Re: [R] Efficient Way to gather data from various files

2012-10-02 Thread Jeff Newmiller
File operations are not vectorizable. About the only thing you can do for the 
iterating through files part might be to use lapply instead of a for loop, but 
that is mostly a style change.

Once you have read the dbf files there will probably be vector functions you 
can use (quantile). Off the top of my head I don't know a function that tells 
you which value corresponds to a particular quantile, but you can probably sort 
the data with order(), find the value whose ecdf is just below your target with 
which.max, and look at the row number of that value.

x - rnorm(11)
names(x) - seq(x)
xs - x[order(x)]
Row90 - as.numeric(names (xs)[0.9=seq(xs)/length(xs))])

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Sam Asin asin@gmail.com wrote:

Hello,

Sorry if this process is too simple for this list.  I know I can do it,
but
I always read online about how when using R one should always try to
avoid
loops and use vectors.  I am wondering if there exists a more R
friendly
way to do this than to use for loops.

I have a dataset that has a list of IDs.  Let's call this dataset
Master

Each of these IDs has an associated DBF file.  The DBF files each
have
the same title, and they are each located in a directory path that
includes, as one of the folder names, the ID.

These DBF files have 2 columns of interest.  One is the run number
the
other is the statistic.  I'm interested in the median and 90th
percentile
of the statistic as well as their corresponding run numbers. 
Ultimately,
I want a table that consists of

ID Run_50th Stat_50 Run_90 Stat_90
1AB  5102010 3 144376
1AC  399 6 9

etc.

Where I currently have a dataset that has

ID
1AB
1AC

etc.

And there are several DBF files that are in folders i.e.
folder1/1AC/folder2/blah.dbf

This dbf looks like

run   Stat

1  10
2  10
3  99
4  1000
5  1
6   99
7  1
8 10
9 10
1010
11 100


I know i could do this with a loop, but I can't see the efficient, R
way.
 I was hoping that you experienced R programmers could give me some
pointers on the most efficient way to achieve this result.

Sam

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Re: [R] patch download?

2012-10-03 Thread Jeff Newmiller
That is silly. The primary difference is that certain bugfix patches have been 
applied.  A side-effect of this first patched status is that unintentional 
bugs may have been introduced, so unless you have already encountered a bug 
that was patched there then you are likely better off sticking with the 
standard release. That said, I have had no qualms about using a patched version 
when I encountered a bug that got fixed.
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Poggio, John jpog...@ku.edu wrote:

Is the primary difference between the current version download (e.g.,
2.15.1) and the patched download of these R files that the patched
files have NOT been verified?  Please confirm and thanks!

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Re: [R] Get Cran R to use system proxy settings

2012-10-03 Thread Jeff Newmiller
Off-topic here. Try https://stat.ethz.ch/mailman/listinfo/r-sig-debian.

I also strongly recommend that you be more specific about what you have tried 
and how it didn't work as you expected when you post there.
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Sachinthaka Abeywardana sachin.abeyward...@gmail.com wrote:

Hi all,

I am using Ubuntu and I am having quite a bit of difficulty trying to
get R to use the system proxy settings. Any suggestions?

Thanks,
Sachin

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Re: [R] Text to Speech In R

2012-10-04 Thread Jeff Newmiller
Why do you think this is something R should be able to do?
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Alex Zhang alex.zh...@ymail.com wrote:

Dear all,

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

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

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

- Alex
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Re: [R] Text to Speech In R

2012-10-04 Thread Jeff Newmiller
Please don't shout.

Speech synthesis is associated in my mind with sound generation, which is 
rather platform-specific. R is a platform agnostic research tool. R can be 
interfaced with many kinds of external libraries, so many things are possible. 
However, they aren't necessarily simple, and hiding operating system 
differences can take a lot of work. Your expectation that this be simple is 
what seems unwarranted to me.
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Alex Zhang alex.zh...@ymail.com wrote:

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

- Alex



 From: Jeff Newmiller jdnew...@dcn.davis.ca.us
To: Alex Zhang alex.zh...@ymail.com; Alex Zhang
alex.zh...@ymail.com; r-help@r-project.org r-help@r-project.org 
Sent: Thursday, October 4, 2012 8:46 AM
Subject: Re: [R] Text to Speech In R
 
Why do you think this is something R should be able to do?
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Sent from my phone. Please excuse my brevity.

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

Dear all,

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

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

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

- Alex
    [[alternative HTML version deleted]]





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Re: [R] (no subject)

2012-10-04 Thread Jeff Newmiller
Please use a relevant subject line.

You can easily download source code for all CRAN packages and read it yourself.
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Chia-Yi Chiu chia-yi.c...@gse.rutgers.edu wrote:

Hi

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

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Re: [R] what exactly is the dim of data set yarn in package pls?

2012-10-06 Thread Jeff Newmiller
Perhaps what is wrong is that you need to learn when to use the str() function.

str(yarn)
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C W tmrs...@gmail.com wrote:

Hi list,
I am looking at the data yarn in package, I don't understand what is
dimension of this data set.

I did the following:
 library(pls)

 data(yarn)

 dim(yarn)

[1] 28  3

 head(yarn)

  NIR.1   NIR.2   NIR.3   NIR.4   NIR.5   NIR.6   NIR.7   NIR.8   NIR.9
NIR.10  NIR.11

1 3.06630 3.08610 3.10790 3.09720 2.99790 2.82730 2.62330 2.40390
2.19310
2.00580 1.83790

2 3.06750 3.08570 3.09580 3.06920 2.98180 2.84080 2.67600 2.50590
2.35060
2.22300 2.11920

3 3.07500 3.09660 3.09160 3.02880 2.88490 2.68850 2.47640 2.26940
2.08240
1.91950 1.77470

4 3.08280 3.09730 3.10100 3.07350 2.99130 2.87090 2.73920 2.61020
2.5
2.42370 2.37740

5 3.10290 3.10340 3.08480 3.02280 2.89270 2.71590 2.53840 2.37640
2.23970
2.13460 2.05340

6 3.08150 3.08490 3.04870 2.93050 2.73230 2.50890 2.29440 2.09950
1.93280
1.79250 1.66930

There were 270 columns, I only copy pasted the first 11 columns.  But
either way, this is NOT 3 columns.

Could anyone let me know what is wrong here?

Thanks in advance,
Mike

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Re: [R] Robust regression for ordered data

2012-10-07 Thread Jeff Newmiller
This does not appear to be a question about R. You should post in a list or 
forum dedicated to discussing statistics theory, such as 
stats.stackoverflow.com.
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Eiko Fried tor...@gmail.com wrote:

I have two regressions to perform - one with a metric DV (-3 to 3), the
other with an ordered DV (0,1,2,3).

Neither normal distribution not homoscedasticity is given. I have a two
questions:

(1) Some sources say robust regression take care of both lack of normal
distribution and heteroscedasticity, while others say only of normal
distribution. What is true?
(2) Are there ways of using robust regressions with ordered data, or is
that only possible for metric DVs?

Thanks
Torvon

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Re: [R] Robust regression for ordered data

2012-10-07 Thread Jeff Newmiller
I don't know about the topic of your question. Have you used the RSiteSearch 
function to research it yourself?
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Eiko Fried tor...@gmail.com wrote:

Thank you Jeff! Please ignore the first of my two questions then, and
apologies for not making it clear that my second question was about R.

(2) Are there ways of using robust regressions with ordered data ...
in
R?

Thank you


On 7 October 2012 18:26, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:

 This does not appear to be a question about R. You should post in a
list
 or forum dedicated to discussing statistics theory, such as
 stats.stackoverflow.com.

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 Eiko Fried tor...@gmail.com wrote:

 I have two regressions to perform - one with a metric DV (-3 to 3),
the
 other with an ordered DV (0,1,2,3).
 
 Neither normal distribution not homoscedasticity is given. I have a
two
 questions:
 
 (1) Some sources say robust regression take care of both lack of
normal
 distribution and heteroscedasticity, while others say only of normal
 distribution. What is true?
 (2) Are there ways of using robust regressions with ordered data, or
is
 that only possible for metric DVs?
 
 Thanks
 Torvon
 
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Re: [R] Removing header from a matrix

2012-10-07 Thread Jeff Newmiller
You need to clarify what you have and what you want. Please use the dput() 
function to create a reproducible example that we can enter into R to make 
suggestions about. [1]

There are a couple of possible things that could be going on here, and what you 
have given so far is ambiguous.

[1] 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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Dereje Bacha d_ba...@yahoo.com wrote:

I have three column vectors (X1, X2, X3).�
��� X1� X2� X3 
�� 20�� 25� 40
�� 100 90� 80
I want to put them as one matrix of dimention 2 by 3, �but remove
headers(X1,X2,X3) from the matrix. I wrote
as follows

U-cbind (X1,X2,X3)

the headers are there. I need help please. Thanks
Dereje
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Re: [R] other way of making a table?

2012-10-09 Thread Jeff Newmiller
Use factors?
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Jessica Streicher j.streic...@micromata.de wrote:

I'm making tables for prediction results of classifiers (2 classes)
that show the usual numbers, true positives, false positives, etc

I used the command

table(predictedLabels,realLabels)

to make those.

I just had a case though ,where one of the label vectors had only one
class in it. This will result in only half a table.

Compare:
x-c(1,1,1,0,0)
y-c(1,1,1,0,1)
table(x,y)

to 

x-c(1,1,1,0,0)
y-c(1,1,1,1,1)
table(x,y)

I want the second one to still have all 4 cases (second column all
zeros then).

Any easy solutions?

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Re: [R] How to write out this regression equation in R?

2012-10-09 Thread Jeff Newmiller
But sprintf is itself vectorized. If you give it vectors, it returns vectors. 
So you could obtain that apply-result more efficiently by passing a bunch of 
column vectors of data. There happens to be a convenient object called a data 
frame that holds a bunch of similar-length vectors.

DF - data.frame( m=c(1,2), b=c(-3,4) )
result - sprintf( y=(%d)*x+(%d), DF$m, DF$b)
cat(paste(result, collapse=\n))

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S Ellison s.elli...@lgcgroup.com wrote:

 

 How to make R write out: 
 
 Balance = 2 + 3 * IntGDP + 5 * IntUnemployment + 0.3 * d1
 
 from the table below:
 
 Balance Intercept   IntGDP  GDPNum  IntUnemployment 
 IntInflationd1  d2  d3
 3   2   3   5  
 0.3  0   0 


Maybe ?sprintf would help?

And if you wrap that in a function that takes a vector, using apply()
on the table would give you one string per row,

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Re: [R] RGL package surface plot

2012-10-09 Thread Jeff Newmiller
You should help us per the posting guide reproducible sample data, code you 
have tried that you thought would work, why you think it did not, output of 
sessionInfo function...

Note that you can learn a lot by doing this exercise, your problem may not be 
where you think it is (or where we might guess it is)

See 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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ariklee eric.kra...@respec.com wrote:

I'm completely new at R... I have sinkhole survey data (lat, long, and
elevation) and have been trying to plot a rotatable 3d plot for several
hours... cannot get it right. Examples I see tend to be grid data (one
elevation value per grid cell); however, my data are more random (known
elevations at known (but random) x-y positions). Data format is tab
separated, i.e.:

lat   long   elev
3443   2
36.832.54 6.2
...etc...

please help!

arik



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Re: [R] Optimx Package Error

2012-10-10 Thread Jeff Newmiller
This is operator error. Do not attempt to optimize over an infinite range.
---
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nserdar snes1...@hotmail.com wrote:

estimate- optimx(init.par,Linn,gr=NULL,method= L-BFGS-B,
hessian=TRUE,
control =
list(trace=1),lower=c(0,0,0,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf),upper=c(1,1,1,Inf,Inf,Inf,Inf,Inf,Inf,Inf))
fn is  Linn 
Function has  10  arguments
par[ 1 ]:  0   ? 0   ? 1 In Bounds   
par[ 2 ]:  0   ? 0   ? 1 In Bounds   In Bounds  
par[ 3 ]:  0   ? 0   ? 1 In Bounds   In Bounds   In Bounds   
 
par[ 4 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds
par[ 5 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds   In Bounds   
par[ 6 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds   In Bounds   In Bounds  
par[ 7 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds   In Bounds   In Bounds   In Bounds 
par[ 8 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds   In Bounds   In Bounds   In Bounds   In Bounds
par[ 9 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds  
In
Bounds   In Bounds   In Bounds   In Bounds   In Bounds   In Bounds   
par[ 10 ]:  -Inf   ? 1   ? Inf In Bounds   In Bounds   In Bounds 
 In
Bounds   In Bounds   In Bounds   In Bounds   In Bounds   In Bounds   In
Bounds  
Error in optimx(init.par, Linn, gr = NULL, method = L-BFGS-B, hessian
=
TRUE,  : 
  Cannot evaluate function at initial parameters

Please let me know how to deal with this problem.

Regards,
Serdar



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Re: [R] Contacting Delphi ??

2012-10-10 Thread Jeff Newmiller
Its suddenness is only as sudden as your input that triggered it. Its oblique 
nature is a reflection of the nature of your input.

http://lmgtfy.com/?q=R+contacting+Delphi+oracle+is+unavailable
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Christian Hoffmann c-w.hoffm...@sunrise.ch wrote:

What does the sudden appearance of Contacting Delphi ..the oracle 
is unavailable.
We apologize for any inconvenience. mean? A bug? It appears at
plotting.

Thanks

Christian

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Re: [R] Converting factors to bounded random numerical data

2012-10-10 Thread Jeff Newmiller
Please don't double post.
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KoopaTrooper ncoop...@tulane.edu wrote:

I have a data set (a) with three columns (X,Y,Z). The first 2 columns
are
numeric. The third (Z) is a factor with three levels A,B,C. I want to
turn
each A into a different random number between 1 and 4, each B into a
different random number between 5 and 8, etc.

I tried this:

a$Z-ifelse(a$Z==L,sample(1:4,1),ifelse(a$Z==M,sample(5:9,1),ifelse(a$Z==U,sample(10:12,1),)

and it almost worked but changed all the A's into the same random
number.
I need a different random number for each A. Ideas?

Thanks, 



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Re: [R] Columns and rows

2012-10-11 Thread Jeff Newmiller
Data frames don't have unequal column or row lengths within a single object. 
Your terminology is obscuring your intent. Perhaps you should read the posting 
guide for guidance on posting here and then generate a reproducible example. 
Since you seem to be having difficulty with the how, at least provide sample 
input and desired output following the recommendations here:

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

Are you aware of

?cbind
?rbind
?merge

?
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Santana Sarma aimanusa...@gmail.com wrote:

Hi,

Could you please advice some easy way to do the following for a
dataframe
(header=F) having unequal column-  row- length.

1. Combine/stack/join contents from -
  a) multiple rows into one column.
  b) multiple columns into one row.

2. Stack contents from multiple columns (or, rows) into one column (or,
row).

Thank you.

Cheers,

Santana

   [[alternative HTML version deleted]]

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Re: [R] Columns and rows

2012-10-11 Thread Jeff Newmiller
Better, but you didn't do your reading.  Seriously... learn to make it 
reproducible... you will get quicker help from more skilled R users. 
Also, post in text, not HTML.


Note that data frames are not really the right data structure for doing 
the kinds of operations you are describing below, since you assume that 
rows and columns are essentially interchangeable.  Lists of vectors might 
be, but only if you were reading from a data format like XML that 
reflected that structure.  Matrices have a uniform 2d structure with a 
transpose operation defined, and can contain NA values. Also, vectors are 
not described as column or row vectors in R ... that is Matlab-speak. 
Vectors have no direction, just sequence.  Matrices are folded vectors 
(vectors with row and column dimension attributes).  The as.vector() 
function forgets the row and column attributes of a matrix (or other 
object). The matrix() function adds those attributes, effectively 
folding the vector up into 2d.


Data frames are lists of vectors with the same length, where each vector 
can have a different type of data (e.g. string, numeric, logical, factor). 
They are printed, read from files and written to files with the vectors as 
columns (even though vectors themselves have no notion of being columns). 
Thus, working with one row at a time is much slower than working with one 
column at a time, and since each element of a row can be a different type 
of data it doesn't make sense to mash them up the way you want to.


I suggest you print and use the str() function liberally on the objects 
and expressions below to understand how they work.


#reproducible data
DF - read.csv( text=
Names,'Col x','Col y','Col z'
rowName1,A,E,H
rowName2,B,F,I
rowName3,C,G,J
rowName4,D,,K
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

#solution
mat - as.matrix( DF )
v1 - as.vector( na.omit( as.vector( t( mat ) ) ) )
v2 - as.vector( na.omit( as.vector( mat ) ) )
ans1a - matrix( v1, ncol=1 )
ans1b - matrix( v2, nrow=1 )
ans2a - matrix( v2, ncol=1 )
ans2b - matrix( v1, nrow=1 )


On Fri, 12 Oct 2012, Santana Sarma wrote:


Hi,

Trying to give an example here. 
Say, I have read in a .csv file using read.csv (), and the file contains the
following info. 


Names Col x Col y Col z
rowName1 A E H
rowName2 B F I
rowName3 C G J
rowName4 D K



Now, this is what is required: 

1. Combine/stack/join contents from - 
              a) multiple rows into one column. 


That is:  

A
E
H
B
F
I
C
G
J
D
K

              b) multiple columns into one row.

A B C D E F G H I J K

    
2. Stack contents from 

A) multiple columns into one column.

A
B
C
D
E
F
G
H
I
J
K


B) Multiple rows into one row.

A E H B F I C G J D 


Thank you.

Cheers,
Santana




On Fri, Oct 12, 2012 at 1:32 PM, David Winsemius dwinsem...@comcast.net
wrote:

  On Oct 11, 2012, at 5:55 PM, Santana Sarma wrote:

   Hi,
  
   Could you please advice some easy way to do the following for
  a dataframe
   (header=F) having unequal column-  row- length.
  
   1. Combine/stack/join contents from -
                a) multiple rows into one column.
                b) multiple columns into one row.
  
   2. Stack contents from multiple columns (or, rows) into one
  column (or,
   row).

Could _you_ please produce an example.

Dataframes do not have headers. They do have column names and column
names are required.

--
David Winsemius, MD
Alameda, CA, USA






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Re: [R] Problem with which function

2012-10-12 Thread Jeff Newmiller
See R FAQ 7.31.
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max u...@maxmetzger.de wrote:

Hej,

i need the which() funktion to find the positions of an entry in a
matrix.

the entries i'm looking for are : seq(begin,end,0.01) and there are no
empty spaces
i'm searching in the right range.

so i was looking for the results R can find and i recieved this answer.

for (l in
1:length(qr)){print(c(l,nstu[l,1],nsto[l,1],nsto[l,1]-nstu[l,1],(nsto[l,1]-nstu[l,1])*100,which(sp1[,1]==nstu[l,1]),(nsto[l,1]-nstu[l,1])*100+which(sp1[,1]==nstu[l,1]),which(r[,1]==nstu[l,1]),(nsto[l,1]-nstu[l,1])*100+which(r[,1]==nstu[l,1])))}
[1] 1 0 0 0 0
[1] 2 0 0 0 0
[1] 3 0 0 0 0
[1]4.00   87.34   87.970.63   63.00 2491.00 2554.00 2491.00
2554.00
[1]5.00   86.73   88.461.73  173.00 2430.00 2603.00
[1]6.00   86.22   88.772.55  255.00 2379.00 2634.00
[1]7.00   85.77   89.013.24  324.00 2334.00 2658.00
[1]8.00   85.36   89.223.86  386.00 2293.00 2679.00
[1]9.0   85.0   89.44.4  440.0 2257.0 2697.0
[1]   10.00   84.68   89.574.89  489.00 2225.00 2714.00 2225.00
2714.00
[1]   11.00   84.38   89.735.35  535.00 2195.00 2730.00
[1]   12.00   84.11   89.885.77  577.00 2168.00 2745.00
[1]   13.00   83.86   90.026.16  616.00 2143.00 2759.00
[1]   14.00   83.64   90.156.51  651.00 2121.00 2772.00
[1]   15.00   83.43   90.286.85  685.00 2100.00 2785.00 2100.00
2785.00
[1]   16.00   83.24   90.407.16  716.00 2081.00 2797.00
[1]  17.00  83.07  90.52   7.45 745.00
[1]   18.00   82.91   90.637.72  772.00 2048.00 2820.00
[1]   19.00   82.76   90.757.99  799.00 2033.00 2832.00
[1]   20.00   82.62   90.868.24  824.00 2019.00 2843.00 2019.00
2843.00
[1]   21.00   82.49   90.978.48  848.00 2006.00 2854.00
[1]   22.00   82.37   91.088.71  871.00 1994.00 2865.00 1994.00
2865.00
[1]   23.00   82.26   91.198.93  893.00 1983.00 2876.00
[1]   24.00   82.16   91.319.15  915.00 1973.00 2888.00
[1]   25.00   82.07   91.429.35  935.00 1964.00 2899.00
[1]   26.00   81.98   91.539.55  955.00 1955.00 2910.00
[1]   27.00   81.90   91.649.74  974.00 1947.00 2921.00 1947.00
2921.00
[1]   28.00   81.83   91.759.92  992.00 1940.00 2932.00
[1]   29.00   81.76   91.87   10.11 1011.00 1933.00 2944.00
[1]   30.00   81.69   91.98   10.29 1029.00 1926.00 2955.00
[1]   31.00   81.63   92.09   10.46 1046.00 1920.00 2966.00
[1]   32.00   81.57   92.20   10.63 1063.00
[1]   33.00   81.50   92.31   10.81 1081.00 1907.00 2988.00
[1]   34.00   81.44   92.43   10.99 1099.00 1901.00 3000.00
[1]   35.00   81.38   92.54   11.16 1116.00 1895.00 3011.00
[1]   36.00   81.31   92.65   11.34 1134.00 1888.00 3022.00
[1]   37.00   81.25   92.76   11.51 1151.00 1882.00 3033.00
[1]   38.00   81.18   92.87   11.69 1169.00 1875.00 3044.00 1875.00
3044.00
[1]   39.00   81.12   92.98   11.86 1186.00 1869.00 3055.00 1869.00
3055.00
[1]   40.00   81.05   93.10   12.05 1205.00 1862.00 3067.00
[1]   41.00   80.99   93.21   12.22 1222.00 1856.00 3078.00
[1]   42.00   80.92   93.32   12.40 1240.00 1849.00 3089.00
[1]   43.00   80.86   93.43   12.57 1257.00 1843.00 3100.00
[1]   44.00   80.79   93.54   12.75 1275.00
[1]   45.00   80.73   93.66   12.93 1293.00 1830.00 3123.00
[1]   46.00   80.66   93.77   13.11 1311.00 1823.00 3134.00
[1]   47.00   80.60   93.88   13.28 1328.00 1817.00 3145.00
[1]   48.00   80.53   93.99   13.46 1346.00 1810.00 3156.00
[1]   49.00   80.47   94.10   13.63 1363.00 1804.00 3167.00
[1]   50.00   80.40   94.22   13.82 1382.00 1797.00 3179.00 1797.00
3179.00
[1]   51.00   80.34   94.33   13.99 1399.00 1791.00 3190.00 1791.00
3190.00
[1]   52.00   80.27   94.44   14.17 1417.00 1784.00 3201.00
[1]   53.00   80.20   94.55   14.35 1435.00 1777.00 3212.00
[1]   54.00   80.14   94.66   14.52 1452.00 1771.00 3223.00
[1]   55.00   80.07   94.77   14.70 1470.00 1764.00 3234.00
[1]   56.00   80.01   94.89   14.88 1488.00 1758.00 3246.00
[1]   57.00   79.94   95.00   15.06 1506.00 1751.00 3257.00
[1]   58.00   79.87   95.11   15.24 1524.00 1744.00 3268.00 1744.00
3268.00
[1]   59.00   79.81   95.22   15.41 1541.00 1738.00 3279.00
[1]   60.00   79.74   95.32   15.58 1558.00 1731.00 3289.00
[1]   61.00   79.67   95.43   15.76 1576.00 1724.00 3300.00
[1]   62.00   79.61   95.54   15.93 1593.00 1718.00 3311.00
[1]   63.00   79.54   95.65   16.11 1611.00
[1]   64.00   79.47   95.76   16.29 1629.00 1704.00 .00
[1]   65.00   79.41   95.87   16.46 1646.00 1698.00

Re: [R] Filtering few column from one data frame based on another data frame

2012-10-13 Thread Jeff Newmiller
I suggest you go back to the Introduction to R document supplied with the 
software and read about indexing.

In particular, pay attention to indexing into a matrix or data frame using 
data1[rowspec,columnspec] syntax. There are three kinds of row and column 
specification types, and any of the the three can be used to solve your problem.
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siddu479 onlyfordigitalst...@gmail.com wrote:

Dear R-Users and experts,
This is my first post in this forum.

I have two csv files file1 and file2.
file1(many rows and columns) is read into dataframe *data1*
File2 is read into dataframe *data2* which have only one column of data
which contains the column names that need to be removed from dataframe
*data1*columns and update the *data1 *data frame.
Example:
data1 -read.csv(file1...,header=T)
Sno,Data_1,Data_2,Data_3
1,2,3,4

data2- read.csv(file2,..header=F)
Data_2
Data_3

The output *data1.new* data frame  should be like this

Sno,Data_1
1,2

So I request the R code to accomplish this task.

Regards
Sidda




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Re: [R] stop at error point in for-loop

2012-10-14 Thread Jeff Newmiller
options(error = recover)
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mrzung mrzun...@gmail.com wrote:

hi all:

Following is simple example,

for(i in 1:1000){
a-function(i){
#some function that has an error
}
}

What I want to know is the way to find the error point in for-loop,
What i
makes the error.

is there any way to solve it instead of debugging and finding an error
manually?

Thanks,





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Re: [R] transforming a .csv file column names as per a particular column rows using R code

2012-10-14 Thread Jeff Newmiller

Not clear to me what your successive rows indicate in your output.

However, you may get something similar to what you want using the reshape2 
library:


library(reshape2)

dta1 - read.csv( text=
Tool,Step_Number,Data1,Data2
A,1,0,1
A,2,3,2
A,3,2,3
B,1,3,2
B,2,1,2
B,3,3,2
)

# put into long form
dta1l - melt(dta1,c(Tool,Step_Number))
# re-cast to wide form
# assuming first row is for tool A, second row is for tool B, etc.
dta2 - dcast(dta1l,Tool~variable+Step_Number)


On Sun, 14 Oct 2012, siddu479 wrote:


Hello all,
I have a .csv file like below.
Tool,Step_Number,Data1,Data2... etc up to 100 columns.
A,1,0,1
A,2,3,1
A,3,2,1
.
.
B,1,3,2
B,2,1,2
B,3,3,2
.
.
.. so on upto 50 rows
where the column *Tool* has distinct steps in second column
*Step_Number*,but both have same entries in Step_Number column.
I want the output like below.

Tool_1,Data1_1,Data2_1,Tool_2,Data1_2,Data2_2,Tool_3,Data1_3,Data2_3... so
on
A,0,1,A,3,1,A,2,1
B,3,2,B,1,2,B,3,2
..
so on.

basically I am transposing entire data based on a specific column row values
and renaming the column headers.
I have a shell script based on awk which can do this task, but the script is
taking exceptionally higher processing time.

So I am looking for a script in R which can save the time.
Please revert to me if the problem description is not clear.

Regards
Sidda




-
Sidda
Business Analyst Lead
Applied Materials Inc.

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Re: [R] how to show iterations

2012-10-14 Thread Jeff Newmiller
Optimization is not something R does... this is something implemented by 
specific functions that may be in the base install or in add-on packages, 
but you have not indicated how you are currently approaching your problem.


If you are using the optim() function, then I refer you to the fine 
manual under the control argument. Type


?optim

at the R command line.

For advice on how to communicate clearly on the topic of R, I recommend 
that you also read 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

before posting again.

On Sun, 14 Oct 2012, sffarooqi wrote:


Hi,
I am new to R, and am working on some optimization problems - I was
wondering if there was a way that I could show all the iterations in R -i.e.
showing for each iteration, what the iteration is, how much the function is,
what the norm of the gradient would be...
Any help is greatly appreciated!
thank you!!




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Re: [R] Date Math

2012-10-14 Thread Jeff Newmiller
There are a few ways.  The xts package has a lag function. So does 
zoo.  Pay careful attention to the conventions used for specifying 
relative time in these various packages. You can also infill your missing 
data to create a regularly-spaced time series.  There is no shortage of 
web information about this topic... you can start at the Time Series task 
view on CRAN or just use a search engine.


On Sun, 14 Oct 2012, Noah Silverman wrote:


Hello,


I have a time series object (xts) that I iterate over in a loop.  Works fine.

My challenge is that I want to be able to reference other entries in the series 
by math.  i.e.  For today's observation, what were the last 5 observations?  If 
indexed numerically, it is trivial, but I can figure out how to do this with 
dates.

This is slightly more difficult as there may not be an observation for every 
day.  So I might want the last 5 that exist in the table, not the last 5 
calendar days.

ideally, it would be something like this.

observations[ index(today)-5:today,  ]

However that obviously fails.

Ideas?



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UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095

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Re: [R] Is there any R package that contains Rusboost based on Adaboost.m2?

2012-10-14 Thread Jeff Newmiller

May I suggest you use a more appropriate tool for your search:

RSiteSearch(multiclass classification)

On Sun, 14 Oct 2012, Carlos Andrade wrote:


Hi,

I have been searching everywhere for an implementation of those algorithms,
but I have only observed them in Matlab and on the literature.

I noticed a package called 'ada' in CRAN but it is not for multi class. I
would be happy with just Adaboost.m2, Smoteboost over adaboost.m2 or any
other combination that could account for imbalanced multiclass
classification problems.

Thanks!

Carlos Andrade
http://carlosandrade.co

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Re: [R] Is there any R package that contains Rusboost based on Adaboost.m2?

2012-10-14 Thread Jeff Newmiller
Asking questions and giving answers about R is what this mailing list is about. 
It is not really a theory list. If you understand the theory you can search 
among the libraries yourself. But when the first search I make brings up what 
to me looks like the answer to your question, then you either did not search 
before posting or you did not clarify what you were really looking for properly.
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Carlos Andrade carlosvia...@gmail.com wrote:

Dear Jeff Newmiller,

Thanks for the suggestion I did not know the website but the problem is
more dry than simple multiclass classification. In fact, there are even
few
papers addressing the problem. I was in hope that if not yet available
in
CRAN someone from the R community could point out any 'pre release'
version
of it. I hope it is inappropriate to ask such questions here.

Thanks,

Carlos Andrade
http://carlosandrade.co



2012/10/14 Jeff Newmiller jdnew...@dcn.davis.ca.us

 May I suggest you use a more appropriate tool for your search:

 RSiteSearch(multiclass classification)


 On Sun, 14 Oct 2012, Carlos Andrade wrote:

  Hi,

 I have been searching everywhere for an implementation of those
 algorithms,
 but I have only observed them in Matlab and on the literature.

 I noticed a package called 'ada' in CRAN but it is not for multi
class. I
 would be happy with just Adaboost.m2, Smoteboost over adaboost.m2 or
any
 other combination that could account for imbalanced multiclass
 classification problems.

 Thanks!

 Carlos Andrade
 http://carlosandrade.co

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Re: [R] Date Math

2012-10-14 Thread Jeff Newmiller
?embed
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Noah Silverman noahsilver...@ucla.edu wrote:

Jeff,

My understanding is that the lag command will lag an entire time
series.  That isn't what I'm looking for.

I just want, for example, today, and 5 entries back.  

for exmple:

iter - '2011-05-18'

observations[iter]   # works fine, returns the row at that date.

index(observations[iter[) # just returns the iter back

index(observations[iter]) - 5  # returns 2011-05-17 23:59:57 PDT, so
it subtracted 3 seconds.

really, I want to find:  iter- 5 days.



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Smart Media Corp.



On Oct 14, 2012, at 10:29 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:

 There are a few ways.  The xts package has a lag function. So does
zoo.  Pay careful attention to the conventions used for specifying
relative time in these various packages. You can also infill your
missing data to create a regularly-spaced time series.  There is no
shortage of web information about this topic... you can start at the
Time Series task view on CRAN or just use a search engine.
 
 On Sun, 14 Oct 2012, Noah Silverman wrote:
 
 Hello,
 
 
 I have a time series object (xts) that I iterate over in a loop. 
Works fine.
 
 My challenge is that I want to be able to reference other entries in
the series by math.  i.e.  For today's observation, what were the last
5 observations?  If indexed numerically, it is trivial, but I can
figure out how to do this with dates.
 
 This is slightly more difficult as there may not be an observation
for every day.  So I might want the last 5 that exist in the table, not
the last 5 calendar days.
 
 ideally, it would be something like this.
 
 observations[ index(today)-5:today,  ]
 
 However that obviously fails.
 
 Ideas?
 
 
 
 --
 Noah Silverman, M.S.
 UCLA Department of Statistics
 8117 Math Sciences Building
 Los Angeles, CA 90095
 
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Re: [R] Chopping a two column data frame by rows into a three dimensional array.

2012-10-15 Thread Jeff Newmiller
aperm(array(dat,c(10,10,2)),c(1,3,2))
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Hans Thompson hans.thomps...@gmail.com wrote:

If I have a two column data frame like:

 dat - cbind(x=c(1:100),y=c(100:1))

How can I create an array that splits every ten rows of that data frame
into a third dimension of an array so that:

 newarray[,,1]

,,1

x   y
1   100
2   99
3   98
...   ...
10  91


,,2

xy
11  90
12  89
...   ...


...

Thanks.

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Re: [R] time extraction and normalization

2012-10-15 Thread Jeff Newmiller
There are multiple ways. For lack of your example, I would suggest multiplying 
by 24.

Please post the previous context of the thread if you must post from Nabble.
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york8866 yu_y...@hotmail.com wrote:

Thanks,

The code gives the numbers in days, how can I adjust the code to
directly
get the numbers in hours?

I tried units but it did not work. 

Thanks1



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Re: [R] Windows 7 R (32/64bit) running under cygwin: package not found

2012-10-16 Thread Jeff Newmiller
Probably because when you run it from Cygwin the R_LIBS variable does not point 
to the user and install library directories. I don't know how Rgui knows where 
they are (registry?) but you can look in the .Library and .Library.site 
variables to see the results. 

In a case like this, posting your sessionInfo() for each case is highly 
recommended.
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Aldi Kraja a...@wustl.edu wrote:

Hi,
Using R 2.15.1 on Windows 7. Have installed both versions 32 and 64bit.
In both of them among others I have installed a package rgenoud
When I open R gui of 32bit and write library(rgenoud) it responds by 
showing a functional rgenoud version  5.7-8. The same it does on Rgui 
64bit.

Now I am working in cygwin (v. 1.12.4.0) with xwin. Normally before
when 
I had installed a package, I only had to call the library with the name

of the package and R will find the right one to load.

Now when I apply R CMD BATCH script1.R out1.txt, under cygwin the first

thing it reports:

R version 2.15.1 (2012-06-22) -- Roasted Marshmallows
After some generalities it reports
 library(rgenoud)
Error in library(rgenoud) : there is no package called 'rgenoud'
Execution halted

So my question is why under cygwin in a batch mode, it does not find
the 
installed package, which is already installed in my laptop's R?

Thank you in advance,

Aldi

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Re: [R] how to concatenate factor vectors?

2012-10-18 Thread Jeff Newmiller
mood - factor(c(blue, sunny))
skycolor - factor(c(azure,blue,teal)

If factors are not defined with levels specifications, automatic merging should 
never be allowed. The fact that read.table automatically generates factors 
using default levels is why I nearly always import using as.is=TRUE, perform QC 
and combining of sources using strings, and only then convert to factor.

If you HAVE defined your factors using explicit levels definitions, you should 
have no trouble combining them.
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Sam Steingold s...@gnu.org wrote:

 * Bert Gunter thagre.ore...@trar.pbz [2012-10-17 23:21:44 -0700]:

 However, Is level 5 in 'a' the same as level 5 in 'b' ?

yes, of course.
would anyone want to _different_ factors with identical string
representations?!

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Re: [R] system.time question

2012-10-20 Thread Jeff Newmiller
You asked several questions.
Elapsed: yes
User + System = CPU: yes
Finally: You have to look at the load and/or cpu core count. Unless you setup 
your code to take advantage of multiple cores, R runs on a single core.
Also: Do you really need to ask that question?

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Mark Leeds marklee...@gmail.com wrote:

Hi : I looked at the help for system.time but I still have the
following
question. Can someone explain the output following output
of system.time :

 user  system  elapsed
12399.681  5632.352   56935.647

Here's my take based on the fact that I was doing ps -aux | grep R off
and
on and the total amount of CPU minutes that
got allotted before the job ended was about 5 hours and the total
actual
time that the job took was about 15 hours.

Does elapsed = total actual time job taken ? That seems to be the case
or a
strange coincidence.

Does user + system = CPU time from ps -aux | grep R ? That seems to be
the
case also or a weird coincidence.

Finally, why can't the CPU get a higher percentage ? It's seems like
it's
always around 30% which would make sense since
5 is ~ 30% of 15 hours.

Also, assuming my take above is correct, when talking about timing of
algorithms, in this case, does one say the job took 5 hours or 15 hours
?
I'm trying to see how fast an algorithm is compared to others and I'm
not
sure what the standard is.  I'm on fedora 16.0 and using R 2.15.
Thanks.

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Re: [R] Error in integrate(integrand, 0, Inf) : non-finite function value

2012-10-21 Thread Jeff Newmiller
That is an intrinsic part of nonlinear optimization. Choose your starting point 
wisely.
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stats12 ska...@gmail.com wrote:

Hi,

Thank you for your comment. I worked on the code again and was able to
make
it work. The only problem I am having right now is that nlm depends on
the
initial value.  

When the initial value is 1, I get the following estimates
0.1230414 19.6271029

when it is 2, I get the following
29.46874 20.01679



d-matrix(c(1,1,0,0,0,0,0,0,2,1,0,0,1,1,0,1,2,2,1,0),nrow=10,ncol=2) 
h-matrix(runif(20,0,1),10) 
delta-matrix(c(2,1,0,1,0,1,0,0,2,1,0,0,1,1,1,1,0,2,1,0),nrow=10,ncol=2)


out-vector(numeric,length(1:2)) 
integ-vector(numeric,length(1:10)) 

for (k in 1:2){ 
ll-function(p){ 
cmh-delta[,k]*(h[,k]*log(0.5))+p*log(gamma(1+1/p)) 
for(s in 1:10){ 
integrand-function(x)
x^d[s,k]*exp(-x*gamma(1+1/p))^p*p*x^(p-1)*exp(-x*h[s,k]) 
integ-integrate(integrand,0,Inf)$value 
return(integ)
}

lik-sum(cmh+log(integ)) 
-lik 
} 
initial-c(1) 
t-nlm(ll,initial) 
out[k]-t$estimate 
} 

est-as.vector(out)








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Re: [R] [Vars] package: impulse response function

2012-10-21 Thread Jeff Newmiller
Read the posting guide. We have no way to know what you are doing unless you 
show the list readers with sample data and code.

If you are wondering about Vector Autoregressive models, I don't claim any 
specific experience with them, but [1] says they are linear, and I do know that 
linear models respond in proportion to the magnitude of the input so you don't 
learn anything useful by changing the magnitude of the input.

I have used the Kronecker delta (impulse) function before, and I have never 
heard of anyone trying to ascribe a standard deviation to it... only impulse 
magnitude.

[1] http://en.m.wikipedia.org/wiki/Vector_autoregression
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Maria Grigoryeva mm.grigory...@gmail.com wrote:

Hello,

I'm doing a research on the impulse responses in VAR models and I'm
having
troubles in interpretation of R results.
My question is what is the shock of impulse variable that is produced
to
obtain the response? Is it one-standard-deviation positive shock? If it
is
so how can I obtain  the responses on other types of shocks, say, 10%
negative shock?

Thanks a lot,

Marion

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Re: [R] daylight

2012-10-23 Thread Jeff Newmiller
library(maptools)
?sunriset
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bambus sonja_...@hotmail.com wrote:

hi there,
does anyone know how to calculate the amount of daylight on every day
of the
year in R? I mean the time between sunrise and sunset.

thanks



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Re: [R] incomplete final line found by readTableHeader

2012-10-23 Thread Jeff Newmiller
Keep in mind that this is NOT a homework support list... you are supposed to 
use the support provided by your educational institution if you are in a 
course. Read the posting guide mentioned in the footer of every email.

FWIW, as described your problem is with Excel, and this is not an Excel support 
list either. I suggest that you edit the csv file with a text editor (e.g. 
NotePad) to make sure the correct number of commas are on each line of the data 
file.
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Shane2012 wenbo...@live.cn wrote:

Hello, 

I am trying to read in an Excel file that I saved as a .csv so I can
analyze 
my assignment data! I am getting really frustrated because this is what
I 
keep getting: 
Warning message:
In read.table(CityData.CSV, sep = /, header = T) :
  incomplete final line found by readTableHeader on 'CityData.CSV'

I have open the file and make sure click return after the last line,
then
save the file.

I try to use kinds of read data frame methods, such as 

read.table  

read.csv   or 
my.files - list.files(path)
for(i in my.files) 
{ 
   nam - paste(CityData, substr(i, 14, 15), sep ='')
   assign(nam, read.csv(i))
}
-
They all failed.

Can anyone offer some help? Thanks a lot!  



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Re: [R] Results not dispalying in R console

2012-10-24 Thread Jeff Newmiller
wow. That sounds like a problem.

For a more constructive response, read the posting guide and post specific 
examples and explain how the output you get is not what you expect. Also, the 
output of sessionInfo() is likely to be helpful, and if you use a Mac, or 
Debian/other Linux, one of the platform-specific email lists might be a more 
appropriate forum.
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arunkumar akpbond...@gmail.com wrote:

Hi

I'm not getting any results in R console when i run commands. I
reinstalled
R but the results are same



-
Thanks in Advance
Arun
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Re: [R] Merge matrices with different column names

2012-10-25 Thread Jeff Newmiller
If they have the same number of rows, you can use cbind() to create one object 
to write out.
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Charles Determan Jr deter...@umn.edu wrote:

A general question that I have been pursuing for some time but have set
aside.  When finishing some analysis, I can have multiple matrices that
have specific column names.  Ideally, I would like to combine these
separate matrices for a final output as a csv file.

A generic example:

Matrix 1
var1A  var1B  var1C
x  x   x
x  x   x

Matrix 2
var2A  var2B  var2C
x  x   x
x  x   x

I would like a final exportable matrix or dataframe or whichever format
is
most workable.

Matrix 3
var1A  var1B  var1C
x  x   x
x  x   x

var2A  var2B  var2C
x  x   x
x  x   x

However, no matter which function I try reports an error that the
column
names are not the same.

Any insights would be appreciated.
Thanks as always,
Charles

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Re: [R] SQL via RODBC

2012-10-26 Thread Jeff Newmiller
That would be very implementation-specific, and ODBC is generic in its own way. 
No, you must run one query at a time in general, and deal with the results 
using the procedural language.

Keep in mind that you have to pick a back-end database to work with, and for 
creating the database you may have tools available that can directly process 
that sql file once.  Then you can use a small number of queries from within R 
to retrieve data at will. If the file contains data queries whose output you 
want to process in R, they have to be handled one-by-one.

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Dr. Alireza Zolfaghari ali.zolfagh...@gmail.com wrote:

Hi list,
Is there a way to use sqlQuery function where there is a sql file (ie.
sample.sql)? I just want to mention that in my sql file there are some
comment lines (starting with --). This means that if I paste all the
lines
in the sql file, I'll come up with a long string that most part of it
is
commneted (after commnet sign --).

thanks
Arvin

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Re: [R] asking about R Code

2012-10-26 Thread Jeff Newmiller
You seem to be quite lost. You should execute one statement at a time, and 
troubleshoot from there. I suspect that your attempts to get your file in the 
right place went wrong, or the file is the wrong type of file for that function 
(regardless of the name you gave it).

As for giving you an explanation about read.pnm, I would rather not. There is 
documentation available on any library function using the ? command in R, e.g.

?read.pnm

You should not have to replace files with other files in order to read 
them... you should be able to specify another name within the quotes. You may 
learn a bit about your current directory by using the getwd() function and the 
list.files() function. Once you have successfully read in the data, you can 
learn more about the object by following the useful hyperlinks in the 
documentation for read.pnm.

I highly recommend reading the Introduction to R pdf that is supplied with your 
R software, and then reading the posting guide mentioned at the bottom of any 
email on this list. If your project is of an academic nature, the posting guide 
will remind you that this is not the place to ask homework questions... that is 
for you to obtain from your educational institution.
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Ellen Praman ellen.praman...@yahoo.com wrote:

Hi, my name is Ellen. I want to ask you about R Code.
I got a code for extracting a pixel value, but I can't compile it..�
It is said Error in is.data.frame(x) : object 'lena' not found

Here is the original full code:

library(pixmap)
lena - read.pnm(oldlennablur.pgm)
write.table(lena@grey,mylenna, quote=FALSE, row.names = FALSE,
col.names= FALSE)
plot(pixmapGrey(as.matrix(read.table(mylenn

do you know why i got the error? i have change the�oldlennablur.pgm
with the one i have and also with the specified path on my computer..|
And would you like give me an explanation about read.pnm. I need to
know what result will i get using this syntax.
Does this syntax give us a 256x256 matrix if we input the 256x256
pixels image?
I will use the data for my project, so I need to know how and what the
data i would get..
Thank you very much..�
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Re: [R] Openbugs- Array Index

2012-10-26 Thread Jeff Newmiller
You seem to be lost. Try 
http://mathstat.helsinki.fi/openbugs/community/CommBBS.html
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SuzieK suzanne.f.ke...@hotmail.co.uk wrote:

Hi, 

I'm working on the codes below however every time I run them when they
get
to OpenBUGS I keep getting the error message: array index is greater
than
array upper bound for hab.

Any help would be greatly appreciated,

Suzie

Codes:
ungulate - read.csv(file.choose ()) #ungulate
ungulate -
as.matrix(ungulate);colnames(ungulate)-NULL;rownames(ungulate)-NULL
habitat - read.csv(file.choose ())   #ungulate habitat
habitat - habitat[,3]
site.data - read.csv(file.choose ()) #ungulate
site
site - site.data$SiteId
visit - site.data$visit
date - site.data$date
date - matrix(date,nrow=4,ncol=5,byrow=TRUE)
S - dim(ungulate)[1]   # set dimension forungulate i.e. number of
species
m - 6 # number of augmented species
G - read.csv(file.choose ()) # ungulate group
G - G[,2]
g - rep(NA,length=m)
G - c(G,g)
g - length(table(G))# number of groups

 # habitat id
Hab - cbind(seq(1,4),habitat)
hab - matrix(NA,nrow=3,ncol=2) #make a matrix with 4 rows and two
colums
for the four sites with 2 different habitat types has NA values??
for(i in 1:2){
hab[i,] - Hab[habitat==i,1]
}
# arrange bird data
S - dim(ungulate)[1]# number of collected species in our case 
Y - array(NA,dim=c(S,4,5))  
for(i in 1:S){
Y[i,,] - matrix(ungulate[i,],nrow=4,ncol=5,byrow=TRUE)   
}
AY - array(NA,dim=c((S+m),4,5))# Y of augmented
for(i in 1:4){
y - matrix(0,nrow=m,ncol=5)  
AY[,i,] - rbind(Y[,i,],y)  
}
## bugs code
library(R2OpenBUGS)
sink(ungulate.txt)
cat(
model{

# hyperparameters
# habitat effects for each functional group
for(i in 1:g){  # number of functional
group
for(j in 1:2){   # number of
habitat
type
mu.h[i,j] ~ dnorm(0,0.0001) I(-2,2)  # filling mu.h with values
based
on a  random distribution
 
sigma.h[i,j] ~ dunif(0,6)
tau.h[i,j] - 1/(sigma.h[i,j]*sigma.h[i,j])
}
}

# detectability
mu.r ~ dnorm(0,0.0001) I(-3,3)
sigma.r ~ dunif(0,6)
tau.r - 1/(sigma.r*sigma.r)

psi ~ dunif(0,1)# inclusion rate that generates wi

# proportion of number of species among groups
for(i in 1:g){
prop[i] ~ dgamma(1,1)
prob[i] - prop[i]/sum(prop[])
 }

for(i in 1:(S+m)){
r[i] ~ dnorm(mu.r,tau.r) I(-2,2)# generating parameters related to
detectability
p[i] - 1/(1+exp(-(r[i])))# individual-level detection probability
w[i] ~ dbern(psi)# indicator variable whether each species is exposed
to
sampling or not
G[i] ~ dcat(prob[1:g])# group identity
for(h in 1:2){# habitat effects
habitat.eff[i,h] ~ dnorm(mu.h[G[i],h],tau.h[G[i],h]) I(-2,2)
 }
for(j in 1:4){# fitting process
# ecological process model
lambda[i,j] -  exp(habitat.eff[i,habitat[j]])
Z[i,j] ~ dpois(lambda[i,j])# latent abundance of each species at each
site
at each visit
A[i,j] - Z[i,j]*w[i]# latent abundance only for species exposed to
sampling
for(v in 1:5){
# detection process model
AY[i,j,v] ~ dbin(p[i],A[i,j])
}
}
# group identity (indicator variable)
G1[i] - equals(G[i],1)
G2[i] - equals(G[i],2)
 }

for(i in 1:(S+m)){
for(j in 1:4){
A1[i,j] - A[i,j]*G1[i]
A2[i,j] - A[i,j]*G2[i]
O[i,j] - step(A[i,j]-1)# latent occupancy of each species for each
site
O1[i,j] - O[i,j]*G1[i]
O2[i,j] - O[i,j]*G2[i]
 }
 }

for(j in 1:4){
AB0[j] - sum(A[,j])
AB1[j] - sum(A1[,j])
AB2[j] - sum(A2[,j])
SpR0[j] - sum(O[,j])
SpR1[j] - sum(O1[,j])
SpR2[j] - sum(O2[,j])
}

for(i in 1:2){
for(j in 1:4){
HabAB0[i,j] - AB0[hab[i,j]]
HabAB1[i,j] - AB1[hab[i,j]]
HabAB2[i,j] - AB2[hab[i,j]]
HabSpR0[i,j] - SpR0[hab[i,j]]
HabSpR1[i,j] - SpR1[hab[i,j]]
HabSpR2[i,j] - SpR2[hab[i,j]]
}   
HAB0[i] - mean(HabAB0[i,])
HAB1[i] - mean(HabAB1[i,])
HAB2[i] - mean(HabAB2[i,])
HSpR0[i] - mean(HabSpR0[i,])
HSpR1[i] - mean(HabSpR1[i,])
HSpR2[i] - mean(HabSpR2[i,])
}

R - sum(w[1:(S+m)])# estimating unknown number of species that occupy
any
sites 

}
,fill=TRUE)
sink()

data - list(m,S,
habitat,
G,g,
#date,
hab,
AY)
inits - function()list(
mu.h=matrix(rnorm(g*2),nrow=2),sigma.h=matrix(runif(g*2),nrow=2),
mu.r=rnorm(1),sigma.r=runif(1),
r=rnorm(S+m),
prop=runif(n=g,min=0,max=5),
Z=array(rpois(n=(S+m)*4,lambda=2),dim=c((S+m),4)),
w=rep(1,(S+m)),psi=runif(1))   
parameters - c(
mu.h,sigma.h,
habitat.eff,
mu.r,sigma.r,
r,
R,psi,
A,
O,
HAB0,HAB1,HAB2,
HSpR0,HSpR1,HSpR2
)

out - bugs(data,inits,parameters,ungulate.txt

Re: [R] using match-type function to return correctly ordered data from a dataframe

2012-10-27 Thread Jeff Newmiller
Have you actually read

?%in%

?

Although a valuable tool, not all answers are most effectively obtained by 
Googling.

Also, your repeated assertions that the answers are not maintained in order are 
poorly framed. They DO stay in order according to the zipcode database order. 
That said, your desire for numeric indexes is only as far away as your help 
file.
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Markus Weisner r...@themarkus.com wrote:

I am regularly running into a problem where I can't seem to figure out
how
maintain correct data order when selecting data out of a dataframe. 
The
below code shows an example of trying to pull data from a dataframe
using
ordered zip codes.  My problem is returning the pulled data in the
correct
order.  This is a very simple example, but it illustrates a regular
problem
that I am running into.

In the past, I have used fairly complicated solutions to pull this off.
There has got to be a more simple and straightforward method ...
probably
some function that I missed in all my googling.

Thanks in advance for anybody's help figuring this out.
~Markus


### Function Definitions ###

# FUNCTION #1 (returns wrong order)
getLatitude1 = function(myzips) {

  # load libraries and data
  library(zipcode)
  data(zipcode)

  # get latitude values
 mylats = zipcode[zipcode$zip %in% myzips, latitude] #problem is that
this code does not maintain order

  # return data
  return(mylats)
}

# FUNCTION #2 (also returns wrong order)
getLatitude2 = function(myzips) {

  # load libraries and data
  library(zipcode)
  data(zipcode)

  # convert myzips to DF
  myzips = as.data.frame(as.character(myzips))

  # merge in zipcode data based on zip
  results = merge(myzips, zipcode[,c(zip, latitude)], by.x =
as.character(myzips), by.y=zip, all.x=TRUE)

  # return data
  return(results$latitude)
}


### Code ###

# specify a set of zip codes
myzips = c(74432, 72537, 06026, 01085, 65793)

# create a DF
myzips.df = data.frame(zip=myzips, latitude=NA, longitude=NA)

# look at data to determine what should be returned and in what order
library(zipcode)
data(zipcode)
zipcode[zipcode$zip %in% myzips,]

# test function #1 (function definition below)
myzips.df$latitude = getLatitude1(myzips.df$zip) #returns wrong order

# test function #2 (function definition below)
myzips.df$latitude = getLatitude2(myzips.df$zip) #also returns wrong
order



# need myzips %in% zipcode$zip to return array/df indices rather than
logical

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Re: [R] help install RExcel

2012-10-27 Thread Jeff Newmiller
For RExcel support, see http://rcom.univie.ac.at, where they have various 
support options including a mailing list.
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tt tiziano.fiore_1...@libero.it wrote:

Hello,
I just installed RExcel but I have some problems. I am writing you the
set
of commands that I used: 

 install.packages(RExcelInstaller, rcom, rsproxy)
 library(rcom)
 comRegisterRegistry()
 library(RExcelInstaller)
 installstatconnDCOM()
 installRExcel()

But when I select RStart from add-ins I see this written No R server
configured. Please set Rserver to work with RExcel. I am using excel
2010.
Can you help me? Thanks



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Re: [R] keep average values and delete duplicate rows

2012-10-28 Thread Jeff Newmiller
Please read the posting guide... this is not a homework help line.
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fuckecon iamsta...@gmail.com wrote:

Hello experts,

I am sorry that my subject line is confusing, because I am confused as
nuts.
Let me take a shot at explaining what I am trying to do.

I have a data set of log GDP, education, democracy index, and a whole
bunch
of variables for every country from 1950 to  2000. Each country
accounts for
10 observations with each observation representing the mean GDP for
each 5
year interval.

Example:

Country  log GDP yr
Canada 9.115211 1950
Canada 9.205848 1955
Canada 9.247975 1960
Canada 9.429002 1965
Canada 9.554069 1970
Canada 9.719351 1975
Canada 9.851376 1980
Canada 9.937892 1985
Canada 10.01457 1990
Canada 10.04093 1995
Canada 10.20005 2000
USA   9.278241950
USA   9.389681955
USA   9.415136  1960
USA   9.594625  1965
USA   9.702071970
USA   9.800418  1975
USA   9.968131980
USA   10.07001  1985
USA   10.18331  1990
USA   10.25446  1995
USA   10.41312000

For log GDP:

I want to create a new object in R with one line for each country and 
the
average log GDP from the 10 5yr interval observations. With the subset
I
want to then create a table with 3 columns and 4 rows.

(I have no idea how to write the codes to create the new object. Friend
said
something about conditional median.)

Columns
1) All countries
2) High income countries
3) Low income countries

Rows
1) Democracy index
2) Log GDP
3) Obs
4) Countries

To create a high and low income columns, I am using the median as the
boundary. (i.e. high income for gdp  median of the mean for each
country,
low income for gdp = median of the mean for each country.)

I hope someone can understand what I am writing here and help me out
with
it.

Thanks so much!





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Re: [R] Having some Trouble Data Structures

2012-10-28 Thread Jeff Newmiller
Search on ragged array. 

My preferred approach is to use a data frame with one row per effector that 
repeats the per-ID information. If that occupies too much memory, you can setup 
another data frame with one row per ID and refer to that information as using 
lapply and subset the effectors data as needed. The plyr package is also useful 
for such processing.
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Benjamin Ward (ENV) b.w...@uea.ac.uk wrote:

Hi All,

I'm trying to run a simulation of host-pathogen evolution based around
individuals.
What I need to have is a dataframe or table of some description -
describing all the individuals of a pathogen population (so far I've
implemented this as a matrix):

 ID No_of_Effectors   Effectors (Sequences)
  [1,] 0001  3   ##   3 Random Numbers ##

There will be many such rows for many individuals. They have something
called effectors, the number of which is randomly generated, so say you
get 3 in the No_of_Effectors column. Then I make R generate 3 numbers
from between 1 and 10,000, this gives me three numerical
representations of genes. These numbers will be compared to a similar
data structure of the host individuals who have their immune genes with
similar numbers.

My problem is that obviously I can't stick 3 numbers in one cell of
the matrix (I've tried) :

Pathogen_Individuals[1,3] - c(2,3,4)
Error in Pathogen_Individuals[1, 3] - c(345, 567, 678) :
  number of items to replace is not a multiple of replacement length

In future I'm also going to have more variables such as whether a gene
is expressed. Such information may require a matrix in itself -
something like:


Effector ID Sequence  Expressed?
 [1,] 0001  345,567,678   1 (or 0).

Is there a way then I can put more than one value in the cell like a
list of values, or a way to put objects in a cell of a data frame,
matrix or table etc. Almost an inception deal - data structures nested
in a data structure? If I search for things like insert list into
matrix I get results like how to turn one into another, which is not
what I think I need to be doing.

I have been considering having several data structures not nested in
each other, something like for every individual create a new matrix
object with the name Effectors_[Individual_ID] and some how get my
simulation loops operating on those objects but I find it hard to see
how to tell R all of those matrices are to be included in an operation,
as you can all lines of a data frame for example with for loops.
This is strange for me because this model was written in a macro-code
for another program which handles data in a different format and layout
to R.

My problem is I think, each individual in the model has many variables
- in this case representations of genes. So I'm having trouble getting
my head about this.

Hopefully someone more experienced will be able to offer advice or a
solution, it will be very appreciated.

Many Thanks,
Ben Ward (ENV, UEA  The Sainsbury Lab, JIC).

P.S. I have searched previous queries to the list, and I'm not sure but
this may be useful for relevant:


Have you thought of using a list?

 a - matrix(1:10, nrow=2)
 b - 1:5
 x - list(a=a, b=b)
 x
$a
 [,1] [,2] [,3] [,4] [,5]
[1,]13579
[2,]2468   10

$b
[1] 1 2 3 4 5

 x$a
 [,1] [,2] [,3] [,4] [,5]
[1,]13579
[2,]2468   10
 x$b
[1] 1 2 3 4 5

oliveoil and yarn datasets have been mentioned.





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Re: [R] PROBLEMS INSTALLING LIFEMETRICS

2012-10-28 Thread Jeff Newmiller
If your goal is to use Excel with R, or if you are having difficulty getting 
Lifehacker to work with Excel, perhaps you should go to 
http://rcom.univie.ac.at/. You don't need Lifehacker to use Excel with R, 
though (as you stated your problem). That just needs Statconn (which is off 
topic here).
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mary marisa_cipri...@hotmail.it wrote:

I am trying to install lifemetrics to use excel with R. Is there anyone
who
knows lifemetrics and can help me to install it?



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Re: [R] find the Best-ticker

2012-10-29 Thread Jeff Newmiller
This is not a question about R, but rather about quantitative financial 
analysis theory that could be applied using any of many different computational 
tools. As such this question is off-topic here.
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sheenmaria sheenmar...@gmail.com wrote:

i need to find the best ticker from the group of some tickers.?
i also need to know on what basis we calculate the best ticker?
   i have some idea about the if the risk rate low, or the market price
high we can say the ticker is best.
   but i dont know is it true.


Anyone can help me .

Thank you



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Re: [R] changing date by +/- days

2012-10-29 Thread Jeff Newmiller
I try to use difftimes as much as possible.  For example,

oneday - as.difftime( 24, units=hours )
sometime - 10 * oneday

You might also find the lubridate package useful.
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chuck.01 charliethebrow...@gmail.com wrote:

got it, sorry!

structure(1243792800, class = c(POSIXct, POSIXt), tzone =
GMT)-(60*60*24*10)




chuck.01 wrote
 Hi, 
 Let say I have a date variable like this:
 
 structure(1243792800, class = c(POSIXct, POSIXt), tzone = GMT)
 [1] 2009-05-31 18:00:00 GMT
 
 How can I make it 10 days prior, so:
 [1] 2009-05-21 18:00:00 GMT
 
 I'm randomly selecting dates from a list  and want a  second value a
set #
 of days either before or after. 
 
 Thanks ahead of time.





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Re: [R] Interaction

2012-10-29 Thread Jeff Newmiller
I thought addition was commutative. I am puzzled by your question.
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Silvano Cesar da Costa silv...@uel.br wrote:

Hi,

I'm fitting a model with 3 variables: A, B and SNP.
The response variable is Y.

I would like fit the following model, in this order:

Y ~ A + B + A*B + SNP

In general, the R sets of the form:

Y ~ A + B + SNP + A*B

How do this?


-
Silvano Cesar da Costa

Universidade Estadual de Londrina
Centro de Ciências Exatas
Departamento de Estatística

Fone: (43) 3371-4346

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Re: [R] Error unary operator

2012-10-29 Thread Jeff Newmiller
a) not reproducible, read the posting guide
b) there is actually a forum dedicated to ggplot2 where you would get better 
answers
c) I have not had luck providing an expression as the data argument to ggplot.
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Jean Jang jeanwaij...@gmail.com wrote:

Hi R - listers,

I am receiving an error. Does anyone know what this means? J

ggplot(subset(foo, Rayos != Rayos.NA), aes(x=HTL, y=DevelopIndex,
colour=TotalEggs)) +geom_point() +geom_jitter() +  
facet_grid(Aeventexhumed ~ Rayos)
+ geom_smooth(method=lm, fill=NA) + ylim(c(0, 7))


Error in +geom_smooth(method = lm, fill = NA) :
   invalid argument to unary operator
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Re: [R] box() doesnt work

2012-10-30 Thread Jeff Newmiller
Not without reproducible code and data.
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penguins cat...@bas.ac.uk wrote:

Hi,

when plotting a graphic i find that the surrounding box disappears if I
adjust the margins with par(mar=..). Ive tried reassigning it with
box() but
it doesnt seem to make any difference. Does anyone know a way to
overcome
this?

Thanks in advance 



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Re: [R] sqlFetch doesn't read the whole objects.

2012-10-30 Thread Jeff Newmiller
Not reproducible, not even close. Other than saying 600 is not a magic limit in 
R, not much help can be given.
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happyji e...@bcm.edu wrote:

Hi everyone,

I connected a data in FileMaker server with odbcConnect.
When I call the data CANCERS using sqlFetch, it looks okay. 
However, the number of obs was different with the actual number.
If I read the same data from Microsoft Access, there are 656 obs. in it
but
srt() shows me 600 obs.

Does anyone know why this happened and how to read all obs. from R?

Thank you in advance.

EJ


 psb-odbcConnect(PSB,uid=XXX,pwd=XXX)
 cancers-sqlFetch(psb,CANCERS)
 str(cancers)
'data.frame':   600 obs. of  5 variables:
 $ study_id   : int  7 152 11 12 3 14 16 22 23 24 ...
$ cancer_type: Factor w/ 121 levels  Breast cancer ,..: 15 24 65 NA
45 16
73 46 23 66 ...
$ cancer_age : Factor w/ 70 levels 17,1968,1983,..: NA 14 31 NA
58 NA
NA 38 NA 35 ...
 $ cancer_recs: num  656 656 656 656 656 656 656 656 656 656 ...
$ cancer_year: Factor w/ 43 levels 1/2003,12/05,..: 34 34 NA 34 NA
34
26 NA 34 34 ...




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Re: [R] POSIXct date missing time component

2012-10-31 Thread Jeff Newmiller
Rather than altering your data, it is better to use the same date time format 
at all times, and specify it explicitly when converting to or from character.
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chuck.01 charliethebrow...@gmail.com wrote:

I did  not read that close enough; thanks again. 




arun kirshna wrote
 Hi,
 
 ?as.POSIXct()
 format: character string giving a date-time format as used by
   ‘strptime’.
 
 ?strptime()
 format: A character string.  The default for the ‘format’ methods is
   ‘%Y-%m-%d %H:%M:%S’ if any component has a time component
   which is not midnight, and ‘%Y-%m-%d’ otherwise.  If
   ‘options(digits.secs)’ is set, up to the specified number
   of digits will be printed for seconds.
 
 So, may be it is better to add a sec and later subtract it.
 
  dates.mine2-dates.mine+1
 dates.mine2[3]
 #[1] 2009-05-22 00:00:01 GMT
  dates.mine3-dates.mine2-1
  dates.mine3[3]
 #[1] 2009-05-22 GMT
 A.K.
 
 
 
 
 
 
 - Original Message -
 From: chuck.01 lt;

 CharlieTheBrown77@

 gt;
 To: 

 r-help@

 Cc: 
 Sent: Tuesday, October 30, 2012 5:21 PM
 Subject: Re: [R] POSIXct  date missing time component
 
 Um, OK. 
 My dates have times, they are 00:00:00 (i.e. midnight)
 I'll just add a fraction of a second my dates and go with it.
 
 Thanks for the reply. 
 
 
 
 
 arun kirshna wrote
 HI,
 Please check this link
 (http://rss.acs.unt.edu/Rdoc/library/base/html/as.POSIXlt.html). 
 Dates without times are treated as being at midnight UTC. 
 
 May be you can try this (if it doesn't create additional problems):
  dates.mine2-dates.mine+1
 dates.mine2
 #[1] 2009-05-21 23:30:01 GMT 2009-05-21 23:45:01 GMT
 #[3] 2009-05-22 00:00:01 GMT 2009-05-22 00:15:01 GMT
  dates.mine2[3]
 #[1] 2009-05-22 00:00:01 GMT
 str(dates.mine2)
  #POSIXct[1:4], format: 2009-05-21 23:30:01 2009-05-21 23:45:01
...
 A.K.
 
 
 
 
 
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Re: [R] mergeing a large number of large .csvs

2012-11-02 Thread Jeff Newmiller
I would first confirm that you need the data in wide format... many algorithms 
are more efficient in long format anyway, and rbind is way more efficient than 
merge.

If you feel this is not negotiable, you may want to consider sqldf. Yes, you 
need to learn a bit of SQL, but it is very well integrated into R.
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Benjamin Caldwell btcaldw...@berkeley.edu wrote:

Dear R help;
I'm currently trying to combine a large number (about 30 x 30) of large
.csvs together (each at least 1 records). They are organized by
plots,
hence 30 X 30, with each group of csvs in a folder which corresponds to
the
plot. The unmerged csvs all have the same number of columns (5). The
fifth
column has a different name for each csv. The number of rows is
different.

The combined csvs are of course quite large, and the code I'm running
is
quite slow - I'm currently running it on a computer with 10 GB ram,
ssd,
and quad core 2.3 ghz processor; it's taken 8 hours and it's only  75%
of
the way through (it's hung up on one of the largest data groupings now
for
an hour, and using 3.5 gigs of RAM.

I know that R isn't the most efficient way of doing this, but I'm not
familiar with sql or C. I wonder if anyone has suggestions for a
different
way to do this in the R environment. For instance, the key function now
is
merge, but I haven't tried join from the plyr package or rbind from
base.
I'm willing to provide a dropbox link to a couple of these files if
you'd
like to see the data. My code is as follows:


#multmerge is based on code by Tony cookson,
http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/;
The function takes a path. This path should be the name of a folder
that
contains all of the files you would like to read and merge together and
only those files you would like to merge.

multmerge = function(mypath){
filenames=list.files(path=mypath, full.names=TRUE)
datalist = try(lapply(filenames,
function(x){read.csv(file=x,header=T)}))
try(Reduce(function(x,y) {merge(x, y, all=TRUE)}, datalist))
}

#this function renames files using a fixed list and outputs a .csv

merepk - function (path, nf.name) {

output-multmerge(mypath=path)
name - list(x, y, z, depth, amplitude)
try(names(output) - name)

write.csv(output, nf.name)
}

#assumes all folders are in the same directory, with nothing else there

merge.by.folder - function (folderpath){

foldernames-list.files(path=folderpath)
n- length(foldernames)
setwd(folderpath)

for (i in 1:n){
path-paste(folderpath,foldernames[i], sep=\\)
 nf.name - as.character(paste(foldernames[i],.csv, sep=))
merepk (path,nf.name)
 }
}

folderpath - yourpath

merge.by.folder(folderpath)


Thanks for looking, and happy friday!



*Ben Caldwell*

PhD Candidate
University of California, Berkeley

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Re: [R] mergeing a large number of large .csvs

2012-11-03 Thread Jeff Newmiller
On the absence of any data examples from you per the posting guidelines, I will 
refer you to the help files for the melt function in the reshape2 package.  
Note that there can be various mixtures of wide versus long... such as a wide 
file with one date column and columns representing all stock prices and all 
trade volumes. The longest format would be what melt gives (date, column name, 
and value) but an in-between format would have one distinct column each for 
dollar values and volume values with a column indicating ticker label and of 
course another for date.

If your csv files can be grouped according to those with similar column 
types, then as you read them in you can use cbind( csvlabel=somelabel, 
csvdf) to distinguish it and then rbind those data frames together to create an 
intermediate-width data frame. When dealing with large amounts of data you will 
want to minimize the amount of reshaping you do, but it would require knowledge 
of your data and algorithms to say any more.
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Benjamin Caldwell btcaldw...@berkeley.edu wrote:

Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:

 I would first confirm that you need the data in wide format... many
 algorithms are more efficient in long format anyway, and rbind is way
more
 efficient than merge.

 If you feel this is not negotiable, you may want to consider sqldf.
Yes,
 you need to learn a bit of SQL, but it is very well integrated into
R.

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 Benjamin Caldwell btcaldw...@berkeley.edu wrote:

 Dear R help;
 I'm currently trying to combine a large number (about 30 x 30) of
large
 .csvs together (each at least 1 records). They are organized by
 plots,
 hence 30 X 30, with each group of csvs in a folder which corresponds
to
 the
 plot. The unmerged csvs all have the same number of columns (5). The
 fifth
 column has a different name for each csv. The number of rows is
 different.
 
 The combined csvs are of course quite large, and the code I'm
running
 is
 quite slow - I'm currently running it on a computer with 10 GB ram,
 ssd,
 and quad core 2.3 ghz processor; it's taken 8 hours and it's only 
75%
 of
 the way through (it's hung up on one of the largest data groupings
now
 for
 an hour, and using 3.5 gigs of RAM.
 
 I know that R isn't the most efficient way of doing this, but I'm
not
 familiar with sql or C. I wonder if anyone has suggestions for a
 different
 way to do this in the R environment. For instance, the key function
now
 is
 merge, but I haven't tried join from the plyr package or rbind from
 base.
 I'm willing to provide a dropbox link to a couple of these files if
 you'd
 like to see the data. My code is as follows:
 
 
 #multmerge is based on code by Tony cookson,
 

http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/
 ;
 The function takes a path. This path should be the name of a folder
 that
 contains all of the files you would like to read and merge together
and
 only those files you would like to merge.
 
 multmerge = function(mypath){
 filenames=list.files(path=mypath, full.names=TRUE)
 datalist = try(lapply(filenames,
 function(x){read.csv(file=x,header=T)}))
 try(Reduce(function(x,y) {merge(x, y, all=TRUE)}, datalist))
 }
 
 #this function renames files using a fixed list and outputs a .csv
 
 merepk - function (path, nf.name) {
 
 output-multmerge(mypath=path)
 name - list(x, y, z, depth, amplitude)
 try(names(output) - name)
 
 write.csv(output, nf.name)
 }
 
 #assumes all folders are in the same directory, with nothing else
there
 
 merge.by.folder - function (folderpath){
 
 foldernames-list.files(path=folderpath)
 n- length(foldernames)
 setwd(folderpath)
 
 for (i in 1:n){
 path-paste(folderpath,foldernames[i], sep

Re: [R] Having some Trouble Data Structures

2012-11-04 Thread Jeff Newmiller
Please keep mail threads on the mailing list. Please follow the posting 
guidelines and provide a sample of data and desired outcome.
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Benjamin Ward (ENV) b.w...@uea.ac.uk wrote:

Hi,
Thank you very much for your reply - how you prefer, is how my
supervisor implemented the layout in Minitab, however I was unsure of
how to get R to do this repeating ID behaviour and how to know that in
a for loop going through individual 1 to say 10, I want it to: 

Randomly sample a number from a distribution for the number of
effectors (I can do this but with runif), 

Then put one value in a cell of the Effector column and repeat the ID
for each effector row. I'm also then left wondering when I do for loops
then that use ID, will it go and apply operations row by row, or ID by
ID - for example in the immunology part I would need a loop to check
individual by individual if any of the effectors it has means death in
the host, in which case all instances of - say ID 1 would need to be
deleted.

Would you be able to provide an example chunk of how you accomplish
this with your preferred approach, if you have the time?

Thanks,
Ben W.


From: Jeff Newmiller [jdnew...@dcn.davis.ca.us]
Sent: 28 October 2012 15:27
To: Benjamin Ward (ENV); r-help@r-project.org
Subject: Re: [R] Having some Trouble Data Structures

Search on ragged array.

My preferred approach is to use a data frame with one row per effector
that repeats the per-ID information. If that occupies too much memory,
you can setup another data frame with one row per ID and refer to that
information as using lapply and subset the effectors data as needed.
The plyr package is also useful for such processing.
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Benjamin Ward (ENV) b.w...@uea.ac.uk wrote:

Hi All,

I'm trying to run a simulation of host-pathogen evolution based around
individuals.
What I need to have is a dataframe or table of some description -
describing all the individuals of a pathogen population (so far I've
implemented this as a matrix):

 ID No_of_Effectors   Effectors
(Sequences)
  [1,] 0001  3   ##   3 Random Numbers ##

There will be many such rows for many individuals. They have something
called effectors, the number of which is randomly generated, so say
you
get 3 in the No_of_Effectors column. Then I make R generate 3 numbers
from between 1 and 10,000, this gives me three numerical
representations of genes. These numbers will be compared to a similar
data structure of the host individuals who have their immune genes
with
similar numbers.

My problem is that obviously I can't stick 3 numbers in one cell of
the matrix (I've tried) :

Pathogen_Individuals[1,3] - c(2,3,4)
Error in Pathogen_Individuals[1, 3] - c(345, 567, 678) :
  number of items to replace is not a multiple of replacement length

In future I'm also going to have more variables such as whether a gene
is expressed. Such information may require a matrix in itself -
something like:


Effector ID Sequence  Expressed?
 [1,] 0001  345,567,678   1 (or
0).

Is there a way then I can put more than one value in the cell like a
list of values, or a way to put objects in a cell of a data frame,
matrix or table etc. Almost an inception deal - data structures nested
in a data structure? If I search for things like insert list into
matrix I get results like how to turn one into another, which is not
what I think I need to be doing.

I have been considering having several data structures not nested in
each other, something like for every individual create a new matrix
object with the name Effectors_[Individual_ID] and some how get my
simulation loops operating on those objects but I find it hard to see
how to tell R all of those matrices are to be included in an
operation

Re: [R] Excluding fixed number of rows from calculation while summarizing using ddply() function.

2012-11-05 Thread Jeff Newmiller
This is not a free do-my-work-for-me forum, this is a help-you-learn-R forum. 
If you study the solution you have been given until you understand how it 
works, and combine that with an understanding of indexing from the Introduction 
to R document that comes with R, you should be able to complete your code the 
way you want it. Don't abuse the generosity of your helpers.
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siddu479 onlyfordigitalst...@gmail.com wrote:

Hi Arun,

Thanks for your reply but your script is removing only one row( first
row
and last row) for each Unique and StepNo combination and calculating
mean
for the rest of rows.
For below data , your script removing the #'s rows perfectly.
But in reality I may need to ignore *say first 10 rows and last 20 rows
for
each Unique and StepNo combination.
* for statistics calculation.
 Unique StepNo Data1 Data2
1  A  1 4 5  #Your script removing this row
successfully.
2  A  1 5 6
3  A  1 7 8
4  A  1 3 4  
5  A  1 1 1  #Your script removing this row
successfully.
6  B  1 2 4  #Your script removing this row
successfully.
7  B  1 3 5   
8  B  1 4 5
9  B  1 5 6
10B  1 6 7  #Your script removing this row
successfully.

Can you modify your script to get my requirement like below (making it
generic, here *N=2*, removing first 2 lines and last 2 lines..
*sometimes I
may have two numbers N1  N2 (no.of rows need to be removed from and
top and
bottom respectively*)

 Unique StepNo Data1 Data2
1  A  1 4 5  #Ignore this
2  A  1 5 6  #Ignore this
3  A  1 7 8
4  A  1 3 4  #Ignore this
5  A  1 1 1  #Ignore this
6  B  1 2 4  #Ignore this
7  B  1 3 5  #Ignore this
8  B  1 4 5
9  B  1 5 6 #Ignore this
10B  1 6 7 #Ignore this

and then calculate the statistics using ddply.

I hope my problem statement is much clear now.







-
Sidda
Business Analyst Lead
Applied Materials Inc.

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Re: [R] Aggregate data frame across columns

2012-11-07 Thread Jeff Newmiller
The best way to get an answer is to provide sample data and desired results 
(computed by hand or by any other available means). See 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example.
 

In the vague question begets a vague answer department, I would use melt from 
the reshape package, and make some columns that contain unique key values for 
each time interval, and then use aggregate or ddply from the plyr package.
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Keith Weintraub kw1...@gmail.com wrote:

Folks,
  I have a data frame with columns 200401, 200402, ..., 201207, 201208.

These represent years/months. What would be the best way to sum these
columns by year? What about by quarter?

Thanks for your time,
KW

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Re: [R] save/load and package namespaces

2012-11-07 Thread Jeff Newmiller
Stop being surprised. Loaded packages are not part of envir (whatever that 
is), nor are they part of the global environment.  You have to reload any 
packages needed separately from the load call.
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Jamie Olson inspired2apa...@gmail.com wrote:

Could someone explain to me what namespaces are loaded/saved when
objects
are saved?

Specifically, I'm using this:
 save(list = ls(all.names = TRUE, envir = envir), file = name, envir =
envir)
to save out everything from an environment.

Later, loading it on another machine, I'm surprised to see the load
failing
for being unable to load certain packages.  Could anyone help me
understand
why this happens?
Jamie Olson

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Re: [R] date and time coding question

2013-05-16 Thread Jeff Newmiller
I stand by my previous statement: you should avoid thinking of difftime objects 
as having units. At the point when you learn to regard the units argument as 
being associated with the numeric values you pass TO as.difftime or that you 
extract FROM a difftime object with as.numeric, then you will understand 
POSIXct arithmetic.
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MacQueen, Don macque...@llnl.gov wrote:

The difftime() function has a units argument, so it seems kind of
obvious
to me to think of objects of class 'difftime' as having units. And
indeed
they do, stored as an attribute:

 str(difftime( Sys.time()+3700, Sys.time(), units='min'))
Class 'difftime'  atomic [1:1] 61.7
  ..- attr(*, units)= chr mins

 str(difftime( Sys.time()+3700, Sys.time(), units='sec'))
Class 'difftime'  atomic [1:1] 3700
  ..- attr(*, units)= chr secs

The numeric values are already in the specified units, prior to any
printing or other further use of the result.


?difftime says,

-quote-
If 'units = auto', a suitable set of units is chosen, the
largest possible (excluding 'weeks') in which all the absolute
differences are greater than one.


-end quote-

So it's pretty clear that the units are chosen when the difference is
calculated, not later. Inspection of the difftime() function source
code
confirms this. One can also inspect print.difftime() and see that the
numeric value is used as is, without any unit conversion.


Which returns me to my original point: when using diff() on a POSIXt
object, I don't know how to specify the units, whereas with difftime()
I
do. If there is a way to specify units using diff(), I'd love to know
about it!


-Don

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Re: [R] Comma separated vector

2013-05-17 Thread Jeff Newmiller
You are wrong... since the SQL query you wish to create is itself a string.

Of course, you cannot send a SQL fragment such as you used as an example, so be 
sure to form a complete, syntactically correct SQL statement before giving it 
to your database query function.

Oh, and if you need more assistance with this topic, then you should probably 
post in the R-sig-DB mailing list. Please also keep in mind that the actual SQL 
syntax used for your specific database is basically off-topic in any R forum, 
so you may need other help resources as well.
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Manta mantin...@libero.it wrote:

Hi all,

I have a vector of numbers, and to be able to pass it to RMySQL and use
the
IN clause I need to have this vector to be a single list numeric and
comma
separated.

I saw the post below but it is about strings, which I do not need (I
cannot
pass strings in this SQL query, I need something like ' where ASSETT in
(1,2,3,4,5)'

http://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector


Any clue?



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Re: [R] as.vector with mode=list and POSIXct

2013-05-20 Thread Jeff Newmiller
I don't know what you plan to do with this list, but lists are quite a bit less 
efficient than fixed-mode vectors, so you are likely losing a lot of 
computational speed by using this list. I don't hesitate to use simple data 
frames (lists of vectors), but processing lists is on par with for loops, not 
vectorized computation. It may still support a simpler model of computation, 
but that is an analyst comprehension benefit rather than a computational 
efficiency benefit.
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Alexandre Sieira alexandre.sie...@gmail.com wrote:

I was trying to convert a vector of POSIXct into a list of POSIXct,
However, I had a problem that I wanted to share with you.

Works fine with, say, numeric:


 v = c(1, 2, 3)
 v
[1] 1 2 3
 str(v)
 num [1:3] 1 2 3
 l = as.vector(v, mode=list)
 l
[[1]]
[1] 1

[[2]]
[1] 2

[[3]]
[1] 3

 str(l)
List of 3
 $ : num 1
 $ : num 2
 $ : num 3

If you try it with POSIXct, on the other hand…


 v = c(Sys.time(), Sys.time())
 v
[1] 2013-05-20 18:02:07 BRT 2013-05-20 18:02:07 BRT
 str(v)
 POSIXct[1:2], format: 2013-05-20 18:02:07 2013-05-20 18:02:07
 l = as.vector(v, mode=list)
 l
[[1]]
[1] 1369083728

[[2]]
[1] 1369083728

 str(l)
List of 2
 $ : num 1.37e+09
 $ : num 1.37e+09

The POSIXct values are coerced to numeric, which is unexpected.

The documentation for as.vector says: The default method handles 24
input types and 12 values of type: the details of most coercions are
undocumented and subject to change. It would appear that treatment for
POSIXct is either missing or needs adjustment.

Unlist (for the reverse) is documented to converting to base types, so
I can't complain. Just wanted to share that I ended up giving up on
vectorization and writing the two following functions:


unlistPOSIXct - function(x) {
  retval = rep(Sys.time(), length(x))
  for (i in 1:length(x)) retval[i] = x[[i]]
  return(retval)
}

listPOSIXct - function(x) {
  retval = list()
  for (i in 1:length(x)) retval[[i]] = x[i]
  return(retval)
}

Is there a better way to do this (other than using *apply instead of
for above) that better leverages vectorization? Am I missing something
here?

Thanks!




-- 
Alexandre Sieira
CISA, CISSP, ISO 27001 Lead Auditor

The truth is rarely pure and never simple.
Oscar Wilde, The Importance of Being Earnest, 1895, Act I



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Re: [R] Lattice, ggplot, and pointsize

2013-05-21 Thread Jeff Newmiller
That is like complaining that your hammer does not fit these newfangled Philips 
screws.

These are different tools. Do not expect them to interoperate.
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Milan Bouchet-Valat nalimi...@club.fr wrote:

Le mardi 21 mai 2013 à 23:30 +1000, Duncan Mackay a écrit :
 Hi
 
 See par.settings in xyplot
 
 Things are also controlled by
 trellis.par.get()
 to see values
 trellis.par.set()
 
 eg
 xyplot(~Freq|Year, data = sheep2,
 groups   = farm,
 par.settings = list(strip.background = list(col =
transparent),
 axis.text = list(cex = 0.75),
 par.xlab.text = list(cex = 0.80),
 par.ylab.text = list(cex = 0.80)) ,
...)
 
 HTH
Thanks, but that's not really my question. I've already found the way
to
change text size. What I'm wondering is whether something could be done
so that the pointsize argument that is passed to graphical devices has
an effect on Lattice and ggplot2 plots.


Regards


 Duncan
 
 Duncan Mackay
 Department of Agronomy and Soil Science
 University of New England
 Armidale NSW 2351
 Email: home: mac...@northnet.com.au
 
 
 
 At 21:18 21/05/2013, you wrote:
 Hi!
 
 When inserting R plots into a document using odfWeave, I fought for
a
 while to get Lattice plots use the same text size as base plots. I
 eventually discovered that specifying a point size via e.g.
 svg(pointsize=10) has no effect on Lattice plots. One needs to
adjust
 the size manually via:
 trellis.par.set(fontsize=list(text=10, points=8))
 
 This is also developed for both Lattice and ggplot2 by this blog
post:
 http://gforge.se/2013/03/exporting-plain-lattice-or-ggplot/
 
 So I am wondering whether is a by-design limitation or whether this
 could be improved. I find it very useful to be able to adapt text
size
 to the output device instead of changing plotting parameters for
each
 plotting system (especially when you change the resolution of PNG
 output, or move from one output device to another).
 
 Thanks in advance
 
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Re: [R] as.vector with mode=list and POSIXct

2013-05-21 Thread Jeff Newmiller
I recommend that you not plan on waiting for the hash package to be redesigned 
to meet your expectations. Also, your response to discovering this feature of 
the hash package seems illogical.

From a computer science perspective, the hash mechanism is an implementation 
trick that is intended to improve lookup speed. It does not actually represent 
a fundamental data structure like a vector or a set does. You can always put 
your keys in a vector and search through them (e.g. vector indexing by string) 
to get an equivalent data retrieval. If the hash package is not improving the 
speed of your data access, adding an extra layer of data structure is hardly 
an appropriate solution.

Why are you not using normal vectors or data frames and accessing with string 
or logical indexing?

If you are avoiding vectors because they seem slow in loops, perhaps you just 
need to preallocate the vectors you will store your results in before your loop 
to regain acceptable speed. Or, perhaps the duplicated() or merge() functions 
could save you from this mess of incremental data processing.
---
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Alexandre Sieira alexandre.sie...@gmail.com wrote:

You are absolutely right.

I am storing POSIXct objects into a hash (from the hash package).
However, if I try to get them out as a vector using the values()
function, they are unclassed. And that breaks my (highly vectorized)
code. Take a look at this:


 h = hash()
 h[[a]] = Sys.time()
 str(h[[a]])
 POSIXct[1:1], format: 2013-05-20 16:54:28
 str(values(h))
 Named num 1.37e+09
 - attr(*, names)= chr a


I have reported this to the hash package maintainers. In the meantime,
however, I am storing, for each key, a list containing a single
POSIXct. Then, when I extract all using values(), I get a list
containing all POSIXct entries with class preserved. 


 h = hash()
 h[[a]] = list( Sys.time() )
 h[[b]] = list( Sys.time() )
 h[[c]] = list( Sys.time() )
 values(h)
$a
[1] 2013-05-21 09:54:03 BRT

$b
[1] 2013-05-21 09:54:07 BRT

$c
[1] 2013-05-21 09:54:11 BRT

 str(values(h))
List of 3
 $ a: POSIXct[1:1], format: 2013-05-21 09:54:03
 $ b: POSIXct[1:1], format: 2013-05-21 09:54:07
 $ c: POSIXct[1:1], format: 2013-05-21 09:54:11


However, the next thing I need to do is a min() over that list, so I
need to convert the list into a vector again.

I agree completely with you that this is horrible for performance, but
it is a temporary workaround until values() is fixed.

-- 
Alexandre Sieira
CISA, CISSP, ISO 27001 Lead Auditor

The truth is rarely pure and never simple.
Oscar Wilde, The Importance of Being Earnest, 1895, Act I
On 20 de maio de 2013 at 19:40:14, Jeff Newmiller
(jdnew...@dcn.davis.ca.us) wrote:
I don't know what you plan to do with this list, but lists are quite a
bit less efficient than fixed-mode vectors, so you are likely losing a
lot of computational speed by using this list. I don't hesitate to use
simple data frames (lists of vectors), but processing lists is on par
with for loops, not vectorized computation. It may still support a
simpler model of computation, but that is an analyst comprehension
benefit rather than a computational efficiency benefit.  
---
 
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---
 
Sent from my phone. Please excuse my brevity.  

Alexandre Sieira alexandre.sie...@gmail.com wrote:  

I was trying to convert a vector of POSIXct into a list of POSIXct,  
However, I had a problem that I wanted to share with you.  
  
Works fine with, say, numeric:  
  
  
 v = c(1, 2, 3)  
 v  
[1] 1 2 3  
 str(v)  
 num [1:3] 1 2 3  
 l = as.vector(v, mode=list)  
 l  
[[1]]  
[1] 1  
  
[[2]]  
[1] 2  
  
[[3]]  
[1] 3  
  
 str(l)  
List of 3  
 $ : num 1  
 $ : num 2  
 $ : num 3  
  
If you try it with POSIXct, on the other hand…  
  
  
 v = c(Sys.time(), Sys.time())  
 v  
[1] 2013-05-20 18:02:07 BRT 2013-05-20 18:02:07 BRT  
 str(v)  
 POSIXct[1:2], format: 2013-05-20 18:02:07 2013-05-20 18:02:07  
 l = as.vector(v, mode=list)  
 l  
[[1]]  
[1] 1369083728  
  
[[2]]  
[1] 1369083728  
  
 str(l)  
List of 2  
 $ : num 1.37e+09  
 $ : num 1.37e+09

Re: [R] as.vector with mode=list and POSIXct

2013-05-22 Thread Jeff Newmiller
My perception of illogic was in your addition of more data structure complexity 
when faced with this difficulty. R has best performance when calculations are 
pushed into simple typed vectors where precompiled code can handle the majority 
of the work. These are simpler structures, not more complex structures. It 
seems like you are fighting the natural paradigm for working in R and holding 
fast to your ideas about how things should be rather than dealing with how 
they are by introducing lists rather than working with vectors or data frames.
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--- 
Sent from my phone. Please excuse my brevity.

Alexandre Sieira alexandre.sie...@gmail.com wrote:

Hi, Jeff.

Thanks for your thoughtful suggestions.

I do not plan to wait for the hash package to be redesigned to meet my
expectations. As a matter of fact, I have:

a) Submitted a report of unexpected behavior in hash::values, which the
package maintainer quickly replied to and said would examine. 
b) Designed (with the help of this list) and implemented a workaround
in the form of wrapping the POSIXct objects in lists, which has my
program working correctly for now. 

If the hash package is updated and the  workaround is no longer
necessary, then I'll reverse this change. Otherwise, I'll look more
deeply into my alternatives which might involve maintaining this
workaround permanently, or analyzing alternative architectures.

The hash package is a beautiful piece of code that is working perfectly
for me in many situations. Even with the list wrapping around the
POSIXct objects, it is meeting my performance requirements much better
than the alternatives I tested. So I'd rather not completely
re-engineer working complex code without a very good reason.

However, I would like to respectfully disagree with you that my
reaction to hash::values behavior was illogical. I don't want to start
a flame war or anything, so let's try to keep the discussion civil. :)

See, a hash table (or a queue, or a stack, or an R vector) is a data
structure that works as a container. You insert objects and you get
them back according to the specificities of each data structure (stacks
will have a FILO ordering, queues will have FIFO ordering, hashes will
maintain key/value pairs, and so).

It is completely unreasonable to insert an object of class X into a
container, and then get it back altered in a way that is not part of
the 'contract' behind the data structure. If I assign X to key K on a
hash, however I choose to ask the hash for the value associated with
key K back, I should get exactly X as a response. I believe most
computer scientists would agree that to be self-evident.

And that is to be expected by reading hash::values documentation:

   Extract values from a hash object. This is a pseudo- accessor method
that returns hash values (without keys) as a vector if possible, a list
otherwise.


Moreover, it has this to say about non-primitive types:

   If the values are of different types or of a complex class than a
named list is returned. 


It never says it will unclass objects, or coerce them into primitive
types. Hence the 'contract' implies I will get back what I inserted,
unaltered, either in a vector or a list. And that is provably not what
is happening. I would have been ok with a vector of POSIXct or a named
list containing the POSIXct values, but instead I am getting a numeric
vector.

I understand R is based on S, and that OOP concepts were introduced
later into its history. However, one of the key concepts in OOP is
encapsulation - as an outside entity you do not get to see the internal
implementation of a class, you interact with it exclusively through its
published interface (method, public member variables, etc). 

I cannot find any justification as for why an object losing its class
unintentionally is ever acceptable, as it violates the concept of
encapsulation. That is essentially what's happening if I look up
several keys using values(). So this violates the encapsulation of the
POSIXct class, as I am exposed to its internal numeric value. Moreover,
it breaks the method-dispatch of R functions that know to treat
POSIXct values differently. All of a sudden, the POSIXct objects I
inserted are being treated, for example, by format as numeric instead
of being dispatched to format.Date as expected.

So I don't think my reaction to this issue was illogical at all. Hope
you'll agree now that I've explained myself a little better. :)

-- 
Alexandre

Re: [R] group data based on row value

2013-05-22 Thread Jeff Newmiller
dat$group - cut( dat$Var, breaks=c(-Inf,0.1, 0.6,Inf))
levels(dat$group) - LETTERS[1:3]

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Ye Lin ye...@lbl.gov wrote:

hey, I want to divide my data into three groups based on the value in
one
column with group name.

dat:

Var
0
0.2
0.5
1
4
6

I tried:

dat - cbind(dat, group=cut(dat$Var, breaks=c(0.1,0.6)))

But it doesnt work, I want to group those 0.1 as group A, 0.1-0.6 as
group
B, 0.6 as group C

Thanks for your help!

   [[alternative HTML version deleted]]

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Re: [R] sample(c(0, 1)...) vs. rbinom

2013-05-22 Thread Jeff Newmiller
You seem to be building an elaborate structure for testing the reproducibility 
of the random number generator. I suspect that rbinom is calling the random 
number generator a different number of times when you pass prob=0.5 than 
otherwise.
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Michael Hannon jm_han...@yahoo.com wrote:

Greetings.  My wife is teaching an introductory stat class at UC
Davis.  The
class emphasizes the use of simulations, rather than mathematics, to
get
insight into statistics, and R is the mandated tool.   A student in the
class
recently inquired about different approaches to sampling from a
binomial
distribution.  I've appended some code that exhibits the idea, the gist
of
which is that using sample(c(0, 1), ...) and rbinom(...) should give
equivalent results.

The surprising (to me) result is that the two approaches DO give the
same
result, EXCEPT when the probability is exactly 0.5.  See Appendix A for
the
code and Appendix B for the output.  I don't think this issue is
system-dependent, but I've put my session information in Appendix C.

Another wrinkle in this is that if I omit the prob parameter from the
call
to sample, meaning to take the default value of 0.5, the two methods DO
give
the same result.

Any thoughts about this?  Thanks.

--Mike

Appendix A: some R code that exhibits the problem
=

ppp - seq(0, 1, by = 0.01)

result - do.call(rbind, lapply(ppp, function(p) {
    set.seed(1)
    sampleRes - sample(c(0, 1), size = 1, replace = TRUE,
    prob=c(1-p, p))
    
    set.seed(1)
    rbinomRes - rbinom(1, size = 1, prob = p)
    
    data.frame(prob = p, equivalent = all(sampleRes == rbinomRes))
    
}))

result


Appendix B: the output from the R code
==

    prob equivalent
1   0.00   TRUE
2   0.01   TRUE
3   0.02   TRUE
4   0.03   TRUE
5   0.04   TRUE
6   0.05   TRUE
7   0.06   TRUE
8   0.07   TRUE
9   0.08   TRUE
10  0.09   TRUE
11  0.10   TRUE
12  0.11   TRUE
13  0.12   TRUE
14  0.13   TRUE
15  0.14   TRUE
16  0.15   TRUE
17  0.16   TRUE
18  0.17   TRUE
19  0.18   TRUE
20  0.19   TRUE
21  0.20   TRUE
22  0.21   TRUE
23  0.22   TRUE
24  0.23   TRUE
25  0.24   TRUE
26  0.25   TRUE
27  0.26   TRUE
28  0.27   TRUE
29  0.28   TRUE
30  0.29   TRUE
31  0.30   TRUE
32  0.31   TRUE
33  0.32   TRUE
34  0.33   TRUE
35  0.34   TRUE
36  0.35   TRUE
37  0.36   TRUE
38  0.37   TRUE
39  0.38   TRUE
40  0.39   TRUE
41  0.40   TRUE
42  0.41   TRUE
43  0.42   TRUE
44  0.43   TRUE
45  0.44   TRUE
46  0.45   TRUE
47  0.46   TRUE
48  0.47   TRUE
49  0.48   TRUE
50  0.49   TRUE
51  0.50  FALSE
52  0.51   TRUE
53  0.52   TRUE
54  0.53   TRUE
55  0.54   TRUE
56  0.55   TRUE
57  0.56   TRUE
58  0.57   TRUE
59  0.58   TRUE
60  0.59   TRUE
61  0.60   TRUE
62  0.61   TRUE
63  0.62   TRUE
64  0.63   TRUE
65  0.64   TRUE
66  0.65   TRUE
67  0.66   TRUE
68  0.67   TRUE
69  0.68   TRUE
70  0.69   TRUE
71  0.70   TRUE
72  0.71   TRUE
73  0.72   TRUE
74  0.73   TRUE
75  0.74   TRUE
76  0.75   TRUE
77  0.76   TRUE
78  0.77   TRUE
79  0.78   TRUE
80  0.79   TRUE
81  0.80   TRUE
82  0.81   TRUE
83  0.82   TRUE
84  0.83   TRUE
85  0.84   TRUE
86  0.85   TRUE
87  0.86   TRUE
88  0.87   TRUE
89  0.88   TRUE
90  0.89   TRUE
91  0.90   TRUE
92  0.91   TRUE
93  0.92   TRUE
94  0.93   TRUE
95  0.94   TRUE
96  0.95   TRUE
97  0.96   TRUE
98  0.97   TRUE
99  0.98   TRUE
100 0.99   TRUE
101 1.00   TRUE

Appendix C: Session information
===

 sessionInfo()
R version 3.0.0 (2013-04-03)
Platform: x86_64-redhat-linux-gnu (64-bit)

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

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

Re: [R] Could graph objects be stored in a two-dimensional list?

2013-05-23 Thread Jeff Newmiller
You could use lists of lists, and index them with vectors.

a - list()
a[[1]] - list()
a[[2]] - list()
a[[c(1,1)]] - g11
a[[c(1,2)]] - g12
a[[c(2,1)]] - g21
a[[c(2,2)]] - g22
print(a[[c(2,1)]])

but this seems like an inefficient use of memory because your indexed data is 
stored more compactly than the graph object is. I would index the data and 
generate the graph object on the fly when I wanted to see it.
---
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--- 
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jpm miao miao...@gmail.com wrote:

Hi,

I have a few graph objects created by some graphic package (say,
ggplot2,
which I use frequently). Because of the existent relation between the
graphs, I'd like to index them in two dimensions as p[1,1], p[1,2],
p[2,1],
p[2,2] for convenience.

To my knowledge, the only data type capable of storing graph objects
(and
any R object) is list, but unfortunately it is available in only one
dimension. Could the graphs be stored in any two-dimensional data type?

  One remedy that comes to my mind is to build a function f so that
f(1,1)=1
f(1,2)=2
f(2,1)=3
f(2,2)=4
With functions f and f^{-1} (inverse function of f) , the
two-dimensional
indices could be mapped to and from a set of one-dimensional indices,
and
the functions are exactly the way R numbers elements in a matrix. Does
R
have this built-in function for a m by n matrix or more generally,
m*n*p
array? (I know this function is easy to write, but just want to make
sure
whether it exists already)

   Thanks,

Miao

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Re: [R] ggmap, hexbin and length distortion in Lat/long

2013-05-24 Thread Jeff Newmiller
I would suggest asking this on R-sig-geo.

https://stat.ethz.ch/mailman/listinfo/r-sig-geo
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Mike Bock mb...@environcorp.com wrote:

I am working with spatial data in ggmap, generally with great success.
I have a huge data set with the coordinates in NAD 83 UTM Zone 11
(meters).  To map the data the coordinates were converted to Lat Long
in GIS prior to use in R and ggmap/ggplot. I am using hexagonal binning
to aggregate the data :
#create bins and calculate stats

hb-hexbin(DF$lon,DF$lat,xbins=80,IDs=TRUE)
hb.avg-hexTapply(hb,DF$Res,mean,na.rm=TRUE)
hb.mx-hexTapply(hb,DF$Res,max,na.rm=TRUE)
hb.p80-hexTapply(hb,DF$Res,quantile,prob=0.80,na.rm=TRUE)
#create df for ggplot
hx_dat - data.frame(hcell2xy(hb), count = hb@count,
xo = hb@xcm, yo = hb@ycm, Mean=hb.avg,Max=hb.mx,
p80=hb.p80)

#Base Map
#BBox is the bounding box
Base-get_map(BBox,source='google')
m_hx-ggmap(Base,legend = bottom,
base_layer=ggplot(aes(x=x,y=y),data=hx_dat))
#Map of means
a-0.55
hc-'grey60'

m_hx+geom_hex(aes(x = x, y = y, fill = Mean),
  color = hc, ,alpha=a,stat = identity) +
  scale_fill_gradientn(Mean,colours=rev(rainbow(4)),trans='sqrt')

...and so on for other stats
I can also run statistical analyses on hx_dat.
By creating hexbins based on lat/long it seems there will be distortion
due to the differences in length of a degree at different locations on
the earth's surface. What is the most efficient way to eliminate this
distortion? Should I run hexbin in NAD83 and convert the x/y
coordinates to Lat Long? Can I get ggmap to convert the baselayer to
NAD84 and just do everything in NAD(my preferred option)?
I have tried converting Lat Long to NAD84 and back but the coordinates
are coming up in the eastern Pacific and not in California, so I am
missing something and I am not sure that is the best way to solve the
problem anyway. Thanks in advance, any help is greatly appreciated
Mike

Michael J. Bock, PhD | Senior Manager
mb...@environcorp.com







This message contains information that may be
confidenti...{{dropped:8}}

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Re: [R] Parameters estimation for extreme value models

2013-05-26 Thread Jeff Newmiller
Well, then, you had better get busy and stop posting here. To learn why, read 
the Posting Guide. Some pointers:

a) No homework help here.
b) No posting in HTML.
c) This list is for questions about R, not statements about your needs.

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assaedi76 assaedi76 assaed...@yahoo.com wrote:

Thanks in advance R
users

�

I have time series data
and I need to estimate the parameters involved in three different
models for
generalized extreme values

�

Model 1: �a, b,
c�are constants.

�

Model 2:�a(t)=B0+B1 t,
but� b, c�are constants

�

Model 3:�c(t)=
Exp(B0+B1 t) �but�a, b�are constants

�

Where a, b�and�c�are location, scale and
shape parameter respectively; t is time.







Regards�
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Re: [R] curiosity: next-gen x86 processors and FP32?

2013-05-26 Thread Jeff Newmiller
I am no HPC expert, but I have been computing for awhile.

There are already many CPU-specific optimizations built into most compilers 
used to compile the R source code. Anyone sincerely interested in getting work 
done today should get on with their work and hope that most of the power of new 
processors gets delivered the same way.

The reason single precision is so uncommon in many computing environments is 
that numerical errors propagate much faster with single precision. I don't 
expect the typical R user to want to perform detailed uncertainty analysis 
every time they set up a computation to decide whether it can be computed with 
sufficient accuracy using SP.

Most speed problems I have encountered have been related to memory (swapping, 
fragmentation) and algorithm inefficiency, not CPU speed.
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ivo welch ivo.we...@anderson.ucla.edu wrote:

dear R experts:

although my question may be better asked on the HPC R mailing list, it
is really about something that average R users who don't plan to write
clever HPC-optimized code would care about: is there a quantum
performance leap on the horizon with CPUs?

like most R average non-HPC users, I want to stick mostly to
mainstream R, often with library parallel but that's it.  I like R to
be fast and effortless.  I don't want to have to rewrite my code
greatly to take advantage of my CPU.  the CUDA forth-and-back on the
memory which requires code rewrites makes CUDA not too useful for me.
in fact, I don't even like setting up computer clusters.  I run code
only on my single personal machine.

now, I am looking at the two upcoming processors---intel haswell (next
month) and amd kaveri (end of year).  does either of them have the
potential to be a quantum leap for R without complex code rewrites?
I presume that any quantum leaps would have to come from R using a
different numerical vector engine.   (I tried different compiler
optimizations when compiling R (such as AVX) on the 1-year old i7-27*,
but it did not really make a difference in basic R benchmarks, such as
simple OLS calculations.  I thought AVX would provide a faster vector
engine, but something didn't really compute here.  pun intended.)

I would guess that haswell will be a nice small evolutionary step
forward.  5-20%, perhaps.  but nothing like a factor 2.

[tomshardware details how intel FP32 math is 4 times as fast as double
math on the i7 architecture.  for most of my applications, a 4 times
speedup at a sacrifice in precision would be worth it.  R seems to use
only doubles---even as.single is not even converting to single, much
less inducing calculations to be single-precision.  so I guess this is
a no-go.  correct?? ]

kaveri's hUMA on the other hand could be a quantum leap.  kaveri could
have the GPU transparently offer common standard built-in vector
operations that we use in R, i.e., improve the speed of many programs
without the need for a rewrite, by a factor of 5?  hard to believe,
but it would seem that AMD actually beat Intel for R users.  a big
turnaround, given their recent deemphasis of FP on the CPU.
(interestingly, the amd-built Xbox One and PS4 processors were also
reported to have  hUMA.)

worth waiting for kaveri?   anything I can do to drastically speed up
R on intel i7 by going to FP32?

regards,

/iaw

Ivo Welch (ivo.we...@gmail.com)

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Re: [R] How sum all possible combinations of rows, given 4 matrices

2013-05-27 Thread Jeff Newmiller
I expect the answer to involve manipulating indices. But why do you need to do 
this? This looks suspiciously like homework, and there is a no-homework policy 
on this list (see the Posting Guide).
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Estigarribia, Bruno estig...@email.unc.edu wrote:

Hello all,

I have 4 matrices with 3 columns each (different number of rows
though). I
want to find a function that returns all possible 3-place vectors
corresponding to the sum by columns of picking one row from matrix 1,
one
from matrix 2, one from matrix 3, and one from matrix 4. So basically,
all
possible ways of picking one row from each matrix and then sum their
columns to obtain a 3-place vector.
Is there a way to use expand.grid and reduce to obtain this result? Or
am
I on the wrong track?
Thank you,
Bruno
PS:I believe I have given all relevant info. I apologize in advance if
my
question is ill-posed or ambiguous.

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Re: [R] adding class attributes to strings: works in a loop, but not directly

2013-05-29 Thread Jeff Newmiller
a) You cannot assign attributes to literal values such as BICY. You must 
assign them to variables, as in
x - BICY
class(x) - AONmode

b) You cannot give the separate elements of a vector their own separate 
attributes unless the vector is of mode list. Ordinarily, attributes apply to 
the whole vector. (When the vector is of mode list then each element may be a 
vector on its own with its own attributes.)
y - c(FOOT,BICY)
class(y) - AONmode
z - vector( list, 2 )
z[[1]] - FOOT
class( z[[1]] ) - AONmode
z[[2]] - BICY
class( z[[2]] ) - AONmode

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Franckx Laurent laurent.fran...@vito.be wrote:

Dear all

I try to assign class attributes to strings. Depending on the approach
I use, it works (including method dispatch) or fails.

Let me clarify with an example.

Let:

 AONtptmodelist - c(FOOT,BICY)


When I assign class attributes directly to these strings, it fails:

 class(BICY) - AONmode
Error in class(BICY) - AONmode :
  target of assignment expands to non-language object

However, when I assign the attributes in a loop, it does work:

 for(tptmode in AONtptmodelist) {
+ class(tptmode) - AONmode
+ cat(The value of is.object(tptmode) for  , tptmode , is:  ,
is.object(tptmode) , .\n)
+ cat(The class of  , tptmode , is:  , class(tptmode) , .\n)
+ }
The value of is.object(tptmode) for  FOOT is:  TRUE .
The class of  FOOT is:  AONmode .
The value of is.object(tptmode) for  BICY is:  TRUE .
The class of  BICY is:  AONmode .

Moreover, within this loop, method dispatch works correctly.

However, when I start a new loop over the same vector, I get:

 for(tptmode in AONtptmodelist) {
+ cat(The value of is.object(tptmode) for  , tptmode , is:  ,
is.object(tptmode) , .\n)
+ cat(The class of  , tptmode , is:  , class(tptmode) , .\n)
+ }
The value of is.object(tptmode) for  FOOT is:  FALSE .
The class of  FOOT is:  character .
The value of is.object(tptmode) for  BICY is:  FALSE .
The class of  BICY is:  character .


I find this troublesome for two reasons. First, I do not understand the
difference between the two approaches. Why can I assign a class
attribute to a string when it is called in a loop, but not directly?
And why is the class attribution not permanent? Second, the problem was
concealed until now precisely because I assigned the attributes in a
link. However, I would like to centralise my class definition in one
single place in my code, thus outside the loop where the methods are
called.



Laurent Franckx, PhD
VITO NV
Boeretang 200, 2400 MOL, Belgium
Tel. + 32 14 33 58 22
Skype: laurent.franckx
laurent.fran...@vito.be
Visit our website: www.vito.be/english and http://www.vito.be/transport











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Re: [R] write table in ascii

2013-05-30 Thread Jeff Newmiller
I would bet the problem is end-of-line convention. This is a very old problem 
of text file conversion that has many solutions, none of which should be 
addressed by R. Rather, you should use any of the many standard file transfer 
mechanisms that understand EOL issues in text files, rather than transferring 
the data as if it were binary data. Email is one such mechanism, FTP is 
another. Many text editors can also fix this for you. It is possible to fix it 
in R, but that is a very bad approach. Google is your friend.
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Gary Dong pdxgary...@gmail.com wrote:

I am working in Windows system. But a software I am using requires data
input in ascii (.dat, delimiters can be tabs or spaces). I was able to
save
the data into .dat (ascii) from SPSS, but it keeps causing problems.
Thank
you!

Gary


On Thu, May 30, 2013 at 3:51 PM, Sarah Goslee
sarah.gos...@gmail.comwrote:

 On Thu, May 30, 2013 at 6:35 PM, Gary Dong pdxgary...@gmail.com
wrote:
  Dear R users,
 
  I have a data set in .csv and I hope to convert it to .dat (ascii)
so it
  can work in an UNIX environment. Anyone can help me? Thank you!

 Please expand. Why can't you work with csv in a UNIX environment?

 I do it all the time.

 Sarah

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


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Re: [R] Strange behaviour of R graphics copied to PowerPoint

2013-06-03 Thread Jeff Newmiller
I have not seen this particular problem, but I have seen other problems and I 
tend to export bitmaps or pdf files as a result.

Note that a reproducible example is usually required to to obtain help on this 
list, and posting in HTML format is bad because it mutilates example code, so 
fix your email client. Please read the Posting Guide for more etiquette tips.
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Erling Johan Frøysa erling.fro...@gmail.com wrote:

Hello,

I am using R to create graphics, especially to plot time series charts.
These charts are then copied as metafiles (for best quality) to a
PowerPoint
presentation and then saved to PDF (via the Save As dialog).

Attached is two pictures. The first picture shows how my chart looks
like in
the R Graphics window, and the second picture shows how the chart
becomes
after saving it to PDF.

 http://r.789695.n4.nabble.com/file/n4668522/R.png

 http://r.789695.n4.nabble.com/file/n4668522/Rppt.png

As you can see. After saving the metafile to PDF via PowerPoint, some
straight lines appears (it seems like all of the lines has the same
origin
in the upper left corner and ends somewhere on the times series line).
This
happens in both plot() and ggplot(). The problem appears more often
when
using daily data in my time series. With monthly data the problem don't
exist.

Have anyone experienced this before? Do you think the problem is
related to
R or to Powerpoint?

Thanks all,

E

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Re: [R] Error code from optim - NEW_X what does it mean?

2013-06-03 Thread Jeff Newmiller
Your email Is missing a reproducible example... very sub-optimal, likely to be 
ignored.
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Adler, Avraham avraham.ad...@guycarp.com wrote:

Hello.

Does anyone know what the error code NEW_X means in optim? Here is
what the return looks like:

   $counts
   function gradient
302  302

   $convergence
   [1] 1

   $message
   [1] NEW_X

I have searched and found a couple of mentions without responses:

https://stat.ethz.ch/pipermail/r-help/2011-September/290722.html  
September 2011
https://stat.ethz.ch/pipermail/r-devel/2013-May/066731.html   
Full disclosure, this is my (unanswered) question on R-devel


I can only find one instance of the call in `optim.c` and I don't
understand what it is doing.

Thank you,

Avraham Adler




 

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Re: [R] How to write a loop in R to select multiple regression model and validate it ?

2013-06-04 Thread Jeff Newmiller
This doesn't look like a task you have acquired through a real-life problem... 
it looks like homework. There is a stated no-homework policy in the Posting 
Guide (please read it), since you should be using the resources provided along 
with your educational environment (teaching assistants, tutors, office 
hours...), and we don't know whether the help we provide would be considered 
cheating.
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beginner pa...@nottingham.ac.uk wrote:

I would like to run a loop in R. I have never done this before, so I
would be
very grateful for your help !

1. I have a sample set: 25 objects. I would like to draw 1 object from
it
and use it as a test set for my future external validation. The
remaining 24
objects I would like to use as a training set (to select a model). I
would
like to repeat this process until all 25 objects are used as a test
set. 

2. For each of the training sets I would like to run the following
code:


library(leaps)
forward - regsubsets(Y ~.,data = training, method = forward,
nbest=1) 
backward - regsubsets(Y ~.,data = training, method = backward,
nbest=1)
stepwise - regsubsets(Y ~., data = training, method = seqrep,
nbest=1)
exhaustive - regsubsets(Y ~.,data = training, method = forward,
nbest=1)
summary(forward)
summary(backward)
summary(stepwise)
summary(exhaustive)

I would like R programme to select the best model (with the highest
adjusted
R2) using each of the selection methods, so there are 4 final best
models
(e.g. the best model selected with forward selection, the best model
selected with backward selection and so on...). 

 
Afterwards I would like to perform internal cross validation of all 4
selected models and choose 1 out of 4 which has the lowest average mean
squared error (MSE). I used to do it using the code below:

library(DAAG)
val.daag-CVlm(df=training, m=1, form.lm=formula(Y ~ X1+X2+X3))
val.daag-CVlm(df=training, m=1, form.lm=formula(Y ~ X1+X2+X4))
val.daag-CVlm(df=training, m=1, form.lm=formula(Y ~ X3+X4+X5))
val.daag-CVlm(df=training, m=1, form.lm=formula(Y ~ X4+X5+X7))

For the best selected model (the lowest MSE) I would like to perform an
external validation on 1 object left on the site at the beginning of
the
study (please refer to point 1.).

3. And loop again using different training and test set 


I hope that you could help me with this. 

If you have any suggestions how to select the best model and perform
validation more efficiently, I would be happy to hear about that.

Thank you !



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Re: [R] Loop FOR with histogram() from lattice

2013-06-05 Thread Jeff Newmiller
If you have read the Posting Guide, you will know that you should have read ALL 
the FAQs before posting your question.

http://cran.r-project.org/doc/FAQ/R-FAQ.html

That said, this particular FAQ applies to any library that depends on grid 
graphics, including lattice and ggplot2.
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Xochitl CORMON xochitl.cor...@ifremer.fr wrote:

Hi Jim,

Thank you a lot. Is it a FAQ concerning lattice or FOR loop in general?

Regards,

Xochitl C.


Le 05/06/2013 10:55, Jim Holtman a écrit :
 This is an FAQ.  you have to explicitly 'print' the histogram:

 print(histogram(~ Mpool[[2]]$LngtClas | Mpool[[2]]$SpCode, type =
count, col = lightgrey, xlab= LngtClas, main = paste(Length
distribution per species for Mpool, 2, sep = _)))

 Sent from my iPad

 On Jun 5, 2013, at 4:37, Xochitl CORMONxochitl.cor...@ifremer.fr 
wrote:

 Hi all,

 I'm encountering a problem I do not understand on my data:

 library (lattice)

 Mpool1- Table[Table$Subarea %in% c(52E9, 51E9),]
 Mpool2- Table[Table$Subarea %in% c(53F0, 52F0),]
 Mpool3- Table[Table$Subarea %in% c(51F0, 50F0),]
 Mpool4- Table[Table$Subarea %in% c(51F1, 52F1),]

 Mpool- list(Mpool1, Mpool2, Mpool3, Mpool4)


 histogram(~ Mpool[[2]]$LngtClas | Mpool[[2]]$SpCode, type = count,
col = lightgrey, xlab= LngtClas, main = paste(Length distribution
per species for Mpool, 2, sep = _))

  This part works perfectly and I obtain the graph reprensenting
Mpool2 length class count per species.
  Now when I want to automatize this with a for loop nothing is
plotted.

 for (i in c(2)){
 windows()
 histogram(~ Mpool[[i]]$LngtClas | Mpool[[i]]$SpCode, type = count,
col = lightgrey, xlab= LngtClas, main = paste(Length distribution
per species for Mpool, i, sep = _))
 print (i)
 }

 ### Running this loop I obtained  windows filled grey (no plot drawn
at all) but the print (i) print a 2 as expected. I really dont
understand what's wrong with the loop. There is no error message and no
notification in R. You can find enclosed my data in txt file.

 Thank you very much for any help,

 Xochitl C.

   

 Xochitl CORMON
 +33 (0)3 21 99 56 84

 Doctorante en sciences halieutiques
 PhD student in fishery sciences

   

 IFREMER
 Centre Manche Mer du Nord
 150 quai Gambetta
 62200 Boulogne-sur-Mer

   

 Table.txt
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Re: [R] Fortunes nomination (was: colour coded dotchart)

2013-06-09 Thread Jeff Newmiller
Oh, Lord, give me strength, for I am about to read an R Help file! Give me 
insight, for I must parse the words of statisticians forced by R CMD check to 
follow the way of the Literate Programmer! And please, Oh Lord, give me the 
wisdom to recall these tribulations when I set foot upon that road, that I may 
mark the path I blaze clearly for the novitiate to follow yet my words not 
become as obscure as those I read!

Commentary:
There is both truth and fiction in Rolf's advice, and for the truth I applaud 
the nomination, but for the fiction I fear the picture it paints in the 
unsuspecting mind.

On the one hand, it seems odd to ask that the R user believe that meaning is 
there because this is software, not the Bible. On the other hand, some people 
seem to think they should not have to study to use this tool properly, so 
perhaps some faith will help them dig a little harder.

Writing documentation is hard, and R CMD check at least makes package 
developers write SOMETHING down. That doesn't mean that it verifies that it is 
complete or clearly written. You can drag the horse to water but you cannot 
make them drink. Fortunately, the author of the code often does write just what 
you need to use it... even if it is a bit terse.

There is also a bit of transformation that happens as one learns how R code 
works, that makes the obscure documentation more intuitive. What the beginner 
does not seem to realize is that documentation that they would think is perfect 
would likely have to be customized to fit their specific set of educational 
holes, but would bore others who knew all that stuff. The solution is for 
certain conventions and shorthand to be used, and to let the user search as 
deep as they need to get their answer. Also, it is common for teenagers to say 
they will never be like their parents, yet later find themselves doing just 
that.

It is worth reminding new users that the maintainer() function is the key to 
helping improve R documentation, since what beginners think of as R is not a 
monolithic product but rather is the product of many individuals. Having the 
package maintainer be the gateway for such changes is a bit random, since 
different maintainers have different time and skills for the job, but 
communicating with the right person (rather than expecting them to read every 
message on R- help) is in general the best practice for fixing problems in 
packages. (Faulty documentation is indeed a problem, but a constructive 
suggestion for fixing it is far more likely to bring about change than either 
throwing of stones or abasing oneself before the guru.)
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Achim Zeileis achim.zeil...@uibk.ac.at wrote:

On Sun, 9 Jun 2013, Duncan Murdoch wrote:

 On reading the help pages,

Thanks, online on R-Forge now :-)

Best,
Z

 On 13-06-08 9:32 PM, Rolf Turner wrote:
 ...

 You need to get the hang of reading the online help.  The
information
 required is actually there in ?dotchart --- it's just tersely and
obscurely
 expressed.  A certain degree of optimism is required.  You need to
 ***believe*** that the information is there; then ask yourself What
 could they possibly mean by what they have written that would tell
 me what I need to know?.

 Duncan Murdoch

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Re: [R] Fwd: Problem with ODBC connection

2013-06-10 Thread Jeff Newmiller
Given the resounding silence, I would venture to guess that no-one here is 
interested in troubleshooting ODBC connections to Excel. The problem is most 
likely in the ODBC driver for Excel (not in R or RODBC), and Excel is NOT a 
database (so any data format problem is unlikely to be detected).
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Christofer Bogaso bogaso.christo...@gmail.com wrote:

Any response please? Was my question not clear to the list? Please let
me
know.

Thanks and regards,

-- Forwarded message --
From: Christofer Bogaso bogaso.christo...@gmail.com
Date: Sat, Jun 8, 2013 at 9:39 PM
Subject: Re: Problem with ODBC connection
To: r-help r-help@r-project.org


Hello All,

My previous post remains unanswered probably because the attachment was
not
working properly.

So I am re-posting it again.

My problem is in reading an Excel-2003 file through ODBC connection
using
RODBC package. Let say I have this Excel file:

http://www.2shared.com/document/HS3JeFyW/MyFile.html


I saved it in my F: drive and tried reading the contents using RODBC
connection:

 library(RODBC)
 MyData - sqlFetch(odbcConnectExcel(f:/MyFile.xls), )
 head(MyData, 30)


However it looks that the second column (with header 's') is not read
properly.

Can somebody here explain this bizarre thing? Did I do something wrong
in
reading that?

Really appreciate if someone could point out anything what might go
wrong.

Thanks and regards,


On Fri, Jun 7, 2013 at 4:46 PM, Christofer Bogaso 
bogaso.christo...@gmail.com wrote:

 Hello again,

 I am having problem with ODBC connection using the RODBC package.

 I am basically trying to read the attached Excel-2003 file using
RODBC
 package. Here is my code:

  head(sqlFetch(odbcConnectExcel(d:/1.xls), ), 30);
 odbcCloseAll()
Criteria  s  d fd  ffd1
 f1fd2f2 fd3 f3 F12 F13 F14 F15 F16
F17
 F18 F19 F20
 1 a NA NA NA NA 0.
 0.27755576 -0.00040332321NA  NA NA 
NA
  NA  NA  NA  NA  NA  NA  NA  NA
 2 s NA  0 NA NA 0.
 0.  0.000NA  NA NA 
NA
  NA  NA  NA  NA  NA  NA  NA  NA
 3 d NA  0 NA NA 0.01734723
 0.06938894  0.2775558  5.00  NA NA 
NA
  NA  NA  NA  NA  NA  NA  NA  NA
 4 f NA NA NA NA NA
 NA NA -4.25  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 5 f NA  0 NA NA 0.
 0.  0.000 -1.53  NA NA 
NA
  NA  NA  NA  NA  NA  NA  NA  NA
 6 f NA NA NA NA NA
 NA  0.000  0.00  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 7 f NA NA NA NA NA
 NA  0.000NA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 8 f NA  0 NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 9 f NA  0 NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 10f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 11f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 12f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA
 13f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA 
NA
  NA  NA  NA

 Here you see the data in second column could not read at all.

 Can somebody point me if I did something wrong?

 Thanks and regards,


   [[alternative HTML version deleted]]

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Re: [R] Bytes to Numeric/Float conversion

2013-06-11 Thread Jeff Newmiller
I recommend the hexView package for setting up such conversions.
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David Winsemius dwinsem...@comcast.net wrote:


On Jun 11, 2013, at 9:01 AM, Bikash Agrawal wrote:

 Is there any packages available in R, that can convert Bytes array to
Float.
 Using rJava we can do it. But it is kind of slow. Is there any R
 specific packages.
 I am having problem converting my bytes array to floating point.
 Could any one help me with this problem.

There is a raw data type that is designed to hold bytes than can be
indexed.

?raw

`scan` can read files of type raw

?scan

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Re: [R] Big, complex, well-structured .R file for demonstration?

2013-06-12 Thread Jeff Newmiller
Just because you have an editor that can let you see the organization within 
the file does not mean the code itself is well-structured. If you do put a lot 
of code in one file, you will be more likely in your next project that builds 
on this one to load code you do not need (bloat), and that is a very practical 
defect in the structure of the current project. Regardless of any arguments you 
can think of to the contrary, that is why single large files with otherwise 
well-structured code are uncommon.
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Thorsten Jolitz tjol...@gmail.com wrote:

Greg Snow 538...@gmail.com writes:

 Some would argue that big and well structured are not compatible.
 Part
 of structuring a project well is knowing when and how to break it
into
 smaller pieces, so those authors who are best at creating well
structured R
 code will often split it between several small files rather than one
big
 file.

As Emacs Org-mode has proven for text files, this structuring into
smaller pieces can be done in one single file too (that is structured
as
a hierarchical outline tree) an this can be even more convenient than
to
deal with many small files. But otherwise I agree with you, its much
better to split a file up before it becomes a growing mess.

 On Tue, Jun 11, 2013 at 9:06 AM, Thorsten Jolitz tjol...@gmail.com
wrote:


 Hi List,

 I'm looking for a rather big, but well structured R file that
contains
 as much of R language features as possible (i.e. that uses a lot of
the
 functionality described in the 'R Reference Card' and, if possible,
S4
 classes too).

 I want to check some code I wrote against such a file and use it for
 demonstration purposes. However, most .R files I find out there are
 rather short without much structure.

 Any links to candidate (open source) files would be appreciated.

 --
 cheers,
 Thorsten

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Re: [R] Remove levels

2013-06-13 Thread Jeff Newmiller
Please read the Posting Guide, which among other things points out that you 
should be posting in plain text format, not HTML (which tends to corrupt 
example R code).

Then please explain why your problem is not addressed by the below referenced 
section of the R Inferno. You may need to read [1] for advice on providing a 
reproducible example per the Posting Guide so we can follow your argument 
clearly.

[1] 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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Shane Carey careys...@gmail.com wrote:

Nope, but thanks


On Thu, Jun 13, 2013 at 1:11 PM, Albin Blaschka 
albin.blasc...@standortsanalyse.net wrote:



 Am 13.06.2013 14:02, schrieb Shane Carey:

  I have a dataframe consisting of factors in one column. Im trying to
 remove
 certain levels using the following code:
 toBeRemoved1-which(DATA$**UnitName_1==lake)
 DATA-DATA[-toBeRemoved1,]

 However it will not remove the level lake


 Hello!

 Is this a part of the R Inferno?
 See The R Inferno from Patrick Burns, specially Chapter 8.2.4
dropping
 factor levels , page 83


http://www.burns-stat.com/**pages/Tutor/R_inferno.pdfhttp://www.burns-stat.com/pages/Tutor/R_inferno.pdf

 HTH,
 Albin


 --
 | Albin Blaschka, Mag.rer.nat.
 | Etrichstrasse 26, A-5020 Salzburg
 | * www.albinblaschka.info * www.thinkanimal.info *
 | - It's hard to live in the mountains, hard but not hopeless!

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Re: [R] R session freezes when I try to save a new script

2013-06-14 Thread Jeff Newmiller
You will probably need to read the Posting Guide and follow its recommendations 
to get a constructive response. Things like posting in text instead of HTML and 
providing the output of sessionInfo go a long way toward speeding up the 
troubleshooting process.

You can also begin the process of identifying what your in-memory objects look 
like by stopping your script just before you run the function or click on a 
menu that hangs, and providing to us the output of the ls() function in your 
next email. You may have redefined an object with an important name. 
Alternatively, you may have a windows permissions issue (which would be outside 
the scope of this mailing list.

Ideally you would generate a self- contained reproducible example, but if that 
is not possible then you will have to live with guesses as responses. (The dump 
or dput functions may let you share the problematic memory configuration.)
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anferg...@aol.com wrote:

All:


Recently my R session freezes when I try to open a file or save a new
script after I have run existing scripts.  The session freezes so that
I can no longer click on any windows within the R session -- including
other scripts that are open or the R console.  (I hear the ding sound
when I try to click on anything.)  I have looked at the Task Manager on
the computer and R is still running (and is still responsive according
to the Task Manager), but I cannot do anything within R.  I have tried
pressing Esc and that does not help.


Any suggestions would be greatly appreciated!  I am using a 32 bit
Windows 7 computer with 4gb ram and 250gb hard drive.


Thanks in advance for any suggestions!


~Amanda Ferguson

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Re: [R] Widows 8

2013-06-15 Thread Jeff Newmiller

On Sat, 15 Jun 2013, Chet Seligman wrote:


Can anyone confirm that R runs on Widows 8?


You can : http://lmgtfy.com/?q=site%3Astat.ethz.ch+%22windows+8%22


Thanks,
Chet Seligman



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Please read the Posting Guide, and asi it requests post in plain text.


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Re: [R] Optimization of a function using optim

2013-06-15 Thread Jeff Newmiller
Not unless you read the Posting Guide, stop posting in HTML mail format, and 
provide a reproducible example.
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Graham McDannel graham.mcdan...@gmail.com wrote:

I am attempting to optimize a function I have developed using optim.

I am getting the below error message:

Error in n  1: 'n' is missing

Could some one provide some additional clarity regarding this message
and
what it entails, as well as, how to rectify this issue.

Thanks

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Re: [R] R for Chrome OS

2013-06-16 Thread Jeff Newmiller
You should probably try to study a bit more about what ChromeOS is about before 
asking questions like that. The real question you should have asked is whether 
R is offered as SAAS in the cloud. (I don't think so.) You can setup an 
instance of Linux in the cloud and run R there. RStudio Server may also be 
helpful. If you don't like paying for your use of R by the minute, then you may 
want to purchase your own server or reconsider using ChromeOS at all.

Also, please post here in plain text rather than using HTML email as the 
Posting Guide requests.
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Kevin Chavers kevinchav...@gmail.com wrote:

Hello,

I have heard that Chrome OS is Linux based, so I am wondering if there
is
anyway to use R on a Google Chromebook using one of your Linux
packages. If
not, do you plan on making a version of R that is compatible with
chromebooks or cloud based?

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Re: [R] help with string split in R

2013-06-17 Thread Jeff Newmiller
?sub
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CompBiol ATL along...@gmail.com wrote:

#I want to remove '_alpha' in a vector of strings
myInput = c( afg3_alpha, alg12_alpha, dbp3_alpha,  elp4_alpha,
 fob1_alpha,  gpr1_alpha)

#my current solution
tmpsplit = strsplit(myInput, '_')
tmp = NA
for (item in tmpsplit){
  tmp = c(tmp, item[1])
}
results1  = tmp[-1];
#this is what I need, but I want a better solution.

#Here is what I come up with, but it is pretty awkward
tmp3 = lapply(myInput, FUN=function(x){strsplit(x, '_')[[1]][1]})
results2 = unlist(tmp3)

#Please help me revise the above loop. Thanks, -- Hong

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Re: [R] BINARY traits GENETIC CORRELATION

2013-06-20 Thread Jeff Newmiller
Re-posting, particularly without referencing your earlier post, is bad 
mailing-list etiquette. Posting in HTML is particularly frowned upon here also. 
Nor is this a statistical methods support forum... it is about R. I for one am 
finding your question very jargonish and obscure. If your question is 
straightforward in the genetics domain then there might be more responsiveness 
in the Bioconductor mailing list. Otherwise, you may need to frame your 
question more directly in terms of the R language, with an example starting 
data set and expected results [1].

You are also expected to read the Posting Guide mentioned at the end of every 
R-help email.

[1] 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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Vallejo, Roger roger.vall...@ars.usda.gov wrote:

Dear R Users,
I have always used ASReml, MTDFREML, SAS, etc. to estimate genetic
correlation (Rg) between two continuous quantitative traits (also
continuous and binary trait) but I have never used R package to
estimate Rg. However, I use R package for many of my other statistical
analysis needs. The R package is a great tool that has advanced to a
level that now we can estimate whole genome-enabled breeding values in
genomic selection research. So, with the aim of moving out of those
packages (listed above) which are either expensive or have little
support, I would like to know if we can estimate Rg between two binary
traits (disease status: alive vs. dead) with the R package.

My data: we have 100 full-sib (FS) families, and two random samples
(each with n= 200 FS fish) from each FS family were evaluated for
resistance response to two different bacterial diseases, separately. It
implies that both traits are not recorded in the same individual; each
trait is recorded in different groups of full-sibs (random sampled from
a FS family). So using this family relationship (full-sibs), I would
like to estimate the Rg between these two disease resistance traits; of
course I would like to estimate Rg with the R package. I will
appreciate having directions on which R package to use if any, or
combination of R functions needed to use to calculate the Rg (between
two binary traits; disease survival traits).
Thank you very much in advance for your help.
Roger


Roger L. Vallejo, Ph.D.
U.S. Department of Agriculture, ARS, NCCCWA
Voice:  (304) 724-8340 Ext. 2141
Email:  roger.vall...@ars.usda.govmailto:roger.vall...@ars.usda.gov
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Re: [R] R 2.11 on Mac OS X

2013-06-24 Thread Jeff Newmiller
RStudio questions belong on the RStudio support website. It is fine software, 
but it is off-topic here.
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Norman Jessup njes...@tpg.com.au wrote:

David,

Thank you.  You are correct - I was inadvertently accessing an old 
version.  Deleting the old and re-installing means I can fire up R 
V3.0.1 which runs fine.

However, I still can't get R Studio to start up, receivng the following

message:


ERROR r error 4 (R code execution error) [errormsg=Error in
identical(call[[1L]], quote(doTryCatch)) :

7 arguments passed to .Internal(identical) which requires 5

,
code=local(source(/Applications/RStudio.app/Contents/Resources/R/Tools.R,
   local=TRUE, echo=FALSE, verbose=FALSE, encoding='UTF-8'))]; OCCURRED
AT: core::Error r::exec::anonymous
   namespace::evaluateExpressions(SEXP, SEXP, SEXP *, sexp::Protect *)
/Users/rstudio/rstudio/src/cpp/r/RExec.cpp:145



Even though I re-installed the latest version 0.97.551.  I wonder if 
there is an old Library or other support file that was not properly 
replaced?

Thank you

Norman Jessup
 On Jun 21, 2013, at 7:27 PM, Norman Jessup wrote:

 Hello,

 I've recently upgraded to R 2.11.1 on Mac OS X 10.8.4.   Now when I
start R up I get the following message:

 Error in identical(call[[1L]], quote(doTryCatch)) :
   7 arguments passed to .Internal(identical) which requires 5
 Error in normalizePath(dirname(pkgpath), /, TRUE) :
   3 arguments passed to .Internal(normalizePath) which requires 1
 cannot find system Renviron

 I get a similar message with user-defined functions ( i.e X
arguments passed when Y defined ) though the functions appear to work.
This problem is also encountered when Rstudio fires up and so it cannot
run now.

 I did find a post that suggested it may be due to R accessing an
old, possibly 32 bit library (I used to have 32 and 64 bit R installed
and they both ran without trouble).  Possibly I need to completely
clean out the installation and start again?  but I'm not sure precisely
where the R support files are stored on Macs.  Can anyone give me a
pointer and/or suggest an alternative fix?
 R 2.11.1 is a rather archaic version. The current version is 3.0.1.
You seems to have skipped major versions 2.12, 2.13,, 2.14 2.15. I
doubt that OSX 10.8.4 was available when 2.11.1 was compiled. There is
a mailing list for MacOS versions of R but I doubt there will be much
interest in supporting version 2.11.1 on OSX 10.8.4. I suggest you
install instead version 3.0.1



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Re: [R] R CMD BATCH Unicode

2013-06-26 Thread Jeff Newmiller
Just because the subject mentions R doesn't mean it is on topic here. This is 
more related to Windows than R. I recommend studying windows documentation for 
awhile. A quick search turned up a number of discussions on the web, including 
http://stackoverflow.com/questions/1035388/unicode-output-on-windows-command-line.
---
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Sent from my phone. Please excuse my brevity.

Ned Harding n...@alteryx.com wrote:

So just to clarify - there is no way to use R CMD BATCH on windows with
Unicode?  Any advice of how to use R in a batch mode with Unicode
inputs and outputs?

Ned.

-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Prof Brian Ripley
Sent: Wednesday, June 26, 2013 12:45 AM
To: r-help@r-project.org
Subject: Re: [R] R CMD BATCH Unicode

On 25/06/2013 20:35, Ned Harding wrote:
 Just to clarify: The encoding didn't come through in the email. 
print(éíôåëëåãåâáè) is meant to be a bunch of random greek
characters.

In that case the message is likely correct.  You failed to give us the
'at a minimum information' required by the posting guide, but you can
only have input scripts in the locale encoding (and there are no UTF-8
locales on Windows).  So unless you were in a Greek locale, the
re-encoding should have failed.

 Ned.

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Ned Harding
 Sent: Tuesday, June 25, 2013 11:35 AM
 To: r-help@r-project.org
 Subject: [R] R CMD BATCH Unicode

 Hey,

 I am looking for some help using Unicode with R CMD BATCH on windows.
In particular I would like my input and output files to be UTF-8
encoded.  My command line looks like this:

 r CMD BATCH --encoding=UTF-8 in.txt out.txt

 in.txt is utf-8 encoded and contains:

 print(éíôåëëåãåâáè)

 out.txt gets:

 + ERROR: re-encoding failure from encoding 'UTF-8'

 What is the proper way to specify encoding on the command line?

 Thanks in advance,

 Ned.

  [[alternative HTML version deleted]]

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Re: [R] R CMD BATCH Unicode

2013-06-27 Thread Jeff Newmiller
Well, I admit that I don't mess with this stuff much, but it worked fine for me 
in a simple test as long as I viewed the output with an editor or console that 
understood UTF-8, so I dispute your assertion that this is a problem internal 
to R. (I needed no special arguments to R for it to work either.)
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Ned Harding n...@alteryx.com wrote:

I have no problems with the windows command line.  I don't need any
Unicode there.  It really is an internal R question because of the way
R is reading and writing the input and output files.  

Ned.

-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us] 
Sent: Wednesday, June 26, 2013 11:06 AM
To: Ned Harding; Prof Brian Ripley; r-help@r-project.org
Subject: Re: [R] R CMD BATCH Unicode

Just because the subject mentions R doesn't mean it is on topic here.
This is more related to Windows than R. I recommend studying windows
documentation for awhile. A quick search turned up a number of
discussions on the web, including
http://stackoverflow.com/questions/1035388/unicode-output-on-windows-command-line.
---
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.

Ned Harding n...@alteryx.com wrote:

So just to clarify - there is no way to use R CMD BATCH on windows
with 
Unicode?  Any advice of how to use R in a batch mode with Unicode 
inputs and outputs?

Ned.

-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Prof Brian Ripley
Sent: Wednesday, June 26, 2013 12:45 AM
To: r-help@r-project.org
Subject: Re: [R] R CMD BATCH Unicode

On 25/06/2013 20:35, Ned Harding wrote:
 Just to clarify: The encoding didn't come through in the email. 
print(éíôåëëåãåâáè) is meant to be a bunch of random greek 
characters.

In that case the message is likely correct.  You failed to give us the

'at a minimum information' required by the posting guide, but you can 
only have input scripts in the locale encoding (and there are no UTF-8

locales on Windows).  So unless you were in a Greek locale, the 
re-encoding should have failed.

 Ned.

 -Original Message-
 From: r-help-boun...@r-project.org
 [mailto:r-help-boun...@r-project.org] On Behalf Of Ned Harding
 Sent: Tuesday, June 25, 2013 11:35 AM
 To: r-help@r-project.org
 Subject: [R] R CMD BATCH Unicode

 Hey,

 I am looking for some help using Unicode with R CMD BATCH on
windows.
In particular I would like my input and output files to be UTF-8 
encoded.  My command line looks like this:

 r CMD BATCH --encoding=UTF-8 in.txt out.txt

 in.txt is utf-8 encoded and contains:

 print(éíôåëëåãåâáè)

 out.txt gets:

 + ERROR: re-encoding failure from encoding 'UTF-8'

 What is the proper way to specify encoding on the command line?

 Thanks in advance,

 Ned.

 [[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.

Please do, and note what it says about HTML mail, too.

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Re: [R] Data Package Query

2013-06-27 Thread Jeff Newmiller
data is a base function (so you should not have to load any packages to use 
it), and it is not capitalized. Depending on the particular data set you want 
the data function to load, you may need to load the package that contains that 
data set.

A common error by beginners (which may or may not be your problem in this case) 
is to create a variable called data. Unfortunately this hides the function 
named data and from that time forward that R session doesn't work when you 
type example code that uses the data function. If this is your problem, the 
best solution is to restart R and do your analysis from the beginning using a 
different variable name than data. (This is why keeping your working code in 
a separate text file is standard operating procedure.)

Note that I am guessing here... the Posting Guide asks you to give a 
reproducible example so we don't have to guess what you have done.
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Yasmine Refai y_re...@hotmail.com wrote:

Hello,

When i type in the below syntax:
Data (name of the data set)

I get an error message specifying that the data package is not found.

Please note that i installed all packages having the word data 
included in them and loaded all these packages.

Please advice.

Regards!
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Re: [R] [SPAM?] Re: Data Package Query

2013-06-27 Thread Jeff Newmiller
Please reply to all so the thread stays on the mailing list for all to see (and 
correct if wrong information is given).

Regarding your code below, either you are missing a library statement to load a 
package that contains this trial data, or you are intending to read in your 
own data set but are not yet aware of how that is done.

I could find no package containing a data set named trial (but I might have 
missed one). 

If you have some data of your own, then the data function is almost certainly 
not the right tool for the job. A PDF document called R Data Import/Export is 
supplied with R that can get you started. There are also various intro books 
available.
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Yasmine Refai y_re...@hotmail.com wrote:

Unfortunately, i have no variable called data. The below is the
syntax i am running:
data(trial)
fit-logistf(data=trial, y=x1+x2+x3)

Sorry for the disturbance.

Regards!

On Jun 27, 2013, at 9:47 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:

 data is a base function (so you should not have to load any
packages to use it), and it is not capitalized. Depending on the
particular data set you want the data function to load, you may need to
load the package that contains that data set.
 
 A common error by beginners (which may or may not be your problem in
this case) is to create a variable called data. Unfortunately this
hides the function named data and from that time forward that R
session doesn't work when you type example code that uses the data
function. If this is your problem, the best solution is to restart R
and do your analysis from the beginning using a different variable name
than data. (This is why keeping your working code in a separate text
file is standard operating procedure.)
 
 Note that I am guessing here... the Posting Guide asks you to give a
reproducible example so we don't have to guess what you have done.

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Live...
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Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#. 
rocks...1k

---

 Sent from my phone. Please excuse my brevity.
 
 Yasmine Refai y_re...@hotmail.com wrote:
 
 Hello,
 
 When i type in the below syntax:
 Data (name of the data set)
 
 I get an error message specifying that the data package is not
found.
 
 Please note that i installed all packages having the word data 
 included in them and loaded all these packages.
 
 Please advice.
 
 Regards!
 __
 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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Re: [R] Data Package Query

2013-06-28 Thread Jeff Newmiller
You need to learn to execute one statement at a time in order to debug this 
yourself. Copy and paste is your friend. Hint: I already told you that the data 
function is inappropriate if the data does not come from a package.

You should be learning to use the str(), head(), and ls() functions to explore 
your R in-memory environment, and use the built-in help system with the 
question mark (?str) or the help.search() and RSiteSearch() functions.
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Yasmine Refai y_re...@hotmail.com wrote:

hello,
 
please advice what is wrong at the below syntax:
Trial-read.table(Trial.txt,header=TRUE)
Trial
save.image(file=Trial.RData)
data(Trial)
fit-logistf(data=Trial, y~x1+x2)

 
and here is the error I get:
Warning message:
In data(Trial) : data set ‘Trial’ not found

 
regards,
yasmine

 
 Date: Fri, 28 Jun 2013 10:29:21 +1200
 From: rolf.tur...@xtra.co.nz
 To: jdnew...@dcn.davis.ca.us
 CC: y_re...@hotmail.com; r-help@r-project.org
 Subject: Re: [R] Data Package Query
 
 On 28/06/13 04:47, Jeff Newmiller wrote:
 
  SNIP
  A common error by beginners (which may or may not be your problem
in this case) is to create a variable called data. Unfortunately this
hides the function named data and from that time forward that R
session doesn't work when you type example code that uses the data
function.
 
  SNIP
 
 This is simply not true.  I believe it *used* to be true, sometime 
 wa back,
 but hasn't been true for years.  The R language is much cleverer now.
 
 If there
 is a function melvin() somewhere on the search path and also a data
object
 melvin (earlier on the search path) then doing
 
  melvin(whatever)
 
 will correctly call the function melvin() with no complaints.  The R 
 language
 can tell by the parentheses that you mean the *function* melvin and

 not the
 data object melvin.
 
 E.g.
 
  data - 42
  require(akima)
  akima
  Error: object 'akima' not found
  data(akima)  # No error message, nor nothin'!
  akima
  # The data set akima is displayed.
 
 All that being said it is ***BAD PRACTICE***, just in terms of 
 comprehensibility
 and avoiding confusion, to give a data set set the same name as a
function
 (either built in, or one of your own).
 
  fortune(dog)
 
 is relevant.
 
  cheers,
 
  Rolf Turner
 


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Re: [R] Bhattacharyya in R

2013-07-01 Thread Jeff Newmiller
There is a built-in guide to syntax called the help system. If you have read it 
you should be able to be mitre specific with your questions.

I had never heard of this function. Your example code is not reproducible, nor 
does it indicate what result you got from running it.

I used

RSiteSearch(bhattacharya)

and there appears to be a function bhattacharyya.matrix in the fpc package. The 
syntax indicated in the help for that function mentions wanting input matrices, 
not vectors. It also mentions another function bhattacharyya.dist which does 
accept vectors, but it does not accept the misclassification.bound argument.

You should read the help file and ask more specific questions and provide 
reproducible examples per the Posting Guide recommendations.
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Dzu dizem.ue...@alumni.fh-aachen.de wrote:

Dear R-user,
I am trying to apply bhattacharyya-distance function to my data. Did
anybody
use it before ?

My code is the following

#Bhattacharya Distance measure
#a and b are vectors 
a - (1,2,3,4,2,2,2,2,2,2,2,1,4,5,6,-1,-1,-1,-1,-1,-3,-3,-3)
b -
(1.1,1.1,1.2,1.2,1.2,1.2,1.2,2.1,2.1,2.2,2.2,2,0,0,0,0,2,2,2,2,2,3.1,3.1)

dist - bhattacharyya.matrix(a,b, misclassification.bound = TRUE)
plot(dist)


Could somebody give me a guide on the syntax ?

Thanks
Dizem





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Re: [R] Lee carter model

2013-07-01 Thread Jeff Newmiller
The Posting Guide warns you to not post in HTML format (it messes up R code 
examples). It also warns you to supply reproducible examples (we should be able 
to paste your code into a fresh R session and see essentially what you see). 

The warning message means you have given the function a factor where it 
expected a numeric value (max does not work on factors). I suggest that you 
learn how to use the str() function on your data so you know that you have 
imported it successfully before you give it to analysis functions. You may need 
to review the Introduction to R and R Data Import/Export PDF files that 
come with R if you don't know what a factor is or how to import numeric data.
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ntamjo achille ntamjo2...@yahoo.fr wrote:

Hi everybody,

I'm running codes for for a Lee carter modeling. I compute the
demogdata function in the package demography.�
Age is a class of factor

Base-demogdata(data=x,pop=y,ages=AGE,years=YEAR,type=mortality,label=City,name=Hommes,lambda=1)
�
Now, I try to run codes for lee carter model, and i always get the
following message:
Error in Summary.factor(c(1L, 2L, 24L, 35L, 46L, 57L, 68L, 79L, 90L,
101L, �:�
� max not meaningful for factors

Please can you help to understand what means this message.�

Many Thanks

Achille

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Re: [R] Outer function in R

2013-07-02 Thread Jeff Newmiller
Read the Posting Guide. The example you provide below is not reproducible [1], 
so we cannot tell what you're giving to the outer() function.

[1] 
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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Dzu dizem.ue...@alumni.fh-aachen.de wrote:

Dear members 

I am trying to apply the function kl.dist (Kullback-Leibler Distance
measure) to multiple matrixes.

I tried the following : 

veckldist - Vectorize(kl.dist)
distancematrix - outer (matrix1,matrix2, veckldist)


But the code is complaining that the list of the object does not match.
The
lengths of my matrixes are same 

How could I fix the error?

Thanks





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Re: [R] What package to use to download pictures and its description from server?

2013-07-02 Thread Jeff Newmiller
Wouldn't this be highly dependent on how the pictures and descriptions were 
formatted online, as well as what you planned to do with them? I generally find 
that a web browser is quite sufficient for my needs.
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C W tmrs...@gmail.com wrote:

Hi R community,
What package would you recommend to download pictures and descriptions
of the pictures?

I have looked at package XML and RCurl so far.  Is this what everyone
uses?

Thanks,
Mike

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Re: [R] acf question...

2013-07-02 Thread Jeff Newmiller
Then you are posting in the wrong forum, since this is a forum about getting R 
to do things for which you already understand the theory.

As to the results you are getting, I highly recommend reading the details 
section of ?ccf.

BTW The Posting Guide indicates that you should post in text format because 
HTML mutilates R code. This is a setting in your email program.
---
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.

Venkatesh Nagarajan goca...@yahoo.com wrote:

I am trying to understand lagged correlations. 
�
x= 1:100; 
y = c(rep(NA,40), 1:60)ccf(x = x, y = y, lag.max=100,
na.action=na.pass, type = correlation) 
�
I was hoping to see max cor at lag = 40. But I am not. What am I doing
wrong?
�
Thanks
VN
   [[alternative HTML version deleted]]





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PLEASE do read the posting guide
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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.


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