmail.com wrote:
 Try this:

  merge(mydata, cbind(reference, group = rep(unique(mydata$group), each
 = nrow(reference))), all = TRUE)

 On Mon, Apr 4, 2011 at 2:24 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 To clarify just in case, here is the result I am trying to get:

 mydate  group   values
 12/29/2008      Group1  0.453466522
 1/5/2009        Group1  NA
 1/12/2009       Group1  0.416548943
 1/19/2009       Group1  2.066275155
 1/26/2009       Group1  2.037729638
 2/2/2009        Group1  -0.598040483
 2/9/2009        Group1  1.658999227
 2/16/2009       Group1  -0.869325211
 12/29/2008      Group2  NA
 1/5/2009        Group2  NA
 1/12/2009       Group2  NA
 1/19/2009       Group2  0.375284194
 1/26/2009       Group2  0.706785401
 2/2/2009        Group2  NA
 2/9/2009        Group2  2.104937151
 2/16/2009       Group2  2.880393978



 On Mon, Apr 4, 2011 at 1:09 PM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!
 I have my data frame mydata (below) and data frame reference -
 that contains all the dates I would like to be present in the final
 data frame.
 I am trying to merge them so that the the result data frame contains
 all 8 dates in both subgroups (i.e., Group1 should have 8 rows and
 Group2 too). But when I merge it it's not coming out this way. Any
 hint would be greatly appreciated!
 Dimitri

 mydata-data.frame(mydate=rep(seq(as.Date(2008-12-29), length = 8,
 by = week),2),
 group=c(rep(Group1,8),rep(Group2,8)),values=rnorm(16,1,1))
 (reference);(mydata)
 set.seed(1234)
 out-sample(1:16,5,replace=F)
 mydata-mydata[-out,]; dim(mydata)
 (mydata)

 # reference contains the dates I want to be present in the final data 
 frame:
 reference-data.frame(mydate=seq(as.Date(2008-12-29), length = 8, by
 = week))

 # Merging:
 new.data-merge(mydata,reference,by=mydate,all.x=T,all.y=T)
 new.data-new.data[order(new.data$group,new.data$mydate),]
 (new.data)
 # my new.data contains only 7 rows in Group 1 and 4 rows in Group 2


 --
 Dimitri Liakhovitski
 Ninah Consulting




 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.com

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 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] A fortunes candidate?

2011-04-05 Thread Bert Gunter

A fortunes candidate?

On Tue, Apr 5, 2011 at 3:59 PM, David Winsemius dwinsem...@comcast.net wrote:

... snipped

 Tested solutions offered when reproducible examples are provided. 





-- 
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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[R] Error in match.names(clabs, names(xi))

2011-04-05 Thread Quan Zhou

Hi Guys,
I have this part of a program:
library(survival)
Gastric - cbind.data.frame(Gp=c(rep(1,45),rep(0,45)),  ### 2nd gp 0
time=c(1,63,105,129,182,216,250,262,301,301,342,354,356,358,
380,383, 383,388,394,408,460,489,499,523,524,535,562,569,675,676,
748,778,786,797,955,968,1000,1245,1271,1420,1551,1694,2363,2754,2950,
17,42,44,48,60,72,74,95,103,108,122,144,167,170,183,185,193,195,197,
208,234,235,254,307,315,401,445,464,484,528,542,547,577,580,795,855,
1366,1577,2060,2412,2486,2796,2802,2934,2988), Dth=c(rep(1,43),
0,0, rep(1,39), rep(0,6)))
CoxG0 - coxph(Surv(time,Dth) ~ Gp, Gastric)
srvGastA - survfit(Surv(Gastric$time,Gastric$Dth)~1) ## 88 distinct times
#Gastric$time is all the time points either death or largest obervation
time.
#srvGastA$time is all the unique times
newGas - data.frame(start=0, stop=1, Dth=1, Ploidy=1, tim=0)
#newGas - r(0,1,1,1,0)
for (i in 2:90) {
timind - match(Gastric$time[i],srvGastA$time)
tmpmat - array(0, dim=c(timind,5))#build an array with dim('index',5)
tmpmat[,4] - rep(Gastric[i,1], timind)#fourth column, return i's group
tmpmat[timind,3] - Gastric$Dth[i]
tmpmat[,1] - if(timind1) c(0,srvGastA$time[1:(timind-1)]) else 0
tmpmat[,2] - srvGastA$time[1:timind]
newGas - rbind(newGas,tmpmat) }

I found when include the last line in the for loop. the error will jump
out. But I do not know how to fix it. I initialize newGas without names. it
does not work either
It would be great if anyone knows how to fix the problem. Thanks a lot.
BR
Quan

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Re: [R] A fortunes candidate?

2011-04-05 Thread John Kane

Sounds more like an advertisement  A bit like the old TV story Paladin, have 
gun, will travel. 

--- On Tue, 4/5/11, Bert Gunter gunter.ber...@gene.com wrote:

 From: Bert Gunter gunter.ber...@gene.com
 Subject: [R] A fortunes candidate?
 To: David Winsemius dwinsem...@comcast.net
 A fortunes candidate?

 On Tue, Apr 5, 2011 at 3:59 PM, David Winsemius dwinsem...@comcast.net
 wrote:

  Tested solutions offered when reproducible examples are
 provided. 

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Re: [R] Error in match.names(clabs, names(xi))



On Apr 5, 2011, at 8:28 PM, Quan Zhou wrote:


Hi Guys,
I have this part of a program:
library(survival)
Gastric - cbind.data.frame(Gp=c(rep(1,45),rep(0,45)),  ### 2nd gp 0
   time=c(1,63,105,129,182,216,250,262,301,301,342,354,356,358,
   380,383, 383,388,394,408,460,489,499,523,524,535,562,569,675,676,

748,778,786,797,955,968,1000,1245,1271,1420,1551,1694,2363,2754,2950,

17,42,44,48,60,72,74,95,103,108,122,144,167,170,183,185,193,195,197,

208,234,235,254,307,315,401,445,464,484,528,542,547,577,580,795,855,

   1366,1577,2060,2412,2486,2796,2802,2934,2988), Dth=c(rep(1,43),
   0,0, rep(1,39), rep(0,6)))
CoxG0 - coxph(Surv(time,Dth) ~ Gp, Gastric)
srvGastA - survfit(Surv(Gastric$time,Gastric$Dth)~1) ## 88 distinct  
times
#Gastric$time is all the time points either death or largest  
obervation

time.
#srvGastA$time is all the unique times
newGas - data.frame(start=0, stop=1, Dth=1, Ploidy=1, tim=0)
#newGas - r(0,1,1,1,0)
for (i in 2:90) {
   timind - match(Gastric$time[i],srvGastA$time)
   tmpmat - array(0, dim=c(timind,5))#build an array with  
dim('index',5)
   tmpmat[,4] - rep(Gastric[i,1], timind)#fourth column, return i's  
group

   tmpmat[timind,3] - Gastric$Dth[i]
   tmpmat[,1] - if(timind1) c(0,srvGastA$time[1:(timind-1)]) else 0
   tmpmat[,2] - srvGastA$time[1:timind]
   newGas - rbind(newGas,tmpmat) }


You haven't told us what you expect (or even what you are trying to  
do),  but if you add an assignment of colnames before the final  
rbind()-ing of a data.frame and a matrix you get some sort of result:


.
+tmpmat[,2] - srvGastA$time[1:timind] ; colnames(tmpmat) -  
names(newGas)

+newGas - rbind(newGas,tmpmat) }
 str(newGas)
'data.frame':   3988 obs. of  5 variables:
 $ start : num  0 0 1 17 42 44 48 60 0 1 ...
 $ stop  : num  1 1 17 42 44 48 60 63 1 17 ...
 $ Dth   : num  1 0 0 0 0 0 0 1 0 0 ...
 $ Ploidy: num  1 1 1 1 1 1 1 1 1 1 ...
 $ tim   : num  0 0 0 0 0 0 0 0 0 0 ...




I found when include the last line in the for loop. the error will  
jump
out. But I do not know how to fix it. I initialize newGas without  
names.


Good luck with trying to make a data.frame without column names.


it
does not work either
It would be great if anyone knows how to fix the problem. Thanks a  
lot.

BR
Quan



--
David Winsemius, MD
West Hartford, CT

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[R] Pulling strings from a Flat file

2011-04-05 Thread Kalicin, Sarah

Hi,

I have a flat file that contains a bunch of strings that look like this. The 
file was originally in Unix and brought over into Windows:

E123456E234567E345678E456789E567891E678910E. . . .
Basically the string starts with E and is followed with 6 numbers. One 
string=E123456, length=7 characters. This file contains 10,000's of these 
strings. I want to separate them into one vector the length of the number of 
strings in the flat file, where each string is it's on unique value.

cc-c(7,7,7,7,7,7,7)
 aa- file(Master,r, raw=TRUE)
 readChar(aa, cc, useBytes = FALSE)
[1] E123456  \nE23456 7\nE3456 78\nE456 789\nE56 7891\nE6 78910\nE
 close(aa)
 unlink(Master)

The biggest issue is I am getting \n added into the string, which I am not sure 
where it is coming from, and splices the strings. Any suggestions on getting 
rid of the /n and create an infinite sequence of 7's for the string length for 
the cc vector? Is there a better way to do this?

Sarah



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Re: [R] Pulling strings from a Flat file



On Apr 5, 2011, at 7:48 PM, Kalicin, Sarah wrote:


Hi,

I have a flat file that contains a bunch of strings that look like  
this. The file was originally in Unix and brought over into Windows:


E123456E234567E345678E456789E567891E678910E. . . .
Basically the string starts with E and is followed with 6 numbers.  
One string=E123456, length=7 characters. This file contains 10,000's  
of these strings. I want to separate them into one vector the length  
of the number of strings in the flat file, where each string is it's  
on unique value.


cc-c(7,7,7,7,7,7,7)

aa- file(Master,r, raw=TRUE)
readChar(aa, cc, useBytes = FALSE)
[1] E123456  \nE23456 7\nE3456 78\nE456 789\nE56  
7891\nE6 78910\nE

close(aa)
unlink(Master)


 txt - E123456E234567E345678E456789E567891E678910E
# You could use readLines to bring in from the file
# and assign to a character vector for work in R.

 gsub((E[[:digit:]]{6}), \\1\n, txt)
[1] E123456\nE234567\nE345678\nE456789\nE567891\nE678910\nE
# Seems to be working properly

 ?scan

 scan(textConnection(gsub((E[[:digit:]]{6}), \\1\n, txt)),  
what=character)

Read 7 items
[1] E123456 E234567 E345678 E456789 E567891 E678910 E

You might be able to use read.table or variants.



The biggest issue is I am getting \n added into the string, which I  
am not sure where it is coming from, and splices the strings. Any  
suggestions on getting rid of the /n and create an infinite sequence  
of 7's for the string length for the cc vector? Is there a better  
way to do this?


Sarah



David Winsemius, MD
West Hartford, CT

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Re: [R] Arrangement of Lattice Histograms - Top to bottom and then left to right?