date:20120314

yes but giving error
 t-read.table(file.choose())
 t-ts(t)
 plot(t)
Hit Return to see next plot: 
 ht-HoltWinters(t)
Error in decompose(ts(x[1L:wind], start = start(x), frequency = f),
seasonal) : 
  time series has no or less than 2 periods

what should i do? 

--
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http://r.789695.n4.nabble.com/how-to-find-best-model-of-time-series-tp4469296p4471097.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Lazarus Mramba

Dear all,

I have a large matrix with about 2500 variables, and 100 rows.

I would like to calculate the means of the every 10 variables starting from
1:2500  and saving the results as a vector or matrix.
How can I do that?
Alternatively, How can I create 250 subset matrices in the order of
variables 1:2500 in groups of 10 from the single matrix which had initially
2500 variables ?
I guess I have to use a loop, but I can't figure out how.

Any help will be appreciated,

Regards,
Lazarus

[[alternative HTML version deleted]]

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[R] Apply a loop containing a function on a list

2012-03-14 Thread ali_protocol

Hi all,
I want to do this:

B.list$aa= (a loop containing My.fun acting on the reults of second
function on a A.list$aa))
or, overally
B.list$aa = function (A.list$aa)
B.list and A.list has many sublists aa, ab and Is there a way I can
apply the function and loop on all sublists of A.list and get B.list?


Thanks in advance.

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Re: [R] reshaping a dataset for a network

2012-03-14 Thread Marco Guerzoni


Thank you fro the reply.
I managed to arrive till here, then I would like to have it in matrix 
where the $1 $2...$5 are the first column.


Il 3/13/2012 8:15 PM, William Dunlap ha scritto:

Is the following what you want?
 a- c(1,2,3,4,4,4,5,5)
 b- c(11,7,4,9,8,3,12,4)
 split(b, a)
   $1
   [1] 11

   $2
   [1] 7

   $3
   [1] 4

   $4
   [1] 9 8 3

   $5
   [1] 12  4


Here I did

c - split(b, a)
d - do.call(rbind,c)

and I get

  [,1] [,2] [,3]
1   11   11   11
2777
3444
4983
5   124   12


Instead of what I would like:

  [,1] [,2] [,3]
1   11
27
34
4983
5   124

A solution could be rbind.fill , which does not seem to work with list.


thanks
Marco Guerzoni,
Department of Economics
University of Turin




Note that your df-cbind(a,b) produces a matrix, not the data.frame
that your df suggests you want.  Use df-data.frame(a,b) to make
a data.frame.  Then you could do with(df, split(a,b)) to operate on
the a and b in the data.frame df.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf
Of Marco Guerzoni
Sent: Tuesday, March 13, 2012 10:51 AM
To: r-help@r-project.org
Subject: [R] reshaping a dataset for a network

dear all,
apologizes for bothering with a probably stupid question but I really
don' t know how to proceed.

I have a dataset which look like df

a- c(1,2,3,4,4,4,5,5)
b- c(11,7,4,9,8,3,12,4)
df-cbind(a,b)

I would like to have one which looks like this:

a
1 11
2 7
3 4
4 9 8 3
5 12 4

a are vertex of a network, b the edges. In the data the lenght of a is
about 5

I read several posts about reshape, reshape2, split, ldply but I
couldn't manage to do it. The problem seems to be that the is not a real
panel.

Any help would be really appreciated,
my best regards
Marco

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Re: [R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Tsjerk Wassenaar

Hi Lazarus,

Checkout arrays (?array). You can cast your matrix to an array of
submatrices, and calculate the means per block using apply.

Cheers,

Tsjerk

On Mar 14, 2012 9:25 AM, Lazarus Mramba lmra...@gmail.com wrote:

Dear all,

I have a large matrix with about 2500 variables, and 100 rows.

I would like to calculate the means of the every 10 variables starting from
1:2500  and saving the results as a vector or matrix.
How can I do that?
Alternatively, How can I create 250 subset matrices in the order of
variables 1:2500 in groups of 10 from the single matrix which had initially
2500 variables ?
I guess I have to use a loop, but I can't figure out how.

Any help will be appreciated,

Regards,
Lazarus

   [[alternative HTML version deleted]]

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

[[alternative HTML version deleted]]

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[R] Partial Canonical Correlation Analysis Help

2012-03-14 Thread Proitsi, Petroula

Hello,

I am trying to perform partial Canonical Correlation (CCA) in R. I have 
calculated the CCA between two vectors  X  and Y and now I want to control 
for Z.
I know I have to modify the cc and rcc functions to include Z but what I am 
doing does not seem to be working...
Below I have the original code and my modified code

Original cc:
function (X, Y) 
{
Xnames = dimnames(X)[[2]]
Ynames = dimnames(Y)[[2]]
ind.names = dimnames(X)[[1]]
res = rcc(X, Y, 0, 0)
return(res)
}

I changed this  to:  

function (X, Y, Z) 
{
Xnames = dimnames(X)[[2]]
Ynames = dimnames(Y)[[2]]
Znames = dimnames(Z) [[2]]
   ind.names = dimnames(X)[[1]]
res = rcc2(X, Y, Z, 0, 0, 0)
return(res)
}

and then for rcc

Original rcc:

function (X, Y, lambda1, lambda2) 
{
Xnames - dimnames(X)[[2]]
Ynames - dimnames(Y)[[2]]
ind.names - dimnames(X)[[1]]
Cxx - var(X, na.rm = TRUE, use = pairwise) + diag(lambda1, 
ncol(X))
Cyy - var(Y, na.rm = TRUE, use = pairwise) + diag(lambda2, 
ncol(Y))
Cxy - cov(X, Y, use = pairwise)
res - geigen(Cxy, Cxx, Cyy)
names(res) - c(cor, xcoef, ycoef)
scores - comput(X, Y, res)
return(list(cor = res$cor, names = list(Xnames = Xnames, 
Ynames = Ynames, ind.names = ind.names), xcoef = res$xcoef, 
ycoef = res$ycoef, scores = scores))
}

I know I have to calculate Czz Czx and Czy but do I have to calculate Cxy.z 
(partial covariance)?
and I am not sure what to add in the res- geigen etc
when I add Czz and/or Czx Czy I get an error - I have also left scores and 
return ans original since I am not sure what to include 

modified rcc2

function (X, Y, Z,  lambda1, lambda2, lambda3) 
{
Xnames - dimnames(X)[[2]]
Ynames - dimnames(Y)[[2]]
Znames - dimnames(Z)[[2]]
ind.names - dimnames(X)[[1]]
Cxx - var(X, na.rm = TRUE, use = pairwise) + diag(lambda1, 
ncol(X))
Cyy - var(Y, na.rm = TRUE, use = pairwise) + diag(lambda2, 
ncol(Y))
Czz - var(Z, na.rm = TRUE, use = pairwise) + diag(lambda3, 
ncol(Z))
Cxy - cov(X, Y, use = pairwise)
Czx - cov(X, Z, use = pairwise)
Czy - cov(Y, Z, use = pairwise)
# Cxy.z?
res - geigen(Cxy, Cxx, Cyy, Czz, Czx, Czy) # doesnt work
names(res) - c(cor, xcoef, ycoef, zcoef)
scores - comput(X, Y, Z, res)
return(list(cor = res$cor, names = list(Xnames = Xnames, 
Ynames = Ynames, Znames=Znames, ind.names = ind.names), xcoef = 
res$xcoef, 
ycoef = res$ycoef, zcoef = res$zcoef,  scores = scores))
}


Any help would be really appreciated..I am completely new to this and lost.

many thanks

Joe

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Re: [R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Petr Savicky

On Wed, Mar 14, 2012 at 03:22:39AM -0400, Lazarus Mramba wrote:
 Dear all,
 
 I have a large matrix with about 2500 variables, and 100 rows.
 
 I would like to calculate the means of the every 10 variables starting from
 1:2500  and saving the results as a vector or matrix.
 How can I do that?
 Alternatively, How can I create 250 subset matrices in the order of
 variables 1:2500 in groups of 10 from the single matrix which had initially
 2500 variables ?
 I guess I have to use a loop, but I can't figure out how.

Hi.

Try the following. I will use smaller parameters for simplicity.

  # a matrix 3 times 20
  a - matrix(1:60, nrow=3, ncol=20)
  a
  
   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] 
[,14]
  [1,]147   10   13   16   19   22   2528313437
40
  [2,]258   11   14   17   20   23   2629323538
41
  [3,]369   12   15   18   21   24   2730333639
42
   [,15] [,16] [,17] [,18] [,19] [,20]
  [1,]434649525558
  [2,]444750535659
  [3,]454851545760
  
  #combine each 5 consecutive columns to a single column
  dim(a) - c(15, 4)
  a
  
[,1] [,2] [,3] [,4]
   [1,]1   16   31   46
   [2,]2   17   32   47
   [3,]3   18   33   48
   [4,]4   19   34   49
   [5,]5   20   35   50
   [6,]6   21   36   51
   [7,]7   22   37   52
   [8,]8   23   38   53
   [9,]9   24   39   54
  [10,]   10   25   40   55
  [11,]   11   26   41   56
  [12,]   12   27   42   57
  [13,]   13   28   43   58
  [14,]   14   29   44   59
  [15,]   15   30   45   60
  
  # compute column means
  colMeans(a)
  
  [1]  8 23 38 53

Hope this helps.

Petr Savicky.

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Re: [R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Tsjerk Wassenaar

Hi Lazarus,

Sorry for my brevity before. I sent it from a phone.

# Generate dummy data matrix
x - matrix(1:100,nrow=5)

# Rearrange into array; ten groups of two columns
y - array(x,c(5,2,10))

# Average columns for each submatrix
apply(y,c(1,3),mean)

# Average per submatrix
apply(y,3,mean)

Hope it helps,

Tsjerk
On Wed, Mar 14, 2012 at 10:01 AM, Petr Savicky savi...@cs.cas.cz wrote:
 On Wed, Mar 14, 2012 at 03:22:39AM -0400, Lazarus Mramba wrote:
 Dear all,

 I have a large matrix with about 2500 variables, and 100 rows.

 I would like to calculate the means of the every 10 variables starting from
 1:2500  and saving the results as a vector or matrix.
 How can I do that?
 Alternatively, How can I create 250 subset matrices in the order of
 variables 1:2500 in groups of 10 from the single matrix which had initially
 2500 variables ?
 I guess I have to use a loop, but I can't figure out how.

 Hi.

 Try the following. I will use smaller parameters for simplicity.

  # a matrix 3 times 20
  a - matrix(1:60, nrow=3, ncol=20)
  a

       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] 
 [,14]
  [1,]    1    4    7   10   13   16   19   22   25    28    31    34    37    
 40
  [2,]    2    5    8   11   14   17   20   23   26    29    32    35    38    
 41
  [3,]    3    6    9   12   15   18   21   24   27    30    33    36    39    
 42
       [,15] [,16] [,17] [,18] [,19] [,20]
  [1,]    43 46 49 52 55    58
  [2,]    44    47    50    53    56    59
  [3,]    45    48    51    54    57    60

  #combine each 5 consecutive columns to a single column
  dim(a) - c(15, 4)
  a

        [,1] [,2] [,3] [,4]
   [1,]    1   16   31   46
   [2,]    2   17   32   47
   [3,]    3   18   33   48
   [4,]    4   19   34   49
   [5,]    5   20   35   50
   [6,]    6   21   36   51
   [7,]    7   22   37   52
   [8,]    8   23   38   53
   [9,]    9   24   39   54
  [10,]   10   25   40   55
  [11,]   11   26   41   56
  [12,]   12   27   42   57
  [13,]   13   28   43   58
  [14,]   14   29   44   59
  [15,]   15   30   45   60

  # compute column means
  colMeans(a)

  [1]  8 23 38 53

 Hope this helps.

 Petr Savicky.

 __
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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.



-- 
Tsjerk A. Wassenaar, Ph.D.

post-doctoral researcher
Molecular Dynamics Group
* Groningen Institute for Biomolecular Research and Biotechnology
* Zernike Institute for Advanced Materials
University of Groningen
The Netherlands

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Re: [R] Idea/package to linearize a curve along the diagonal?

2012-03-14 Thread Yvonnick Noel




I am trying to normalize some data. First I fitted a principal curve
(using the LCPM package), but now I would like to apply a
transformation so that the curve becomes a straight diagonal line on
the plot.  The data used to fit the curve would then be normalized by
applying the same transformation to it.


It is unclear to me what you mean by diagonal but I suspect what 
you're looking for is to locate projected points onto the unfolded curve.


That is exactly what coordinates on the principal curve would give you.

Sorry if I misunderstood your point,

Yvonnick Noel
University of Brittany,
Rennes, France

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[R] [R-pkgs] diveMove 1.3.1

2012-03-14 Thread Sebastian P. Luque

Version 1.3.1 of diveMove is now available on CRAN.  Changes since
previous version (1.2.9) are:


   o Improved formatting of code in vignette.  Figure resolution reduced to
 satisfy package check requirements.

   o At least R 2.13.0 is required.

   o Dives occurring in trivial wet periods are now identified.

   o Improved validity checking for 'TDRcalibrate' objects.

   o 'TDRcalibrate' accessor methods now explicitly check and report
 that requested dives exist.

   o plotTDR() now conditionally generates initial plot, after preparing
 Tcl/Tk widgets, to accomodate for changes in R 2.14.2. Legend is
 plotted only if there is at least one level in the phase factor.


Cheers,

-- 
Sebastian P. Luque, Ph.D.
Department of Biological Sciences
University of Manitoba
http://www.ucs.mun.ca/~sluque

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Re: [R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Tsjerk Wassenaar

Hi Lazarus,

You should use apply, not sapply.

 a =matrix(1:12, nrow=2,ncol=6,byrow=TRUE)
 b = array(a,c(2,2,3));b
 colMeans(b)

 f1=function(x) sum(x)/length(row(a.df[,1:2]))
 y3 - apply(b, 1, f1)

 It only gives one matrix with only two values, I expected 3 values.

To apply the function on each submatrix, you have to apply it over the
third index:

y3 - apply(b, 3, f1)

Cheers,

Tsjerk



-- 
Tsjerk A. Wassenaar, Ph.D.

post-doctoral researcher
Molecular Dynamics Group
* Groningen Institute for Biomolecular Research and Biotechnology
* Zernike Institute for Advanced Materials
University of Groningen
The Netherlands

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Re: [R] how to find best model of time series?giving error

2012-03-14 Thread Berend Hasselman


On 14-03-2012, at 07:22, sagarnikam123 wrote:

 yes but giving error
 t-read.table(file.choose())
 t-ts(t)
 plot(t)
 Hit Return to see next plot: 
 ht-HoltWinters(t)
 Error in decompose(ts(x[1L:wind], start = start(x), frequency = f),
 seasonal) : 
  time series has no or less than 2 periods
 
 what should i do? 

Read HoltWinters help thoroughly.
Your timeseries should have a frequency  1.

So how about t - ts(t, start=??, frequency=4)

Berend

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Re: [R] Help with plot Grouped Bar Plot by using R

2012-03-14 Thread Bert Gunter

Please stop asking such basic questions on this list and start reading
An Intro to R (ships with R) or any other beginning R tutorial. It
seems to me to be impolite to pester this list for beginners'
instructions without first having made at least a minimal effort to to
familiarize yourself with R basics (in this case, basic plotting
procedures).

-- Bert

On Tue, Mar 13, 2012 at 9:36 PM, R_beginner_starter
cjyxiao...@gmail.com wrote:
 Thanks a lot, Jim.

 I get my desired output by using your R script at the end
 Yesterday was my mistake

 Apart from that, do you have any idea to plot a line graph/chart?
 My Input file is shown as below:

 Range   Data1   Data2
 1       38      37
 2       38      37
 3       38      37
 4       37      37
 5       37      37
 6       37      37
 7       37      37
 8       37      37
 9       37      37
 10      37      37
 11      37      37
 12      37      37
 13      37      37
 14      37      37
 15      37      37
 16      37      37
 17      37      37
 18      37      37
 19      37      37
 20      37      37
 21      37      36
 22      37      36
 23      37      36
 24      37      36
 25      37      36
 26      37      36
 27      36      36
 28      36      36
 29      36      36
 30      36      36
 31      36      36
 32      36      36
 33      36      36
 34      36      36
 35      36      36
 36      36      36
 37      36      36
 38      36      36
 39      36      36
 40      36      36

 Thanks for your further notice and advice :)

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Help-with-plot-Grouped-Bar-Plot-by-using-R-tp4448762p4471020.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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

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Re: [R] Creating 250 submatrices from a large single matrix with 2500 variables using loops

2012-03-14 Thread Bert Gunter

??
Petr's solution is shorter and for large problems considerably faster...

-- Bert

On Wed, Mar 14, 2012 at 2:08 AM, Tsjerk Wassenaar tsje...@gmail.com wrote:
 Hi Lazarus,

 Sorry for my brevity before. I sent it from a phone.

 # Generate dummy data matrix
 x - matrix(1:100,nrow=5)

 # Rearrange into array; ten groups of two columns
 y - array(x,c(5,2,10))

 # Average columns for each submatrix
 apply(y,c(1,3),mean)

 # Average per submatrix
 apply(y,3,mean)

 Hope it helps,

 Tsjerk
 On Wed, Mar 14, 2012 at 10:01 AM, Petr Savicky savi...@cs.cas.cz wrote:
 On Wed, Mar 14, 2012 at 03:22:39AM -0400, Lazarus Mramba wrote:
 Dear all,

 I have a large matrix with about 2500 variables, and 100 rows.

 I would like to calculate the means of the every 10 variables starting from
 1:2500  and saving the results as a vector or matrix.
 How can I do that?
 Alternatively, How can I create 250 subset matrices in the order of
 variables 1:2500 in groups of 10 from the single matrix which had initially
 2500 variables ?
 I guess I have to use a loop, but I can't figure out how.

 Hi.

 Try the following. I will use smaller parameters for simplicity.

  # a matrix 3 times 20
  a - matrix(1:60, nrow=3, ncol=20)
  a

       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] 
 [,14]
  [1,]    1    4    7   10   13   16   19   22   25    28    31    34    37   
  40
  [2,]    2    5    8   11   14   17   20   23   26    29    32    35    38   
  41
  [3,]    3    6    9   12   15   18   21   24   27    30    33    36    39   
  42
       [,15] [,16] [,17] [,18] [,19] [,20]
  [1,]    43 46 49 52 55    58
  [2,]    44    47    50    53    56    59
  [3,]    45    48    51    54    57    60

  #combine each 5 consecutive columns to a single column
  dim(a) - c(15, 4)
  a

        [,1] [,2] [,3] [,4]
   [1,]    1   16   31   46
   [2,]    2   17   32   47
   [3,]    3   18   33   48
   [4,]    4   19   34   49
   [5,]    5   20   35   50
   [6,]    6   21   36   51
   [7,]    7   22   37   52
   [8,]    8   23   38   53
   [9,]    9   24   39   54
  [10,]   10   25   40   55
  [11,]   11   26   41   56
  [12,]   12   27   42   57
  [13,]   13   28   43   58
  [14,]   14   29   44   59
  [15,]   15   30   45   60

  # compute column means
  colMeans(a)

  [1]  8 23 38 53

 Hope this helps.

 Petr Savicky.

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 * Groningen Institute for Biomolecular Research and Biotechnology
 * Zernike Institute for Advanced Materials
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Re: [R] Problem installing RMySQL package!

2012-03-14 Thread S Ellison

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of AGGARWAL, 
 NEERAJ (NEERAJ)
 Sent: 14 March 2012 05:14
 To: r-help@r-project.org
 Subject: [R] Problem installing RMySQL package!

 I am facing issues while installing RMySQL package on 
 windows(32 bit) installation of R-Project.
 I am getting the following warning messages.

 Warning messages:
 1: running command 'C:/PROGRA~1/R/R-213~1.2/bin/i386/R CMD 
 INSTALL -l C:/Program Files/R/R-2.13.2/library   
 D:\DOCUME~1\nagga002\LOCALS~1\Temp\RtmpLZndlE/downloaded_packa
 ges/RMySQL_0.9-3.tar.gz' had status 1
 2: In install.packages(RMySQL, type = source) :
   installation of package 'RMySQL' had non-zero exit status

 Is RTools something different from RProject. 
Read 'R installation and administration' in your R Help system. That will tell 
you what R Tools is/are and where to get them.

In relation to accessing a MySQL database, you could also consider installing 
RODBC (which installs from binary unsinf install.packages()), registering your 
MySQL database as an ODBC data source in Windows and then using RODBC's query 
facilities. That certainly works for ordinary SELECTS, INSERTS and UPDATES, and 
does not require RMySQL. 

S***
This email and any attachments are confidential. Any use...{{dropped:8}}

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Re: [R] Apply a loop containing a function on a list

2012-03-14 Thread Milan Bouchet-Valat

Le mercredi 14 mars 2012 à 00:51 -0700, ali_protocol a écrit :
 Hi all,
 I want to do this:
   
   B.list$aa= (a loop containing My.fun acting on the reults of second
 function on a A.list$aa))
 or, overally  
   B.list$aa = function (A.list$aa)
 B.list and A.list has many sublists aa, ab and Is there a way I can
 apply the function and loop on all sublists of A.list and get B.list?
Please provide a small example, you can easily create two lists to
illustrate your point. It's hard to help without that, it would require
us mocking up an example for you, and we might get it wrong.


Cheers

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[R] Creating polygons from scattered points

2012-03-14 Thread Louise Mair

Hello,

I have a distribution dataset for species consisting of xy coordinates at
the 1km resolution, with only presence data. So a simplified example of a
species distribution might be:

y - rbind(as.integer(rnorm(100,50,20)), as.integer(rnorm(200,100,30)),
as.integer(rnorm(100,180,15)))
x - rbind(as.integer(rnorm(200,50,20)), as.integer(rnorm(200,100,20)),
as.integer(rnorm(100,200,15)))
plot(y~x)

I would like to create polygons for each species distribution, where if an
island is present (as I have tried to show in the example), it would be a
seperate polygon, and the jagged edges of coastlines etc are maintained. I
have spent ages trying to find a package that will allow me to convert
scattered point distributions to polygons but haven't found anything that
works, the functions I have found require the data already to be in the
format where the only xy coordinates present are the outline of the
polygon.

Can anyone please recommend a function I can use here, or suggest a way of
extracting the outline points? I have tried this manually but cannot seem
to write a code that will effectively take account of jagged edges and
islands.

Thanks very much for your help,

Louise.

[[alternative HTML version deleted]]

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[R] Moore-Penrose Generalized determinant?

2012-03-14 Thread Sean O'Riordain

Is there a function in R to calculate the generalized determinant of a
singular matrix? - similar to the ginv() used to compute the generalized
inverse.

I can't seem to find any R related posts at all.

Thanks in advance,
Sean O'Riordain
Trinity College Dublin

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Re: [R] multi-histogram plotting

It's easier to do log plots as

plot(h$density, log = L)

where L is one of x, y, or xy as you choose. This is well documented.

Michael

On Wed, Mar 14, 2012 at 12:02 AM, Sam Steingold s...@gnu.org wrote:
 * David Winsemius qjvafrz...@pbzpnfg.arg [2012-03-13 17:53:14 -0400]:
 On Mar 13, 2012, at 5:33 PM, Sam Steingold wrote:
 I can, of course, plot log(h$density), but then the number labels will
 be wrong.

 You could try apply a log transform to the appropriate component of
 the h object and using barplot to display the results.

 that's what I said above: plot log(h$density).
 However, the ordinate will be labeled with log values, not the original
 values. how do I get the log ticks on the ordinate?


 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 
 11.0.11004000
 http://www.childpsy.net/ http://jihadwatch.org http://www.memritv.org
 http://dhimmi.com http://memri.org http://pmw.org.il http://truepeace.org
 Profanity is the one language all programmers know best.

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] reshaping a dataset for a network

You can't have empty spots like that in an array. One choice would
be to fill them with NAs:

library(plyr)
do.call(rbind.fill.matrix,lapply(split(b,a), t))

Michael

On Wed, Mar 14, 2012 at 4:01 AM, Marco Guerzoni marco.guerz...@unito.it wrote:
 Thank you fro the reply.
 I managed to arrive till here, then I would like to have it in matrix where
 the $1 $2...$5 are the first column.

 Il 3/13/2012 8:15 PM, William Dunlap ha scritto:

 Is the following what you want?
     a- c(1,2,3,4,4,4,5,5)
     b- c(11,7,4,9,8,3,12,4)
     split(b, a)
   $1
   [1] 11

   $2
   [1] 7

   $3
   [1] 4

   $4
   [1] 9 8 3

   $5
   [1] 12  4


 Here I did

 c - split(b, a)
 d - do.call(rbind,c)

 and I get

  [,1] [,2] [,3]
 1   11   11   11
 2    7    7    7
 3    4    4    4
 4    9    8    3
 5   12    4   12


 Instead of what I would like:

  [,1] [,2] [,3]

 1   11
 2    7
 3    4
 4    9    8    3
 5   12    4

 A solution could be rbind.fill , which does not seem to work with list.


 thanks
 Marco Guerzoni,
 Department of Economics
 University of Turin




 Note that your df-cbind(a,b) produces a matrix, not the data.frame
 that your df suggests you want.  Use df-data.frame(a,b) to make
 a data.frame.  Then you could do with(df, split(a,b)) to operate on
 the a and b in the data.frame df.

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf
 Of Marco Guerzoni
 Sent: Tuesday, March 13, 2012 10:51 AM
 To: r-help@r-project.org
 Subject: [R] reshaping a dataset for a network

 dear all,
 apologizes for bothering with a probably stupid question but I really
 don' t know how to proceed.

 I have a dataset which look like df

 a- c(1,2,3,4,4,4,5,5)
 b- c(11,7,4,9,8,3,12,4)
 df-cbind(a,b)

 I would like to have one which looks like this:

 a
 1 11
 2 7
 3 4
 4 9 8 3
 5 12 4

 a are vertex of a network, b the edges. In the data the lenght of a is
 about 5

 I read several posts about reshape, reshape2, split, ldply but I
 couldn't manage to do it. The problem seems to be that the is not a real
 panel.

 Any help would be really appreciated,
 my best regards
 Marco

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 PLEASE do read the posting guide
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Re: [R] Problem installing RMySQL package!

In addition to Dr. Ellison's sound advice, as the link says: follow
the provided link, go to a CRAN mirror
(http://cran.r-project.org/mirrors.html), click download for windows
and Rtools will be available.

Michael

On Wed, Mar 14, 2012 at 1:14 AM, AGGARWAL, NEERAJ (NEERAJ)
neeraj.aggar...@alcatel-lucent.com wrote:
 I am facing issues while installing RMySQL package on windows(32 bit) 
 installation of R-Project.
 I am getting the following warning messages.

 Warning messages:
 1: running command 'C:/PROGRA~1/R/R-213~1.2/bin/i386/R CMD INSTALL -l 
 C:/Program Files/R/R-2.13.2/library   
 D:\DOCUME~1\nagga002\LOCALS~1\Temp\RtmpLZndlE/downloaded_packages/RMySQL_0.9-3.tar.gz'
  had status 1
 2: In install.packages(RMySQL, type = source) :
  installation of package 'RMySQL' had non-zero exit status

 It seems many people faced the issue and some got it resolved manipulating 
 with some jars.
 I got one solution from - 
 http://stackoverflow.com/questions/4785933/adding-rmysql-package-to-r-fails/5236954#5236954


 The basic process is described 
 herehttp://biostat.mc.vanderbilt.edu/wiki/Main/RMySQL, but there are 
 several hints, So I will describe the whole solution(please change the R 
 version and paths if needed):
 1.               Install latest RTools from 
 herehttp://www.murdoch-sutherland.com/Rtools/
 2.               install MySQL or header and library files of mysql
 3.               create or edit file C:\Program 
 Files\R\R-2.12.1\etc\Renviron.site and add line likeMYSQL_HOME=C:/mysql (path 
 to your mysql files)
 4.               copy libmysql.lib from mysql/lib to mysql/lib/opt to meet 
 dependencies.
 5.               copy libmysql.dll to C:\Program Files\R\R-2.12.1\bin or to 
 windows/system32 directory.
 6.               run install.packages('RMySQL',type='source') and wait while 
 compilation will end.

 This worked for me on Windows 7 64 bit, so there should be no problems with 
 32 bit versions

 II tried the above steps, but I am still facing problems.
 Is RTools something different from RProject. I did not find anything by the 
 name RTools at the given link?

 Regards,
 Neeraj Aggarwal


        [[alternative HTML version deleted]]

 __
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Re: [R] Creating polygons from scattered points

This might be a question for the R-SIG-Geo or R-SIG-Ecology lists.

Best,
Michael

On Wed, Mar 14, 2012 at 7:13 AM, Louise Mair lm...@york.ac.uk wrote:
 Hello,

 I have a distribution dataset for species consisting of xy coordinates at
 the 1km resolution, with only presence data. So a simplified example of a
 species distribution might be:

 y - rbind(as.integer(rnorm(100,50,20)), as.integer(rnorm(200,100,30)),
 as.integer(rnorm(100,180,15)))
 x - rbind(as.integer(rnorm(200,50,20)), as.integer(rnorm(200,100,20)),
 as.integer(rnorm(100,200,15)))
 plot(y~x)

 I would like to create polygons for each species distribution, where if an
 island is present (as I have tried to show in the example), it would be a
 seperate polygon, and the jagged edges of coastlines etc are maintained. I
 have spent ages trying to find a package that will allow me to convert
 scattered point distributions to polygons but haven't found anything that
 works, the functions I have found require the data already to be in the
 format where the only xy coordinates present are the outline of the
 polygon.

 Can anyone please recommend a function I can use here, or suggest a way of
 extracting the outline points? I have tried this manually but cannot seem
 to write a code that will effectively take account of jagged edges and
 islands.

 Thanks very much for your help,

 Louise.

        [[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.

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Re: [R] Problem installing RMySQL package!

2012-03-14 Thread Gabor Grothendieck

On Wed, Mar 14, 2012 at 1:14 AM, AGGARWAL, NEERAJ (NEERAJ)
neeraj.aggar...@alcatel-lucent.com wrote:
 I am facing issues while installing RMySQL package on windows(32 bit) 
 installation of R-Project.
 I am getting the following warning messages.

 Warning messages:
 1: running command 'C:/PROGRA~1/R/R-213~1.2/bin/i386/R CMD INSTALL -l 
 C:/Program Files/R/R-2.13.2/library   
 D:\DOCUME~1\nagga002\LOCALS~1\Temp\RtmpLZndlE/downloaded_packages/RMySQL_0.9-3.tar.gz'
  had status 1
 2: In install.packages(RMySQL, type = source) :
  installation of package 'RMySQL' had non-zero exit status

 It seems many people faced the issue and some got it resolved manipulating 
 with some jars.
 I got one solution from - 
 http://stackoverflow.com/questions/4785933/adding-rmysql-package-to-r-fails/5236954#5236954


 The basic process is described 
 herehttp://biostat.mc.vanderbilt.edu/wiki/Main/RMySQL, but there are 
 several hints, So I will describe the whole solution(please change the R 
 version and paths if needed):
 1.               Install latest RTools from 
 herehttp://www.murdoch-sutherland.com/Rtools/
 2.               install MySQL or header and library files of mysql
 3.               create or edit file C:\Program 
 Files\R\R-2.12.1\etc\Renviron.site and add line likeMYSQL_HOME=C:/mysql (path 
 to your mysql files)
 4.               copy libmysql.lib from mysql/lib to mysql/lib/opt to meet 
 dependencies.
 5.               copy libmysql.dll to C:\Program Files\R\R-2.12.1\bin or to 
 windows/system32 directory.
 6.               run install.packages('RMySQL',type='source') and wait while 
 compilation will end.

 This worked for me on Windows 7 64 bit, so there should be no problems with 
 32 bit versions

 II tried the above steps, but I am still facing problems.
 Is RTools something different from RProject. I did not find anything by the 
 name RTools at the given link?

The above description is outdated if you are using the latest version of RMySQL.

Install Rtools from http://cran.r-project.org/bin/windows/Rtools and
then look at the Windows section of the Installation Info link at
http://cran.r-project.org/package=RMySQL (which is the installation
information that comes with the package itself).



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

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Re: [R] Creating polygons from scattered points

2012-03-14 Thread Keith Jewell

No time to really think about this, but:

a) to convert scattered point distributions to polygons you might look at 
convex hulls; e.g. convhulln {geometry}
b) to identify islands some kind of cluster analysis

Hope that helps a little.

KJ

Louise Mair lm...@york.ac.uk wrote in message 
news:camkzt57pep1zkpgm2dg5tn6r0p-_o5tf+wsnurxdohf8zvm...@mail.gmail.com...
 Hello,

 I have a distribution dataset for species consisting of xy coordinates at
 the 1km resolution, with only presence data. So a simplified example of a
 species distribution might be:

 y - rbind(as.integer(rnorm(100,50,20)), as.integer(rnorm(200,100,30)),
 as.integer(rnorm(100,180,15)))
 x - rbind(as.integer(rnorm(200,50,20)), as.integer(rnorm(200,100,20)),
 as.integer(rnorm(100,200,15)))
 plot(y~x)

 I would like to create polygons for each species distribution, where if an
 island is present (as I have tried to show in the example), it would be a
 seperate polygon, and the jagged edges of coastlines etc are maintained. I
 have spent ages trying to find a package that will allow me to convert
 scattered point distributions to polygons but haven't found anything that
 works, the functions I have found require the data already to be in the
 format where the only xy coordinates present are the outline of the
 polygon.

 Can anyone please recommend a function I can use here, or suggest a way of
 extracting the outline points? I have tried this manually but cannot seem
 to write a code that will effectively take account of jagged edges and
 islands.

 Thanks very much for your help,

 Louise.

 [[alternative HTML version deleted]]


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[R] How to extend a slot of a class?

2012-03-14 Thread Marco Smolla

Hej hej,

is there a way to extend the SpatialPointsDataFrame data slot?This is the 
structure of an object of it: str(coord)
Formal class 'SpatialPointsDataFrame' [package sp] with 5 slots
  ..@ data   :'data.frame': 214 obs. of  2 variables:
  .. ..$ location.long: num [1:214] -79.8 -79.8 -79.8 -79.8 -79.8 ...
  .. ..$ location.lat : num [1:214] 9.16 9.16 9.16 9.16 9.16 ...
  ..@ coords.nrs : num(0) 
  ..@ coords : num [1:214, 1:2] -79.8 -79.8 -79.8 -79.8 -79.8 ...
  .. ..- attr(*, dimnames)=List of 2
  .. .. ..$ : NULL
  .. .. ..$ : chr [1:2] location.long location.lat
  ..@ bbox   : num [1:2, 1:2] -79.84 9.16 -79.84 9.17
  .. ..- attr(*, dimnames)=List of 2
  .. .. ..$ : chr [1:2] location.long location.lat
  .. .. ..$ : chr [1:2] min max
  ..@ proj4string:Formal class 'CRS' [package sp] with 1 slots
  .. .. ..@ projargs: chr NA

data is a data.frame including the information of long and lat location. I 
would like to have there a third information: time (but as POSIXct). Is there 
an elegant way to do this?

Best,
marco
[[alternative HTML version deleted]]

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Re: [R] ROC Analysis

2012-03-14 Thread Camille Leclerc

Hi,
Pascal, I tried to use your method which functions but the values of the
abscissa are not good. When I talked about “value”, I wanted mean the values
of my starting dataset that I have classified into two classes positive and
negative classes, such as this :

Value  Status  TPF   TNF
10001   
487 0   
852 1   
927 1   
799 1   
369 0   
…   …   

And for each value of my dataset, I want the tpf and tnf. 
With your method, I have an abscissa which has values between 0 and 1.

All the best,
Camille



-
--
Camille Leclerc, Master student
Lab ESE, UMR CNRS 8079
Univ Paris-Sud
Bat 362
F-91405  Orsay Cedex FRANCE
--
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[R] plm function

2012-03-14 Thread Ieva Sriubaitė

Dear Sir/ Madam,

I am writing about the panel data for my bachelor degree.
I would really appreciate if You could help dealing with R functions.
I am trying to estimate the panel data lm model with plm function. When i
include 3dummy variables into the regression it dont appear in the sumarry
of the model, but when i estimate a simple lm model it appears.
Why is it so? What should i do to estimate the statistics for those dummy
variables?

Thank You.
Ieva

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Re: [R] beginner's loop issue

2012-03-14 Thread aledanda

Thank you all for your helpful comments!
I solved my problem by creating an empty matrix before the loop and
adjusting the loop itself, this is the code:

size - dim(input)
out - matrix('',nrow =size[1], ncol =  9)

for (i in 1:nrow(input)) {
  
  out[i,1:3] - unlist(MyFunction(input[i,1],input[i,2], input[i,3]) 
out[i,4:6] - unlist(MyFunction(input[i,5],input[i,7], input[i,6]) 
  out[i,7:9] - unlist(MyFunction(input[i,8],input[i,10], input[i,9]) 

}

It's probably not the most efficient way of doing it but it gives me the
desired output matrix... 
Thanks for pointing me to this pdf R inferno pdf, I had an old version of
it..
I'm, starting to use R again after long time and this is a very useful
guide!

All the best

Ale




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Re: [R] Sum results in a matrix

2012-03-14 Thread RMSOPS

thank you

   It is working, a question in the wake of the array.

with the following code I am creating a data frame to store the data without
repeating the code is working.
   The question is the best way to do this process in R

tab-NULL
for(i in 1: nrow(res4))
{
  for(j in i:nrow(res4))
  {
#print(paste(i,-,j,-,res4[i,j]))
temp-data.frame(i,j,res4[i,j])
tab-rbind(tab,temp)
  }
}

 tab
   i j res4.i..j.
1  1 1  0
2  1 2 21
3  1 3  0
4  1 4  0
5  1 5  0
6  1 6  0
7  1 7  0
8  2 2  0

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Re: [R] reshaping a dataset for a network

2012-03-14 Thread Marco Guerzoni


thank you very much
I had managed with

df - data.frame(a,b)
m - lapply(split(df, df$a), function(x) x$b)
n - max(sapply(m, length))
a - t(sapply(m, function (x) c(x, rep(NA, n - length(x)

 but your solution is much more elegant.
best regards
Marco


Il 3/14/2012 1:54 PM, R. Michael Weylandt ha scritto:

You can't have empty spots like that in an array. One choice would
be to fill them with NAs:

library(plyr)
do.call(rbind.fill.matrix,lapply(split(b,a), t))

Michael

On Wed, Mar 14, 2012 at 4:01 AM, Marco Guerzonimarco.guerz...@unito.it  wrote:

Thank you fro the reply.
I managed to arrive till here, then I would like to have it in matrix where
the $1 $2...$5 are the first column.

Il 3/13/2012 8:15 PM, William Dunlap ha scritto:


Is the following what you want?
   a- c(1,2,3,4,4,4,5,5)
   b- c(11,7,4,9,8,3,12,4)
   split(b, a)
   $1
   [1] 11

   $2
   [1] 7

   $3
   [1] 4

   $4
   [1] 9 8 3

   $5
   [1] 12  4


Here I did

c- split(b, a)
d- do.call(rbind,c)

and I get

  [,1] [,2] [,3]
1   11   11   11
2777
3444
4983
5   124   12


Instead of what I would like:

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

1   11
27
34
4983
5   124

A solution could be rbind.fill , which does not seem to work with list.


thanks
Marco Guerzoni,
Department of Economics
University of Turin




Note that your df-cbind(a,b) produces a matrix, not the data.frame
that your df suggests you want.  Use df-data.frame(a,b) to make
a data.frame.  Then you could do with(df, split(a,b)) to operate on
the a and b in the data.frame df.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf
Of Marco Guerzoni
Sent: Tuesday, March 13, 2012 10:51 AM
To: r-help@r-project.org
Subject: [R] reshaping a dataset for a network

dear all,
apologizes for bothering with a probably stupid question but I really
don' t know how to proceed.

I have a dataset which look like df

a- c(1,2,3,4,4,4,5,5)
b- c(11,7,4,9,8,3,12,4)
df-cbind(a,b)

I would like to have one which looks like this:

a
1 11
2 7
3 4
4 9 8 3
5 12 4

a are vertex of a network, b the edges. In the data the lenght of a is
about 5

I read several posts about reshape, reshape2, split, ldply but I
couldn't manage to do it. The problem seems to be that the is not a real
panel.

Any help would be really appreciated,
my best regards
Marco

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[R] not saving plot with resizable size?

i want to save plots in batch mode(100 plots in single code) in resizable
mode
which format should i use,i m working on windows

 savePlot(zatka,type=pdf,device=dev.cur())
Error in savePlot(zatka, type = pdf, device = dev.cur()) : 
  can only copy from 'windows' devices

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[R] R course: London, Monday 18th - Friday 22nd June 2012.

2012-03-14 Thread Leonard Schalkwyk



r-help readers may be interested in our week-long summer school
'Introduction to R'.  Further details including links to the 
course materials from last year and the application form are at:  

http://www.kcl.ac.uk/schools/summerschool/si/sgdp/course2/

The course is presented by the MRC Social, Genetic and Developmental 
Psychiatry Research Centre at the Institute of Psychiatry in London (UK).  
Because the emphasis is on fundamentals, the course may be of interest to people
from diverse fields. There are practical sessions paired with each lecture.

Leo Schalkwyk
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[R] Real-Time data transfer from Excel to R

2012-03-14 Thread burcy

Hi All
I receive through DDE ,real time data from an external supplier on an Excel
2003 sheet.
I use R as platform to make backtest trading and prepare trade.
My question :   Is existing a solution to transfer real-time data from Excel
to R? Such transfer keeping  the streaming condition.
I don't intend to open an account to IB. And I don't want make backtest on
Excel,using R as calculator

Thanks for help

Burcy


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Re: [R] sort list

2012-03-14 Thread sybil kennelly

I appreciate the reading Thank you. May i ask one final question. If i have:

matrix:
 var1var2 var3
 cell1x   x x
 cell2x   x x
 cell3x   x x

 cell4

 .
 .
 .
 .
 cell100

and:

vector1 - c(cell1, cell5,cell19, cell50, cell70)

your_data$mycells - factor(your_data$cells %in% vector1, c(Special,
NotSpecial))

So my output will be something like:

[25] SpecialSpecialSpecialSpecialSpecialSpecial
  [31] SpecialNotSpecial NotSpecial NotSpecial NotSpecial NotSpecial
  [37] NotSpecial NotSpecial NotSpecial NotSpecial

is there a way to plot the data so that my Special cells are plotted on
top of my not special cells. The reason is my data may have 1 not
special points,and i may have 5 special cells, I find I'm not able to see
where they are on my plot because they are being covered by my not special
cells :(

I have been looking around for  order of factors plotted , 'order of
levels, order of factor levels, is this on the right track or can it
even be done?

Syb




On Wed, Mar 14, 2012 at 2:04 PM, sybil kennelly sybilkenne...@gmail.comwrote:

 I appreciate the reading Thank you. May i ask one final question. If i
 have:

 matrix:
  var1var2 var3
  cell1x   x x
  cell2x   x x
  cell3x   x x
 
  cell4
 
  .
  .
  .
  .
  cell100

 and:

 vector1 - c(cell1, cell5,cell19, cell50, cell70)

 your_data$mycells - factor(your_data$cells %in% vector1, c(Special,
 NotSpecial))

 So my output will be something like:

 [25] SpecialSpecialSpecialSpecialSpecialSpecial
   [31] SpecialNotSpecial NotSpecial NotSpecial NotSpecial NotSpecial
   [37] NotSpecial NotSpecial NotSpecial NotSpecial

 is there a way to plot the data so that my Special cells are plotted on
 top of my not special cells. The reason is my data may have 1 not
 special points,and i may have 5 special cells, I find I'm not able to see
 where they are on my plot because they are being covered by my not special
 cells :(

 I have been looking around for  order of factors plotted , 'order of
 levels, order of factor levels, is this on the right track or can it
 even be done?

 Syb

 On Tue, Mar 13, 2012 at 12:29 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:

 On Tue, Mar 13, 2012 at 5:15 AM, sybil kennelly sybilkenne...@gmail.com
 wrote:
  Thanks Josh. I'm quite new, just wondering re:factor levels?
 
  In this example (shamelessly stolen from the internet):
 
  schtyp
 
  [1] 0 0 1 0 0 0 1 0 1 0 1 1 1 1 0 0 1 1 1 0
 
  schtyp.f - factor(schtyp, labels = c(private, public))
 
  schtyp.f
 
  [1] private private public private private private public private public
  [10] private public public public public private private public public
 
  [19] public private
 
 
  Levels: private public
 
 
 
  in my data i have a table:
 
  var1var2 var3
  cell1x   x x
  cell2x   x x
  cell3x   x x
 
  cell4
 
  .
  .
  .
  .
  cell100
 
 
  and i have a subset of those cells that are interesting to me as a list
 of
  data
  list1 = [cell1, cell5,cell19, cell50, cell70]
 
  is it possible to create (similar to above):
 
  schtyp.f - factor(schtyp, labels = c(special, normal))

 Sure.  Again, probably better to have cells of interest in a vector,
 not a list a la:

 list1 - c(cell1, cell5,cell19, cell50, cell70)

 your_data$mycells - factor(your_data$cells %in% list1, c(Special,
 NotSpecial))

 basically compares the cells to those in your list and returns
 TRUE/FALSE, which is then converted to a factor, labeled, and stored.
 If you are just starting, some background reading will help.  Here are
 some suggestions:

 1) Go here: http://www.burns-stat.com/pages/tutorials.html and read
 the tutorials for R -- Beginning (this should not take more than 1
 day).
 2) Sit down and read:
 http://cran.r-project.org/doc/manuals/R-intro.pdf through Appendix A
 (for now you can probably skip the rest of the appendices).  That will
 probably take another entire day or so.
 3) Head back to Patrick Burn's website:
 http://www.burns-stat.com/pages/tutorials.html and read the
 intermediate guide, The R Inferno (1-3 days depending if you can read
 for 8 hours straight or not)

 Cheers,

 Josh

 
  so that when i plot this data, i can color the items in list1 as one
 color
  (eg all the special cells are red), and the rest of the items as a
 second
  color (eg all the other cells are black/blue)?
 
 
  Syb
 
 
 
  On Tue, Mar 13, 2012 at 11:48 AM, Joshua Wiley jwiley.ps...@gmail.com
  wrote:
 
  Hi Sybil,
 
  You cannot turn a list into a factor.  You could do:
 
  cell_data -c('cell1','cell2')
  factor_list - factor(cell_data)
 
  or if you already have a list, unlist() or as.vector() may convert it
  into a vector that you can then convert to a factor.
 
  Cheers,
 
  Josh
 
  On Tue, Mar 13, 2012 at 4:29 AM, sybil kennelly 
 sybilkenne...@gmail.com

[R] geom_plot creates Area Instead Of Lines

2012-03-14 Thread kyleR

Hi,

I have created a large (?) data frame.  I have one variable to plot, lets
call it X. further X has been observed under conditions Y, Z. And the
observation took place at times T (0:10e5). Now my data frame D consists of
x,y,z,t each corresponding to the combination of X, Y, Z and T.  I would
like to create different plots and use this:

h - ggplot(D, aes(y=x, x=t))
h + geom_line(data=subset(subset(D,y=someThing),z=thingSome))
+geom_line(data=subset(subset(D,y=someThingElse),z=ThingSomeElse));

Now the issue is that T is a vector of repeated sequences from 0 to 10e5
(something like 1 2 3 1 2 3 1 2 3).  This way X corresponds to the correct
time of measurement FOR MANY REPEATED MEASUREMENTS.  

From the above code I get something very similar to what i want – only that
the individual lines for X vs T which  I was expecting seem to be just one
coloured area between Xmax and Xmin for each T.  

http://r.789695.n4.nabble.com/file/n4471583/Rpot.png 

What I would have liked is something like a cloud of lines, similar to what
I get when I convert the data into a matrix (why do I not just use a matrix? 
I come from MATLAB and this seems natural, however, my data is large and a
data frame seems to be an advantageous way to handle that).

On  side note, generating this “area” plot takes AGES, idk where I got so
inefficient in my approach or whether this is unavoidable, but this is
really … upsetting.


Any help is very much appreciated.

K

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[R] list factoring

2012-03-14 Thread sybil kennelly

Hi Guys, this is actually a thread of emails, but for some reason, even
though i am a member, it's withholding my email so i said i would try it
this route instead!...

I appreciate the reading Thank you. If i have:

matrix:
 var1var2 var3
 cell1x   x x
 cell2x   x x
 cell3x   x x

 cell4

 .
 .
 .
 .
 cell100

and:

vector1 - c(cell1, cell5,cell19, cell50, cell70)

your_data$mycells - factor(your_data$cells %in% vector1, c(Special,
NotSpecial))

So my output will be something like:

[25] SpecialSpecialSpecialSpecialSpecialSpecial  
  [31] SpecialNotSpecial NotSpecial NotSpecial NotSpecial NotSpecial
  [37] NotSpecial NotSpecial NotSpecial NotSpecial

is there a way to plot the data so that my Special cells are plotted on
top of my not special cells. The reason is my data may have 1 not
special points,and i may have 5 special cells, I find I'm not able to see
where they are on my plot because they are being covered by my not special
cells :(

I have been looking around for  order of factors plotted , 'order of
levels, order of factor levels, is this on the right track or can it even
be done?

Syb

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[R] logical test not functioning correctly on zoo series...what the hell?

Here's the raw data I'm working with (will be available temporarily):

http://dl.dropbox.com/u/41922443/dataout_2471_843.csv

Next, here's the code I want to impose (error that I'm seeing SHOULD
reproduce on your shell when script is executed...btw using Ubuntu Linux
11.10 if that makes a difference):

http://pastebin.com/LDNw8UT0

The issue specifically is the last line of code in which I want to print out
any value greater than 0. The outputs that I'm getting, however, are
printing out 0's down the columns. I've tried various methods assuming
various theories, read the R manual via ? for different possible
solutions, Googled stuff, tried the ifelse function which produces the same
error, tried creating logical vectors to get a better visual of what the
process is, tried doing all of the above on a small sample data set (which
actually works correctly for some reason, but doesn't work on the actual raw
data), and basically have done a lot of research and trial and error before
posting on here. Help is much appreciated. Thank you.

P.S. I hope I asked this correctly on the forum (putting reproducible code
that reproduces the error, etc.). If I made some noob mistake, I apologize
in advance so please don't be mad at me. I will clarify if necessary.

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[R] random effect of species in lme on slope taking into account two variables

2012-03-14 Thread Bhogal, Harkiran

Hi!



Just wondering how you would change this code so i can look at the random 
effect of species on slopes of effect of logM (body mass) and K (temperature) 
together.

I did this so far:

model1-lme(logSSP~logM + K,random=~1|species,data=data1) #random effect of spp 
on intercept

model2-lme(logSSP~logM + K,random=~K|species,data=data1) #random effect of spp 
on slope K
model3-lme(logSSP~logM + K,random=~logM|species,data=data1) #random effect of 
spp on slope logM



The one I now want is:

model4-lme(logSSP~logM + K,random=~logM|species,K|species,data=data1) #random 
effect of spp on both slopes of logM and K

# but this code doesn't work so how do i change the code to have a random 
effect of species on slopes of logM and K together???



Please help :(

Harkiran

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[R] how to calculate AIC value from below values using time series model

t 
-0.15264004 
0.056076439 
-0.07276116 
-0.00917326 
-0.02069089 
-0.00416232 
-0.07225855 
-0.02654577 
-0.06131410 
-0.09380202 
0.057414014 
-0.05239976 
0.014397612 
0.016145161 
-0.00670587 
0.018696335 
0.036943654 
-0.02450233 
0.031161705 
0.006513503 
-0.02892329 
-0.00831519 
-0.00877744 
-0.00634399 
-0.02612019 
-0.02531800 
-0.01435533 
0.011148840 
-0.01893775 
0.029859128 
0.029878797 
-0.00125987 
0.031404385 
0.035127606 
-0.00191775 
0.059797202 
-0.03268047 
-0.06026960 
-0.02216465 
-0.08145612 
-0.02772806 
-0.03171683 
-0.02842562 
-0.11807898 
-0.01457311 
-0.12612482 
0.409631265 
-0.06375234 

if i have 200 such t variable (which have time observations ranging from
100-400)  want to select best model,how should i select it?

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Re: [R] how to find best model of time series? calculate aic value, if i have

yes,but i want AIC value to calculate for below value, using frequency =1 in
ts() function
t 
-0.15264004 
0.056076439 
-0.07276116 
-0.00917326 
-0.02069089 
-0.00416232 
-0.07225855 
-0.02654577 
-0.06131410 
-0.09380202 
0.057414014 
-0.05239976 
0.014397612 
0.016145161 
-0.00670587 
0.018696335 
0.036943654 
-0.02450233 
0.031161705 
0.006513503 
-0.02892329 
-0.00831519 
-0.00877744 
-0.00634399 
-0.02612019 
-0.02531800 
-0.01435533 
0.011148840 
-0.01893775 
0.029859128 
0.029878797 
-0.00125987 
0.031404385 
0.035127606 
-0.00191775 
0.059797202 
-0.03268047 
-0.06026960 
-0.02216465 
-0.08145612 
-0.02772806 
-0.03171683 
-0.02842562 
-0.11807898 
-0.01457311 
-0.12612482 
0.409631265 
-0.06375234 


in short, i want to know which is best model if i have this type of
vectors(t) 200,each having 400 observations at respective time i.e.
1,2,3,4.400


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Re: [R] Rolling regressions with sample extended one period at a time

2012-03-14 Thread pie'

Hi,

I took a look at the documentation concerning rollapply and it doesnt seem
to be the command I need. Just to be more clear, on an initial sample of 80
observations I want to perform a probit regression and save the fitted
probabilities in a vector. Subsequently another probit is run on the sample
1-81 with relative save of the fitted probabilities. And so again for the
sample 1-82, 1-83, 1-84 etc... through the end of the sample. Does rollapply
allow me to do this?

P.

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Re: [R] not saving plot with resizable size?

Just plot directly to your desired output device and don't try to copy.

pdf(outfilename.pdf)

# lots of plots

dev.off()

and everything will come out in a single pdf

Michael

On Wed, Mar 14, 2012 at 6:44 AM, sagarnikam123 sagarnikam...@gmail.com wrote:
 i want to save plots in batch mode(100 plots in single code) in resizable
 mode
 which format should i use,i m working on windows

 savePlot(zatka,type=pdf,device=dev.cur())
 Error in savePlot(zatka, type = pdf, device = dev.cur()) :
  can only copy from 'windows' devices

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[R] aggregate

2012-03-14 Thread Alaios

Dear all I am having a vector with large length and I would like to ask you if 
I can aggregate the values by constant sized windows. For example for the 
following vector, I would like to take 30 points until the end
and find their mean.


 myData-seq(1:10)

 c(mean(myData[1:30]),mean(myData[31:60])) #...and so one until the end
[1] 15.5 45.5

I have searched in the R documentation and I found the aggregate but it seems 
to operate on data.frames. It also has this by argument where I tried to set it 
to 30 but it expects there a list rather than a numeric value.

Could you please help me ?
I would like to thank you in advance for your help

B.R
Alex

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[R] where I can find more color names or color definition?

2012-03-14 Thread Jie Tang

hi everyone .
  Now I want to draw several lines in one frame.And it seems needs more
colors except for blue red,black .Where can i found these color name or
define some new color ?thank you .

-- 
TANG Jie
Email: totang...@gmail.com
Tel: 0086-2154896104
Shanghai Typhoon Institute,China

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[R] gam - Y axis probability scale with confidence/error lines

2012-03-14 Thread Ben quant

Hello,

How do I plot a gam fit object on probability (Y axis) vs raw values (X
axis) axis and include the confidence plot lines?

Details...

I'm using the gam function like this:
l_yx[,2] = log(l_yx[,2] + .0004)
fit - gam(y~s(x),data=as.data.frame(l_yx),family=binomial)

And I want to plot it so that probability is on the Y axis and values are
on the X axis (i.e. I don't want log likelihood on the Y axis or the log of
my values on my X axis):

xx - seq(min(l_yx[,2]),max(l_yx[,2]),len=101)
plot(xx,predict(fit,data.frame(x=xx),type=response),type=l,xaxt=n,xlab=Churn,ylab=P(Top
Performer))
at - c(.001,.01,.1,1,10)  # -- I'd also like to generalize
this rather than hard code the numbers
axis(1,at=log(at+ .0004),label=at)

So far, using the code above, everything looks the way I want. But that
does not give me anything information on variability/confidence/certainty.
How do I get the dash plots from this:
plot(fit)
...on the same scales as above?

Related question: how do get the dashed values out of the fit object so I
can do 'stuff' with it?

Thanks,

Ben

PS - thank you Patrick for your help previously.

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Re: [R] Sum results in a matrix

2012-03-14 Thread Petr Savicky

On Wed, Mar 14, 2012 at 02:28:22AM -0700, RMSOPS wrote:
 thank you
 
It is working, a question in the wake of the array.
 
 with the following code I am creating a data frame to store the data without
 repeating the code is working.
The question is the best way to do this process in R
 
 tab-NULL
 for(i in 1: nrow(res4))
 {
   for(j in i:nrow(res4))
   {
 #print(paste(i,-,j,-,res4[i,j]))
 temp-data.frame(i,j,res4[i,j])
 tab-rbind(tab,temp)
   }
 }

Hi.

Try the following.

  # create a matrix
  res4 - matrix(1:12, nrow=3, ncol=4)
  d - dim(res4)
  ind - expand.grid(i=1:d[1], j=1:d[2])
  cbind(ind, res4=c(res4))

 i j res4
  1  1 11
  2  2 12
  3  3 13
  4  1 24
  5  2 25
  6  3 26
  7  1 37
  8  2 38
  9  3 39
  10 1 4   10
  11 2 4   11
  12 3 4   12

Hope this helps.

Petr Savicky.

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Re: [R] Moore-Penrose Generalized determinant?

2012-03-14 Thread JLucke

If the matrix is singular, the determinant of the matrix and its M-P 
inverse are both zero.




Sean O'Riordain sean...@acm.org 
Sent by: r-help-boun...@r-project.org
03/14/2012 07:41 AM

To
r-help@r-project.org
cc

Subject
[R] Moore-Penrose Generalized determinant?






Is there a function in R to calculate the generalized determinant of a
singular matrix? - similar to the ginv() used to compute the generalized
inverse.

I can't seem to find any R related posts at all.

Thanks in advance,
Sean O'Riordain
Trinity College Dublin

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[R] help with expression

2012-03-14 Thread Thomas Hoffmann

Hi all,

I still fail to plot an axis title with the following expression:

plot(0,xlab=expression('(SOC [' * kgm^{-2} * '])' * ^{-2}))

the xlab should look like:   (SOC [kgm^2])^0.25

with an out bracket and a superscript.


Thanks for your advice.

Thomas


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[R] How to use a saved SVM model from e1071

2012-03-14 Thread Thomas Coffin

Hello,

I have an SVM model previously calibrated using libsvm R implementation from
the e1071 package. 
I would like to use this SVM to predict values, from a Java program.
I first tried to use jlibsvm and the standard java implementation of
libsvm, without success.
Thus, I am now considering writing data in files from my Java code, calling
an R program to predict values, then gather the predicted values in Java.

The problem is that I do not know how to re-use the model saved using the
write.svm() function, since there is no read.svm() function.
I read the following message in the mailing archive, stating that I may use
the save and load methods built in R :
http://www.mail-archive.com/r-help@r-project.org/msg64428.html

Still, I am not sure how to pre-process the data and to post-process the
results.
Since write.svm() writes .scale and .ysale files as well as an svm file, I
figure that the scaling data is included in the svm object.

Does that mean that I do not have to worry about scaling my data and
unscaling the results provided by the predict function on a model reloaded
using save/load ?
I am asking this because I previously succeeded in loading the svm model
from libsvm in Java, but the results using unscaled data were obviously
wrong.

Thanks in advance,
Thomas

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Re: [R] aggregate

2012-03-14 Thread Petr Savicky

On Wed, Mar 14, 2012 at 07:46:40AM -0700, Alaios wrote:
 Dear all I am having a vector with large length and I would like to ask you 
 if I can aggregate the values by constant sized windows. For example for the 
 following vector, I would like to take 30 points until the end
 and find their mean.
 
 
  myData-seq(1:10)
 
  c(mean(myData[1:30]),mean(myData[31:60])) #...and so one until the end
 [1] 15.5 45.5

Hi.

Try the following.

  myData - 1:100
  gr - ceiling((1:length(myData))/30)
  c(tapply(myData, gr, FUN=mean))

 1234 
  15.5 45.5 75.5 95.5 

Hope this helps.

Petr Savicky.

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Re: [R] where I can find more color names or color definition?

apropos(color)
?colors
colors()

Michael

On Wed, Mar 14, 2012 at 10:55 AM, Jie Tang totang...@gmail.com wrote:
 hi everyone .
  Now I want to draw several lines in one frame.And it seems needs more
 colors except for blue red,black .Where can i found these color name or
 define some new color ?thank you .

 --
 TANG Jie
 Email: totang...@gmail.com
 Tel: 0086-2154896104
 Shanghai Typhoon Institute,China

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Re: [R] how to find best model of time series? calculate aic value, if i have

2012-03-14 Thread Berend Hasselman


On 14-03-2012, at 13:00, sagarnikam123 wrote:

 yes,but i want AIC value to calculate for below value, using frequency =1 in
 ts() function
 t 
 -0.15264004 
 0.056076439 
 -0.07276116 
 -0.00917326 
 -0.02069089 
 -0.00416232 
 -0.07225855 
 -0.02654577 
 -0.06131410 
 -0.09380202 
 0.057414014 
 -0.05239976 
 0.014397612 
 0.016145161 
 -0.00670587 
 0.018696335 
 0.036943654 
 -0.02450233 
 0.031161705 
 0.006513503 
 -0.02892329 
 -0.00831519 
 -0.00877744 
 -0.00634399 
 -0.02612019 
 -0.02531800 
 -0.01435533 
 0.011148840 
 -0.01893775 
 0.029859128 
 0.029878797 
 -0.00125987 
 0.031404385 
 0.035127606 
 -0.00191775 
 0.059797202 
 -0.03268047 
 -0.06026960 
 -0.02216465 
 -0.08145612 
 -0.02772806 
 -0.03171683 
 -0.02842562 
 -0.11807898 
 -0.01457311 
 -0.12612482 
 0.409631265 
 -0.06375234 
 
 
 in short, i want to know which is best model if i have this type of
 vectors(t) 200,each having 400 observations at respective time i.e.
 1,2,3,4.400
 
You seem to be changing what you want.
First you wanted seasonal decomposition, you got answers and now you've changed 
the goal(post).

Well my final answer is look at the forecast package. Maybe auto.arima is what 
you want.

Berend

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Re: [R] where I can find more color names or color definition?

2012-03-14 Thread Jorge I Velez

Hi,

One way would be to type

colors()

and choose your favorite(s)  ;-)

HTH,
Jorge.-


On Wed, Mar 14, 2012 at 10:55 AM, Jie Tang  wrote:

 hi everyone .
  Now I want to draw several lines in one frame.And it seems needs more
 colors except for blue red,black .Where can i found these color name or
 define some new color ?thank you .

 --
 TANG Jie
 Email: totang...@gmail.com
 Tel: 0086-2154896104
 Shanghai Typhoon Institute,China

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


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Re: [R] where I can find more color names or color definition?



On Mar 14, 2012, at 10:55 AM, Jie Tang wrote:


hi everyone .
 Now I want to draw several lines in one frame.And it seems needs more
colors except for blue red,black .Where can i found these color name  
or

define some new color ?thank you .


When you have a question, your first thought should be ... what will  
happen if I type :


?colors


--

David Winsemius, MD
West Hartford, CT

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Re: [R] Moore-Penrose Generalized determinant?

2012-03-14 Thread Berend Hasselman


On 14-03-2012, at 12:41, Sean O'Riordain wrote:

 Is there a function in R to calculate the generalized determinant of a
 singular matrix? - similar to the ginv() used to compute the generalized
 inverse.
 
 I can't seem to find any R related posts at all.

Is this what you want:

http://en.wikipedia.org/wiki/Pseudo-determinant

Berend

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Re: [R] Moore-Penrose Generalized determinant?

2012-03-14 Thread Ranjan Maitra

Since the OP says Generalized determinant (I do not really know
what that means), it is different from the usual determinant perhaps?

Is it the product of the nonzero eigenvalues?

Ranjan

On Wed, 14 Mar 2012 10:52:14 -0400 jlu...@ria.buffalo.edu wrote:

If the matrix is singular, the determinant of the matrix and its M-P
inverse are both zero.

Sean O'Riordain sean...@acm.org
Sent by: r-help-boun...@r-project.org
03/14/2012 07:41 AM

To
r-help@r-project.org
cc

Subject
[R] Moore-Penrose Generalized determinant?

Is there a function in R to calculate the generalized determinant of a
singular matrix? - similar to the ginv() used to compute the generalized
inverse.

I can't seem to find any R related posts at all.

Thanks in advance,
Sean O'Riordain
Trinity College Dublin

--
View this message in context:
http://r.789695.n4.nabble.com/Moore-Penrose-Generalized-determinant-tp4471629p4471629.html

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Re: [R] Questing on fitting Baseline category Logit model

2012-03-14 Thread Michael Friendly

Not sure why VGAM::vglm doesn't work here, but most likely it is the 
small  zero counts cited on the page you quote below.  This data set

is very sparse.  You should communicate with the author of VGAM.

You can fit this model with nnet::multinom instead, something like

library(nnet)
# multinomial logit model
(mod1 - multinom(food ~ lake+size+sex, data=Alligator, weights=count))

 multinom(food ~ lake+size+sex, data=Alligator, weights=count)
# weights:  35 (24 variable)
initial  value 352.466903
iter  10 value 270.397070
iter  20 value 268.958046
final  value 268.932740
converged
Call:
multinom(formula = food ~ lake + size + sex, data = Alligator,
weights = count)

Coefficients:
(Intercept) lakeHancock lakeOklawaha lakeTrafford sizesmall 
 sexmale
fish 1.70178892  -0.57524030.5503569  -1.23679067 0.7303298 
0.60639521
invert   0.53452560  -2.35574511.4635491  -0.08096449 2.0665999 
0.14342792
other   -0.01957203   0.19139190.5764707   0.32102428 1.0209285 
0.35382356
reptile -1.15700455   0.55392633.0803954   1.82403973 0.1733300 
-0.02116283


Residual Deviance: 537.8655
AIC: 585.8655



On 3/14/2012 4:15 AM, Christofer Bogaso wrote:

Dear all,

I am facing some problem with how to fit a Baseline category Logit
model with R. Basically I am considering famous  Alligator data as
discussed by Agresti. This data can also be found here:

https://onlinecourses.science.psu.edu/stat504/node/174
(there is also an accompanying R file, however the underlying R code
could not load the data properly!!!)

Below are the stuffs what I have done so far:

My_Data- structure(list(Number = c(7L, 4L, 1L, 0L, 0L, 0L, 0L, 1L, 5L,
2L, 16L, 3L, 3L, 0L, 2L, 1L, 2L, 2L, 3L, 3L, 2L, 13L, 2L, 7L,
0L, 6L, 0L, 0L, 1L, 0L, 3L, 0L, 9L, 1L, 1L, 0L, 0L, 1L, 2L, 0L,
3L, 8L, 7L, 6L, 1L, 6L, 0L, 3L, 1L, 5L, 2L, 0L, 4L, 1L, 1L, 0L,
1L, 0L, 4L, 0L, 13L, 9L, 10L, 0L, 0L, 0L, 2L, 1L, 2L, 2L, 3L,
8L, 9L, 1L, 1L, 0L, 0L, 0L, 1L, 1L), Food = structure(c(2L, 3L,
5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L,
1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L,
4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L,
2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L,
3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L, 2L, 3L, 5L, 1L, 4L), .Label = c(Bird,
Fish, Invertebrate, Other, Reptile), class = factor),
 Size = structure(c(2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L), .Label = c(Large,
 Small), class = factor), Sex = structure(c(2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L), .Label = c(Female, Male), class = factor),
 Lake = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L,
 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(George,
 Hancock, Oklawaha, Trafford), class = factor)), .Names =
c(Number,
Food, Size, Sex, Lake), row.names = c(NA, 80L), class = data.frame)


library(VGAM)
vglm(Food~Size+Sex+Lake, data = My_Data, fam=multinomial, weights = Number)



However I am getting following error:

Error in if (max(abs(ycounts - round(ycounts)))  smallno)
warning(converting 'ycounts' to integer in @loglikelihood) :
   missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12
2: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12
3: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12
4: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12
5: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12
6: In checkwz(wz, M = M, trace = trace, wzepsilon = control$wzepsilon) :
   96 elements replaced by 1.819e-12

Can somebody points me why I am getting this error?

Thanks for you help




--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University  Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele StreetWeb:   http://www.datavis.ca

Re: [R] How to use a saved SVM model from e1071

2012-03-14 Thread Steve Lianoglou

Hi Thomas,

On Wed, Mar 14, 2012 at 11:00 AM, Thomas Coffin
thomas.cof...@artelys.com wrote:
 Hello,

 I have an SVM model previously calibrated using libsvm R implementation from
 the e1071 package.
 I would like to use this SVM to predict values, from a Java program.
 I first tried to use jlibsvm and the standard java implementation of
 libsvm, without success.
 Thus, I am now considering writing data in files from my Java code, calling
 an R program to predict values, then gather the predicted values in Java.

 The problem is that I do not know how to re-use the model saved using the
 write.svm() function, since there is no read.svm() function.
 I read the following message in the mailing archive, stating that I may use
 the save and load methods built in R :
 http://www.mail-archive.com/r-help@r-project.org/msg64428.html

 Still, I am not sure how to pre-process the data and to post-process the
 results.
 Since write.svm() writes .scale and .ysale files as well as an svm file, I
 figure that the scaling data is included in the svm object.

 Does that mean that I do not have to worry about scaling my data and
 unscaling the results provided by the predict function on a model reloaded
 using save/load ?
 I am asking this because I previously succeeded in loading the svm model
 from libsvm in Java, but the results using unscaled data were obviously
 wrong.

I think you'll find it helpful (and enlightening) to peruse the source
code of the svm stuffs in e1071.

Start with the `svm.default` function: you'll see where the scale
attributes are calculated (and stored) in the returned object.

Then take a peak at the `predict.svm` function. You'll find if and
when any scaling is performed on the `newdata` object you are trying
to predict labels on.

In short, things should just work if you save/load the svm object
you've previously learned as long as your `newdata` object maintains
the same structure your training data had (number and type of features
(and column names if you're using the formula interface, I guess)).

HTH,
-steve

-- 
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] logical test not functioning correctly on zoo series...what the hell?



On Mar 14, 2012, at 7:57 AM, knavero wrote:


Here's the raw data I'm working with (will be available temporarily):

http://dl.dropbox.com/u/41922443/dataout_2471_843.csv


Nothing appears.



Next, here's the code I want to impose (error that I'm seeing SHOULD
reproduce on your shell when script is executed...btw using Ubuntu  
Linux

11.10 if that makes a difference):

http://pastebin.com/LDNw8UT0

The issue specifically is the last line of code in which I want to  
print out

any value greater than 0.


You're missing a comma between 0 and ] in the last line if your goal  
is to retrieve the rows that satisfy that condition (and if the  
condition makes any sense). Haven't tested the rest of your code,  
though.


?Extract



The outputs that I'm getting, however, are
printing out 0's down the columns. I've tried various methods assuming
various theories, read the R manual via ? for different possible
solutions, Googled stuff, tried the ifelse function which produces  
the same
error, tried creating logical vectors to get a better visual of what  
the
process is, tried doing all of the above on a small sample data set  
(which
actually works correctly for some reason, but doesn't work on the  
actual raw
data), and basically have done a lot of research and trial and error  
before

posting on here. Help is much appreciated. Thank you.

P.S. I hope I asked this correctly on the forum (putting  
reproducible code

that reproduces the error, etc.).


If there is an error then you should post the complete error message.


If I made some noob mistake, I apologize
in advance so please don't be mad at me. I will clarify if necessary.



David Winsemius, MD
West Hartford, CT

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[R] How to test the statistical significance of the difference of two univariate Linear Regression betas?

How to test the statistical significance of the difference of two
univariate Linear Regression betas?

Hi all,

There are two samples of data: D1 and D2.

On data D1 we do a univariate Linear Regression and get the coefficient
beta1.

On data D2 we do a univariate Linear Regression and get the coefficient
beta2.

How do I test the statistical significance of (beta1-beta2)?

Could you please recommend packages/commands in R for doing this?

Thanks a lot!

[[alternative HTML version deleted]]

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Re: [R] where I can find more color names or color definition?

Here's a function I stole from a blog post somewhere (I forgot where
and I'm pressed for time, but I'm sure google can recover it) -- it
might be helpful to you:

`col.wheel` - function(str, nearby=3, cex=0.75) {
cols - colors()
hsvs - rgb2hsv(col2rgb(cols))
srt - order(hsvs[1,], hsvs[2,], hsvs[3,])
cols - cols[srt]
ind - grep(str, cols)
if (length(ind) 1) stop(no colour matches found,
call.=FALSE)
#s.ind - ind
#if (nearby1) for (i in 1:nearby) {
#   s.ind - c(s.ind, ind+i, ind-i)
#}
ind - sort(unique(as.vector(outer(ind, -nearby:nearby, +
ind - ind[ind = length(cols)]
cols - cols[ind]
pie(rep(1, length(cols)), labels=cols, col=cols, cex=cex)
cols
}

Behold, a good use of the bad ink/info ratio of a pie chart!

Michael


On Wed, Mar 14, 2012 at 11:06 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 apropos(color)
 ?colors
 colors()

 Michael

 On Wed, Mar 14, 2012 at 10:55 AM, Jie Tang totang...@gmail.com wrote:
 hi everyone .
  Now I want to draw several lines in one frame.And it seems needs more
 colors except for blue red,black .Where can i found these color name or
 define some new color ?thank you .

 --
 TANG Jie
 Email: totang...@gmail.com
 Tel: 0086-2154896104
 Shanghai Typhoon Institute,China

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[R] How to use ggplot to do the binned quantile plots(one type of scatter plot)?

How to use ggplot to do the binned quantile plots(one type of scatter plot)?

Hi all,

I have done scatter plot: plot(x, y).

Now I wanted to do binned quantile plots... can ggplot2 help me?

For example, we bin x data into 10 bins.

For each bin, we draw the 10 deciles of the corresponding y data in that
bin as points/dots.

And then accross all bins, we would like to connect the corresponding
decile points/dots together(something like equi-quantile or equi-decile
curves)...

How do I do that in R or ggplot2? Is there an existing function/command
that can do this?

Thanks a lot!

[[alternative HTML version deleted]]

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

[R] extract values for the dates and a new column

2012-03-14 Thread RMSOPS


#add column Start
Hello

Good Afternoon I am trying to create a new column in the data frame, the
target is based on a column of dates to extract a value of life and add a
new column to data frame

f-x.res$T-x.res$duration
Dataset$StarEvent-(as.character.POSIXt(f))


x.sub$T is:
[1] 2006-10-31 20:10:35 EST 2006-10-31 20:11:08 EST 2006-10-31 20:11:21
EST
 [4] 2006-10-31 20:12:16 EST 2006-10-31 20:13:20 EST 2006-10-31
20:13:28 EST

x$duration
159
234
345
35
544


Error in `$-.data.frame`(`*tmp*`, StarEvent, value = c(2006-10-31
20:08:03,  : 
  replacement has 18 rows, data has 103)

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Re: [R] reshaping a dataset for a network

2012-03-14 Thread William Dunlap

I see you have a solution, but why do you want the
results of split() in matrix form?  E.g., does it make
a nicer printout, is it needed to interface with other
R functions, is it needed to interface with other
software?

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: Marco Guerzoni [mailto:marco.guerz...@unito.it]
 Sent: Wednesday, March 14, 2012 1:02 AM
 To: William Dunlap
 Cc: r-help@r-project.org
 Subject: Re: [R] reshaping a dataset for a network
 
 Thank you fro the reply.
 I managed to arrive till here, then I would like to have it in matrix
 where the $1 $2...$5 are the first column.
 
 Il 3/13/2012 8:15 PM, William Dunlap ha scritto:
  Is the following what you want?
   a- c(1,2,3,4,4,4,5,5)
   b- c(11,7,4,9,8,3,12,4)
   split(b, a)
 $1
 [1] 11
 
 $2
 [1] 7
 
 $3
 [1] 4
 
 $4
 [1] 9 8 3
 
 $5
 [1] 12  4
 
 Here I did
 
 c - split(b, a)
 d - do.call(rbind,c)
 
 and I get
 
[,1] [,2] [,3]
 1   11   11   11
 2777
 3444
 4983
 5   124   12
 
 
 Instead of what I would like:
 
[,1] [,2] [,3]
 1   11
 27
 34
 4983
 5   124
 
 A solution could be rbind.fill , which does not seem to work with list.
 
 
 thanks
 Marco Guerzoni,
 Department of Economics
 University of Turin
 
 
 
  Note that your df-cbind(a,b) produces a matrix, not the data.frame
  that your df suggests you want.  Use df-data.frame(a,b) to make
  a data.frame.  Then you could do with(df, split(a,b)) to operate on
  the a and b in the data.frame df.
 
  Bill Dunlap
  Spotfire, TIBCO Software
  wdunlap tibco.com
 
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
 Behalf
  Of Marco Guerzoni
  Sent: Tuesday, March 13, 2012 10:51 AM
  To: r-help@r-project.org
  Subject: [R] reshaping a dataset for a network
 
  dear all,
  apologizes for bothering with a probably stupid question but I really
  don' t know how to proceed.
 
  I have a dataset which look like df
 
  a- c(1,2,3,4,4,4,5,5)
  b- c(11,7,4,9,8,3,12,4)
  df-cbind(a,b)
 
  I would like to have one which looks like this:
 
  a
  1 11
  2 7
  3 4
  4 9 8 3
  5 12 4
 
  a are vertex of a network, b the edges. In the data the lenght of a is
  about 5
 
  I read several posts about reshape, reshape2, split, ldply but I
  couldn't manage to do it. The problem seems to be that the is not a real
  panel.
 
  Any help would be really appreciated,
  my best regards
  Marco
 
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Re: [R] How to use ggplot to do the binned quantile plots(one type of scatter plot)?



On Mar 14, 2012, at 11:33 AM, Michael wrote:

How to use ggplot to do the binned quantile plots(one type of  
scatter plot)?


More like a quantile regression plot.



Hi all,

I have done scatter plot: plot(x, y).

Now I wanted to do binned quantile plots... can ggplot2 help me?

For example, we bin x data into 10 bins.

For each bin, we draw the 10 deciles of the corresponding y data in  
that

bin as points/dots.

And then accross all bins, we would like to connect the corresponding
decile points/dots together(something like equi-quantile or equi- 
decile

curves)...

How do I do that in R or ggplot2? Is there an existing function/ 
command

that can do this?


I'm not aware of it if there is. But I agree that it can be a very  
informative display of data.  I do such plots on data structures  
generated using tapply with age, sex categories and quantile() to  
generate a value. It gets kind of hairy with the list structure that  
results but if you have a more simple data situation it will probably  
be more straightforward, and you should post it.



[[alternative HTML version deleted]]


And you should learn to post in plain text


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


And you should read the Posting Guide.


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



And I already said that.

--

David Winsemius, MD
West Hartford, CT

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Re: [R] help with expression

2012-03-14 Thread ilai

On Wed, Mar 14, 2012 at 8:56 AM, Thomas Hoffmann
hoffm...@giub.uni-bonn.de wrote:
 Hi all,

 I still fail to plot an axis title with the following expression:

 plot(0,xlab=expression('(SOC [' * kgm^{-2} * '])' * ^{-2}))

 the xlab should look like:   (SOC [kgm^2])^0.25

 with an out bracket and a superscript.


Like this ?
plot(0,xlab=expression(paste('(SOC[',kgm^-2,'])'^0.25,sep='')))


 Thanks for your advice.

 Thomas


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Re: [R] Real-Time data transfer from Excel to R

2012-03-14 Thread Richard M. Heiberger

Yes.

please look at RExcel.  You can download it from rcom.univie.ac.at
The wiki page there has many papers discussing similar projects (follow the
Literature link in the left column).  Followup should be on the rcom
mailing list.

Rich


On Wed, Mar 14, 2012 at 7:12 AM, burcy dupuy.dang...@free.fr wrote:

 Hi All
 I receive through DDE ,real time data from an external supplier on an Excel
 2003 sheet.
 I use R as platform to make backtest trading and prepare trade.
 My question :   Is existing a solution to transfer real-time data from
 Excel
 to R? Such transfer keeping  the streaming condition.
 I don't intend to open an account to IB. And I don't want make backtest on
 Excel,using R as calculator

 Thanks for help

 Burcy


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Re: [R] logical test not functioning correctly on zoo series...what the hell?

The outputs that I'm getting, however, are
printing out 0's down the columns. I've tried various methods assuming
various theories, read the R manual via ? for different possible
solutions, Googled stuff, tried the ifelse function which produces
the same
error, tried creating logical vectors to get a better visual of what
the
process is, tried doing all of the above on a small sample data set
(which
actually works correctly for some reason, but doesn't work on the
actual raw
data), and basically have done a lot of research and trial and error
before
posting on here. Help is much appreciated. Thank you.

P.S. I hope I asked this correctly on the forum (putting
reproducible code
that reproduces the error, etc.).
... [show rest of quote]

If there is an error then you should post the complete error message.

Hm, technically there is no error message? It just prints zeros (the wrong
output) as a result of explicitly making the logical condition 0. But
anyway, I'll be sure to copy paste the error output next time. Thanks for
the advice.

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Re: [R] How to use ggplot to do the binned quantile plots(one type of scatter plot)?

2012-03-14 Thread Clint Bowman

?quantile on the individual bins, make your deciles, then plot the ten 
series as usual with your x values at the midpoint of the bins.


Clint BowmanINTERNET:   cl...@ecy.wa.gov
Air Quality Modeler INTERNET:   cl...@math.utah.edu
Department of Ecology   VOICE:  (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600

USPS:   PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Wed, 14 Mar 2012, David Winsemius wrote:



On Mar 14, 2012, at 11:33 AM, Michael wrote:

How to use ggplot to do the binned quantile plots(one type of scatter 
plot)?


More like a quantile regression plot.



Hi all,

I have done scatter plot: plot(x, y).

Now I wanted to do binned quantile plots... can ggplot2 help me?

For example, we bin x data into 10 bins.

For each bin, we draw the 10 deciles of the corresponding y data in that
bin as points/dots.

And then accross all bins, we would like to connect the corresponding
decile points/dots together(something like equi-quantile or equi-decile
curves)...

How do I do that in R or ggplot2? Is there an existing function/command
that can do this?


I'm not aware of it if there is. But I agree that it can be a very 
informative display of data.  I do such plots on data structures generated 
using tapply with age, sex categories and quantile() to generate a value. It 
gets kind of hairy with the list structure that results but if you have a 
more simple data situation it will probably be more straightforward, and you 
should post it.



[[alternative HTML version deleted]]


And you should learn to post in plain text

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


And you should read the Posting Guide.


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



And I already said that.

--

David Winsemius, MD
West Hartford, CT

__
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Re: [R] extract values for the dates and a new column

2012-03-14 Thread Jeff Newmiller

The error seems clear enough to me... you are trying to mix vectors of 
different lengths where one length is not a multiple of the other length.

Your discussion is pretty confused though, referring to possible data frames x, 
x.sub, x.res and Dataset. You might benefit from reviewing ?str and ?dput for 
future posts.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

RMSOPS ricardosousa2...@clix.pt wrote:


#add column Start
Hello

Good Afternoon I am trying to create a new column in the data frame,
the
target is based on a column of dates to extract a value of life and add
a
new column to data frame

f-x.res$T-x.res$duration
Dataset$StarEvent-(as.character.POSIXt(f))


x.sub$T is:
[1] 2006-10-31 20:10:35 EST 2006-10-31 20:11:08 EST 2006-10-31
20:11:21
EST
 [4] 2006-10-31 20:12:16 EST 2006-10-31 20:13:20 EST 2006-10-31
20:13:28 EST

x$duration
159
234
345
35
544


Error in `$-.data.frame`(`*tmp*`, StarEvent, value = c(2006-10-31
20:08:03,  : 
  replacement has 18 rows, data has 103)

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Re: [R] logical test not functioning correctly on zoo series...what the hell?

 Here's the raw data I'm working with (will be available temporarily):

 http://dl.dropbox.com/u/41922443/dataout_2471_843.csv

Nothing appears. 

^ Clicking on the link should prompt you to download a csv file and save it
somewhere in your HDD.

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Re: [R] MANOVA and Extra Sums-of-Squares Tests

2012-03-14 Thread chris33

Hi John,

Thanks again.  That looks like an easy and convenient approach.  Regards,

Chris

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Re: [R] help with expression



On Mar 14, 2012, at 11:53 AM, ilai wrote:


On Wed, Mar 14, 2012 at 8:56 AM, Thomas Hoffmann
hoffm...@giub.uni-bonn.de wrote:

Hi all,

I still fail to plot an axis title with the following expression:

plot(0,xlab=expression('(SOC [' * kgm^{-2} * '])' * ^{-2}))

the xlab should look like:   (SOC [kgm^2])^0.25

with an out bracket and a superscript.



Like this ?
plot(0,xlab=expression(paste('(SOC[',kgm^-2,'])'^0.25,sep='')))


If so, then this is more simple:

plot(0,xlab=expression('(SOC['*kgm^-2*'])'^0.25))  # using * as a  
spaceless connector


And as far as I know, unlike the base R paste function, there is no  
sep argument in the plotmath paste function. Look what happens with:


plot(0,xlab=expression(paste('(SOC[',kgm^-2,'])'^0.25,sep='#')))

# not used as a separator above, but rather is appended at end

And:

plot(0,xlab=expression(paste('(SOC[',kgm^-2,'])'^0.25))) # no default  
space


--
David Winsemius, MD
West Hartford, CT

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Re: [R] logical test not functioning correctly on zoo series...what the hell?



On Mar 14, 2012, at 11:47 AM, knavero wrote:

 Here's the raw data I'm working with (will be available  
temporarily):


http://dl.dropbox.com/u/41922443/dataout_2471_843.csv


Nothing appears. 

^ Clicking on the link should prompt you to download a csv file and  
save it

somewhere in your HDD.


Nope. I just see a blank screen.



pastedGraphic.pdf
Description: Adobe PDF document




--

David Winsemius, MD
West Hartford, CT

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Re: [R] Amount of memory occupied by data type

2012-03-14 Thread Folkes, Michael

Thanks David for the details and pointer to bitops functions. Buried a
bit deep that was. 
I like to think the memory constraints of win xp keeps my code lean and
efficient.  RAM is like a suburban garage, the bigger it is, the more
useless junk people stuff in there.

Michael Folkes


-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net] 
Sent: March 13, 2012 5:25 PM
To: Folkes, Michael
Cc: r-help@r-project.org
Subject: Re: [R] Amount of memory occupied by data type


On Mar 13, 2012, at 7:02 PM, Folkes, Michael wrote:

 Hello all,
 I was under the (false?) assumption that an object that is class 
 logical, would take up less memory than an object with class integer.

Nope.

 Below am I correctly showing this is not the case?

 This was an attempt to reduce memory usage.

I think there is a package that will do bitwise operations. Yep... all
we needed to do is look:

http://finzi.psych.upenn.edu/R/library/bitops/html/00Index.html


 I'm dealing with two large
 arrays (could be integers).  Their contents are the exact same, but 
 one has NA's in random locations.  I thought instead of having the 
 second array as an integer, it could be logical and the TRUE vs FALSE 
 could be used to update data in the first array.  (but even this idea 
 may be weak if I just end up with a third temporary array...)

You probably would since any assignment is going to create a copy. And
even having a bitwise logical option wouldn't necessarily help since the
indexing would be of necessity either integer or logical (both 8 bit
values).

 I'm running win xp sp3, R version 2.14.1 (2011-12-22).

  31-bit addressing constraints as well?  (That's so last decade.) You
aren't making life easy for yourself are you.

-- 

David Winsemius, MD
West Hartford, CT

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Re: [R] How to test the statistical significance of the difference of two univariate Linear Regression betas?

2012-03-14 Thread Mark Leeds

Hi: I'm jot sure if it's exactly what you want but check out Hotelling's
paper from 1940. It
should be in the archives because I answered this question before ( not
from you ).
If you can't find it, I'll find the title  actually, here's the title:

Hotelling, The Selection of Variates For Use in Prediction with Some
Comments on the
General Problem of Nuisance Parameters, Annals of Mathematical Statistics,
11,
271-283.




On Wed, Mar 14, 2012 at 11:21 AM, Michael comtech@gmail.com wrote:

 How to test the statistical significance of the difference of two
 univariate Linear Regression betas?

 Hi all,

 There are two samples of data: D1 and D2.

 On data D1 we do a univariate Linear Regression and get the coefficient
 beta1.

 On data D2 we do a univariate Linear Regression and get the coefficient
 beta2.

 How do I test the statistical significance of (beta1-beta2)?

 Could you please recommend packages/commands in R for doing this?

 Thanks a lot!

[[alternative HTML version deleted]]

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


[[alternative HTML version deleted]]

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Re: [R] gam - Y axis probability scale with confidence/error lines

2012-03-14 Thread Patrick Breheny

The predict() function has an option 'se.fit' that returns what you are 
asking for.  If you set this equal to TRUE in your code:


pred - predict(fit,data.frame(x=xx),type=response,se.fit=TRUE)

will return a list with two elements, 'fit' and 'se.fit'.  The pointwise 
confidence intervals will then be


pred$fit + 1.96*se.fit
pred$fit - 1.96*se.fit

for 95% confidence intervals (replace 1.96 with the appropriate quantile 
of the normal distribution for other confidence levels).


You can then do whatever stuff you want to do with them, including 
plot them.


--Patrick

On 03/14/2012 10:48 AM, Ben quant wrote:

Hello,

How do I plot a gam fit object on probability (Y axis) vs raw values (X
axis) axis and include the confidence plot lines?

Details...

I'm using the gam function like this:
l_yx[,2] = log(l_yx[,2] + .0004)
fit- gam(y~s(x),data=as.data.frame(l_yx),family=binomial)

And I want to plot it so that probability is on the Y axis and values are
on the X axis (i.e. I don't want log likelihood on the Y axis or the log of
my values on my X axis):

xx- seq(min(l_yx[,2]),max(l_yx[,2]),len=101)
plot(xx,predict(fit,data.frame(x=xx),type=response),type=l,xaxt=n,xlab=Churn,ylab=P(Top
Performer))
at- c(.001,.01,.1,1,10)  #-- I'd also like to generalize
this rather than hard code the numbers
axis(1,at=log(at+ .0004),label=at)

So far, using the code above, everything looks the way I want. But that
does not give me anything information on variability/confidence/certainty.
How do I get the dash plots from this:
plot(fit)
...on the same scales as above?

Related question: how do get the dashed values out of the fit object so I
can do 'stuff' with it?

Thanks,

Ben

PS - thank you Patrick for your help previously.

[[alternative HTML version deleted]]

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--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky

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Re: [R] logical test not functioning correctly on zoo series...what the hell?

You're missing a comma between 0 and ] in the last line if your goal
is to retrieve the rows that satisfy that condition (and if the
condition makes any sense). Haven't tested the rest of your code,
though.

?Extract

P.S. I hope I asked this correctly on the forum (putting
reproducible code
that reproduces the error, etc.).
... [show rest of quote]

If there is an error then you should post the complete error message.

About to post the error because it's still showing up. I understand the ,
is supposed to act as an (x, y) or (i, j) separator such that [,1] would be
column 1 similar to how rawData$KW.ch..1..set.0. looks at the same column.
Let me know if you're able to retrieve the csv file from the dropbox link.
Again, it should prompt you to download it. I would've put a smaller sample
size, but the logical condition seemed to work on a small scale version
unfortunately, thus being unable to reproduce the problem.

--
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Re: [R] reshaping a dataset for a network

2012-03-14 Thread Marco Guerzoni


Il 3/14/2012 4:43 PM, William Dunlap ha scritto:

I see you have a solution, but why do you want the
results of split() in matrix form?  E.g., does it make
a nicer printout, is it needed to interface with other
R functions, is it needed to interface with other
I need matrix to export the data in to pajek. I don't any other way to 
have them like this and save them as text file.

Probably is my lack of knowledge...


software?

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com



-Original Message-
From: Marco Guerzoni [mailto:marco.guerz...@unito.it]
Sent: Wednesday, March 14, 2012 1:02 AM
To: William Dunlap
Cc: r-help@r-project.org
Subject: Re: [R] reshaping a dataset for a network

Thank you fro the reply.
I managed to arrive till here, then I would like to have it in matrix
where the $1 $2...$5 are the first column.

Il 3/13/2012 8:15 PM, William Dunlap ha scritto:

Is the following what you want?
   a- c(1,2,3,4,4,4,5,5)
   b- c(11,7,4,9,8,3,12,4)
   split(b, a)
$1
[1] 11

$2
[1] 7

$3
[1] 4

$4
[1] 9 8 3

$5
[1] 12  4

Here I did

c- split(b, a)
d- do.call(rbind,c)

and I get

[,1] [,2] [,3]
1   11   11   11
2777
3444
4983
5   124   12


Instead of what I would like:

[,1] [,2] [,3]
1   11
27
34
4983
5   124

A solution could be rbind.fill , which does not seem to work with list.


thanks
Marco Guerzoni,
Department of Economics
University of Turin



Note that your df-cbind(a,b) produces a matrix, not the data.frame
that your df suggests you want.  Use df-data.frame(a,b) to make
a data.frame.  Then you could do with(df, split(a,b)) to operate on
the a and b in the data.frame df.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On

Behalf

Of Marco Guerzoni
Sent: Tuesday, March 13, 2012 10:51 AM
To: r-help@r-project.org
Subject: [R] reshaping a dataset for a network

dear all,
apologizes for bothering with a probably stupid question but I really
don' t know how to proceed.

I have a dataset which look like df

a- c(1,2,3,4,4,4,5,5)
b- c(11,7,4,9,8,3,12,4)
df-cbind(a,b)

I would like to have one which looks like this:

a
1 11
2 7
3 4
4 9 8 3
5 12 4

a are vertex of a network, b the edges. In the data the lenght of a is
about 5

I read several posts about reshape, reshape2, split, ldply but I
couldn't manage to do it. The problem seems to be that the is not a real
panel.

Any help would be really appreciated,
my best regards
Marco

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Re: [R] logical test not functioning correctly on zoo series...what the hell?

Here's the exact error I'm receiving:

http://pastebin.com/mNsPauwk

Tracked each output along the way. Starting to think there's a bug in the
source code. 

--
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[R] Reading and analyzing data with two headers

2012-03-14 Thread jorge Rogrigues

I have data that I would like to analyze in R with the following format:

*,M1,,M2,
*,S1,S2,S1,S2,
p1,m1s1v1,m1s2v1,m2s1v1,m2s2v1
p2,m1s1v2,m1s2v2,m2s1v2,m2s2v2
..,...,...,...,

so I have a method M1 and M2 and each of these have attributes that I would
like to analyze for the different problems
p1,p2...
So I would like to read the table and get some kind of summary like,
For method M1 the average on S1 is ... The avererage on S2 is ...
and the same thing for method M2.

I tried reading using the cvs reading function but it doesn't work the way
I want.

[[alternative HTML version deleted]]

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[R] climate data, Netcdf

2012-03-14 Thread Amen

As you can see:

time  is from 1 to 1460, 1 means they the measure Tem for the globe at the
first 6 hours , 2=after 12 hours .; and so on 1460 *6 = 8760 hours which
equals 1 year
first of all I want to convert all data from kelvin to degree
second I want to convert from 6 hourly  to daily and make it like this

  raw datawhat i want to make

time   temp (k)
1(6horse)290
2  289
3 
4  1 day (min,max,averag)   
one map(file) for the globe
5
6
7
8  2
   
second map(file) for the
. 
;
;
;
1460365  days


so in the end will develop   365  file   fordaily min Tem
 365  file   for 
daily max   Tem
365  file   for 
average  daily Tem

Thanks in advance 

--
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Re: [R] How to test the statistical significance of the difference of two univariate Linear Regression betas?

2012-03-14 Thread peter dalgaard


On Mar 14, 2012, at 16:21 , Michael wrote:

 How to test the statistical significance of the difference of two
 univariate Linear Regression betas?
 
 Hi all,
 
 There are two samples of data: D1 and D2.
 
 On data D1 we do a univariate Linear Regression and get the coefficient
 beta1.
 
 On data D2 we do a univariate Linear Regression and get the coefficient
 beta2.
 
 How do I test the statistical significance of (beta1-beta2)?
 
 Could you please recommend packages/commands in R for doing this?

Well, the large-sample test would be to calculate the se of the difference as 

  se - sqrt(se1^2+se2^2)

and then Z - (beta1-beta2)/se; 2*pnorm(-abs(Z)).

For small samples you might consider joining the two data sets and test for an 
interaction between your predictor and an indicator for D1 or D2. Other options 
are possible, if you know a bit of theory.

 
 Thanks a lot!
 
   [[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
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-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] How to test the statistical significance of the difference of two univariate Linear Regression betas?

2012-03-14 Thread Bert Gunter

1. Is this homework? -- we don't do homework.

2. This is not an R question -- posting to stats.stackexchange.com or
other statistics websites is usually more appropriate for such non-R
statistical questions

3. General approach: Combine all data; model it with both a simpler
(fewer parameter, single slope) model and a more complex (separate
slopes for different groups of data); test difference in models (e.g.
via likelihood ratio which reduces to anova for the normal theory
case).

4. Better approach: Ignore 3, because the test has not been
pre-planned, suitable power has not been designed in, or the
hypothesis is post hoc, etc., so that P values are therefore likely
just nonsense with no sensible interpretation. The correct answer is:
The slopes _are_ different (entropy is on my side here -- a very
powerful ally that one should avoid messing with). The relevant
question then becomes: Are they different enough to matter in your
context (which if it's homework is -- who cares?) .

5. Perhaps the best approach: Ignore the rantings of a grumpy
statistician and consult your local statistical resource for help,
which you clearly need more of then can be provided here.

Cheers,
Bert

On Wed, Mar 14, 2012 at 9:24 AM, Mark Leeds marklee...@gmail.com wrote:
Hi: I'm jot sure if it's exactly what you want but check out Hotelling's
paper from 1940. It
should be in the archives because I answered this question before ( not
from you ).
If you can't find it, I'll find the title actually, here's the title:

Hotelling, The Selection of Variates For Use in Prediction with Some
Comments on the
General Problem of Nuisance Parameters, Annals of Mathematical Statistics,
11,
271-283.

On Wed, Mar 14, 2012 at 11:21 AM, Michael comtech@gmail.com wrote:

How to test the statistical significance of the difference of two
univariate Linear Regression betas?

Hi all,

There are two samples of data: D1 and D2.

On data D1 we do a univariate Linear Regression and get the coefficient
beta1.

On data D2 we do a univariate Linear Regression and get the coefficient
beta2.

How do I test the statistical significance of (beta1-beta2)?

Could you please recommend packages/commands in R for doing this?

Thanks a lot!

[[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.

[[alternative HTML version deleted]]

Bert Gunter
Genentech Nonclinical Biostatistics

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

Re: [R] How to test the statistical significance of the difference of two univariate Linear Regression betas?

Thank you!

How large is large sample?

What about 50 data points in D1 and another 50 data points in D2?

Thanks a lot!

On Wed, Mar 14, 2012 at 11:49 AM, peter dalgaard pda...@gmail.com wrote:


 On Mar 14, 2012, at 16:21 , Michael wrote:

  How to test the statistical significance of the difference of two
  univariate Linear Regression betas?
 
  Hi all,
 
  There are two samples of data: D1 and D2.
 
  On data D1 we do a univariate Linear Regression and get the coefficient
  beta1.
 
  On data D2 we do a univariate Linear Regression and get the coefficient
  beta2.
 
  How do I test the statistical significance of (beta1-beta2)?
 
  Could you please recommend packages/commands in R for doing this?

 Well, the large-sample test would be to calculate the se of the difference
 as

  se - sqrt(se1^2+se2^2)

 and then Z - (beta1-beta2)/se; 2*pnorm(-abs(Z)).

 For small samples you might consider joining the two data sets and test
 for an interaction between your predictor and an indicator for D1 or D2.
 Other options are possible, if you know a bit of theory.

 
  Thanks a lot!
 
[[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.htmlhttp://www.r-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.

 --
 Peter Dalgaard, Professor,
 Center for Statistics, Copenhagen Business School
 Solbjerg Plads 3, 2000 Frederiksberg, Denmark
 Phone: (+45)38153501
 Email: pd@cbs.dk  Priv: pda...@gmail.com










[[alternative HTML version deleted]]

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Re: [R] How to use ggplot to do the binned quantile plots(one type of scatter plot)?

How to stack these subplots horizontally and vertically together in a nice
way?

Thank you!

On Wed, Mar 14, 2012 at 11:04 AM, Clint Bowman cl...@ecy.wa.gov wrote:

?quantile on the individual bins, make your deciles, then plot the ten
series as usual with your x values at the midpoint of the bins.

Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600

USPS: PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Wed, 14 Mar 2012, David Winsemius wrote:

On Mar 14, 2012, at 11:33 AM, Michael wrote:

How to use ggplot to do the binned quantile plots(one type of scatter
plot)?

More like a quantile regression plot.

Hi all,

I have done scatter plot: plot(x, y).

Now I wanted to do binned quantile plots... can ggplot2 help me?

For example, we bin x data into 10 bins.

For each bin, we draw the 10 deciles of the corresponding y data in that
bin as points/dots.

And then accross all bins, we would like to connect the corresponding
decile points/dots together(something like equi-quantile or equi-decile
curves)...

How do I do that in R or ggplot2? Is there an existing function/command
that can do this?

I'm not aware of it if there is. But I agree that it can be a very
informative display of data. I do such plots on data structures generated
using tapply with age, sex categories and quantile() to generate a value.
It gets kind of hairy with the list structure that results but if you have
a more simple data situation it will probably be more straightforward, and
you should post it.

[[alternative HTML version deleted]]

And you should learn to post in plain text

PLEASE do read the posting guide http://www.R-project.org/**
posting-guide.html http://www.r-project.org/posting-guide.html

And you should read the Posting Guide.

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

And I already said that.

David Winsemius, MD
West Hartford, CT

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[[alternative HTML version deleted]]

[R] statistical contrasts on 3-way interaction

2012-03-14 Thread array chip

Hi all,Â 

I was trying to use glht() from multcomp package to construct a contrast on 
interaction term
in a linear model to do some comparisons. I am little uncertain on how to 
construct contrasts on a 3-way interaction containing a continuous variable, 
and hope someone can confirm what I did is correct or wrong:

The linear model has a continuous dependent
variable âyâ, with treatment factor âTrtâ with value 0 and 1, a factor 
variable
âAâ with value 0 and 1, a continuous variable âxâ. 
Â 
A simpler model is:
Â 
set.seed(10)
dat - cbind(y=c(rnorm(10,3),rnorm(10,4),rnorm(10,3.1),rnorm(10,6)),
x=runif(40,5,15), expand.grid(A=rep(factor(0:1),each=10),Trt=factor(0:1)))
Â 
fit - lm(y
~ x + Trt * A,dat)
Â 
My purpose is to test whether treatment effect is significant given
each level of factor A, so I used contrasts:
Â 
library(multcomp)
Â 
K - rbind(c(0,0,1,0,0), c(0,0,1,0,1))
rownames(K) - c('Trt 1-0|A=0','Trt 1-0|A=1')
colnames(K) - names(coef(fit))
K
Â  Â  Â  Â  Â  Â  (Intercept) x Trt1 A1 Trt1:A1
Trt 1-0|A=0 Â  Â  Â  Â  Â  0 0 Â  Â 1 Â 0 Â  Â  Â  0
Trt 1-0|A=1 Â  Â  Â  Â  Â  0 0 Â  Â 1 Â 0 Â  Â  Â  1
Â 
(glht.fit - summary(glht(fit, linfct =
K), test=adjusted(type='none')))
Linear
Hypotheses:
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Estimate Std. Error t value
Pr(|t|)Â Â  
Trt 1-0|A=0 ==
0Â  -0.2720Â Â Â Â  0.3616Â  -0.752Â  0.45701Â Â  
Trt 1-0|A=1 ==
0Â Â  1.0690Â Â Â Â  0.3564Â Â  2.999Â  0.00496 **
Â 
Now I suspect independent variable âxâ may play a role in the treatment
effect at each level of A, so I would like to add in a 3-way interaction
between Trt, A and x:
Â 
fit - lm(y
~ x * Trt * A,dat)
Â 
If my purpose is to test whether treatment is significant at each level
of factor A and certain value of covariate âxâ, for example, when x=10, 
would
following code give me what I wanted?
Â 
K - rbind(c(0,0,1,0,10,0,0,0), c(0,0,1,0,10,0,1,10))
rownames(K) - c('Trt 1-0|A=0 x=10','Trt
1-0|A=1 x=10')
colnames(K) - names(coef(fit))
K
Â  Â  Â  Â  Â  Â  Â  Â  Â (Intercept) x Trt1 A1 x:Trt1 x:A1 Trt1:A1 x:Trt1:A1
Trt 1-0|A=0 x=10 Â  Â  Â  Â  Â  0 0 Â  Â 1 Â 0 Â  Â  10 Â  Â 0 Â  Â  Â  0 Â  Â  
Â  Â  0
Trt 1-0|A=1 x=10 Â  Â  Â  Â  Â  0 0 Â  Â 1 Â 0 Â  Â  10 Â  Â 0 Â  Â  Â  1 Â  Â  
Â  Â 10
Â 
(glht.fit - summary(glht(fit, linfct =
K), test=adjusted(type='none')))
Linear
Hypotheses:
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Estimate Std. Error t
value Pr(|t|)Â Â Â  
Trt 1-0|A=0
x=10 == 0Â  -0.3526Â Â Â Â  0.3254Â  -1.083 0.286731Â Â Â  
Trt 1-0|A=1
x=10 == 0Â Â  1.4621Â Â Â Â  0.3328Â Â  4.394 0.000115 ***
Â 
So the above test was testing whether treatment effect is significant at each 
level of factor A when x=10, am I correct?

Appreciate if someone would confirm this?

Thanks

JohnÂ 
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Re: [R] gam - Y axis probability scale with confidence/error lines

2012-03-14 Thread Ben quant

That was embarrassingly easy. Thanks again Patrick! Just correcting a
little typo to his reply. this is probably what he meant:

pred = predict(fit,data.frame(x=xx),type=response,se.fit=TRUE)
upper = pred$fit + 1.96 * pred$se.fit
lower = pred$fit - 1.96 * pred$se.fit

# For people who are interested this is how you plot it line by line:

plot(xx,pred$fit,type=l,xlab=fd$getFactorName(),ylab=ylab,ylim=
c(min(down),max(up)))
lines(xx,upper,type=l,lty='dashed')
lines(xx,lower,type=l,lty='dashed')

In my opinion this is only important if the desired y axis is different
than what plot(fit) gives you for a gam fit (i.e fit - gam(...stuff...))
and you want to plot the confidence intervals.

thanks again!

Ben

On Wed, Mar 14, 2012 at 10:39 AM, Patrick Breheny
patrick.breh...@uky.eduwrote:

 The predict() function has an option 'se.fit' that returns what you are
 asking for.  If you set this equal to TRUE in your code:

 pred - predict(fit,data.frame(x=xx),**type=response,se.fit=TRUE)

 will return a list with two elements, 'fit' and 'se.fit'.  The pointwise
 confidence intervals will then be

 pred$fit + 1.96*se.fit
 pred$fit - 1.96*se.fit

 for 95% confidence intervals (replace 1.96 with the appropriate quantile
 of the normal distribution for other confidence levels).

 You can then do whatever stuff you want to do with them, including plot
 them.

 --Patrick


 On 03/14/2012 10:48 AM, Ben quant wrote:

 Hello,

 How do I plot a gam fit object on probability (Y axis) vs raw values (X
 axis) axis and include the confidence plot lines?

 Details...

 I'm using the gam function like this:
 l_yx[,2] = log(l_yx[,2] + .0004)
 fit- gam(y~s(x),data=as.data.frame(**l_yx),family=binomial)

 And I want to plot it so that probability is on the Y axis and values are
 on the X axis (i.e. I don't want log likelihood on the Y axis or the log
 of
 my values on my X axis):

 xx- seq(min(l_yx[,2]),max(l_yx[,2]**),len=101)
 plot(xx,predict(fit,data.**frame(x=xx),type=response),**
 type=l,xaxt=n,xlab=Churn**,ylab=P(Top
 Performer))
 at- c(.001,.01,.1,1,10)  #-- I'd also like to generalize
 this rather than hard code the numbers
 axis(1,at=log(at+ .0004),label=at)

 So far, using the code above, everything looks the way I want. But that
 does not give me anything information on variability/confidence/**
 certainty.
 How do I get the dash plots from this:
 plot(fit)
 ...on the same scales as above?

 Related question: how do get the dashed values out of the fit object so I
 can do 'stuff' with it?

 Thanks,

 Ben

 PS - thank you Patrick for your help previously.

[[alternative HTML version deleted]]

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



 --
 Patrick Breheny
 Assistant Professor
 Department of Biostatistics
 Department of Statistics
 University of Kentucky


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Re: [R] gam - Y axis probability scale with confidence/error lines

2012-03-14 Thread Patrick Breheny

Actually, I responded a bit too quickly last time, without really 
reading through your example carefully.  You're fitting a logistic 
regression model and plotting the results on the probability scale.  The 
better way to do what you propose is to obtain the confidence interval 
on the scale of the linear predictor and then transform to the 
probability scale, as in:


x - seq(0,1,by=.01)
y - rbinom(length(x),size=1,p=x)
require(gam)
fit - gam(y~s(x),family=binomial)
pred - predict(fit,se.fit=TRUE)
yy - binomial()$linkinv(pred$fit)
l - binomial()$linkinv(pred$fit-1.96*pred$se.fit)
u - binomial()$linkinv(pred$fit+1.96*pred$se.fit)
plot(x,yy,type=l)
lines(x,l,lty=2)
lines(x,u,lty=2)

--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky




On 03/14/2012 01:49 PM, Ben quant wrote:

That was embarrassingly easy. Thanks again Patrick! Just correcting a
little typo to his reply. this is probably what he meant:

pred = predict(fit,data.frame(x=xx),type=response,se.fit=TRUE)
upper = pred$fit + 1.96 * pred$se.fit
lower = pred$fit - 1.96 * pred$se.fit

# For people who are interested this is how you plot it line by line:

plot(xx,pred$fit,type=l,xlab=fd$getFactorName(),ylab=ylab,ylim=
c(min(down),max(up)))
lines(xx,upper,type=l,lty='dashed')
lines(xx,lower,type=l,lty='dashed')

In my opinion this is only important if the desired y axis is different
than what plot(fit) gives you for a gam fit (i.e fit -
gam(...stuff...)) and you want to plot the confidence intervals.

thanks again!

Ben

On Wed, Mar 14, 2012 at 10:39 AM, Patrick Breheny
patrick.breh...@uky.edu mailto:patrick.breh...@uky.edu wrote:

The predict() function has an option 'se.fit' that returns what you
are asking for.  If you set this equal to TRUE in your code:

pred - predict(fit,data.frame(x=xx),__type=response,se.fit=TRUE)

will return a list with two elements, 'fit' and 'se.fit'.  The
pointwise confidence intervals will then be

pred$fit + 1.96*se.fit
pred$fit - 1.96*se.fit

for 95% confidence intervals (replace 1.96 with the appropriate
quantile of the normal distribution for other confidence levels).

You can then do whatever stuff you want to do with them, including
plot them.

--Patrick


On 03/14/2012 10:48 AM, Ben quant wrote:

Hello,

How do I plot a gam fit object on probability (Y axis) vs raw
values (X
axis) axis and include the confidence plot lines?

Details...

I'm using the gam function like this:
l_yx[,2] = log(l_yx[,2] + .0004)
fit- gam(y~s(x),data=as.data.frame(__l_yx),family=binomial)

And I want to plot it so that probability is on the Y axis and
values are
on the X axis (i.e. I don't want log likelihood on the Y axis or
the log of
my values on my X axis):

xx- seq(min(l_yx[,2]),max(l_yx[,2]__),len=101)

plot(xx,predict(fit,data.__frame(x=xx),type=response),__type=l,xaxt=n,xlab=Churn__,ylab=P(Top
Performer))
at- c(.001,.01,.1,1,10)  #-- I'd also like to
generalize
this rather than hard code the numbers
axis(1,at=log(at+ .0004),label=at)

So far, using the code above, everything looks the way I want.
But that
does not give me anything information on
variability/confidence/__certainty.
How do I get the dash plots from this:
plot(fit)
...on the same scales as above?

Related question: how do get the dashed values out of the fit
object so I
can do 'stuff' with it?

Thanks,

Ben

PS - thank you Patrick for your help previously.

[[alternative HTML version deleted]]


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




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

Re: [R] sort list

What sort of plot are you using? I'm not really clear on what your
data as a whole look like: if you use dput() you can create a
representation and we can work from there.

In addition to the resources Josh recommended, the following sites can
direct you to all sorts of graphical goodies, all of which come with
source code (though at the latter site, they are often a little
complicated):

http://had.co.nz/ggplot2/
http://addictedtor.free.fr/graphiques/

Michael

On Wed, Mar 14, 2012 at 10:13 AM, sybil kennelly
sybilkenne...@gmail.com wrote:
 I appreciate the reading Thank you. May i ask one final question. If i have:

 matrix:
         var1    var2     var3
 cell1    x       x         x
 cell2    x       x         x
 cell3    x       x         x

 cell4

 .
 .
 .
 .
 cell100

 and:

 vector1 - c(cell1, cell5,cell19, cell50, cell70)

 your_data$mycells - factor(your_data$cells %in% vector1, c(Special,
 NotSpecial))

 So my output will be something like:

 [25] Special    Special    Special    Special    Special    Special
  [31] Special    NotSpecial NotSpecial NotSpecial NotSpecial NotSpecial
  [37] NotSpecial NotSpecial NotSpecial NotSpecial

 is there a way to plot the data so that my Special cells are plotted on
 top of my not special cells. The reason is my data may have 1 not
 special points,and i may have 5 special cells, I find I'm not able to see
 where they are on my plot because they are being covered by my not special
 cells :(

 I have been looking around for  order of factors plotted , 'order of
 levels, order of factor levels, is this on the right track or can it
 even be done?

 Syb




 On Wed, Mar 14, 2012 at 2:04 PM, sybil kennelly 
 sybilkenne...@gmail.comwrote:

 I appreciate the reading Thank you. May i ask one final question. If i
 have:

 matrix:
          var1    var2     var3
  cell1    x       x         x
  cell2    x       x         x
  cell3    x       x         x
 
  cell4
 
  .
  .
  .
  .
  cell100

 and:

 vector1 - c(cell1, cell5,cell19, cell50, cell70)

 your_data$mycells - factor(your_data$cells %in% vector1, c(Special,
 NotSpecial))

 So my output will be something like:

 [25] Special    Special    Special    Special    Special    Special
   [31] Special    NotSpecial NotSpecial NotSpecial NotSpecial NotSpecial
   [37] NotSpecial NotSpecial NotSpecial NotSpecial

 is there a way to plot the data so that my Special cells are plotted on
 top of my not special cells. The reason is my data may have 1 not
 special points,and i may have 5 special cells, I find I'm not able to see
 where they are on my plot because they are being covered by my not special
 cells :(

 I have been looking around for  order of factors plotted , 'order of
 levels, order of factor levels, is this on the right track or can it
 even be done?

 Syb

 On Tue, Mar 13, 2012 at 12:29 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:

 On Tue, Mar 13, 2012 at 5:15 AM, sybil kennelly sybilkenne...@gmail.com
 wrote:
  Thanks Josh. I'm quite new, just wondering re:factor levels?
 
  In this example (shamelessly stolen from the internet):
 
  schtyp
 
  [1] 0 0 1 0 0 0 1 0 1 0 1 1 1 1 0 0 1 1 1 0
 
  schtyp.f - factor(schtyp, labels = c(private, public))
 
  schtyp.f
 
  [1] private private public private private private public private public
  [10] private public public public public private private public public
 
  [19] public private
 
 
  Levels: private public
 
 
 
  in my data i have a table:
 
          var1    var2     var3
  cell1    x       x         x
  cell2    x       x         x
  cell3    x       x         x
 
  cell4
 
  .
  .
  .
  .
  cell100
 
 
  and i have a subset of those cells that are interesting to me as a list
 of
  data
  list1 = [cell1, cell5,cell19, cell50, cell70]
 
  is it possible to create (similar to above):
 
  schtyp.f - factor(schtyp, labels = c(special, normal))

 Sure.  Again, probably better to have cells of interest in a vector,
 not a list a la:

 list1 - c(cell1, cell5,cell19, cell50, cell70)

 your_data$mycells - factor(your_data$cells %in% list1, c(Special,
 NotSpecial))

 basically compares the cells to those in your list and returns
 TRUE/FALSE, which is then converted to a factor, labeled, and stored.
 If you are just starting, some background reading will help.  Here are
 some suggestions:

 1) Go here: http://www.burns-stat.com/pages/tutorials.html and read
 the tutorials for R -- Beginning (this should not take more than 1
 day).
 2) Sit down and read:
 http://cran.r-project.org/doc/manuals/R-intro.pdf through Appendix A
 (for now you can probably skip the rest of the appendices).  That will
 probably take another entire day or so.
 3) Head back to Patrick Burn's website:
 http://www.burns-stat.com/pages/tutorials.html and read the
 intermediate guide, The R Inferno (1-3 days depending if you can read
 for 8 hours straight or not)

 Cheers,

 Josh

 
  so that when i plot this data, i can color the items in list1 as one
 color
  (eg all the special

Re: [R] Merging fully overlapping groups

2012-03-14 Thread Jean V Adams

This code performs the same operation in about 1/10th the time on my 
machine.
Give it a try.

look - function(i) {
# look for subsets
dif - m[, i] - m
apply(dif, 2, min)  -0.5
}
nosubsets - function(df) {
# eliminate events that are subsets of other events in terms of 
attendance
m - table(df$B, df$A)
nevents - dim(m)[2]
found - sapply(seq(nevents), look)
diag(found) - FALSE
df[df$A %in% dimnames(m)[[2]][rowSums(found)0.5], ]
}
nosubsets(DF)

Jean



mdvaan wrote on 03/13/2012 10:56:33 PM:

 Hi,
 
 I have data on individuals (B) who participated in events (A). If ALL
 participants in an event are a subset of the participants in another 
event I
 would like to remove the smaller event and if the participants in one 
event
 are exactly similar to the participants in another event I would like to
 remove one of the events (I don't care which one). The following example
 does that however it is extremely slow (and the true dataset is very 
large).
 What would be a more efficient way to solve the problem? I really 
appreciate
 your help. Thanks! 
 
 DF - data.frame(read.table(textConnection(  A  B
 1209569832
 1209551750
 120956734
 1877451750
 1877451733
 187746734
 1877469833
 1926851750
 192686734
 1926851733
 1926865251
 516954441
 516915480
 51693228
 596651733
 596665251
 596668197
 59666734
 596651750
 596669833
 7189135523
 718965251
 718951733
 718969833
 7189135522
 718968197
 71896734
 779751750
 77976734
 779769833
 78666734
 786669833
 786651733
 859651733
 859651750
 859665251
 86776734
 867751750
 867751733
 893668197
 89366734
 893665251
 893651733
 920451750
 920469833
 92046734
 920451733),head=TRUE,stringsAsFactors=FALSE))
 
 data - unique(DF$A)
 for (m in 1:length(data))
{
for (m in 1:length(data))
   {
   tdata - data[-m]
   q - 0
   for (n in 1:length(tdata))
  {
  if (length(which(DF[DF$A == data[m], 2] %in% DF[DF$A == 
 tdata[n], 2] ==
 TRUE)) == length(DF[DF$A == data[m], 2]))
 {
 q - q + 1
 }
  }
   if (q  0)
  {
  data - data[-m]
  m - m - 1
  }
   }
}
 DF - DF[DF$A %in% data,]

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Re: [R] Rolling regressions with sample extended one period at a time