date:20100716

Dear list,

I am running R2.11.1 on 32 bit windows. I am receiving messages as
follows ...
 require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it

The package kohonen was another example.

This failure appears to be fatal and not only affects the package
concerned, but also all its dependents.

Is there anything I can do at my end, or do I have to wait for the
packages to be re-built,

Thanks,

Philip


This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it immediately.
Statements of intent shall only become binding when confirmed in hard copy by 
an authorised signatory.  The contents of this email may relate to dealings 
with other companies within the Detica Limited group of companies.

Detica Limited is registered in England under No: 1337451.

Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England.

[[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.

[R] Question about KLdiv and large datasets

2010-07-16 Thread Ralf B

Hi all,

when running KL on a small data set, everything is fine:

require(flexmix)
n - 20
a - rnorm(n)
b - rnorm(n)
mydata - cbind(a,b)
KLdiv(mydata)

however, when this dataset increases

require(flexmix)
n - 1000
a - rnorm(n)
b - rnorm(n)
mydata - cbind(a,b)
KLdiv(mydata)


KL seems to be not defined. Can somebody explain what is going on?

Thanks,
Ralf

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

[R] filtered back-projection

2010-07-16 Thread Kendralin Freeman

Hi List,

Does R have a package that will do filtered back-projection?  I'm a medical
physicist trying to produce an image of a dose cloud within a phantom using
software other than Matlab and wondered if R has this capability.  I looked
through the imaging pages on the project site but didn't see anything
specific about back-projection.

Thanks!

[[alternative HTML version deleted]]

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

Re: [R] Packages built before R 2.10.0


On 16/07/2010 9:50 AM, Philip Whittall wrote:

Dear list,

I am running R2.11.1 on 32 bit windows. I am receiving messages as
follows ...
 require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it

The package kohonen was another example.

This failure appears to be fatal and not only affects the package
concerned, but also all its dependents.

Is there anything I can do at my end, or do I have to wait for the
packages to be re-built,
  


Just follow the instructions:  re-install the packages and they will be 
fine.  They've already been rebuilt for 2.11.x long ago.


Duncan Murdoch

Thanks,

Philip


This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it immediately.
Statements of intent shall only become binding when confirmed in hard copy by 
an authorised signatory.  The contents of this email may relate to dealings 
with other companies within the Detica Limited group of companies.

Detica Limited is registered in England under No: 1337451.

Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England.

[[alternative HTML version deleted]]

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



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

Re: [R] Packages built before R 2.10.0

Hi Duncan,

Many thanks for the response. I did indeed re-install car and this is
what occurred ...

 utils:::menuInstallLocal()
package 'car' successfully unpacked and MD5 sums checked
 require(candisc)
Loading required package: candisc
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 utils:::menuInstallLocal()
package 'candisc' successfully unpacked and MD5 sums checked
 require(candisc)
Loading required package: candisc
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 
I had checked the .zip files on the CRAN site ahd I do have the current
ones I had also
run update.packages() both as a command and from the console pull down
menu.

I have probably missed something obvious,

Thanks,

Philip
 

-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: 16 July 2010 15:24
To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10.0

On 16/07/2010 9:50 AM, Philip Whittall wrote:
 Dear list,

 I am running R2.11.1 on 32 bit windows. I am receiving messages as 
 follows ...
  require(car)
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it

 The package kohonen was another example.

 This failure appears to be fatal and not only affects the package 
 concerned, but also all its dependents.

 Is there anything I can do at my end, or do I have to wait for the 
 packages to be re-built,
   

Just follow the instructions:  re-install the packages and they will be
fine.  They've already been rebuilt for 2.11.x long ago.

Duncan Murdoch
 Thanks,

 Philip


 This message should be regarded as confidential. If you have received
this email in error please notify the sender and destroy it immediately.
 Statements of intent shall only become binding when confirmed in hard
copy by an authorised signatory.  The contents of this email may relate
to dealings with other companies within the Detica Limited group of
companies.

 Detica Limited is registered in England under No: 1337451.

 Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP,
England.

   [[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.
   

This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it immediately.
Statements of intent shall only become binding when confirmed in hard copy by 
an authorised signatory.  The contents of this email may relate to dealings 
with other companies within the Detica Limited group of companies.

Detica Limited is registered in England under No: 1337451.

Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England.


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

Re: [R] Question about KLdiv and large datasets

2010-07-16 Thread Peter Ehlers


On 2010-07-16 7:56, Ralf B wrote:

Hi all,

when running KL on a small data set, everything is fine:

require(flexmix)
n- 20
a- rnorm(n)
b- rnorm(n)
mydata- cbind(a,b)
KLdiv(mydata)

however, when this dataset increases

require(flexmix)
n- 1000
a- rnorm(n)
b- rnorm(n)
mydata- cbind(a,b)
KLdiv(mydata)


KL seems to be not defined. Can somebody explain what is going on?

Thanks,
Ralf


Ralf,

You can adjust the 'eps=' argument. But I don't know
what this will do to the reliability of the results.

KLdiv(mydata, eps = 1e-7)

  -Peter Ehlers

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

Re: [R] Packages built before R 2.10.0


On 16/07/2010 10:41 AM, Philip Whittall wrote:

Hi Duncan,

Many thanks for the response. I did indeed re-install car and this is
what occurred ...

 utils:::menuInstallLocal()
package 'car' successfully unpacked and MD5 sums checked
 require(candisc)
Loading required package: candisc
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 utils:::menuInstallLocal()
package 'candisc' successfully unpacked and MD5 sums checked
 require(candisc)
Loading required package: candisc
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
 
I had checked the .zip files on the CRAN site ahd I do have the current

ones I had also
run update.packages() both as a command and from the console pull down
menu.

I have probably missed something obvious,
  


I would guess you need to use a more up-to-date mirror.  It's also 
possible that you have somehow messed up the path to your library and 
are installing to one place but loading from another, but normally new 
installs go into the first search position, so I don't see how that 
could be the cause.


Duncan Murdoch

Thanks,

Philip
 


-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: 16 July 2010 15:24

To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10.0

On 16/07/2010 9:50 AM, Philip Whittall wrote:
 Dear list,

 I am running R2.11.1 on 32 bit windows. I am receiving messages as 
 follows ...

  require(car)
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it

 The package kohonen was another example.

 This failure appears to be fatal and not only affects the package 
 concerned, but also all its dependents.


 Is there anything I can do at my end, or do I have to wait for the 
 packages to be re-built,
   


Just follow the instructions:  re-install the packages and they will be
fine.  They've already been rebuilt for 2.11.x long ago.

Duncan Murdoch
 Thanks,

 Philip


 This message should be regarded as confidential. If you have received
this email in error please notify the sender and destroy it immediately.
 Statements of intent shall only become binding when confirmed in hard
copy by an authorised signatory.  The contents of this email may relate
to dealings with other companies within the Detica Limited group of
companies.

 Detica Limited is registered in England under No: 1337451.

 Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP,
England.

[[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.
   


This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it immediately.
Statements of intent shall only become binding when confirmed in hard copy by 
an authorised signatory.  The contents of this email may relate to dealings 
with other companies within the Detica Limited group of companies.

Detica Limited is registered in England under No: 1337451.

Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England.





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

Re: [R] Packages built before R 2.10.0


 I should have added that this problem is not serious under 2.10 as you
only get a Warning and everything works

  require(candisc)
Loading required package: candisc
Loading required package: car
Loading required package: heplots
Warning message:
package 'car' was built under R version 2.9.2 and help will not work
correctly
Please re-install it 
 require(car)
 require(candisc)
 

I am using the mirror from Imperial College London and a check on the
\library\car subdirectory of my
R-2.11.1 installation show that it was updated at the expected time,

Thanks,

Philip


-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: 16 July 2010 15:45
To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10.0

On 16/07/2010 10:41 AM, Philip Whittall wrote:
 Hi Duncan,

 Many thanks for the response. I did indeed re-install car and this is 
 what occurred ...

  utils:::menuInstallLocal()
 package 'car' successfully unpacked and MD5 sums checked
  require(candisc)
 Loading required package: candisc
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  utils:::menuInstallLocal()
 package 'candisc' successfully unpacked and MD5 sums checked
  require(candisc)
 Loading required package: candisc
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  require(car)
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  
 I had checked the .zip files on the CRAN site ahd I do have the 
 current ones I had also run update.packages() both as a command and 
 from the console pull down menu.

 I have probably missed something obvious,
   

I would guess you need to use a more up-to-date mirror.  It's also
possible that you have somehow messed up the path to your library and
are installing to one place but loading from another, but normally new
installs go into the first search position, so I don't see how that
could be the cause.

Duncan Murdoch
 Thanks,

 Philip
  

 -Original Message-
 From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
 Sent: 16 July 2010 15:24
 To: Philip Whittall
 Cc: r-help@r-project.org
 Subject: Re: [R] Packages built before R 2.10.0

 On 16/07/2010 9:50 AM, Philip Whittall wrote:
  Dear list,
 
  I am running R2.11.1 on 32 bit windows. I am receiving messages as 
  follows ...
   require(car)
  Loading required package: car
  Error: package 'car' was built before R 2.10.0: please re-install it
 
  The package kohonen was another example.
 
  This failure appears to be fatal and not only affects the package 
  concerned, but also all its dependents.
 
  Is there anything I can do at my end, or do I have to wait for the 
  packages to be re-built,


 Just follow the instructions:  re-install the packages and they will 
 be fine.  They've already been rebuilt for 2.11.x long ago.

 Duncan Murdoch
  Thanks,
 
  Philip
 
 
  This message should be regarded as confidential. If you have 
  received
 this email in error please notify the sender and destroy it
immediately.
  Statements of intent shall only become binding when confirmed in 
  hard
 copy by an authorised signatory.  The contents of this email may 
 relate to dealings with other companies within the Detica Limited 
 group of companies.
 
  Detica Limited is registered in England under No: 1337451.
 
  Registered offices: Surrey Research Park, Guildford, Surrey, GU2 
  7YP,
 England.
 
  [[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.


 This message should be regarded as confidential. If you have received
this email in error please notify the sender and destroy it immediately.
 Statements of intent shall only become binding when confirmed in hard
copy by an authorised signatory.  The contents of this email may relate
to dealings with other companies within the Detica Limited group of
companies.

 Detica Limited is registered in England under No: 1337451.

 Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP,
England.


   

This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it immediately.
Statements of intent shall only become binding when confirmed in hard copy by 
an authorised signatory.  The contents of this email may relate to dealings 
with other companies within the Detica Limited group of companies.

Detica Limited is registered in England under No: 1337451.

Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP, England.


__
R-help@r-project.org mailing list

[R] read.table input array

2010-07-16 Thread Balpo


 Hello to all!
I am new with R and I need your help.
I'm trying to read a file which contests are similar to this:
ABCTLengths
14.00.001525878918c(1,2,3)
11.00.001716613824c(1,1,4)

So all the columns are numeric values, except Lengths, which is supposed to
be an variable length array of integers.
How can I make R read them as arrays of integers? Or otherwise, convert the
character array to an array of integers.
When I read the file, I do it like this
t1 = read.table(file=paste(./borrar.dat,sep=), header=T, 
colClasses=c(numeric, numeric, numeric, numeric, array))
But the 5th column is treated as an array of characters, and when trying 
to convert it to another class of data, I either
get two strings c(1,2,3) and c(1,1,4) or using a toRaw converter, I 
get the corresponding ASCII ¿? values.
Should the input be modified in order to be able to read it as an array 
of integers?


Thank you for your help.
Balpo

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

[R] assign

2010-07-16 Thread linda.s

Hi.
I am a beginner. What is the difference between:
x-2
and
x=2

Thanks.
Linda

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

Re: [R] Packages built before R 2.10.0


On 16/07/2010 10:51 AM, Philip Whittall wrote:

 I should have added that this problem is not serious under 2.10 as you
only get a Warning and everything works

  require(candisc)
Loading required package: candisc
Loading required package: car
Loading required package: heplots
Warning message:
package 'car' was built under R version 2.9.2 and help will not work
correctly
Please re-install it 
 require(car)

 require(candisc)
 


I am using the mirror from Imperial College London and a check on the
\library\car subdirectory of my
R-2.11.1 installation show that it was updated at the expected time,
  


I just did an install from the UK (London) mirror (the same one?) and it 
works fine.  So I've got no idea what's going wrong for you.  I'd 
suggest deleting the directory holding the problem package (i.e. 
\library\car) and confirming that the error changes to Error in 
library(car) : there is no package called 'car'.  Then do an install using


install.packages(car)

and see if that fixes things.

Duncan Murdoch

Thanks,

Philip


-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: 16 July 2010 15:45

To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10.0

On 16/07/2010 10:41 AM, Philip Whittall wrote:
 Hi Duncan,

 Many thanks for the response. I did indeed re-install car and this is 
 what occurred ...


  utils:::menuInstallLocal()
 package 'car' successfully unpacked and MD5 sums checked
  require(candisc)
 Loading required package: candisc
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  utils:::menuInstallLocal()
 package 'candisc' successfully unpacked and MD5 sums checked
  require(candisc)
 Loading required package: candisc
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  require(car)
 Loading required package: car
 Error: package 'car' was built before R 2.10.0: please re-install it
  
 I had checked the .zip files on the CRAN site ahd I do have the 
 current ones I had also run update.packages() both as a command and 
 from the console pull down menu.


 I have probably missed something obvious,
   


I would guess you need to use a more up-to-date mirror.  It's also
possible that you have somehow messed up the path to your library and
are installing to one place but loading from another, but normally new
installs go into the first search position, so I don't see how that
could be the cause.

Duncan Murdoch
 Thanks,

 Philip
  


 -Original Message-
 From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
 Sent: 16 July 2010 15:24
 To: Philip Whittall
 Cc: r-help@r-project.org
 Subject: Re: [R] Packages built before R 2.10.0

 On 16/07/2010 9:50 AM, Philip Whittall wrote:
  Dear list,
 
  I am running R2.11.1 on 32 bit windows. I am receiving messages as 
  follows ...

   require(car)
  Loading required package: car
  Error: package 'car' was built before R 2.10.0: please re-install it
 
  The package kohonen was another example.
 
  This failure appears to be fatal and not only affects the package 
  concerned, but also all its dependents.

 
  Is there anything I can do at my end, or do I have to wait for the 
  packages to be re-built,


 Just follow the instructions:  re-install the packages and they will 
 be fine.  They've already been rebuilt for 2.11.x long ago.


 Duncan Murdoch
  Thanks,
 
  Philip
 
 
  This message should be regarded as confidential. If you have 
  received

 this email in error please notify the sender and destroy it
immediately.
  Statements of intent shall only become binding when confirmed in 
  hard
 copy by an authorised signatory.  The contents of this email may 
 relate to dealings with other companies within the Detica Limited 
 group of companies.

 
  Detica Limited is registered in England under No: 1337451.
 
  Registered offices: Surrey Research Park, Guildford, Surrey, GU2 
  7YP,

 England.
 
[[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.



 This message should be regarded as confidential. If you have received
this email in error please notify the sender and destroy it immediately.
 Statements of intent shall only become binding when confirmed in hard
copy by an authorised signatory.  The contents of this email may relate
to dealings with other companies within the Detica Limited group of
companies.

 Detica Limited is registered in England under No: 1337451.

 Registered offices: Surrey Research Park, Guildford, Surrey, GU2 7YP,
England.


   


This message should be regarded as confidential. If you have received this 
email in error please notify the sender and destroy it

Re: [R] Packages built before R 2.10.0

Hi Duncan,

That fixed it. Many thanks indeed, I now know what to do if it happens
again,

Philip
 

-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: 16 July 2010 16:00
To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10.0

On 16/07/2010 10:51 AM, Philip Whittall wrote:
  I should have added that this problem is not serious under 2.10 as 
 you only get a Warning and everything works

   require(candisc)
 Loading required package: candisc
 Loading required package: car
 Loading required package: heplots
 Warning message:
 package 'car' was built under R version 2.9.2 and help will not work 
 correctly Please re-install it
  require(car)
  require(candisc)
  

 I am using the mirror from Imperial College London and a check on the 
 \library\car subdirectory of my
 R-2.11.1 installation show that it was updated at the expected time,
   

I just did an install from the UK (London) mirror (the same one?) and it
works fine.  So I've got no idea what's going wrong for you.  I'd
suggest deleting the directory holding the problem package (i.e. 
\library\car) and confirming that the error changes to Error in
library(car) : there is no package called 'car'.  Then do an install
using

install.packages(car)

and see if that fixes things.

Duncan Murdoch
 Thanks,

 Philip


 -Original Message-
 From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
 Sent: 16 July 2010 15:45
 To: Philip Whittall
 Cc: r-help@r-project.org
 Subject: Re: [R] Packages built before R 2.10.0

 On 16/07/2010 10:41 AM, Philip Whittall wrote:
  Hi Duncan,
 
  Many thanks for the response. I did indeed re-install car and this 
  is what occurred ...
 
   utils:::menuInstallLocal()
  package 'car' successfully unpacked and MD5 sums checked
   require(candisc)
  Loading required package: candisc
  Loading required package: car
  Error: package 'car' was built before R 2.10.0: please re-install it
   utils:::menuInstallLocal()
  package 'candisc' successfully unpacked and MD5 sums checked
   require(candisc)
  Loading required package: candisc
  Loading required package: car
  Error: package 'car' was built before R 2.10.0: please re-install it
   require(car)
  Loading required package: car
  Error: package 'car' was built before R 2.10.0: please re-install it
   
  I had checked the .zip files on the CRAN site ahd I do have the 
  current ones I had also run update.packages() both as a command and 
  from the console pull down menu.
 
  I have probably missed something obvious,


 I would guess you need to use a more up-to-date mirror.  It's also 
 possible that you have somehow messed up the path to your library and 
 are installing to one place but loading from another, but normally new

 installs go into the first search position, so I don't see how that 
 could be the cause.

 Duncan Murdoch
  Thanks,
 
  Philip
   
 
  -Original Message-
  From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
  Sent: 16 July 2010 15:24
  To: Philip Whittall
  Cc: r-help@r-project.org
  Subject: Re: [R] Packages built before R 2.10.0
 
  On 16/07/2010 9:50 AM, Philip Whittall wrote:
   Dear list,
  
   I am running R2.11.1 on 32 bit windows. I am receiving messages as

   follows ...
require(car)
   Loading required package: car
   Error: package 'car' was built before R 2.10.0: please re-install 
   it
  
   The package kohonen was another example.
  
   This failure appears to be fatal and not only affects the package 
   concerned, but also all its dependents.
  
   Is there anything I can do at my end, or do I have to wait for the

   packages to be re-built,
 
 
  Just follow the instructions:  re-install the packages and they will

  be fine.  They've already been rebuilt for 2.11.x long ago.
 
  Duncan Murdoch
   Thanks,
  
   Philip
  
  
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Re: [R] assign


On 16/07/2010 10:57 AM, linda.s wrote:

Hi.
I am a beginner. What is the difference between:
x-2
and
x=2
  


It depends on the context.  x - 2 always assigns the value 2 to the 
variable named x.  Typed as an isolated command, that's what x = 2 
does too.  However, when used as an argument to a function, e.g. f(x = 
2), it associates the value 2 to the *argument* matched by x in the 
function call.  This is occasionally confusing, so many people recommend 
that you always use x - 2 when you mean to do an assignment, and 
reserve the use of = for function arguments.


Duncan Murdoch

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Re: [R] question about string handling....

2010-07-16 Thread karena


hey, guys, all these methods work perfectly. thank you!!
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[R] package for rank ordered logit

2010-07-16 Thread Suresh Singh

Is there a package in R that can run rank-ordered logit?

Thanks,
Suresh

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[R] save plot

2010-07-16 Thread linda.s

I made a plot, but after I made a second plot, the previous plot was
gone. How can I save all the plots in a file (I do not manually copy
and paste them one by one)?
Thanks.
Linda

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Re: [R] save plot

We need to know how you're doing this, with a minimal example that we 
can run.  Most graphics devices accept a file or filename argument, so 
that's one way.  If you're using the pdf device, multiple plots will 
create multiple pages in the final output..


linda.s wrote:

I made a plot, but after I made a second plot, the previous plot was
gone. How can I save all the plots in a file (I do not manually copy
and paste them one by one)?
Thanks.
Linda

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[R] Simple question regarding name of column headers

2010-07-16 Thread Addi Wei


names(miceTrainSample)
[1] b_double  KierA2KierFlex  Q_VSA_POS pID50 

In the above code, how do I delete pID50 column to store the resulting
object without indicating column 5.  The code below does the trick, but I
wish to delete the column by specifying -pID50 instead of 5. 

names(miceTrainSample)[-5]
[1] b_double  KierA2KierFlex  Q_VSA_POS

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[R] changing names numeric to a time series

2010-07-16 Thread Erin Hodgess

Dear R People:

What is the best way to change a named numeric (in which the names are
dates) to a time series, please?

 str(y1)
 Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
 - attr(*, names)= chr [1:730] 2006-01-01 2006-01-02
2006-01-03 2006-01-04 ...


Thanks in advance,
Sincerely,
Erin


-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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[R] changing names numeric to a time series - solved

2010-07-16 Thread Erin Hodgess

Here is the original question:

What is the best way to change a named numeric to a time series

 str(y1)
 Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
 - attr(*, names)= chr [1:730] 2006-01-01 2006-01-02
2006-01-03 2006-01-04 ...


y2 - zoo(y1,order=names(y1))

Perfect!

Thanks though,
Erin


-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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Re: [R] package for rank ordered logit

2010-07-16 Thread Tal Galili

Did you try:
library(MASS)
?polr


?


Contact
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www.r-statistics.com (English)
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On Fri, Jul 16, 2010 at 6:15 PM, Suresh Singh singh@osu.edu wrote:

 Is there a package in R that can run rank-ordered logit?

 Thanks,
 Suresh

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Re: [R] Simple question regarding name of column headers

2010-07-16 Thread Ista Zahn

Hi Addi,

On Fri, Jul 16, 2010 at 3:22 PM, Addi Wei addi...@gmail.com wrote:

 names(miceTrainSample)
 [1] b_double  KierA2    KierFlex  Q_VSA_POS pID50

 In the above code, how do I delete pID50 column to store the resulting
 object without indicating column 5.  The code below does the trick, but I
 wish to delete the column by specifying -pID50 instead of 5.

 names(miceTrainSample)[-5]
 [1] b_double  KierA2    KierFlex  Q_VSA_POS

If I understand you correctly, than this code will not do the trick.
All it does is print the column names minus pID50. It does nothing to
miceTrainSample.

Anyway, I have often wished that something like

new.mt.sample - miceTrainSample[, -pID50]

would return miceTrainSample without the pID50 column. Here are three
alternative ways to do it.

# Method 1: Assign NULL to the column
new.mt.sample - miceTrainsSample
new.mt.sample$pID50 - NULL

# Method 2: Use which()
new.mt.sample - miceTrainSample[, - which(names(miceTrainSample == pID50)]

# Method 3: use %in% (the one I usually use)
new.mt.sample - miceTrainSample[, ! names(miceTrainSample) %in% pID50]

Hope it helps,
Ista

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-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] changing names numeric to a time series

On Fri, Jul 16, 2010 at 11:26 AM, Erin Hodgess erinm.hodg...@gmail.com wrote:
 Dear R People:

 What is the best way to change a named numeric (in which the names are
 dates) to a time series, please?

 str(y1)
  Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
  - attr(*, names)= chr [1:730] 2006-01-01 2006-01-02
 2006-01-03 2006-01-04 ...


Try this:

  zoo(unname(y1), as.Date(names(y1)))

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[R] I need help making a data.fame comprised of selected columns of an original data frame.

I must have missed something simple, but still, i don't know what.

I obtained my basic data as follows:

x - sprintf(SELECT m_id,sale_date,YEAR(sale_date) AS
sale_year,WEEK(sale_date) AS sale_week,return_type,0.0001 +
DATEDIFF(return_date,sale_date) AS elapsed_time FROM
`merchants2`.`risk_input` WHERE DATEDIFF(return_date,sale_date) IS NOT
NULL)
moreinfo - dbGetQuery(con, x)

I then made the data frame I want to use as follows:

fun_m_id - function(df)
  if (length(df$elapsed_time)  5) {
rv = fitdist(df$elapsed_time,exp)
rv$mid = df$m_id[1]
rv
  }
aaa - lapply(split(moreinfo,list(moreinfo$m_id),drop = TRUE), fun_m_id)
m_id_default_res - do.call(rbind, aaa)

At this point, each row in m_id_default_res corresponds to one data.frame
produced by fitdist.  When I print it, I get the output I expected.
However, I need to store only some of it into my DB.

And then, because fitdist produces a data frame that includes a lot of info
I don't need to store in the DB, I tried making a new data.frame containing
only the info I need as follows:
ndf = data.frame()
for (i in 1:length(m_id_default_res[,1])) {
  ndf$mid[i] = m_id_default_res$mid[i]
  ndf$estimate[i] = m_id_default_res$estimate[i]
  ndf$sd[i] = m_id_default_res$sd[i]
  ndf$n[i] = m_id_default_res[i]
  ndf$loglik[i] = m_id_default_res$loglik[i]
  ndf$aic[i] = m_id_default_res$aic[i]
  ndf$bic[i] = m_id_default_res$bic[i]
  ndf$chisq[i] = m_id_default_res$chisq[i]
  ndf$chisqpvalue[i] = m_id_default_res$chisqpvalue[i]
  ndf$chisqdf[i] = m_id_default_res$chisqdf[i]
}
ndf

And I get the following error:
Error in `$-.data.frame`(`*tmp*`, n, value = list(0.114752782316094)) :
  replacement has 1 rows, data has 0

I need to either get rid of the columns in m_id_default_res that I don't
need, or I need to copy only those columns I need to a new data.frame.  How
do I do this.  Obviously, doing an element-wise copy, at least as I tried to
do it, doesn't work.

Thanks,

Ted

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Re: [R] discrepancy matrix


I guess your data frame is a little different from the reference, so your
as.logical doesn't work.

attach(Q)

FUN - function(X, Y) {abs(X - Y)}

round(outer(rank(date)[colour==b],rank(date)[colour==g],FUN) +
outer(rank(number)[colour==b],rank(number)[colour==g],FUN))

detach(Q) 

-
A R learner.
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Re: [R] I need help making a data.fame comprised of selected columns of an original data frame.

2010-07-16 Thread Steve Lianoglou

Hi,

First: it's kind of hard to play along w/o some reproducible data. To
that end, you can paste into an email the output of:

dput(moreinfo)

If there are lots of rows in `moreinfo`, just give us the first ~10-20

dput(head(moreinfo, 20))

Anyway:

snip
 At this point, each row in m_id_default_res corresponds to one data.frame
 produced by fitdist.  When I print it, I get the output I expected.
 However, I need to store only some of it into my DB.

 And then, because fitdist produces a data frame that includes a lot of info
 I don't need to store in the DB, I tried making a new data.frame containing
 only the info I need as follows:
 ndf = data.frame()
 for (i in 1:length(m_id_default_res[,1])) {
  ndf$mid[i] = m_id_default_res$mid[i]
  ndf$estimate[i] = m_id_default_res$estimate[i]
  ndf$sd[i] = m_id_default_res$sd[i]
  ndf$n[i] = m_id_default_res[i]
  ndf$loglik[i] = m_id_default_res$loglik[i]
  ndf$aic[i] = m_id_default_res$aic[i]
  ndf$bic[i] = m_id_default_res$bic[i]
  ndf$chisq[i] = m_id_default_res$chisq[i]
  ndf$chisqpvalue[i] = m_id_default_res$chisqpvalue[i]
  ndf$chisqdf[i] = m_id_default_res$chisqdf[i]
 }

Forget the for loop. How about:

ndf - m_id_default[, c('mid, 'estimate', 'sd', 'loglik', 'aic',
'bic', 'chisq', 'chisqpvalue', 'chisqdf')

Having just written that, I see something strange in your for loop.
Specifically this line:

  ndf$n[i] = m_id_default_res[i]

m_id_default_res is a data.frame, right? Why don't you try to see what
`m_id_default_res[1]` returns.

I'm not sure that that's what your error message is coming from, but I
foresee this to be a problem anyway, if I follow your build up code
correctly.

Hope that helps,

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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Re: [R] Simple question regarding name of column headers


Anyway, I have often wished that something like

new.mt.sample - miceTrainSample[, -pID50]

would return miceTrainSample without the pID50 column. Here are three
alternative ways to do it.

# Method 1: Assign NULL to the column
new.mt.sample - miceTrainsSample
new.mt.sample$pID50 - NULL

# Method 2: Use which()
new.mt.sample - miceTrainSample[, - which(names(miceTrainSample == pID50)]

# Method 3: use %in% (the one I usually use)
new.mt.sample - miceTrainSample[, ! names(miceTrainSample) %in% pID50]


As a variation on Method 1...

df - data.frame(a = 1:10, b = 2:11, c = 3:12)
transform(df, a = NULL)

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Re: [R] how to skip a specific value when using apply() function to a matrix?

2010-07-16 Thread Shuhua Zhan

Hello Nikhil and Wu,
Thank you very much for your reply!
What I want is to calculate the student's score column-wise by ignoring the 
specific values such as zeros for example only using c(2,1) in column 8 in the 
tmp1 and generate tmp2. I changed the zeros to NAs and modified my stud fun to 
student as below. So I can ignore these specific values but can not put back 
the NAs to the right position in the resulted matrix.
Any suggestions to put back NAs to their original positions in column 7 and 8 
either to list tmp4 or matrix tmp5?

tmp1[tmp1==0]-NA
student- function(x){
x-x[!is.na(x)]
x-(x-mean(x))/sd(x)
return (x)
}
tmp4-apply(tmp1, 2, student)
tmp4
[[1]]
[1]  1.1296201  0.1613743 -1.2909944  1.1296201 -0.3227486 -0.8068715

[[2]]
[1]  0.000  0.000 -1.4680505 -0.5872202  0.5872202  1.4680505

[[3]]
[1] -0.5207910 -0.4817316 -0.4036130  0.2994548  1.9008870 -0.7942062

[[4]]
[1] -0.8630035 -0.2380699 -0.7737273 -0.5951748  1.5474546  0.9225210

[[5]]
[1] -0.1913482 -0.6944435 -0.7202433  0.5826446  1.7565336 -0.7331431

[[6]]
[1] -0.1132277  0.5661385 -1.4719601 -0.7925939  0.5661385  1.2455047

[[7]]
[1] -0.9561829  0.2390457  1.4342743  0.2390457 -0.9561829

[[8]]
[1]  0.7071068 -0.7071068

tmp5- matrix(unlist(tmp4),nrow=6)
tmp5
   [,1]   [,2]   [,3]   [,4]   [,5]   [,6]   
[,7]   [,8]
[1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
-0.9561829 -0.7071068
[2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
0.2390457  1.1296201
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
1.4342743  0.1613743
[4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
0.2390457 -1.2909944
[5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385 
-0.9561829  1.1296201
[6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047  
0.7071068 -0.3227486

Joshua

- Original Message -
From: Shuhua Zhan sz...@uoguelph.ca
To: r-help@r-project.org
Sent: Thursday, July 15, 2010 11:08:34 PM GMT -05:00 US/Canada Eastern
Subject: [R] how to skip a specific value when using apply() function to a 
matrix?

Hello R experts,
I'd like to studentize a matrix (tmp1) by column using apply() function and 
skip some specific values such as zeros in the example below to tmp2 but not 
tmp3. I used the script below and only can get a matrix tmp3. Could you please 
help me to studentize the matrix (tmp1) without changing the zeros and generate 
a new matrix tmp2?
Thanks,
Joshua

tmp1
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]   1579   49   60320
[2,]   137   10   56   21430
[3,]   102   12   50   19102
[4,]   155   30   52  120240
[5,]   129   71   76  211431
[6,]   11   122   69   18520

tmp2
[1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
-0.9561829  0.000
[2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
0.2390457  0.000
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
0.000  0.7071068
[4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
1.4342743  0.000
[5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385  
0.2390457 -0.7071068
[6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047 
-0.9561829  0.000

tmp3
  [,1]   [,2]   [,3]   [,4]   [,5]   [,6]  [,7] 
  [,8]
[1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
-0.243975 -0.5976143
[2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
0.487950 -0.5976143
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601 
-1.707825  1.7928429
[4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
1.219875 -0.5976143
[5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385  
0.487950  0.5976143
[6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047 
-0.243975 -0.5976143

Here is my script:
stud- function(x){
x-(x-mean(x))/sd(x)
return (x)
}
tmp3-apply(tmp1,2,stud)

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Re: [R] read.table input array

2010-07-16 Thread jim holtman

Here is a way of creating a separate list of variable length vectors
that you can use in your processing:

 # read into a dataframe
 x - read.table(textConnection(ABCTLengths
+ 14.00.001525878918c(1,2,3)
+ 14.00.001525878918c(1,2,6,7,8,3)
+ 14.00.001525878918c(1,2,3,1,2,3,4,5,6,7,9)
+ 14.00.001525878918c(1,2,3)
+ 11.00.001716613824c(1,1,4)), header=TRUE)
 # create a  'list' with the variable length vectors
 # assuming the the Lengths are legal R expressions using 'c'
 x$varList - lapply(x$Lengths, function(a) eval(parse(text=a)))

 x
  A B   C  T  Lengths varList
1 1 4 0.001525879 18 c(1,2,3) 1, 2, 3
2 1 4 0.001525879 18   c(1,2,6,7,8,3)1, 2, 6, 7, 8, 3
3 1 4 0.001525879 18 c(1,2,3,1,2,3,4,5,6,7,9) 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 9
4 1 4 0.001525879 18 c(1,2,3) 1, 2, 3
5 1 1 0.001716614 24 c(1,1,4) 1, 1, 4
 str(x)
'data.frame':   5 obs. of  6 variables:
 $ A  : int  1 1 1 1 1
 $ B  : num  4 4 4 4 1
 $ C  : num  0.00153 0.00153 0.00153 0.00153 0.00172
 $ T  : int  18 18 18 18 24
 $ Lengths: Factor w/ 4 levels c(1,1,4),c(1,2,3),..: 2 4 3 2 1
 $ varList:List of 5
  ..$ : num  1 2 3
  ..$ : num  1 2 6 7 8 3
  ..$ : num  1 2 3 1 2 3 4 5 6 7 ...
  ..$ : num  1 2 3
  ..$ : num  1 1 4


On Fri, Jul 16, 2010 at 10:51 AM, Balpo ba...@gmx.net wrote:
  Hello to all!
 I am new with R and I need your help.
 I'm trying to read a file which contests are similar to this:
 A    B    C    T    Lengths
 1    4.0    0.0015258789    18    c(1,2,3)
 1    1.0    0.0017166138    24    c(1,1,4)

 So all the columns are numeric values, except Lengths, which is supposed to
 be an variable length array of integers.
 How can I make R read them as arrays of integers? Or otherwise, convert the
 character array to an array of integers.
 When I read the file, I do it like this
 t1 = read.table(file=paste(./borrar.dat,sep=), header=T,
 colClasses=c(numeric, numeric, numeric, numeric, array))
 But the 5th column is treated as an array of characters, and when trying to
 convert it to another class of data, I either
 get two strings c(1,2,3) and c(1,1,4) or using a toRaw converter, I get
 the corresponding ASCII ¿? values.
 Should the input be modified in order to be able to read it as an array of
 integers?

 Thank you for your help.
 Balpo

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Re: [R] Simple question regarding name of column headers

2010-07-16 Thread jim holtman

subset(miceTrainSample, select = -plD50)

On Fri, Jul 16, 2010 at 11:22 AM, Addi Wei addi...@gmail.com wrote:

 names(miceTrainSample)
 [1] b_double  KierA2    KierFlex  Q_VSA_POS pID50

 In the above code, how do I delete pID50 column to store the resulting
 object without indicating column 5.  The code below does the trick, but I
 wish to delete the column by specifying -pID50 instead of 5.

 names(miceTrainSample)[-5]
 [1] b_double  KierA2    KierFlex  Q_VSA_POS

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 http://r.789695.n4.nabble.com/Simple-question-regarding-name-of-column-headers-tp2291534p2291534.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] I need help making a data.fame comprised of selected columns of an original data frame.

Hi Steve,

Thanks

Here is a tiny subset of the data:
 dput(head(moreinfo, 40))
structure(list(m_id = c(171, 206, 206, 206, 206, 206, 206, 218,
224, 224, 227, 229, 229, 229, 229, 229, 229, 229, 229, 233, 233,
238, 238, 251, 251, 251, 251, 251, 251, 251, 251, 251, 251, 251,
251, 251, 251, 251, 251, 251), sale_date = c(2008-04-25 07:41:09,
2008-05-09 20:58:12, 2008-09-06 19:51:52, 2008-05-01 21:26:40,
2008-08-06 23:53:17, 2008-05-29 18:44:50, 2008-05-16 16:10:52,
2008-12-30 17:59:54, 2008-11-06 18:15:40, 2008-09-05 17:43:51,
2008-10-31 21:55:52, 2008-04-30 21:30:36, 2008-11-11 00:43:54,
2008-07-24 22:26:29, 2008-10-07 17:57:22, 2008-04-23 20:39:41,
2008-09-08 22:42:12, 2008-11-13 00:09:59, 2008-04-15 22:57:31,
2008-07-05 08:52:58, 2008-10-04 13:17:02, 2008-03-20 23:02:12,
2008-08-08 16:48:42, 2008-06-04 04:31:20, 2008-09-27 07:02:14,
2008-09-08 07:16:39, 2008-09-25 07:09:11, 2008-09-23 07:02:39,
2008-08-09 07:31:46, 2008-09-28 07:02:13, 2008-07-05 07:26:46,
2008-05-11 04:01:55, 2008-06-26 07:46:17, 2008-07-09 07:36:16,
2008-07-21 18:36:44, 2008-10-11 07:01:36, 2008-07-21 19:03:42,
2008-05-07 04:21:23, 2008-10-14 07:07:02, 2008-05-12 04:26:21
), sale_year = c(2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L, 2008L,
2008L, 2008L, 2008L, 2008L, 2008L, 2008L), sale_week = c(16L,
18L, 35L, 17L, 31L, 21L, 19L, 52L, 44L, 35L, 43L, 17L, 45L, 29L,
40L, 16L, 36L, 45L, 15L, 26L, 39L, 11L, 31L, 22L, 38L, 36L, 38L,
38L, 31L, 39L, 26L, 19L, 25L, 27L, 29L, 40L, 29L, 18L, 41L, 19L
), return_type = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1), elapsed_time = c(1e-04, 1e-04, 3.0001, 4.0001,
21.0001, 5.0001, 24.0001, 1.0001, 8.0001, 1e-04, 1e-04, 8.0001,
14.0001, 55.0001, 35.0001, 1e-04, 1e-04, 4.0001, 1e-04, 2.0001,
5.0001, 1e-04, 52.0001, 4.0001, 28.0001, 49.0001, 34.0001, 72.0001,
5.0001, 53.0001, 128.0001, 8.0001, 2.0001, 55.0001, 1.0001, 12.0001,
46.0001, 30.0001, 12.0001, 12.0001)), .Names = c(m_id, sale_date,
sale_year, sale_week, return_type, elapsed_time), row.names = c(1,

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40), class = data.frame)


The full dataset has almost 200,000 observations!  That is why I hadn't
posted the raw data.  And m_id_default_res is even bigger because it
includes all the original data along with the computed stats.


Yes, the following line you pointed out has a typo:

ndf$n[i] = m_id_default_res[i]

It should have been

ndf$n[i] = m_id_default_res$n[i]

Correcting that makes the error go away, but at the end of the loop, ndf is
said to have 0 columns and 0 rows.  That I don't understand.

But your statement (as corrected for the right source name) below does what
I'd intended.
ndf - m_id_default_res[, c('mid', 'estimate', 'sd', 'loglik', 'aic','bic',
'chisq', 'chisqpvalue', 'chisqdf')]

Thanks

Ted


On Fri, Jul 16, 2010 at 12:04 PM, Steve Lianoglou 
mailinglist.honey...@gmail.com wrote:

 Hi,

 First: it's kind of hard to play along w/o some reproducible data. To
 that end, you can paste into an email the output of:

 dput(moreinfo)

 If there are lots of rows in `moreinfo`, just give us the first ~10-20

 dput(head(moreinfo, 20))

 Anyway:

 snip
  At this point, each row in m_id_default_res corresponds to one data.frame
  produced by fitdist.  When I print it, I get the output I expected.
  However, I need to store only some of it into my DB.
 
  And then, because fitdist produces a data frame that includes a lot of
 info
  I don't need to store in the DB, I tried making a new data.frame
 containing
  only the info I need as follows:
  ndf = data.frame()
  for (i in 1:length(m_id_default_res[,1])) {
   ndf$mid[i] = m_id_default_res$mid[i]
   ndf$estimate[i] = m_id_default_res$estimate[i]
   ndf$sd[i] = m_id_default_res$sd[i]
   ndf$n[i] = m_id_default_res[i]
   ndf$loglik[i] = m_id_default_res$loglik[i]
   ndf$aic[i] = m_id_default_res$aic[i]
   ndf$bic[i] = m_id_default_res$bic[i]
   ndf$chisq[i] = m_id_default_res$chisq[i]
   ndf$chisqpvalue[i] = m_id_default_res$chisqpvalue[i]
   ndf$chisqdf[i] = m_id_default_res$chisqdf[i]
  }

 Forget the for loop. How about:

 ndf - m_id_default[, c('mid, 'estimate', 'sd', 'loglik', 'aic',
 'bic', 'chisq', 'chisqpvalue', 'chisqdf')

 Having just written that, I see something strange in your for loop.
 Specifically this line:

   ndf$n[i] = m_id_default_res[i]

 m_id_default_res is a data.frame, right? Why don't you try to see what
 `m_id_default_res[1]` returns.

 I'm not sure that that's what your error message is coming from, but I
 foresee this to be a problem anyway, if I follow your build up code
 correctly.

 Hope that helps,

 --
 Steve Lianoglou
 Graduate Student: Computational Systems

[R] setting up dates

2010-07-16 Thread Erin Hodgess

I have several data sets, which begin early in 2002 and run until
yesterday.  They do not have observations every day.  For example:
 xd1[1:10]
2002-02-25 2002-02-26 2002-02-28 2002-03-01 2002-03-04 2002-03-05 2002-03-07
 7  8  1  9 12  3  5
2002-03-08 2002-03-11 2002-03-12
 7 10  5



I want to set up zoo series which has every day from 2002-01-01
through yesterday.  There should be zeros on the non appearing dates
and the previously seen values on the appearing dates.

I've tried union and merge, but they didn't seem to work correctly.

Thank you in advance,
Sincerely,
Erin


-- 
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com

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[R] Toggle between the various pages for multi-page figures

2010-07-16 Thread mahesh samtani

Hello,

I am a new R user having transitioned over from S-plus recently. I have a
question that is probably very trivial but I am having trouble finding a
solution. In S-plus, graphic pages are created as tabs when multi-page
figures are created. I have shown the R code for xpose.VPC (a function
within library xpose4 for R) where I want the figure from each Strata (STRT)
to displayed on a different page. When I run the code, the pages just zoom
by and I was wondering if I could toggle between the various pages. I can
only export the last of the 8 pages that are created with the code below
(and I wish to export all 8 pages not just the last one):



xpose.VPC (vpc.info=vpc1/vpc_results.csv, vpctab=vpc1/vpctab1,

   PI=NULL, by=STRT, PI.ci=area, PI.real=TRUE,type=n,

   PI.limits=c(0.05,0.95),*layout=c(1,1,8)*,xlb=Time
(week),ylb=Concentration (ng/mL),

   scales = list(x = list(at=seq(0,3360,by=336)  ,
labels=seq(0,20,by=2 )),

 y =
list(at=log(c(.3,1,3,7.5,20,50,150)), labels=c(.3,1,3,7.5,20,50,150)) ))



Please help,

MNS

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Re: [R] how to skip a specific value when using apply() function to a matrix?

2010-07-16 Thread Joshua Wiley

Hello,

This does what you want.  The simple solution is shorter but requires
that there is only one value you wish to exclude (e.g., 0).  The
second works for any number of values you wish to exclude, but is
subsequently longer.  Also there is no need to create your own
function to 'studentize', ?scale will do it for you (as well as simply
center), this has the added benefit of not needing to convert the
results of apply() back to a matrix.

##Create data##
tmp1 - structure(c(15L, 13L, 10L, 15L, 12L, 11L, 7L, 7L, 2L, 5L, 9L,
12L, 9L, 10L, 12L, 30L, 71L, 2L, 49L, 56L, 50L, 52L, 76L, 69L,
60L, 21L, 19L, 120L, 211L, 18L, 3L, 4L, 1L, 2L, 4L, 5L, 2L, 3L,
0L, 4L, 3L, 2L, 0L, 0L, 2L, 0L, 1L, 0L), .Dim = c(6L, 8L))

tmp2 - tmp1 #for use in second solution

##Simple Solution##

tmp1[tmp1 == 0] - NA #change 0s to NAs
tmp1 - apply(tmp1, 2, scale) # 'studentize'
tmp1[is.na(tmp1)] - 0 #change NAs back to 0s

##More General Solution##

#Stores positions of 0s and 1s in origianl matrix
positions - which(tmp2 == 0 | tmp2 == 1)

#Store original values at 'positions'
replacements - tmp2[positions]

#Change these to NAs
tmp2[positions] - NA

#Scale using NAs
tmp2 - apply(tmp2, 2, scale)

#Now replace the NAs with the origianl values
tmp2[positions] - replacements

#Print to screen
tmp1
tmp2

HTH,

Josh

On Fri, Jul 16, 2010 at 9:11 AM, Shuhua Zhan sz...@uoguelph.ca wrote:
 Hello Nikhil and Wu,
 Thank you very much for your reply!
 What I want is to calculate the student's score column-wise by ignoring the 
 specific values such as zeros for example only using c(2,1) in column 8 in 
 the tmp1 and generate tmp2. I changed the zeros to NAs and modified my stud 
 fun to student as below. So I can ignore these specific values but can not 
 put back the NAs to the right position in the resulted matrix.
 Any suggestions to put back NAs to their original positions in column 7 and 8 
 either to list tmp4 or matrix tmp5?

 tmp1[tmp1==0]-NA
 student- function(x){
    x-x[!is.na(x)]
    x-(x-mean(x))/sd(x)
    return (x)
 }
 tmp4-apply(tmp1, 2, student)
 tmp4
 [[1]]
 [1]  1.1296201  0.1613743 -1.2909944  1.1296201 -0.3227486 -0.8068715

 [[2]]
 [1]  0.000  0.000 -1.4680505 -0.5872202  0.5872202  1.4680505

 [[3]]
 [1] -0.5207910 -0.4817316 -0.4036130  0.2994548  1.9008870 -0.7942062

 [[4]]
 [1] -0.8630035 -0.2380699 -0.7737273 -0.5951748  1.5474546  0.9225210

 [[5]]
 [1] -0.1913482 -0.6944435 -0.7202433  0.5826446  1.7565336 -0.7331431

 [[6]]
 [1] -0.1132277  0.5661385 -1.4719601 -0.7925939  0.5661385  1.2455047

 [[7]]
 [1] -0.9561829  0.2390457  1.4342743  0.2390457 -0.9561829

 [[8]]
 [1]  0.7071068 -0.7071068

 tmp5- matrix(unlist(tmp4),nrow=6)
 tmp5
           [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       
 [,7]       [,8]
 [1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
 -0.9561829 -0.7071068
 [2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
 0.2390457  1.1296201
 [3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
 1.4342743  0.1613743
 [4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
 0.2390457 -1.2909944
 [5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385 
 -0.9561829  1.1296201
 [6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047  
 0.7071068 -0.3227486

 Joshua

 - Original Message -
 From: Shuhua Zhan sz...@uoguelph.ca
 To: r-help@r-project.org
 Sent: Thursday, July 15, 2010 11:08:34 PM GMT -05:00 US/Canada Eastern
 Subject: [R] how to skip a specific value when using apply() function to a 
 matrix?

 Hello R experts,
 I'd like to studentize a matrix (tmp1) by column using apply() function and 
 skip some specific values such as zeros in the example below to tmp2 but not 
 tmp3. I used the script below and only can get a matrix tmp3. Could you 
 please help me to studentize the matrix (tmp1) without changing the zeros and 
 generate a new matrix tmp2?
 Thanks,
 Joshua

 tmp1
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
 [1,]   15    7    9   49   60    3    2    0
 [2,]   13    7   10   56   21    4    3    0
 [3,]   10    2   12   50   19    1    0    2
 [4,]   15    5   30   52  120    2    4    0
 [5,]   12    9   71   76  211    4    3    1
 [6,]   11   12    2   69   18    5    2    0

 tmp2
 [1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
 -0.9561829  0.000
 [2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
 0.2390457  0.000
 [3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
 0.000  0.7071068
 [4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
 1.4342743  0.000
 [5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385  
 0.2390457 -0.7071068
 [6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047 
 -0.9561829  0.000

 tmp3
          [,1]       [,2]       [,3]       [,4]       [,5]       [,6]      
 [,7]

Re: [R] package for rank ordered logit

2010-07-16 Thread Suresh Singh

My understanding is that polr will do ordered logit but I am not sure if it
is also suited for rank ordered logit (or is there no such distinction)

I am thinking of following two situations

1. there is an ordered response (say small,medium,large coffee) and each
individual selects one of these options. the order is predetermined i.e we
know which one is small, medium or large and interested in knowing which
option is selected. in this case each choice is independent because
different individuals choose them
2. an individual ranks some or all the options (say three different types of
coffee). We do not know apriori what the order is beforehand - these ranks
are dependent because the same individual selects them

I am calling the 1st situation - ordered logit and 2nd situation - rank
ordered logit.

Are these equivalent situations and is polr suited for both?

Suresh


On Fri, Jul 16, 2010 at 11:41 AM, Tal Galili tal.gal...@gmail.com wrote:
 Did you try:
 library(MASS)
 ?polr

 ?


 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

--




 On Fri, Jul 16, 2010 at 6:15 PM, Suresh Singh singh@osu.edu wrote:

 Is there a package in R that can run rank-ordered logit?

 Thanks,
 Suresh

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Re: [R] how to skip a specific value when using apply() function to a matrix?


Hi szhan,

I think Joshua gives all you wants -- scale is a really good function.

You can also make your own function work by setting an argument na.rm.

tmp1[tmp1==0]-NA 
student- function(x){
x-(x-mean(x,na.rm=T))/sd(x,na.rm=T)
return (x)
}
tmp4-apply(tmp1, 2, student)



-
A R learner.
-- 
View this message in context: 
http://r.789695.n4.nabble.com/how-to-skip-a-specific-value-when-using-apply-function-to-a-matrix-tp2290898p2291678.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] setting up dates

On Fri, Jul 16, 2010 at 12:38 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
 I have several data sets, which begin early in 2002 and run until
 yesterday.  They do not have observations every day.  For example:
 xd1[1:10]
 2002-02-25 2002-02-26 2002-02-28 2002-03-01 2002-03-04 2002-03-05 2002-03-07
         7          8          1          9         12          3          5
 2002-03-08 2002-03-11 2002-03-12
         7         10          5



 I want to set up zoo series which has every day from 2002-01-01
 through yesterday.  There should be zeros on the non appearing dates
 and the previously seen values on the appearing dates.

 I've tried union and merge, but they didn't seem to work correctly.


Try this:

z - zoo(c(7, 8, 1, 9, 12, 3, 5, 7, 10, 5),
   as.Date(c(2002-02-25, 2002-02-26, 2002-02-28, 2002-03-01,
  2002-03-04, 2002-03-05, 2002-03-07, 2002-03-08, 2002-03-11,
  2002-03-12)))

dd - seq(start(z), end(z), day)
merge(z, zoo(,dd), fill = 0)

Please try displaying your data using dput to make it easier to copy
it into R.  For example,
dput(z)

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Re: [R] Access web content from within R

2010-07-16 Thread Marshall Feldman

On 7/16/2010 6:00 AM, r-help-requ...@r-project.org wrote:

Message: 5
Date: Thu, 15 Jul 2010 03:36:21 -0700 (PDT)
From: Bart Joosenbartjoo...@hotmail.com
To:r-help@r-project.org
Subject: [R] Access web content from within R
Message-ID:1279190181074-2289953.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii

Hi,

I have to search in an online db for registered manufacturers of raw
materials.
Can I use R for the following:
I have a list with monograph numbers eg: l- c(198, 731,355)

Now I want to make a dataframe, containing the monograph number and the
information listed under COS:
Certificate holder, certificate number, Status, Type

Is this possible with R?

kind regards

Bart

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You don't describe the format of the database. If it's HTML or XML, the 
scrapeR package may do the trick.

Marsh Feldman

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[R] how to collapse categories or re-categorize variables?

2010-07-16 Thread CC

I am sure this is a very basic question:

I have 600,000 categorical variables in a data.frame - each of which is
classified as 0, 1, or 2

What I would like to do is collapse 1 and 2 and leave 0 by itself,
such that after re-categorizing 0 = 0; 1 = 1 and 2 = 1 --- in
the end I only want 0 and 1 as categories for each of the variables.

Also, if possible I would rather not create 600,000 new variables, if I can
replace the existing variables with the new values that would be great!

What would be the best way to do this?

Thank you!


-- 
Thanks,
CC

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[R] Help with Sink Function

2010-07-16 Thread Addi Wei


iterations - 100
nvars - 4
combined - rbind(scaleMiceTrain, scaleMiceTest)
reducedSample - combined
reducedSample - subset(reducedSample, select = -pID50)
reducedSample - subset(reducedSample, select = -id)
for (i in 1:iterations)
{
miceSample - sample(combined[,-c(1,2)],nvars, replace=FALSE)

miceSample$pID50 - combined$pID50
miceTestSample - miceSample[47:55,]
miceTrainSample - miceSample[1:46,]


fit.kknn - kknn(pID50~., miceTrainSample, miceTestSample)
table(miceTestSample$pID50, fit.kknn$fit)
(fit.train1 - train.kknn(pID50~., miceTrainSample, kmax=15,
kernel=c(rectangular), distance=1))

predictedTrain - predict(fit.train1, miceTrainSample,
miceTrainSample$pID50)
pID50Train - miceTrainSample$pID50
lmTrain - lm(predictedTrain~pID50Train)
slm - summary(lmTrain)
str(slm)
if (i == 1) 
{ 
previousR2 -slm$r.squared
sink(file=R2outputKKNN.txt, append=TRUE)
previousR2  
sink() 
}  
else if(i!=1)
{
currentR2 - slm$r.squared
if (previousR2  currentR2)
{
currentR2 - previousR2 
}   
if (previousR2  currentR2) 
{
sink(file=R2outputKKNN.txt, append=TRUE)
currentR2
sink()  
}
}
}


In my code above, I can't get sink to work.  In summary, I'm trying to write
the first run's R2, which is called previousR2 to file, and then anytime
currentR2  previousR2, I will write currentR2 to file. After running
the code above, my file R2outputKKNN.txt is empty... 

However, just running the code below writes / works fine:  
previousR2 -slm$r.squared
sink(file=R2outputKKNN.txt, append=TRUE)
previousR2  
sink() 
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Re: [R] Help with Sink Function

This is not reproducible, and does not look minimal.  You'll get better 
answers, and probably solve many issues on your own, if you construct 
small examples that illustrate the same problem you're having with your 
real data.


Addi Wei wrote:

iterations - 100
nvars - 4
combined - rbind(scaleMiceTrain, scaleMiceTest)
reducedSample - combined
reducedSample - subset(reducedSample, select = -pID50)
reducedSample - subset(reducedSample, select = -id)
for (i in 1:iterations)
{
miceSample - sample(combined[,-c(1,2)],nvars, replace=FALSE)

miceSample$pID50 - combined$pID50
miceTestSample - miceSample[47:55,]
miceTrainSample - miceSample[1:46,]


fit.kknn - kknn(pID50~., miceTrainSample, miceTestSample)
table(miceTestSample$pID50, fit.kknn$fit)
(fit.train1 - train.kknn(pID50~., miceTrainSample, kmax=15,
kernel=c(rectangular), distance=1))

predictedTrain - predict(fit.train1, miceTrainSample,
miceTrainSample$pID50)
pID50Train - miceTrainSample$pID50
lmTrain - lm(predictedTrain~pID50Train)
slm - summary(lmTrain)
str(slm)
	if (i == 1) 
	{ 
		previousR2 -slm$r.squared

sink(file=R2outputKKNN.txt, append=TRUE)
previousR2  
		sink() 
	}  
	else if(i!=1)

{
currentR2 - slm$r.squared
if (previousR2  currentR2)
{
currentR2 - previousR2  
}   
		if (previousR2  currentR2) 
		{

sink(file=R2outputKKNN.txt, append=TRUE)
currentR2
sink()  
}
}
}


In my code above, I can't get sink to work.  In summary, I'm trying to write
the first run's R2, which is called previousR2 to file, and then anytime
currentR2  previousR2, I will write currentR2 to file. After running
the code above, my file R2outputKKNN.txt is empty... 

However, just running the code below writes / works fine:  
previousR2 -slm$r.squared

sink(file=R2outputKKNN.txt, append=TRUE)
previousR2  
sink()


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Re: [R] Help with Sink Function

2010-07-16 Thread Addi Wei


Sorry about that. Still new to this...  The code below should be
reproducible.All R2 should just be 1, and I should write 1 to
R2outputKKNN.txt 10 timesnothing is happening.  Appreciate the efforts
to help! 

for (i in 1:10)
{
adata = 1:5
bdata = 6:10
lm - lm(adata~bdata)
slm - summary(lm)
str(slm)

   if (i == 1) {   
previousR2 -slm$r.squared
   sink(file=R2outputKKNN.txt, append=TRUE)
   previousR2  
   sink()  }   else if(i!=1)
   {
   currentR2 - slm$r.squared
   if (previousR2  currentR2)
   {
   currentR2 - previousR2
   }  
   if (previousR2  currentR2) {
   sink(file=R2outputKKNN.txt, append=TRUE)
   currentR2
   sink()  
   }
   }
}
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Re: [R] how to collapse categories or re-categorize variables?


Do you want to replace specific values of a data set?

df - sample(c(0,1,2),600,replace=T)
table(df)
df[df==2]-1
table(df)

-
A R learner.
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Re: [R] Mathematica and R

2010-07-16 Thread David Bickel


 Hi Albyn,

Thank you very much for the suggestion.

I managed to install Sage on Windows (via a Linux VM), but I cannot find 
any documentation on how to use R from Sage. Maybe I should use the web 
interface of Sage to avoid having to install R on the VM.


Best regards,
David


On 10-07-14 10:24, Albyn Jones wrote:
Take a look at Sage, which is an open source alternative.  It already 
integrates R  (http://www.sagemath.org)


albyn

Quoting David Bickel davidbickel.com+rh...@gmail.com:

What are some effective ways to leverage the strengths of R and 
Mathematica for the analysis of a single data set?


More specifically, are there any functions that can assist with any 
of the following?

1. Calling an R function from Mathematica.
2. Calling a Mathematica function from R.
3. Using XML or another reliable data format to pass vectors, 
matrices, and/or lists from one environment to the other.


Any advice would be appreciated.

David

--
David R. Bickel, PhD
Associate Professor
Ottawa Institute of Systems Biology
Biochem., Micro. and I. Department
Mathematics and Statistics Department
University of Ottawa
451 Smyth Road
Ottawa, Ontario K1H 8M5

http://www.statomics.com

Office Tel: (613) 562-5800 ext. 8670
Office Fax: (613) 562-5185

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[R] Deleting a variable number of characters from a string

2010-07-16 Thread Davis, Brian

I have a text processing problem I'm hoping someone can help me solve.  This 
issue it this.

 I have a character string in which I need to delete a variable number of 
characters from the string.  The string itself contains the number of 
characters to be deleted.  The number of characters to be deleted is proceeded 
by either a + or a -.

A toy example:

Suppose I have

x-c(A-1CB-2GHX, *+11gAgggTgtgggH)
 x
[1] A-1CB-2GHX   *+11gAgggTgtgggH

What I need as output is
ABX *H

I know I can use gsub to remove the control character and the number portion 
with

gsub((\\-|\\+)([0-9]+), replacement=, x)

However, I can't figure out how to delete the variable number of characters 
after the number portion of the string.

Any ideas?


In case this helps
 sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-pc-mingw32

locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252

Brian

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Re: [R] Creating Enumerated Variables

Hi:

Hadley's solution is certainly preferred here due to its relative
simplicity. I just wanted to correct an error from my earlier post.

On Thu, Jul 15, 2010 at 2:08 PM, Dennis Murphy djmu...@gmail.com wrote:

 Hi:

 I sincerely hope there's an easier way, but one method to get this is as
 follows,
 with d as the data frame name of your test data:

 d - d[order(with(d, Age, School, rev(Grade))), ]


Should be

d - d[with(d, order(Age, School, rev(Grade))), ]


 d$Count - do.call(c, mapply(seq, 1, as.vector(t(with(d, table(Age,
 School))
 d

  d
   ID Age School Grade Count
 1  1  10  198 1
 3  3  10  192 2
 7  7  10  180 3
 8  8  10  179 4
 2  2  10  297 1
 4  4  11  190 1
 5  5  11  180 2
 6  6  11  270 1
 9  9  11  270 2


 The code to get the count is a little convoluted:
 - first, find the frequency table of Age and School, transpose it and
 then unlist into a vector
 - use mapply to generate a sequence for each group from 1 up to its
 group count; mapply()  is necessary to use the counts as a vector argument.
 This returns a list of sequences.
 - do.call() applies a function (here, c) to an input list, yielding the
 vector of groupwise indices we wanted. Basically, it flattens the list. This
 is what we assign to d$Count.

 Side note: I didn't get the correct ordering the first time, but I did the
 second time (2.11.1 64bit, Windows 7).


And now we know why :)

D.


 HTH,
 Dennis


 On Thu, Jul 15, 2010 at 7:45 AM, jdellava jdell...@vcu.edu wrote:


 Hi,

 I am trying to create a variable counting the number of individuals based
 on
 two variables. I am able to do it or one variable, but not two. In SAS I
 was
 able to sort by two variables and use a first. statement to create the
 counts based on both. Here is an example:

 What I have
 ID  Age School  Grade
 1   10  1   98
 2   10  2   97
 3   10  1   92
 4   11  1   90
 5   11  1   80
 6   11  2   70
 7   10  1   80
 8   10  1   79
 9   11  2   70

 What I need
 ID  Age School  Grade   School Count
 1   10  1   98  1
 3   10  1   92  2
 7   10  1   80  3
 8   10  1   79  4
 2   10  2   97  1
 4   11  1   90  1
 5   11  1   80  2
 6   11  2   70  1
 9   11  2   70  2

 I want to create counts of individuals age 10 in school 1 then age 10 in
 school two (the what I need set)

 Anyway to do this?

 Thank you.

 --
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Re: [R] threshold in plot

Hi:

Here's a simple example (divide through by 10 for your case):

x - 1:10
y - rnorm(10, x, s = 0.3)
clr - ifelse(x = 5, 'red', 'blue')
plot(x, y, col = clr)

HTH,
Dennis

On Fri, Jul 16, 2010 at 6:08 AM, azam jaafari azamjaaf...@yahoo.com wrote:

 Hi

 I want to draw a plot from observed and predicted data and also shows
 threshold and data before threshold are identified with different color from
 data after threshold.

 Suppose:
 abserved data are 0 or 1
 predicted data= 0 to 1
 threshold=0.5

 Thanks alot





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Re: [R] Deleting a variable number of characters from a string


Hi Davis, 

Please try ??regex

gsub((\\-|\\+)([0-9]+)(\\w*)(\\w), replacement=\\4, x)

-
A R learner.
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Re: [R] Help with Sink Function

2010-07-16 Thread Matt Shotwell

Your code between calls to sink() does not generate any output. Hence,
nothing will be diverted to the file. To illustrate this point,
consider 

for(i in 1:10) i

This produces no output. However, 

for(i in 1:10) print(i)

produces output as expected.

-Matt

On Fri, 2010-07-16 at 13:34 -0400, Addi Wei wrote:
 Sorry about that. Still new to this...  The code below should be
 reproducible.All R2 should just be 1, and I should write 1 to
 R2outputKKNN.txt 10 timesnothing is happening.  Appreciate the efforts
 to help! 
 
 for (i in 1:10)
 {
   adata = 1:5
   bdata = 6:10
   lm - lm(adata~bdata)
   slm - summary(lm)
 str(slm)
 
if (i == 1) {   
   previousR2 -slm$r.squared
sink(file=R2outputKKNN.txt, append=TRUE)
previousR2  
sink()  }   else if(i!=1)
{
currentR2 - slm$r.squared
if (previousR2  currentR2)
{
currentR2 - previousR2
}  
if (previousR2  currentR2) {
sink(file=R2outputKKNN.txt, append=TRUE)
currentR2
sink()  
}
}
 }
-- 
Matthew S. Shotwell
Graduate Student
Division of Biostatistics and Epidemiology
Medical University of South Carolina
http://biostatmatt.com

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Re: [R] Access web content from within R

2010-07-16 Thread Bart Joosen


Marsh, 

you are absolutely right, I forget the link to the db.
Actually its a webpage for each monograph number:

http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/198
http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/731
...

As I have a list with the numbers of interest (l- c(198, 731,355)), I can
paste the webadress together with the numbers, but then I need to substract
the COS info.

Kind regards

Bart


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Re: [R] how to collapse categories or re-categorize variables?

Hi:

See ? levels. Here's a toy example:

 x - factor(sample(0:2, 10, replace = TRUE))
 x
 [1] 1 2 1 0 2 2 2 2 2 1
Levels: 0 1 2

 levels(x) - c(0, 1, 1)# Change level 2 to 1
 x
 [1] 1 1 1 0 1 1 1 1 1 1
Levels: 0 1

HTH,
Dennis


On Fri, Jul 16, 2010 at 10:18 AM, CC turtysm...@gmail.com wrote:

 I am sure this is a very basic question:

 I have 600,000 categorical variables in a data.frame - each of which is
 classified as 0, 1, or 2

 What I would like to do is collapse 1 and 2 and leave 0 by itself,
 such that after re-categorizing 0 = 0; 1 = 1 and 2 = 1 --- in
 the end I only want 0 and 1 as categories for each of the variables.

 Also, if possible I would rather not create 600,000 new variables, if I can
 replace the existing variables with the new values that would be great!

 What would be the best way to do this?

 Thank you!


 --
 Thanks,
 CC

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[R] Elementary question about computing confidence intervals.

I would have thought this to be relatively elementary, but I can't find it
mentioned in any of my stats texts.

Please consider the following:

library(fitdistrplus)

  fp = fitdist(y,exp);
  rate = fp$estimate;
  sd = fp$sd
  fOneWeek = exp(-rate*7); #fraction that happens within a week - y is
measured in days
  fr = exp(-rate*dt);  #fraction remaining - dt = elapsed time from time of
sample to present
  fh = 1 - fr;  # fraction that occurred from time of sample to present

# assume n = total number that have happened from time of sample to present
  T = n / fh  # t is the total number at y = 0
  NR = fr * T
  NNW = NR * (1 - fOneWeek)

(If you wanted to run this, just populate y with random numbers from an
exponential distribution.)

What I show here simply extracts an estimate and standard deviation from the
data.frame returned by fitdist, and tries to compute a number of integrals.
What I need is the number of events that can be expected next week, next
month, and from now to the end of time.  Unless I have gone senile in my old
age, I have the integrals correct. Please correct me if I missed something.
But what I need help (to refresh my memory - I used to know this way back in
the stone age) to compute the confidence intervals for each of these
integrals.

So I don't bother anyone with similar elementary questions, what web
resource exists that defines confidence intervals for such integrals for
arbitrary distributions?  or does such a resource exist?

Thanks

Ted

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Re: [R] how to collapse categories or re-categorize variables?

2010-07-16 Thread Ista Zahn

Hi,
On Fri, Jul 16, 2010 at 5:18 PM, CC turtysm...@gmail.com wrote:
 I am sure this is a very basic question:

 I have 600,000 categorical variables in a data.frame - each of which is
 classified as 0, 1, or 2

 What I would like to do is collapse 1 and 2 and leave 0 by itself,
 such that after re-categorizing 0 = 0; 1 = 1 and 2 = 1 --- in
 the end I only want 0 and 1 as categories for each of the variables.

Something like this should work

for (i in names(dat)) {
dat[, i]  - factor(dat[, i], levels = c(0, 1, 2), labels =
c(0, 1, 1))
}

-Ista

 Also, if possible I would rather not create 600,000 new variables, if I can
 replace the existing variables with the new values that would be great!

 What would be the best way to do this?

 Thank you!


 --
 Thanks,
 CC

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-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] Dot Plot with Confidence Limits

2010-07-16 Thread Deepayan Sarkar

On Wed, Jul 14, 2010 at 10:12 AM, Neil123 neil.wh...@plants.ox.ac.uk wrote:

 Hi,

 I have the following dataset and I would like to create a dotplot with
 confidence limits:

        CAT1              CAT2          MEAN                 Lower
 Upper
 1         1                     1            0.619                 0.392
 0.845
 2         1                    10           1.774                 1.030
 2.518
 3         1                   100          7.700                 4.810
 10.586
 4         1                  1000         45.536               23.612
 67.392
 5         2                     1            0.500                 0.244
 0.755
 6         2                    10           1.725                 1.109
 2.341
 7         2                   100          15.200               10.924
 19.473
 8         2                  1000         88.200               48.030
 128.369

 I need the data grouped by the two categories (independent variables CAT1 
 CAT2). Each row would be a separate dot, and colour-coded by CAT1.

 I have been able to create a dotplot with the data of just one of the
 CAT1's, but not both together in the same graph...

I am not sure what your question is. It would probably help to have
reproducible code showing the progress you have made so far.

-Deepayan
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[R] invalid factor level, NAs generated

2010-07-16 Thread Jared Stabach

I've seen a few threads about this, but none that seem to answer my problem

I have a list of .txt files in a directory that I am reading into R and row
binding together.  I am using the following code to do so:

# Directory where files are found
my.txt.file.directory - C:/Jared/Data/Kenya/Wildebeest/Tracking_Data

names.of.txt.files -
list.files(my.txt.file.directory,pattern=all_data,ignore.case=TRUE,
full.names=TRUE)
# Print names that meet criteria in directory
names.of.txt.files

# The names.of.txt.files will be a vector with the names of all the txt
files
# Dataset will contain all the data in the directory

wildebeest - NULL
# Run loop
for (i in 1:length(names.of.txt.files))
{
dat - read.table(names.of.txt.files[i],header=FALSE)
# Row bind all data together into a file called 'wildebeest'
wildebeest - rbind.data.frame(wildebeest,dat)
rm(dat)
}

When I run this script, I get an error such as:

18: In `[-.factor`(`*tmp*`, ri, value = c(1714.36, 1711.27,  ... :
  invalid factor level, NAs generated


I think I have identified the problem such that when I identify the
structure of some of the files that I am reading in, columns are labeled as
Factors.  In other files, the same columns are labeled as numeric
values.  Is there a way to assign the data structure to these columns in the
dataframe as they are being read in?  Any other suggestions to why I am
getting this error is appreciated.

Jared

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Re: [R] invalid factor level, NAs generated


snip



I think I have identified the problem such that when I identify the
structure of some of the files that I am reading in, columns are labeled as
Factors.  In other files, the same columns are labeled as numeric
values.  Is there a way to assign the data structure to these columns in the
dataframe as they are being read in?  Any other suggestions to why I am
getting this error is appreciated.



Yes, as ?read.table describes, see the colClasses argument.

However, you should investigate why read.table wants to treat these 
particular columns as factors.  If they are indeed simply consisting of 
all numeric values, there shouldn't be a problem.  The fact that they 
are coming out as factors raises a flag to me ...


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Re: [R] invalid factor level, NAs generated

Hi Jared:

Is it possible you're using read.table with variable names in line 1 but not
using header = TRUE as an argument?

HTH,
Dennis

On Fri, Jul 16, 2010 at 12:41 PM, Erik Iverson er...@ccbr.umn.edu wrote:

 snip



 I think I have identified the problem such that when I identify the
 structure of some of the files that I am reading in, columns are labeled
 as
 Factors.  In other files, the same columns are labeled as numeric
 values.  Is there a way to assign the data structure to these columns in
 the
 dataframe as they are being read in?  Any other suggestions to why I am
 getting this error is appreciated.


 Yes, as ?read.table describes, see the colClasses argument.

 However, you should investigate why read.table wants to treat these
 particular columns as factors.  If they are indeed simply consisting of all
 numeric values, there shouldn't be a problem.  The fact that they are coming
 out as factors raises a flag to me ...


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[R] Sage and R (was: Mathematica and R)

2010-07-16 Thread Karl-Dieter Crisman

Dear David,

I managed to install Sage on Windows (via a Linux VM), but I cannot find any
documentation on how to use R from Sage. Maybe I
should use the web interface of Sage to avoid having to install R on the VM.

In the Sage command line, you can type

sage: r?

to get some very basic info, and

sage: r.[tab]

to get various commands. This part of the interface still needs some
work, though it suffices (particularly via r.eval()) for many needs.
You do NOT need to reinstall R; a full copy of R (with all recommended
packages) is in any recent Sage.

However, based on your earlier emails in the thread, you may want to
start Sage, and then type

sage: notebook()

to launch the web server on your own computer, from which you can get
a Mma-style worksheet and do all R computations in there, by simply
selecting 'r' from the drop-down menu at the top which would currently
read 'sage', or by typing %r at the beginning of each cell you want to
use R in.

You are, however, right that there is very little easy-to-find
documentation on how to use R within Sage, and we VERY much welcome
improvements on this front. The worksheets
http://prep.sagenb.org/home/pub/34/ and
http://sagenb.org/home/pub/2232/ might have a few tidbits for you, and
the thread at
http://groups.google.com/group/sage-support/browse_thread/thread/b8411dfebdb54406/0bca74c09bd4145a?lnk=gstq=r+integration#0bca74c09bd4145a
may as well.

In particular, if anyone wants to help improve this (though again,
quite a bit already works seamlessly), you may be interested in the
following talk (mine) at useR! 2010 next week:
http://user2010.org/abstracts/Crisman.pdf

We definitely need experts, and also really crave feedback on how Sage
can best be of use to the R community (why R is useful to Sage should
be obvious).

Karl-Dieter Crisman

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Re: [R] make an model object (e.g. nlme) available in a user defined function (xyplot related)

2010-07-16 Thread Deepayan Sarkar

On Mon, Jul 12, 2010 at 8:16 AM, Jun Shen jun.shen...@gmail.com wrote:
 Dear Deepayan,

 Thank you for taking the time to look into this issue.

 I have a data object called Data, please find it at the end of the
 message. Then I can run the code below separately in the console.

[...]

 

 Then I have my real function as follows: If I run the code as,

compare.curves(Data=Data)

 The analytical part is working but not the plotting part (Error using
 packet 1, object 'model' not found)

I don't have much to help you, but the problem is not in lattice, but
in formula.nlme. The following modification of your function also
fails:

compare.curves - function(curve='ascending',
   Data=stop('A data object must be specified'),
   parameter='EC50',random.pdDiag=FALSE,
   start.values=c(Emax=100,E0=1,EC50=50,gamma=2),...)
{
mymodel=as.formula('RESP ~ E0+(Emax-E0)*CP**gamma/(EC50**gamma+CP**gamma)')
mod.nlme-nlme(model=mymodel,data=Data,method='REML',
   fixed=Emax+E0+EC50+gamma~1,
   random= if (length(parameter)==1)

eval(substitute(variable~1,list(variable=as.name(parameter
   else {
   variable-as.name(parameter[1])
   for (i in 2:length(parameter)) variable-
   paste(variable,'+',as.name(parameter[i]))
   formula-as.formula(paste(variable,'~1'))
   if (random.pdDiag) list(pdDiag(formula))
   else formula
   },
   groups=~ID,
   start=list(fixed=start.values)
   )
formula(mod.nlme)
}

-Deepayan

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Re: [R] Deleting a variable number of characters from a string

On Fri, Jul 16, 2010 at 1:59 PM, Davis, Brian brian.da...@uth.tmc.edu wrote:
 I have a text processing problem I'm hoping someone can help me solve.  This 
 issue it this.

  I have a character string in which I need to delete a variable number of 
 characters from the string.  The string itself contains the number of 
 characters to be deleted.  The number of characters to be deleted is 
 proceeded by either a + or a -.

 A toy example:

 Suppose I have

 x-c(A-1CB-2GHX, *+11gAgggTgtgggH)
 x
 [1] A-1CB-2GHX       *+11gAgggTgtgggH

 What I need as output is
 ABX *H

 I know I can use gsub to remove the control character and the number portion 
 with

 gsub((\\-|\\+)([0-9]+), replacement=, x)

 However, I can't figure out how to delete the variable number of characters 
 after the number portion of the string.


Using gsubfn in the gsubfn package we match

- the - or + via [-+],
- the digits via \\d+ and
- the remaining characters via [^-+]*

parenthesizing the digits and remaining characters so that they form
back references which are passed to the function as args 1 and 2
respectively.  gsubfn supports a formula notation for functions and the
specified function using that formula notation has arguments d and s
and function body which strips the characters and returns the rest
to be substituted back in:

library(gsubfn)
gsubfn([-+](\\d+)([^-+]*), d + s ~ substring(s, as.numeric(d) + 1), x)
   [1] ABX *H

See http://gsubfn.googlecode.com for more.

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Re: [R] how to skip a specific value when using apply() function to a matrix?

2010-07-16 Thread Shuhua Zhan

Hello Joshua and Wu,
Thank you for your excellent solutions.
Joshua 
- Original Message -
From: Joshua Wiley jwiley.ps...@gmail.com
To: Shuhua Zhan sz...@uoguelph.ca
Cc: r-help@r-project.org
Sent: Friday, July 16, 2010 12:39:22 PM GMT -05:00 US/Canada Eastern
Subject: Re: [R] how to skip a specific value when using apply() function to a  
matrix?

Hello,

This does what you want.  The simple solution is shorter but requires
that there is only one value you wish to exclude (e.g., 0).  The
second works for any number of values you wish to exclude, but is
subsequently longer.  Also there is no need to create your own
function to 'studentize', ?scale will do it for you (as well as simply
center), this has the added benefit of not needing to convert the
results of apply() back to a matrix.

##Create data##
tmp1 - structure(c(15L, 13L, 10L, 15L, 12L, 11L, 7L, 7L, 2L, 5L, 9L,
12L, 9L, 10L, 12L, 30L, 71L, 2L, 49L, 56L, 50L, 52L, 76L, 69L,
60L, 21L, 19L, 120L, 211L, 18L, 3L, 4L, 1L, 2L, 4L, 5L, 2L, 3L,
0L, 4L, 3L, 2L, 0L, 0L, 2L, 0L, 1L, 0L), .Dim = c(6L, 8L))

tmp2 - tmp1 #for use in second solution

##Simple Solution##

tmp1[tmp1 == 0] - NA #change 0s to NAs
tmp1 - apply(tmp1, 2, scale) # 'studentize'
tmp1[is.na(tmp1)] - 0 #change NAs back to 0s

##More General Solution##

#Stores positions of 0s and 1s in origianl matrix
positions - which(tmp2 == 0 | tmp2 == 1)

#Store original values at 'positions'
replacements - tmp2[positions]

#Change these to NAs
tmp2[positions] - NA

#Scale using NAs
tmp2 - apply(tmp2, 2, scale)

#Now replace the NAs with the origianl values
tmp2[positions] - replacements

#Print to screen
tmp1
tmp2

HTH,

Josh

On Fri, Jul 16, 2010 at 9:11 AM, Shuhua Zhan sz...@uoguelph.ca wrote:
 Hello Nikhil and Wu,
 Thank you very much for your reply!
 What I want is to calculate the student's score column-wise by ignoring the 
 specific values such as zeros for example only using c(2,1) in column 8 in 
 the tmp1 and generate tmp2. I changed the zeros to NAs and modified my stud 
 fun to student as below. So I can ignore these specific values but can not 
 put back the NAs to the right position in the resulted matrix.
 Any suggestions to put back NAs to their original positions in column 7 and 8 
 either to list tmp4 or matrix tmp5?

 tmp1[tmp1==0]-NA
 student- function(x){
    x-x[!is.na(x)]
    x-(x-mean(x))/sd(x)
    return (x)
 }
 tmp4-apply(tmp1, 2, student)
 tmp4
 [[1]]
 [1]  1.1296201  0.1613743 -1.2909944  1.1296201 -0.3227486 -0.8068715

 [[2]]
 [1]  0.000  0.000 -1.4680505 -0.5872202  0.5872202  1.4680505

 [[3]]
 [1] -0.5207910 -0.4817316 -0.4036130  0.2994548  1.9008870 -0.7942062

 [[4]]
 [1] -0.8630035 -0.2380699 -0.7737273 -0.5951748  1.5474546  0.9225210

 [[5]]
 [1] -0.1913482 -0.6944435 -0.7202433  0.5826446  1.7565336 -0.7331431

 [[6]]
 [1] -0.1132277  0.5661385 -1.4719601 -0.7925939  0.5661385  1.2455047

 [[7]]
 [1] -0.9561829  0.2390457  1.4342743  0.2390457 -0.9561829

 [[8]]
 [1]  0.7071068 -0.7071068

 tmp5- matrix(unlist(tmp4),nrow=6)
 tmp5
           [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       
 [,7]       [,8]
 [1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
 -0.9561829 -0.7071068
 [2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
 0.2390457  1.1296201
 [3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
 1.4342743  0.1613743
 [4,]  1.1296201 -0.5872202  0.2994548 -0.5951748  0.5826446 -0.7925939  
 0.2390457 -1.2909944
 [5,] -0.3227486  0.5872202  1.9008870  1.5474546  1.7565336  0.5661385 
 -0.9561829  1.1296201
 [6,] -0.8068715  1.4680505 -0.7942062  0.9225210 -0.7331431  1.2455047  
 0.7071068 -0.3227486

 Joshua

 - Original Message -
 From: Shuhua Zhan sz...@uoguelph.ca
 To: r-help@r-project.org
 Sent: Thursday, July 15, 2010 11:08:34 PM GMT -05:00 US/Canada Eastern
 Subject: [R] how to skip a specific value when using apply() function to a 
 matrix?

 Hello R experts,
 I'd like to studentize a matrix (tmp1) by column using apply() function and 
 skip some specific values such as zeros in the example below to tmp2 but not 
 tmp3. I used the script below and only can get a matrix tmp3. Could you 
 please help me to studentize the matrix (tmp1) without changing the zeros and 
 generate a new matrix tmp2?
 Thanks,
 Joshua

 tmp1
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
 [1,]   15    7    9   49   60    3    2    0
 [2,]   13    7   10   56   21    4    3    0
 [3,]   10    2   12   50   19    1    0    2
 [4,]   15    5   30   52  120    2    4    0
 [5,]   12    9   71   76  211    4    3    1
 [6,]   11   12    2   69   18    5    2    0

 tmp2
 [1,]  1.1296201  0.000 -0.5207910 -0.8630035 -0.1913482 -0.1132277 
 -0.9561829  0.000
 [2,]  0.1613743  0.000 -0.4817316 -0.2380699 -0.6944435  0.5661385  
 0.2390457  0.000
 [3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601  
 0.000  0.7071068
 [4,]  1.1296201

Re: [R] Access web content from within R

2010-07-16 Thread Henrique Dallazuanna

Try this:

library(XML)
readHTMLTable(http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/198;,
which = 5)

On Fri, Jul 16, 2010 at 4:08 PM, Bart Joosen bartjoo...@hotmail.com wrote:


 Marsh,

 you are absolutely right, I forget the link to the db.
 Actually its a webpage for each monograph number:

 http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/198
 http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/731
 ...

 As I have a list with the numbers of interest (l- c(198, 731,355)), I can
 paste the webadress together with the numbers, but then I need to substract
 the COS info.

 Kind regards

 Bart


 --
 View this message in context:
 http://r.789695.n4.nabble.com/Access-web-content-from-within-R-tp2289953p2291839.html
 Sent from the R help mailing list archive at Nabble.com.

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

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[R] sqldf modify table

2010-07-16 Thread PeterTucker


Hi - I am something of a newbie and am a little perplexed.  When (trying to)
modify a table I issue the following commands with subsequent errors

sqldf(alter table Korea drop column code, dbname = mydb)
error in statement: near drop: syntax error

or

sqldf(alter table Korea rename column hyr to hyrI, dbname = mydb)
error in statement: near column: syntax error

These are simple commands - am I missing something obvious?  I can retrieve
data from them, and retrieve their table_info

Thanks

Peter
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[R] Mixed Conditional Logit with nested data

2010-07-16 Thread Paul Miller

Hello Everyone,
Â 
This is my first attempt to do something in R. As a precursor to a Willingness 
to Pay analysis, I want to conduct a Mixed Conditional Logit analysis but am 
unsure how to proceed because of some nesting within my data.
Â 
Below is some dataÂ and code that illustrate what Iâm trying to do. The data 
are based on responses to a conjoint survey obtained during pilot testing. In 
the survey, cancer patients are asked to complete several discrete choice 
tasks. In each task, the patient indicates whether they prefer treatment A or 
treatment B. The attributes of treatment AÂ and treatment B generally vary from 
one task to the next. The data also contain some of the predictors that that 
are presumed to influence patient choices (chance of side effects, duration of 
side effects, severity of side effects).
Â 
The mclogit package seems like a possible avenue for conducting my analysis. 
The code below appears to be properly specified and produces some intuitive 
results. Even with only 4 respondents, there appears to be a tendency for 
patients to prefer treatments with a low chance of side effects, a short 
duration of side effects, and a low severity of side effects.
Â 
The only problem is that the current setup doesn't seem to address the fact 
that I have multiple discrete choice tasks within each respondent. It seems to 
me that knowing which option person 1 chose in task 1 might also indicate 
something about their responses in the other conjoint tasks and that my model 
doesn'tÂ account for this lack of statistical independence.
Â 
So I'm wondering if there is any way to perform my analysis while taking this 
non-independence into account.
Â 
Any assistance that people can render will be most appreciated.
Â 
Thanks,
Â 
PaulÂ 
Â 
wtp_string -
Subject,Task,Option,selected,chance_low,chance_medium,chance_high,duration_short,duration_medium,duration_long,severity_low,severity_medium,severity_high
1,1,A,1,0,0,1,1,0,0,1,0,0
1,1,B,0,0,1,0,0,0,1,0,0,1
1,2,A,1,1,0,0,1,0,0,0,0,1
1,2,B,0,0,1,0,0,1,0,0,1,0
1,3,A,0,0,0,1,0,1,0,1,0,0
1,3,B,1,1,0,0,0,0,1,0,1,0
1,4,A,1,0,0,1,0,1,0,1,0,0
1,4,B,0,0,1,0,0,0,1,0,0,1
1,5,A,1,0,1,0,1,0,0,1,0,0
1,5,B,0,1,0,0,0,1,0,0,1,0
1,6,A,1,1,0,0,1,0,0,0,0,1
1,6,B,0,0,0,1,0,0,1,0,1,0
1,7,A,1,1,0,0,0,0,1,1,0,0
1,7,B,0,0,1,0,0,1,0,0,0,1
1,8,A,1,0,0,1,1,0,0,0,1,0
1,8,B,0,0,1,0,0,1,0,0,0,1
1,9,A,1,0,1,0,1,0,0,1,0,0
1,9,B,0,0,0,1,0,1,0,0,1,0
1,10,A,1,1,0,0,1,0,0,0,1,0
1,10,B,0,0,0,1,0,0,1,1,0,0
1,11,A,1,0,1,0,0,1,0,1,0,0
1,11,B,0,0,0,1,1,0,0,0,0,1
1,12,A,1,1,0,0,0,1,0,1,0,0
1,12,B,0,0,1,0,0,0,1,0,1,0
1,13,A,0,0,0,1,0,0,1,0,1,0
1,13,B,1,1,0,0,0,1,0,0,0,1
1,14,A,0,0,0,1,0,0,1,0,0,1
1,14,B,1,0,1,0,1,0,0,1,0,0
1,15,A,1,1,0,0,1,0,0,0,1,0
1,15,B,0,0,0,1,0,1,0,0,0,1
1,16,A,1,1,0,0,0,0,1,1,0,0
1,16,B,0,0,1,0,0,1,0,0,1,0
1,17,A,1,0,1,0,1,0,0,0,1,0
1,17,B,0,0,0,1,0,0,1,1,0,0
1,18,A,1,1,0,0,1,0,0,1,0,0
1,18,B,0,0,0,1,0,0,1,0,0,1
1,19,A,1,1,0,0,1,0,0,0,0,1
1,19,B,0,0,0,1,0,1,0,0,1,0
1,20,A,0,0,0,1,0,0,1,0,0,1
1,20,B,1,0,1,0,1,0,0,1,0,0
1,21,A,1,1,0,0,0,1,0,1,0,0
1,21,B,0,0,1,0,0,0,1,0,0,1
1,22,A,1,0,0,1,1,0,0,1,0,0
1,22,B,0,1,0,0,0,1,0,0,1,0
1,23,A,0,1,0,0,0,1,0,0,0,1
1,23,B,1,0,1,0,0,0,1,0,1,0
1,24,A,1,0,0,1,1,0,0,0,1,0
1,24,B,0,0,1,0,0,0,1,1,0,0
1,25,A,0,1,0,0,0,0,1,1,0,0
1,25,B,1,0,1,0,1,0,0,0,0,1
2,1,A,0,0,0,1,1,0,0,1,0,0
2,1,B,1,0,1,0,0,0,1,0,0,1
2,2,A,0,1,0,0,1,0,0,0,0,1
2,2,B,1,0,1,0,0,1,0,0,1,0
2,3,A,1,0,0,1,0,1,0,1,0,0
2,3,B,0,1,0,0,0,0,1,0,1,0
2,4,A,0,0,0,1,0,1,0,1,0,0
2,4,B,1,0,1,0,0,0,1,0,0,1
2,5,A,1,0,1,0,1,0,0,1,0,0
2,5,B,0,1,0,0,0,1,0,0,1,0
2,6,A,1,1,0,0,1,0,0,0,0,1
2,6,B,0,0,0,1,0,0,1,0,1,0
2,7,A,1,1,0,0,0,0,1,1,0,0
2,7,B,0,0,1,0,0,1,0,0,0,1
2,8,A,1,0,0,1,1,0,0,0,1,0
2,8,B,0,0,1,0,0,1,0,0,0,1
2,9,A,0,0,1,0,1,0,0,1,0,0
2,9,B,1,0,0,1,0,1,0,0,1,0
2,10,A,1,1,0,0,1,0,0,0,1,0
2,10,B,0,0,0,1,0,0,1,1,0,0
2,11,A,1,0,1,0,0,1,0,1,0,0
2,11,B,0,0,0,1,1,0,0,0,0,1
2,12,A,1,1,0,0,0,1,0,1,0,0
2,12,B,0,0,1,0,0,0,1,0,1,0
2,13,A,0,0,0,1,0,0,1,0,1,0
2,13,B,1,1,0,0,0,1,0,0,0,1
2,14,A,1,0,0,1,0,0,1,0,0,1
2,14,B,0,0,1,0,1,0,0,1,0,0
2,15,A,1,1,0,0,1,0,0,0,1,0
2,15,B,0,0,0,1,0,1,0,0,0,1
2,16,A,0,1,0,0,0,0,1,1,0,0
2,16,B,1,0,1,0,0,1,0,0,1,0
2,17,A,0,0,1,0,1,0,0,0,1,0
2,17,B,1,0,0,1,0,0,1,1,0,0
2,18,A,0,1,0,0,1,0,0,1,0,0
2,18,B,1,0,0,1,0,0,1,0,0,1
2,19,A,1,1,0,0,1,0,0,0,0,1
2,19,B,0,0,0,1,0,1,0,0,1,0
2,20,A,0,0,0,1,0,0,1,0,0,1
2,20,B,1,0,1,0,1,0,0,1,0,0
2,21,A,0,1,0,0,0,1,0,1,0,0
2,21,B,1,0,1,0,0,0,1,0,0,1
2,22,A,1,0,0,1,1,0,0,1,0,0
2,22,B,0,1,0,0,0,1,0,0,1,0
2,23,A,0,1,0,0,0,1,0,0,0,1
2,23,B,1,0,1,0,0,0,1,0,1,0
2,24,A,1,0,0,1,1,0,0,0,1,0
2,24,B,0,0,1,0,0,0,1,1,0,0
2,25,A,1,1,0,0,0,0,1,1,0,0
2,25,B,0,0,1,0,1,0,0,0,0,1
3,1,A,1,0,0,1,1,0,0,1,0,0
3,1,B,0,0,1,0,0,0,1,0,0,1
3,2,A,0,1,0,0,1,0,0,0,0,1
3,2,B,1,0,1,0,0,1,0,0,1,0
3,3,A,1,0,0,1,0,1,0,1,0,0
3,3,B,0,1,0,0,0,0,1,0,1,0
3,4,A,1,0,0,1,0,1,0,1,0,0
3,4,B,0,0,1,0,0,0,1,0,0,1
3,5,A,0,0,1,0,1,0,0,1,0,0
3,5,B,1,1,0,0,0,1,0,0,1,0
3,6,A,0,1,0,0,1,0,0,0,0,1
3,6,B,1,0,0,1,0,0,1,0,1,0
3,7,A,1,1,0,0,0,0,1,1,0,0
3,7,B,0,0,1,0,0,1,0,0,0,1

[R] a issue about the qutation mark?

2010-07-16 Thread karena


Following is a function that I wrote (It is working well). It's a simple one,
nothing complicated. The only question that I have is a qutation mark issue,
I guess.
#
funcname - function(trait.file){#line1
setwd('/root/subroot') # line 2
load('imge.RData')# line3
imge - imge[,-c(3)] # line4
imge - imge[complete.cases(imge),]   # line5
trait- read.csv(trait.file) # line6
ngenes - nrow(imge) # line7
nsnp - nrow(trait)  #line 8
for(i in 1:ngenes) { 
  for(j in 1:nsnp) {
if(imge$d1[i]==trait$d1[j]  imge$d2[i]==trait$d2[j]) trait$imgene2[j] 
-
imge$Gene[i]
  else trait$imgene2[j] - NA
   }
}
   return(trait)
}

hypertension - funcname(folder/hyper.csv) # last line
##

So, as we all know, when using read.csv, we need to use qutation mark out
side the filename which we wanna read in. At first, for line 8, I wrote:
read.csv(trait.file), at last line, I wrote:
funcname(folder/hyper.csv), but it did not work in this way. Unless I
changed the code to the current one that I showed above, I couldn't get what
I want.  
So, to be straightforward, I will show the difference:
1) the code not working:
read.csv(trait.file)   #line 8
funcname(folder/hyper.csv)  #last line

2) the code working:
read.csv(trait.file) # line 8
funcname(folder/hyper.csv)   # last line

anyone can tell me why is the difference?

thank you,

karena


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[R] Nested if help

2010-07-16 Thread George Coyle

Hello,

I am trying to find a direct way to write a nested if of sorts to find data
for a specific time range for a specific day (across a range of days) and
have exhausted my abilities with the manuals I have at hand.  I have a good
deal of data of this approximate form:

day   time   price
1  1am5
1   2am7
1  3am 9
1  4am 12
 2  1am5
2   2am7
2  3am 9
2  4am 12
etc

I am reading from a file I loaded and using a while statement.  I then am
trying the direct approach of writing the if:

if(time=2am time =4am) { #specifying desired time range--since a while
loop format I don't think date is necessary and if it is I don't know how to
say this?
time1-time #sets first time in desired interval to base time
px1-px #sets first price in desired interval to base price
if(pxpx1){ #nested if stmt to see if subsequent times are higher than base,
thus replacing base with new max (I am seeking max in the time period in
this instance)
px2-px #px2 would be the new higher maximum
time2-time #time associated with px2
px3-px+1 #price immediately following max
time3-time+1 #time immed. follows max
out-(cur_date,px2,time2,px3,time3) #output the high price/time and the
immediately following price/time
cat(out,\n)}

This code however, does nothing.  Any help would be apreciated.  The manuals
seem to only take one so far.

Thanks,

George

[[alternative HTML version deleted]]

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[R] Troubles with DBI's dbWriteTable in RMySQL

I am feeling rather dumb right now.

I created what I thought was a data.frame as follows:

aaa - lapply(split(moreinfo,list(moreinfo$m_id),drop = TRUE), fun_m_id)
m_id_default_res - do.call(rbind, aaa)
print(==)
m_id_default_res
print(==)
ndf - m_id_default_res[, c('mid', 'estimate', 'sd', 'loglik', 'aic','bic',
'chisq', 'chisqpvalue', 'chisqdf')]
ndf

The data in NDF is perfect, exactly what I expected when I print the
contents as shown in the last statement above.

On the asumption tha tthat is a data frame, I tried

dbWriteTable(con,test1,ndf);

But I received the following error:

Error in function (classes, fdef, mtable)  :
  unable to find an inherited method for function dbWriteTable, for
signature MySQLConnection, character, matrix

Then, on the assumption it is trivial to convert a matrix into a data.frame,
i tried:

dbWriteTable(con,test2,as.data.frame(ndf));

But this produced the following error:
Error in write.table(x, file, nrow(x), p, rnames, sep, eol, na, dec,
as.integer(quote),  :
  unimplemented type 'list' in 'EncodeElement'

The silly, and frustrating, thing is that I used dbWriteTable before, and
that worked adequately.  But that was with a simple data frame (within a for
loop, element by element - res$var[[i]] = expression), not the result of
do.call(rbind(...))  The principle limitation I saw in my previous use of
dbWriteTable is that all fields are given the type 'TEXT', and that it
insists on creating a new table.  What I'd prefer is a kind of bulk interset
that just makes extra records for an existing table.

So, given my past experience with dbWriteTable, it is a question of what
do.call(rbind(..)) did to produce ndf that has the effect that dbWriteTable
doesn't like that data.frame.

So, then, what is the bext way to either get dbWriteTable working (ideally
in a way that works around the limitations I mention above) or to do a bulk
insert into my MySQL table (yes, I already have a table in the relevant
schema with all the right data types for each field, and I load RMySQL at
the start of my program.)  In a worst case, I can live with an insertion one
record at a time.

Thanks

Ted

PS: If it helps, here is the the contents of ndf - as shown by entering
'ndf' at the R prompt:
 ndf
mid estimate   sd   loglikaic  bic  chisq
chisqpvalue   chisqdf
206 206 0.1147528  0.04336918   -22.15483 46.30965 46.25556 4.433502
0.035240131
229 229 0.0736 0.01999671   -56.41179 114.8236 115.5962 195307.1
0 2
251 251 0.074421   0.002171616  -4224.072 8450.144 8455.212 593302.2
0 18
252 252 0.03710208 0.0004556731 -28426.82 56855.65 56862.45 3543373
0 38
253 253 0.01397349 0.0005900857 -2925.179 5852.358 5856.677 283.9848
5.232282e-51  16
254 254 0.09043846 0.01528502   -119.108  240.216  241.7713 23.52441
3.139385e-05  3
255 255 0.05078883 0.0006021373 -28294.38 56590.76 56597.63 1988844
0 35
260 260 0.03392846 0.005499136  -166.5730 335.1461 336.7837 10.83060
0.054844135
268 268 0.05357114 0.01785082   -35.3407  72.6814  72.87863 82995.79
0 2
286 286 0.09321947 0.01987217   -74.20157 150.4031 151.4942 1.698603
0.6372445 3
290 290 0.03841793 0.006584153  -144.8139 291.6277 293.1541 135.8937
2.902434e-29  3
292 292 0.06289269 0.01988338   -37.66325 77.32651 77.6291  143099.8
0 2
297 297 0.01674874 0.004047625  -86.52035 175.0407 175.8739 47.27713
3.034432e-10  3
302 302 0.02878066 0.003876092  -250.1428 502.2857 504.293  9.22447
0.2369393 7
306 306 0.07904849 0.0004164051 -127449.0 254899.9 254908.4 111574416
0 40
307 307 0.01655872 0.001320903  -795.7314 1593.463 1596.513 57.38622
1.127804e-08  10
308 308 0.02631102 0.000884155  -4095.149 8192.298 8197.081 142.8876
3.904898e-20  21
309 309 0.09891599 0.0084501-453.9474 909.8947 912.8147 357135.5
0 8
310 310 0.09332047 0.004580396  -1399.262 2800.524 2804.552 217126
0 13
311 311 0.06378327 0.0005049166 -59848.62 119699.2 119706.9 59481893
0 34
313 313 0.06203001 0.0006486936 -34546.67 69095.34 69102.46 18207698
0 32
316 316 0.173  0.07026985   -25.04100 52.08199 52.38458 18002.22
0 2
317 317 0.04405086 0.0005949207 -22578.44 45158.88 45165.49 8923236
0 33
320 320 0.05747093 0.006634162  -289.2357 580.4714 582.7889 8.641322
0.2794433 7
321 321 0.06365155 0.003692525  -1115.037 2232.073 2235.767 19.10553
0.0860133712
322 322 0.05737672 0.01532991   -54.01363 110.0273 110.6663 9.597753
0.008238998   2
323 323 0.03116934 0.001909146  -1188.573 2379.146 2382.73  109.7663
6.656046e-18  12
324 324 0.03027327 0.0004146385 -23922.15 47846.3  47852.88 47330365
0 32
325 325 0.06047783 0.00922026   -163.6356 329.2711 331.0323 1695781
0 3
326 326 0.05627898 0.0008642285 -16432.57 32867.13 32873.48 3405089
0 29
327 327 0.07052627

[R] Creating symbolic expressions in R

2010-07-16 Thread Erin

Hello,

I'm trying to do some differential equation modeling in R using the
package 'deSolve.'  Briefly, I'm trying to use the law of mass action
(the details of which aren't really important) to structure a vector
of rate equation which will be passed into an ODE function and solved
with associated functions from 'deSolve.'  Roughly what this entails
is scanning through a very large stoichiometric matrix containing
integers which map metabolites to reactions and then assembling a
series of monomials that describe the rates of the reactions.  I'm
running into trouble because R does not like doing numerical
operations on characters, but I need the rate to be in symbolic form.
Below I've described an example.

The input is a matrix like this one, but a lot bigger:

  rxn1 rxn2 rxn3
met1   -2   0   2
met2   -1  -1   0
met30   -1 -1

and the goal is to generate a vector of monomials representing the
rate of each reaction 1 - 3:

rateRxn1 = (met1)^2 * (met2)^1
rateRxn2 = (met2)^1 * (met3)^1
rateRxn3 = (met3)^1

The rates need to be represented symbolically for me to pass into the
ODE solver where the metabolites will be assigned starting values.  I
tried using the functions quote and substitute but unfortunately
they are tough to use within a loop because if I pass in a metabolite
label generated from a loop (e.g. metaboliteNames[i]) it's taken
verbatim in quote(), so I get metaboliteNames[i]^-stoich[i, j]
instead of met1^2.

My goal is to make a package of functions for modeling and simulation
of chemical kinetics and I want to make sure that they are open-source
and freely available to all researchers, so any advice on how to
proceed would be greatly appreciated!

Erin

-- 
Erin Rachael Shellman
The University of Michigan
Bioinformatics PhD Candidate
http://www.erinshellman.com
shell...@umich.edu
(937) 321.1129

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Re: [R] RMySQL package on 64bit R for Windows

2010-07-16 Thread abeSRT


Did you ever find a solution for this?
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Re: [R] sqldf modify table

On Fri, Jul 16, 2010 at 2:46 PM, PeterTucker pthet...@gmail.com wrote:

 Hi - I am something of a newbie and am a little perplexed.  When (trying to)
 modify a table I issue the following commands with subsequent errors

 sqldf(alter table Korea drop column code, dbname = mydb)
 error in statement: near drop: syntax error

 or

 sqldf(alter table Korea rename column hyr to hyrI, dbname = mydb)
 error in statement: near column: syntax error

 These are simple commands - am I missing something obvious?  I can retrieve
 data from them, and retrieve their table_info


SQLite does not support dropping columns. See:

   http://www.sqlite.org/lang_altertable.html

however, sqldf does support the H2 and PostgreSQL databases in
addition to sqlite so you can try one of those if this feature is
important to you.

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Re: [R] Nested if help