Re: [R] Read shortcuts of MS Excel files through R

2013-09-27 Thread Mark Lamias 
Assuming you are trying to read the contents of the Excel files and not the 
shortcut itself, try something link this:

#C:\users\mark\desktop\A1.xlsx.lnk is a shortcut I created on my desktop to an 
xlsx document in another directory
#The windows type command, along with find gets the path to the actual file 
from the shortcut
cmd-'type c:\\users\\mark\\desktop\\A1.xlsx.lnk|find \\|findstr/b 
[a-z]:[]'
XLFileName-shell(cmd, intern=T)
read.xlsx(XLFileName, 1)

I hope this helps.


--Mark J. Lamias




 From: Santosh santosh2...@gmail.com
To: r-help r-help@r-project.org 
Sent: Thursday, September 26, 2013 4:54 PM
Subject: [R] Read shortcuts of MS Excel files through R


Dear Rxperts,

Through Windows OS, I created shortcuts (paste as shortcut) to excel
spreadsheets ( with xlsx as the file extension). I wasn't able to read
the shortcuts through R and using read functions of xlsx package.

exf - a1.xlsx.lnk

 read.xlsx(exf,1)
Error in .jcall(RJavaTools, Ljava/lang/Object;, invokeMethod, cl,  :
  java.lang.IllegalArgumentException: Your InputStream was neither an OLE2
stream, nor an OOXML stream
 read.xlsx2(exf,1)
Error in .jcall(RJavaTools, Ljava/lang/Object;, invokeMethod, cl,  :
  java.lang.IllegalArgumentException: Your InputStream was neither an OLE2
stream, nor an OOXML stream


Would truly appreciate your suggestions.

Regards,
Santosh

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Re: [R] Having trouble converting a dataframe of character vectors to factors

2013-02-21 Thread Mark Lamias 
Great point, William.  I agree your approach is the one to take to preserve 
attributes.

Thanks for following up.





 From: William Dunlap wdun...@tibco.com

elp (r-help@r-project.org) r-help@r-project.org 
Sent: Thursday, February 21, 2013 11:32 AM
Subject: RE: [R] Having trouble converting a dataframe of character vectors to 
factors

 scs2-data.frame(lapply(scs2, factor))

Calling data.frame() on the output of lapply() can result in changing column 
names
and will drop attributes that the input data.frame may have had.  I prefer to 
modify
the original data.frame instead of making a new one from scratch to avoid these 
problems.

Also, calling factor() on a factor will drop any unused levels, which you may 
not want
to do.  Calling as.factor will not.

Compare the following three methods

  f1 - function (dataFrame) {
      dataFrame[] - lapply(dataFrame, factor)
      dataFrame
  }
  f2 - function (dataFrame) {
      dataFrame[] - lapply(dataFrame, as.factor)
      dataFrame
  }
  f3 - function (dataFrame) {
      data.frame(lapply(dataFrame, factor))
  }

on the following data.frame
  x - data.frame(stringsAsFactors=FALSE, check.names=FALSE,
               No/Yes = factor(c(Yes,Yes,Yes), levels=c(No,Yes)),
               Size = ordered(c(Small,Large,Medium), 
levels=c(Small,Medium,Large)),
               Name = c(Adam,Bill,Chuck))
  attr(x, Date) - as.POSIXlt(2013-02-21)


   str(x)
  'data.frame':   3 obs. of  3 variables:
   $ No/Yes: Factor w/ 2 levels No,Yes: 2 2 2
   $ Size  : Ord.factor w/ 3 levels SmallMedium..: 1 3 2
   $ Name  : chr  Adam Bill Chuck
   - attr(*, Date)= POSIXlt, format: 2013-02-21

   str(f1(x)) # drops unused levels
  'data.frame':   3 obs. of  3 variables:
   $ No/Yes: Factor w/ 1 level Yes: 1 1 1
   $ Size  : Ord.factor w/ 3 levels SmallMedium..: 1 3 2
   $ Name  : Factor w/ 3 levels Adam,Bill,..: 1 2 3
   - attr(*, Date)= POSIXlt, format: 2013-02-21
   str(f2(x))
  'data.frame':   3 obs. of  3 variables:
   $ No/Yes: Factor w/ 2 levels No,Yes: 2 2 2
   $ Size  : Ord.factor w/ 3 levels SmallMedium..: 1 3 2
   $ Name  : Factor w/ 3 levels Adam,Bill,..: 1 2 3
   - attr(*, Date)= POSIXlt, format: 2013-02-21
   str(f3(x)) # mangles column names, drops unused levels, drops Date attribute
  'data.frame':   3 obs. of  3 variables:
   $ No.Yes: Factor w/ 1 level Yes: 1 1 1
   $ Size  : Ord.factor w/ 3 levels SmallMedium..: 1 3 2
   $ Name  : Factor w/ 3 levels Adam,Bill,..: 1 2 3

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Mark Lamias
 Sent: Wednesday, February 20, 2013 6:51 PM
 To: Daniel Lopez; R help (r-help@r-project.org)
 Subject: Re: [R] Having trouble converting a dataframe of character vectors 
 to factors
 
 How about this?
 
 scs2-data.frame(lapply(scs2, factor))
 
 
 
 
 
  From: Lopez, Dan lopez...@llnl.gov
 To: R help (r-help@r-project.org) r-help@r-project.org
 Sent: Wednesday, February 20, 2013 7:09 PM
 Subject: [R] Having trouble converting a dataframe of character vectors to 
 factors
 
 R Experts,
 
 I have a dataframe made up of character vectors--these are results from survey
 questions. I need to convert them to factors.
 
 I tried the following which did not work:
 scs2-sapply(scs2,as.factor)
 also this didn't work:
 scs2-sapply(scs2,function(x) as.factor(x))
 
 After doing either of above I end up with
 str(scs2)
 
 chr [1:10, 1:10] very important very important very important very 
 important ...
 
 - attr(*, dimnames)=List of 2
 
   ..$ : NULL
 
   ..$ : chr [1:10] Q1_1 Q1_2 Q1_3 Q1_4 ...
 
 class(scs2)
 matrix
 
 But when I do it one at a time it works:
 scs2$Q1_1-as.factor(scs2$Q1_1)
 scs2$Q1_2- as.factor(scs2$Q1_2)
 
 What am I doing wrong?  How do I accomplish this with sapply or similar 
 function?
 
 Data for reproducibility:
 
 
 scs2-structure(list(Q1_1 = c(very important, very important, very 
 important,
 
 very important, very important, very important, very important,
 
 somewhat important, important, very important), Q1_2 = c(important,
 
 somewhat important, very important, important, important,
 
 very important, somewhat important, somewhat important,
 
 very important, very important), Q1_3 = c(very important,
 
 important, very important, very important, important,
 
 very important, very important, somewhat important, not important,
 
 important), Q1_4 = c(very important, important, very important,
 
 very important, important, important, important, very important,
 
 somewhat important, important), Q1_5 = c(very important,
 
 not important, important, very important, not important,
 
 important, somewhat important, important, somewhat important,
 
 not important), Q1_6 = c(very important, not important,
 
 important, very important, somewhat important, very important,
 
 very important, very important, important, important),
 
     Q1_7 = c

Re: [R] Having trouble converting a dataframe of character vectors to factors

2013-02-20 Thread Mark Lamias 
How about this?

scs2-data.frame(lapply(scs2, factor))





 From: Lopez, Dan lopez...@llnl.gov
To: R help (r-help@r-project.org) r-help@r-project.org 
Sent: Wednesday, February 20, 2013 7:09 PM
Subject: [R] Having trouble converting a dataframe of character vectors to 
factors

R Experts,

I have a dataframe made up of character vectors--these are results from survey 
questions. I need to convert them to factors.

I tried the following which did not work:
scs2-sapply(scs2,as.factor)
also this didn't work:
scs2-sapply(scs2,function(x) as.factor(x))

After doing either of above I end up with
str(scs2)

chr [1:10, 1:10] very important very important very important very 
important ...

- attr(*, dimnames)=List of 2

  ..$ : NULL

  ..$ : chr [1:10] Q1_1 Q1_2 Q1_3 Q1_4 ...

class(scs2)
matrix

But when I do it one at a time it works:
scs2$Q1_1-as.factor(scs2$Q1_1)
scs2$Q1_2- as.factor(scs2$Q1_2)

What am I doing wrong?  How do I accomplish this with sapply or similar 
function?

Data for reproducibility:


scs2-structure(list(Q1_1 = c(very important, very important, very 
important,

very important, very important, very important, very important,

somewhat important, important, very important), Q1_2 = c(important,

somewhat important, very important, important, important,

very important, somewhat important, somewhat important,

very important, very important), Q1_3 = c(very important,

important, very important, very important, important,

very important, very important, somewhat important, not important,

important), Q1_4 = c(very important, important, very important,

very important, important, important, important, very important,

somewhat important, important), Q1_5 = c(very important,

not important, important, very important, not important,

important, somewhat important, important, somewhat important,

not important), Q1_6 = c(very important, not important,

important, very important, somewhat important, very important,

very important, very important, important, important),

    Q1_7 = c(very important, somewhat important, important,

    somewhat important, important, important, very important,

    very important, somewhat important, not important),

    Q2 = c(Somewhat, Very Much, Somewhat, Very Much,

    Very Much, Very Much, Very Much, Very Much, Very Much,

    Very Much), Q3 = c(yes, yes, yes, yes, yes, yes,

    yes, yes, yes, yes), Q4 = c(None, None, None,

    None, Confirmed Field of Study, Confirmed Field of Study,

    Confirmed Field of Study, None, None, None)), .Names = c(Q1_1,

Q1_2, Q1_3, Q1_4, Q1_5, Q1_6, Q1_7, Q2, Q3, Q4

), row.names = c(78L, 46L, 80L, 196L, 188L, 197L, 39L, 195L,

172L, 110L), class = data.frame)


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Re: [R] Recommendation for website to format R code

2013-01-23 Thread Mark Lamias 
Are you only interested in formatting code from copy and pasting to/from email? 
 If you are interested in formatting your code in Latex/PDF/HTML take a look at 
the knitr package:

http://yihui.name/knitr/

Also, you could check out the formatR package:
http://cran.r-project.org/web/packages/formatR/formatR.pdf

--Mark Lamias





 From: C W tmrs...@gmail.com
To: r-help r-help@r-project.org 
Sent: Wednesday, January 23, 2013 9:27 PM
Subject: [R] Recommendation for website to format R code

Hi list,

Could anyone recommend some good website for formatting R code?  For
example, when you copy paste R code to gmail and back to R, it loses its
format, the dash symbol causes errors.

I've had someone used it to format my code here on the list, but can't find
it anymore.

Mike

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Re: [R] A question on list and lapply

2012-12-14 Thread Mark Lamias 
How about 

Dat1 - lapply(subset(Dat, Dat!=P), My_Function)

--Mark





 From: Christofer Bogaso bogaso.christo...@gmail.com
To: r-help@r-project.org 
Sent: Friday, December 14, 2012 1:58 PM
Subject: [R] A question on list and lapply
 
Dear all, let say I have following list:

Dat - vector(list, length = 26)
names(Dat) - LETTERS
My_Function - function(x) return(rnorm(5))
Dat1 - lapply(Dat, My_Function)


However I want to apply my function 'My_Function' for all elements of 'Dat' 
except the elements having 'names(Dat) == P'. Here I have specified the name 
P just for illustration however this will be some name specified by user.

Is there any direct way to achieve this, using 'lapply'?

Thanks for your help.

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Re: [R] list to matrix?

2012-12-04 Thread Mark Lamias 
Try:

matrix(unlist(a), ncol=2, byrow=T)

--Mark Lamias





 From: Sam Steingold s...@gnu.org
To: r-help@r-project.org 
Sent: Tuesday, December 4, 2012 3:09 PM
Subject: [R] list to matrix?

How do I convert a list to a matrix?

--8---cut here---start-8---
list(c(5, 101), c(1e+05, 46), c(15, 31), c(2e+05, 17), 
    c(25, 19), c(3e+05, 11), c(35, 12), c(4e+05, 25), 
    c(45, 19), c(5e+05, 16))
as.matrix(a)
      [,1]    
[1,] Numeric,2
[2,] Numeric,2
[3,] Numeric,2
[4,] Numeric,2
[5,] Numeric,2
[6,] Numeric,2
[7,] Numeric,2
[8,] Numeric,2
[9,] Numeric,2
--8---cut here---end---8---

thanks!

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com
http://jihadwatch.org http://www.PetitionOnline.com/tap12009/ http://memri.org
Rhinoceros has poor vision, but, due to his size, it's not his problem.

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Re: [R] reformatting some data

2012-12-04 Thread Mark Lamias 
It's not clear to me what it is you are attempting to do, as you switch from a 
very specific example to some general example with the vague terms var1 
var2, and var3.

It sounds like you might be trying to do something similar to what would be 
available in the shape package using the melt function.

Try ?melt.data.frame.

--Mark Lamias





 From: Charles Determan Jr deter...@umn.edu
To: r-help@r-project.org 
Sent: Tuesday, December 4, 2012 4:17 PM
Subject: [R] reformatting some data

Hello,

I am trying to reformat some data so that it is organized by group in the
columns.  The data currently looks like this:

       group X3.Hydroxybutyrate X3.Hydroxyisovalerate   ADP
347     4              4e-04                 3e-04                  5e-04
353     3              5e-04                 3e-04                  6e-04
359     4              4e-04                 3e-04                  6e-04
365     4              6e-04                 3e-04                  5e-04
371     4              5e-04                 3e-04                  7e-04
377     2              7e-04                 4e-04                  7e-04

I would like to reformat it so it is like this:

                2          3           4
var1
var2
var3


I realize that there unequal numbers in each group but I would like to
none-the-less if possible.
Here is a subset of the data:

structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate =
c(4e-04,
5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate,
X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L,
365L, 371L, 377L), class = data.frame)

Any insight is truly appreciated,
Regards,
Charles

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Re: [R] reformatting some data

2012-12-04 Thread Mark Lamias 
That should read the reshape package -- not the shape package.  

My apologies.






To: Charles Determan Jr deter...@umn.edu; r-help@r-project.org 
r-help@r-project.org 
Sent: Tuesday, December 4, 2012 4:36 PM
Subject: Re: [R] reformatting some data


It's not clear to me what it is you are attempting to do, as you switch from a 
very specific example to some general example with the vague terms var1 
var2, and var3.

It sounds like you might be trying to do something similar to what would be 
available in the shape package using the melt function.

Try ?melt.data.frame.

--Mark Lamias





 From: Charles Determan Jr deter...@umn.edu
To: r-help@r-project.org 
Sent: Tuesday, December 4, 2012 4:17 PM
Subject: [R] reformatting some data

Hello,

I am trying to reformat some data so that it is organized by group in the
columns.  The data currently looks like this:

       group X3.Hydroxybutyrate X3.Hydroxyisovalerate   ADP
347     4              4e-04                 3e-04                  5e-04
353     3              5e-04                 3e-04                  6e-04
359     4              4e-04                 3e-04                  6e-04
365     4              6e-04           
      3e-04                  5e-04
371     4              5e-04                 3e-04                  7e-04
377     2              7e-04                 4e-04                  7e-04

I would like to reformat it so it is like this:

                2          3           4
var1
var2
var3


I realize that there unequal numbers in each group but I would like to
none-the-less if possible.
Here is a subset of the data:

structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate =
c(4e-04,
5e-04,
 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate,
X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L,
365L, 371L, 377L), class = data.frame)

Any insight is truly appreciated,
Regards,
Charles

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Re: [R] How to read SPSS file in R

2012-12-03 Thread Mark Lamias 
The following works just fine for me, using the built-in SPSS dataset 
nhis2000_subset.sav.  

library(foreign)
a=read.spss(C:\\Program 
Files\\IBM\\SPSS\\Statistics\\21\\Samples\\English\\nhis2000_subset.sav, 
to.data.frame=T)
 names(a)
 [1] STRATUM  PSU  WTFA_SA  SEX  AGE_P    REGION   
SMKNOW   VITANY   VITMUL   HERBSUPP VIGFREQW
[12] MODFREQW STRFREQW DESIREWT MOVE1    LIFT age_cat

You do not need to specify a comma delimiter on your read.spss statement since 
the file is already in a native SPSS dataset format -- not a CSV file.

--Mark Lamias






 From: F86 farad...@gmail.com
To: r-help@r-project.org 
Sent: Monday, December 3, 2012 10:55 AM
Subject: [R] How to read SPSS file in R

Dear R-users, 

I have som troubles with .sav file. How is it possible for us R-users to
read SPSS files. I know that is possible, 


I tried the following: 


 library(foreign)
 Corp-read.spss(/Users/kama/Analysis/Corporation.sav, header=TRUE,
 sep=,)
Error in read.spss(/Users/kama/Analysis/Corporation.sav, header = TRUE,  : 
  unused argument(s) (header = TRUE, sep = ,)
 Corp-read.spss(/Users/kama/Analysis/Corporation.sav)
re-encoding from UTF-8

Any suggestions please? 

Regards, 
Faradj
Stockholm University



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[R] knitr error with Lyx

2012-11-29 Thread Mark Lamias 
R Users,


I just upgraded my version of R from R-2.15.0 to R-2.15.2 and installed the 
latest versions of LyX and MikTex running Windows 7 Ultimate, 64-bit OS.  Prior 
to the upgrade, I was using Lyx with knitr to generate a document with no 
problems.  However, after the upgrade, and using the same LyX document, I'm 
receiving the following error when I attempt to compile the document:

\end{verbatim}
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

I have determined that the error is caused when printing the anova results from 
the anova statement in my R source code, but can't seem to resolve the issue.  
Here is an example code chunk that creates the error:

NonCP1, fig.width=6, fig.height=4, out.width='.8\\linewidth' ,par=FALSE=
#Read in data
y=c( 67, 73, 83, 89, 65, 91, 87, 86, 155, 127, 147, 212, 108, 100, 90, 153, 
140, 142, 121, 150, 33, 8, 46, 54 ) 
temp=as.factor(c(rep(seq(360, 380, 10), each=4), rep(seq(380, 360, -10), 
each=4))) 
coat=as.factor(rep(seq(1, 4), 6)) 
replicate=as.factor(rep(seq(1, 6), each=4)) 
#Obtain Factorial/Incorrect Model 
o=lm(y~temp*coat)
ano=anova(o)
ano
@


Removing the ano=anova(o) or ano lines in the code chunk allows the document to 
compile with no problem.  Does anyone else have this problem or did I do 
something wrong when I migrated to the newer versions?

Thanks, in advance for any help!

Sincerely yours,

Mark J. Lamias
[[alternative HTML version deleted]]

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

Re: [R] knitr error with Lyx

2012-11-29 Thread Mark Lamias 
Thanks, Yihui!

Luckily I kept R-2.15.0 and left it untouched (so I can continue to use that 
for now).  If it helps any, I was able to go back into Lyx and change the path 
to point to R-2.15.0 and I also changed the windows path environment variable 
to point to the old version.  After doing this, LyX worked fine with no problem 
on the code below.  Changing the paths back to the new version  R-2.15.2, 
generates the error below.

If anyone else has any idea how to resolve this, either through R or a 
Lyx/LaTeX fix, I'd be all ears.

Thanks, again for your response, Yihui!

Sincerely yours,

Mark J. Lamias





 From: Yihui Xie x...@yihui.name

Cc: r-help@r-project.org r-help@r-project.org 
Sent: Thursday, November 29, 2012 2:02 PM
Subject: Re: [R] knitr error with Lyx

Quite a few people have had this problem, but since I'm unable to
reproduce it, I'm not exactly sure how to fix it either. A few
references that might be helpful to you:

http://stackoverflow.com/q/12448507/559676
https://github.com/yihui/knitr/issues/413

It is very likely to be a pure LaTeX problem. Letting MikTeX install
the missing LaTeX packages on the fly might solve the problem.

Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA



 R Users,


 I just upgraded my version of R from R-2.15.0 to R-2.15.2 and installed the 
 latest versions of LyX and MikTex running Windows 7 Ultimate, 64-bit OS.  
 Prior to the upgrade, I was using Lyx with knitr to generate a document with 
 no problems.  However, after the upgrade, and using the same LyX document, 
 I'm receiving the following error when I attempt to compile the document:

 \end{verbatim}
 The control sequence at the end of the top line
 of your error message was never \def'ed. If you have
 misspelled it (e.g., `\hobx'), type `I' and the correct
 spelling (e.g., `I\hbox'). Otherwise just continue,
 and I'll forget about whatever was undefined.

 I have determined that the error is caused when printing the anova results 
 from the anova statement in my R source code, but can't seem to resolve the 
 issue.  Here is an example code chunk that creates the error:

 NonCP1, fig.width=6, fig.height=4, out.width='.8\\linewidth' ,par=FALSE=
 #Read in data
 y=c( 67, 73, 83, 89, 65, 91, 87, 86, 155, 127, 147, 212, 108, 100, 90, 153, 
 140, 142, 121, 150, 33, 8, 46, 54 )
 temp=as.factor(c(rep(seq(360, 380, 10), each=4), rep(seq(380, 360, -10), 
 each=4)))
 coat=as.factor(rep(seq(1, 4), 6))
 replicate=as.factor(rep(seq(1, 6), each=4))
 #Obtain Factorial/Incorrect Model
 o=lm(y~temp*coat)
 ano=anova(o)
 ano
 @


 Removing the ano=anova(o) or ano lines in the code chunk allows the document 
 to compile with no problem.  Does anyone else have this problem or did I do 
 something wrong when I migrated to the newer versions?

[[elided Yahoo spam]]

 Sincerely yours,

 Mark J. Lamias
         [[alternative HTML version deleted]]


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

Re: [R] Can R be embedded in html?

2012-11-29 Thread Mark Lamias 
I haven't used it yet, so I can't comment about the stability or 
appropriateness of it, but R.NET looks somewhat promising for this:

http://rdotnet.codeplex.com/

Shiny is another potential way to go.

You could also take a look at David Firth's CGIwithR:  
http://www2.warwick.ac.uk/fac/sci/statistics/staff/academic-research/firth/software/cgiwithr/.

--Mark J. Lamias





 From: Gergely Daróczi gerg...@snowl.net
To: jagat.k.sh...@wellsfargo.com 
Cc: r-help@r-project.org; cha...@bio.ntnu.no 
Sent: Thursday, November 29, 2012 11:34 AM
Subject: Re: [R] Can R be embedded in html?

I grab the opportunity to suggest giving a try `pander` package, which
holds a forked version of `brew` called `Pandoc.brew`.
If you are familiar with `brew`, you could easily create nicely formatted
HTML output automatically beside other formats, example:
http://rapporter.github.com/pander/#examples

But if you want to run those files directly, then RApache is the way to go
with `RHandler brew::brew`, although that would end up in a spaghetti code.
It's better to keep R and HTML separately in the long run.

Best,
Gergely

On Thu, Nov 29, 2012 at 4:41 PM, jagat.k.sh...@wellsfargo.com wrote:

 Try R package 'brew'? From its package description

 Description:        brew implements a templating framework for mixing
                     text and R code for report generation. brew
                     template syntax is similar to PHP, Ruby's erb
                     module, Java Server Pages, and Python's psp module.

 There is also the rApache project and many other useful links mentioned on
 R Web Interfaces on the R FAQ page.

 Jagat

  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of chawla
  Sent: Wednesday, November 28, 2012 9:04 AM
  To: r-help@r-project.org
  Subject: [R] Can R be embedded in html?
 
  Hi
  I have previously used PHP script in HTML to connect website with the
  database and do analysis.
  Is it also possible to execute R scripts within HTML files?
  Basically I want to create an application where user can input data
  such
  as gene list and their expression values, which can be processed by R
  and result be displayed or made available for download.
  If there is some guide on how to do this please give the link.
  Thanks
  Konika
 
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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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] knitr error with Lyx

2012-11-29 Thread Mark Lamias 
Hi, Yihui,

Attached is an HTML Diff report of the two files.  The left pane contains the 
R-2.15.0 file. 

Thanks.

--Mark





 From: Yihui Xie x...@yihui.name
To: Mark Lamias mlam...@yahoo.com 
Cc: r-help@r-project.org r-help@r-project.org 
Sent: Thursday, November 29, 2012 2:43 PM
Subject: Re: [R] knitr error with Lyx
 
That is very helpful! Just to continue debugging, can you save the two
versions of the tex files produced from LyX with different versions of
R and do a diff on them? It sounds like something has changed from R
2.15.0 to 2.15.2.

Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA


On Thu, Nov 29, 2012 at 1:26 PM, Mark Lamias mlam...@yahoo.com wrote:
 Thanks, Yihui!

 Luckily I kept R-2.15.0 and left it untouched (so I can continue to use that
 for now).  If it helps any, I was able to go back into Lyx and change the
 path to point to R-2.15.0 and I also changed the windows path environment
 variable to point to the old version.  After doing this, LyX worked fine
 with no problem on the code below.  Changing the paths back to the new
 version  R-2.15.2, generates the error below.

 If anyone else has any idea how to resolve this, either through R or a
 Lyx/LaTeX fix, I'd be all ears.

 Thanks, again for your response, Yihui!


 Sincerely yours,

 Mark J. Lamias
   C:\Users\Mark\Desktop\Test\2.15.0.tex C:\Users\Mark\Desktop\Test\2.15.2.tex
   \batchmode \batchmode
   \makeatletter \makeatletter
   \def\input@path{{C:/Users/Mark/Desktop//}}
   \def\input@path{{C:/Users/Mark/Desktop//}}
   \makeatother \makeatother
   \documentclass{article}   \documentclass{article}\usepackage{graphicx,
   color}
   \usepackage{graphicx, color}  %% maxwidth is the original width if it is
   less than linewidth
 %% otherwise use linewidth (to make sure the graphics do not exceed the
   margin)
 \makeatletter
 \def\maxwidth{ %
  \ifdim\Gin@nat@width\linewidth
\linewidth
  \else
\Gin@nat@width
  \fi
 }
 \makeatother

 \IfFileExists{upquote.sty}{\usepackage{upquote}}{}
 \definecolor{fgcolor}{rgb}{0.2, 0.2, 0.2}
   \newcommand{\hlnumber}[1]{\textcolor[rgb]{0,0,0}{#1}}%
   \newcommand{\hlnumber}[1]{\textcolor[rgb]{0,0,0}{#1}}%
   \newcommand{\hlfunctioncall}[1]{\textcolor[rgb]{.5,0,.33}{\textbf{#1}}}%
   \newcommand{\hlfunctioncall}[1]{\textcolor[rgb]{0.501960784
   313725,0,0.329411764705882}{\textbf{#1}}}%
   \newcommand{\hlstring}[1]{\textcolor[rgb]{.6,.6,1}{#1}}%
   \newcommand{\hlstring}[1]{\textcolor[rgb]{0.6,0.6,1}{#1}}%
   \newcommand{\hlkeyword}[1]{\textbf{#1}}%
   \newcommand{\hlkeyword}[1]{\textcolor[rgb]{0,0,0}{\textbf{#1}}}%
   \newcommand{\hlargument}[1]{\textcolor[rgb]{.69,.25,.02}{#1}}%
   \newcommand{\hlargument}[1]{\textcolor[rgb]{0.6901960784313
   73,0.250980392156863,0.0196078431372549}{#1}}%
   \newcommand{\hlcomment}[1]{\textcolor[rgb]{.18,.6,.34}{#1}}%
   \newcommand{\hlcomment}[1]{\textcolor[rgb]{0.18039215686274
   5,0.6,0.341176470588235}{#1}}%
   \newcommand{\hlroxygencomment}[1]{\textcolor[rgb]{.44,.48,.7}{#1}}%
   \newcommand{\hlroxygencomment}[1]{\textcolor[rgb]{0.4392156
   8627451,0.47843137254902,0.701960784313725}{#1}}%
   \newcommand{\hlformalargs}[1]{\hlargument{#1}}%
   \newcommand{\hlformalargs}[1]{\textcolor[rgb]{0.69019607843
   1373,0.250980392156863,0.0196078431372549}{#1}}%
   \newcommand{\hleqformalargs}[1]{\hlargument{#1}}%
   \newcommand{\hleqformalargs}[1]{\textcolor[rgb]{0.690196078
   431373,0.250980392156863,0.0196078431372549}{#1}}%
   \newcommand{\hlassignement}[1]{\textbf{#1}}%
   \newcommand{\hlassignement}[1]{\textcolor[rgb]{0,0,0}{\textbf{#1}}}%
   \newcommand{\hlpackage}[1]{\textcolor[rgb]{.59,.71,.145}{#1}}%
   \newcommand{\hlpackage}[1]{\textcolor[rgb]{0.58823529411764
   7,0.709803921568627,0.145098039215686}{#1}}%
   \newcommand{\hlslot}[1]{\textit{#1}}% \newcommand{\hlslot}[1]{\textit{#1}}%
   \newcommand{\hlsymbol}[1]{#1}%
   \newcommand{\hlsymbol}[1]{\textcolor[rgb]{0,0,0}{#1}}%
   \newcommand{\hlprompt}[1]{\textcolor[rgb]{.5,.5,.5}{#1}}%
   \newcommand{\hlprompt}[1]{\textcolor[rgb]{0.2,0.2,0.2}{#1}}%

   \usepackage{color}%

   \newsavebox{\hlnormalsizeboxclosebrace}%
   \newsavebox{\hlnormalsizeboxopenbrace}%
   \newsavebox{\hlnormalsizeboxbackslash}%
   \newsavebox{\hlnormalsizeboxlessthan}%
   \newsavebox{\hlnormalsizeboxgreaterthan}%
   \newsavebox{\hlnormalsizeboxdollar}%
   \newsavebox{\hlnormalsizeboxunderscore}%
   \newsavebox{\hlnormalsizeboxand}%
   \newsavebox{\hlnormalsizeboxhash}%
   \newsavebox{\hlnormalsizeboxat}%
   \newsavebox{\hlnormalsizeboxpercent}%
   \newsavebox{\hlnormalsizeboxhat}%
   \newsavebox{\hlnormalsizeboxsinglequote}%
   \newsavebox{\hlnormalsizeboxbacktick}%

   \setbox\hlnormalsizeboxopenbrace=\hbox{\begin{normalsize}\v
   erb.{.\end{normalsize}}%
   \setbox\hlnormalsizeboxclosebrace=\hbox{\begin{normalsize

Re: [R] binning by frequency

2012-11-27 Thread Mark Lamias 
You might find the binning function in the sm package helpful here.

--Mark Lamias




 From: Santosh santosh2...@gmail.com
To: r-help r-help@r-project.org 
Sent: Tuesday, November 27, 2012 9:59 AM
Subject: [R] binning by frequency

Dear Rxperts,

is there way to identify intervals from continuous data (having some kind
of a pattern) and then pick the value of most frequency?


a1 - round(rnorm(50,mean=0,0.1),2)
a2 - round(rnorm(50,mean=1,0.2),1)
a3 - round(rnorm(50,mean=5,1),0)
a4 - round(rnorm(50,mean=14,4),0)
a5 - round(rnorm(50,mean=30,8),0)

b1 - rbind(a1,a2,a3,a4,a5)

hist(b1,brea=100) # shows intervals and values with varying frequency.

unlike the mean values of a1 a5 above, I don't know the nominal values.
I would like an algorithm to identify intervals and pick the value with
most frequency.

I tried cut, split and was not successful.
Any suggestions/tips are highly welcome.
Thanks and regards,
Santosh

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Re: [R] Help in splitting the records

2012-11-26 Thread Mark Lamias 
In the example below, I don't see any $$$ separators.  Are you sure this is 
supposed to be the separate or just a single dollar sign?  If this is the case, 
you don't specify what is to happen when multiple separators appear next to 
each other.  From your example, it appears that two separators are to be 
treated as one?  




 From: arunkumar akpbond...@gmail.com
To: r-help@r-project.org 
Sent: Monday, November 26, 2012 5:51 AM
Subject: [R] Help in splitting the records

Hi

I have set of records seperated by a separator say $$$ i want to get the
values in a dataframe.

eq

qwer$$12$$qwre
ewrtr$7789$ewwe

I want the output as\

V1       V2    V3
qwer 12 qwre
ewrtr  7789 ewwwe



Please help me






-
Thanks in Advance
        Arun
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] How to Label Cases in Scatterplots?

2012-11-25 Thread Mark Lamias 
You could use something like this:


#Generate 26 simulated data points

y=rnorm(26)
x=1:26
your.labels=letters
#Plot data

plot(x, y, xlim=c(min(x), max(x)+.25), ylim=c(min(y), max(y)+.25))
#Label points

text(x+.15, y+.15, your.labels)



You could also use the textxy function in the calibrate package as show here:  
http://www.r-bloggers.com/r-tutorial-series-labeling-data-points-on-a-plot/.




 From: john55 johannes...@gmail.com
To: r-help@r-project.org 
Sent: Saturday, November 24, 2012 11:04 AM
Subject: [R] How to Label Cases in Scatterplots?

Hi everyone,

i´m trying to graphically display distributions with r and i´m working with
makrodata from the WVS.

the command i´m using is
                                                      
 plot (Makrodata$v11, Makrodata$v12, xlab=Democracy Score Economist,
 ylab= share religious people)


i´m having an additional variable that identifies respectively labels the
cases with its country name.
how can i implement that variable, so it identifies the various cases in the
scatterplot and looks like the spss screenshot i posted below?

http://r.789695.n4.nabble.com/file/n4650650/Unbenannt.gif 

thanks in advance!



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

Re: [R] Weighted least squares

2012-11-21 Thread Mark Lamias 
I'm guessing that you have more elements in your weights vector than you have 
observations?  It's hard to tell without seeing your data.
 
See below:
 
#Generate fake data
y=rnorm(10)
X1=rnorm(10)
X2=rnorm(10)
X3=rnorm(10)
 
#Create a 10 element vector of weights
wts=abs(rnorm(10))
 
#weighted lm works fine
lm(y~X1+X2+X3, weights=wts)
 
#Add an additional element to weight vector
wts=c(wts, abs(rnorm(1)))
#weighted lm throws warning message 
lm(y~X1+X2+X3, weights=wts)
 
--Mark J. Lamias



From: rbowman16 roosevelt.bow...@gmail.com
To: r-help@r-project.org 
Sent: Wednesday, November 21, 2012 9:34 AM
Subject: [R] Weighted least squares

Hi everyone, 

I admit I am a bit of an R novice, and I was hoping someone could help me
with this error message:

Warning message:
In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
  extra arguments weigths are just disregarded.


My equation is:


lm( Y ~ X1 + X2 + X3, weigths = seq(0.1, 1, by = 0.1))



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Re: [R] Deviation standard variance

2012-11-20 Thread Mark Lamias 
What about something like this:

#Create dummy data
mydf=data.frame(matrix(rnorm(1000), ncol=100))

#Obtain summary stats

my.vars=apply(mydf, MARGIN=2, FUN=var)
my.std=sqrt(my.vars)

#Plot summary stats
plot(my.vars, col=red)
points(my.std, col=blue)






 From: Elli ellilti_...@hotmail.com
To: r-help@r-project.org 
Sent: Tuesday, November 20, 2012 1:32 PM
Subject: Re: [R] Deviation standard  variance
 
I have a data set consisting of 364 variables and I need to graph the
standard deviation and variance of each variable as only the function sd ()
or var () is not enough. And I want to know how I can graph this?



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Re: [R] kinitr

2012-11-20 Thread Mark Lamias 

This is how I learned everything about knitr:

http://yihui.name/knitr/

Yihui is great and his site gives  you pretty much all the information you need 
to get you started.



 From: Muhuri, Pradip (SAMHSA/CBHSQ) pradip.muh...@samhsa.hhs.gov
To: 'R. Michael Weylandt' michael.weyla...@gmail.com 
Cc: r-help@r-project.org r-help@r-project.org 
Sent: Tuesday, November 20, 2012 9:10 AM
Subject: Re: [R] kinitr

Dear Michael,

I really appreciated that you have sent me the link info - Jeromy Anglim's 
Blog. I was exactly looking for this kind of resources about R Markdown and 
knitr. All this would be of immense help.

Thank you so much. 

Pradip

Pradip K. Muhuri, PhD
Statistician
Substance Abuse  Mental Health Services Administration
The Center for Behavioral Health Statistics and Quality
Division of Population Surveys
1 Choke Cherry Road, Room 2-1071
Rockville, MD 20857
 
Tel: 240-276-1070
Fax: 240-276-1260
e-mail: pradip.muh...@samhsa.hhs.gov
 
The Center for Behavioral Health Statistics and Quality your feedback.  Please 
click on the following link to complete a brief customer survey:   
http://cbhsqsurvey.samhsa.gov


-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] 
Sent: Tuesday, November 20, 2012 8:36 AM
To: Muhuri, Pradip (SAMHSA/CBHSQ)
Cc: r-help@r-project.org
Subject: Re: [R] kinitr

On Tue, Nov 20, 2012 at 1:57 AM, Muhuri, Pradip (SAMHSA/CBHSQ)
pradip.muh...@samhsa.hhs.gov wrote:
 Hello,

 I am an Intro-level R and ggplot2 user and looking for resources to self 
 teach dynamic report generation in R using knitr. Any advice would be highly 
 appreciated.

http://jeromyanglim.blogspot.co.uk/2012/05/getting-started-with-r-markdown-knitr.html

Michael Weylandt

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Re: [R] Greek letters on title

2012-11-20 Thread Mark Lamias 
Try this:

titleline=expression(paste(Time evolution with , alpha, =0.2))
plot(x, main=titleline,col=orange)

--Mark J. Lamias






 From: bgnumis bgnu...@gmail.com
To: r-help@r-project.org 
Sent: Tuesday, November 20, 2012 3:47 PM
Subject: [R] Greek letters on title

Hi all,

I want to plot a series x and I want to put on title Time evolution with
alpha=0.2, e.g.,

plot(x, main=Time evotion with alpha=0.2,col=orange)

Is there any posibility to put the greek letter and avoid  written the word?

Many thanks in advance.

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Re: [R] Greek letters on title

2012-11-20 Thread Mark Lamias 
May may also find this helpful, especially if you know LaTex:

http://yihui.name/en/2011/04/produce-authentic-math-formulas-in-r-graphics/






To: bgnumis bgnu...@gmail.com; r-help@r-project.org r-help@r-project.org 
Sent: Tuesday, November 20, 2012 6:22 PM
Subject: Re: [R] Greek letters on title

Try this:

titleline=expression(paste(Time evolution with , alpha, =0.2))
plot(x, main=titleline,col=orange)

--Mark J. Lamias






From: bgnumis bgnu...@gmail.com
To: r-help@r-project.org 
Sent: Tuesday, November 20, 2012 3:47 PM
Subject: [R] Greek letters on title

Hi all,

I want to plot a series x and I want to put on title Time evolution with
alpha=0.2, e.g.,

plot(x, main=Time evotion with alpha=0.2,col=orange)

Is there any posibility to put the greek letter and avoid  written the word?

Many thanks in advance.

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Re: [R] Looking for a test of standard normality

2012-11-08 Thread Mark Lamias 
The Anderson-Darling Test and the Shapiro-Wilk test have considerably more 
power.

I'd refer you to a non-R related discussion board for statistics related 
questions.  Perhaps the following might be more useful:

http://stats.stackexchange.com/questions/8184/most-powerful-gof-test-for-normality

Sincerely yours,

Mark J. Lamias






 From: Rolf Turner rolf.tur...@xtra.co.nz
To: Herschtal Alan alan.hersch...@petermac.org 
Cc: r-help@r-project.org 
Sent: Thursday, November 8, 2012 10:16 PM
Subject: Re: [R] Looking for a test of standard normality


Others may correct me, but I cannot imagine any test of standard normality
giving appreciably more power than is given by the Kolmogorov-Smirnov
test.

I also wonder about the point of testing for (standard) normality in the
first place.  There is a quote --- I think it refers to testing for 
heteroscedasticity,
but I believe it applies equally to testing for normality  --- about such 
testing
being analogous to going out of the harbour in a rowing dinghy to see if it's
safe for an ocean liner to put to sea.

    cheers,

        Rolf Turner

On 09/11/12 13:23, Herschtal Alan wrote:
 Dear list members,
 
 I am looking for a goodness of test that will tell me if a sample is
 likely to have come from a standard normal distribution. I can find
 plenty of omnibus tests for normality in the nor.test package, but none
 of them appear to allow me to test against the specific alternative that
 the data are not standard normal. My back up option is to use a
 Kolmogorov-Smirnov test, but my impression is that that is not a very
 powerful test. Any suggestions?

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Re: [R] Autofilling a large matrix in R

2012-10-12 Thread Mark Lamias 
If you are after all the possible percentages with two decimal places, why 
don't you use this:

seq(from=0, to=100, by=.01)/100

I'm not really sure what you are trying to do in terms of rows and columns, 
however.  Can you be a bit more specific on what each row/column is?  
Are you trying to group the numbers so that all the entries in a row add up to 
100% and then, once it does, split the following entries onto the next row 
until they add up to 100%, etc.?
Thanks.




 From: wwreith reith_will...@bah.com
To: r-help@r-project.org 
Sent: Friday, October 12, 2012 10:30 AM
Subject: [R] Autofilling a large matrix in R

I wish to create a matrix of all possible percentages with two decimal place
percision. I then want each row  to sum to 100%. I started with the code
below with the intent to then subset the data based on the row sum. This
works great for 2 or 3 columns, but if I try 4 or more columns the number of
rows become to large. I would like to find a way to break it down into some
kind of for loop, so that I can remove the rows that don't sum to 100%
inside the for loop rather than outside it. My first thought was to take
list from 1:10, 11:20, etc. but that does not get all of the points. 

g-as.matrix(expand.grid(rep(list(1:100), times=3)))

Any thoughts how to split this into pieces?



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

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Re: [R] Barplot problem

2012-04-20 Thread Mark Lamias 
Here is an example you can use:
 #Generate random normal variables to use as dummy data for Fisher's Alpha
alphaBCI-rnorm(100)
 
#Assume you have 5 sites, each with 20 data points
sites-rep(c(Site 1, Site 2, Site 3, Site 4, Site 5), each=20)
 
#Making sites a factor will produce the stratified boxplots
sites-as.factor(sites)
 
#Produce boxplot
boxplot(alphaBCI~sites)



From: André Silva berthol...@gmail.com
To: r-help@r-project.org 
Sent: Friday, April 20, 2012 8:28 PM
Subject: [R] Barplot problem

Hello,
I have little experience with r, so I wonder if someone could help me?
I need to plot a diversity measure (Fisher`s alpha) for different sites. I
calculated the values using the following commands:

Data(BCI)
alphaBCI - fisher.alpha(BCI)

I need boxplots showing the data variation between sites. Then I tried:
boxplot(as.data.frame(alphaBCI))

However, it does not appear the box plots for different sites but only the
boxplots for alpha, se, df. Residual and code?

Do anyone know the right expression?

Thank you very much,

Bert

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Re: [R] how to calculate a variance and covariance matrix for a vector

2012-03-12 Thread Mark Lamias 
cov(A) will compute a covariance matrix of the columns of a matrix A.
 
cov(a, b) will compute a covariance of the vectors a and b.
 
here are some examples:
 
#create random vectors
a=rnorm(10)
b=rnorm(10)
 
 
 a
 [1] -0.3110384 -2.6201009 -1.3078118  1.3384743  2.4032381 -0.7245201
 [7] -0.2274577 -1.4619624  2.2165118  1.3621709
 
 b
 [1] -1.1648495 -2.3013411  1.5335541 -0.2165172  1.1157774  1.0360323
 [7] -0.8309283 -0.3403514  0.3640117 -0.6497384

#create a matrix
A=cbind(a,b)  
 
 A
               a          b
 [1,] -0.3110384 -1.1648495
 [2,] -2.6201009 -2.3013411
 [3,] -1.3078118  1.5335541
 [4,]  1.3384743 -0.2165172
 [5,]  2.4032381  1.1157774
 [6,] -0.7245201  1.0360323
 [7,] -0.2274577 -0.8309283
 [8,] -1.4619624 -0.3403514
 [9,]  2.2165118  0.3640117
[10,]  1.3621709 -0.6497384
 
cov(a,b)
 
Covariance
 cov(a,b)
[1] 0.7481056
 
#Covariance matrix
cov(A)
 
 cov(A)
          a         b
a 2.8529990 0.7481056
b 0.7481056 1.3910519
 
By the way, the covariance of two scalars will be zero, since the scalars are 
constants.
 
 
I hope this helps.
 
Sincerely yours,
 
Mark J. Lamias



From: huang jialin huangps...@gmail.com
To: Doran, Harold hdo...@air.org 
Cc: r-help@r-project.org r-help@r-project.org 
Sent: Monday, March 12, 2012 12:08 PM
Subject: Re: [R] how to calculate a variance and covariance matrix for a vector

Doran,

Thanks for your advice.

Sincerely,
Jialin Huang


On Mon, Mar 12, 2012 at 10:56 AM, Doran, Harold hdo...@air.org wrote:

 Always start with the help page. But, the best way to get help on this
 list is to construct some toy data. Even if imperfect, people will work to
 help you. Maybe show us some of your actual data.

 ** **

 Of course, if a and b1 are really scalars, I don’t think there is such a
 thing as the covariance between two scalars

 ** **

 ** **

 * *

 ** **

 *From:* huang jialin [mailto:huangps...@gmail.com]
 *Sent:* Monday, March 12, 2012 11:53 AM
 *To:* Doran, Harold
 *Cc:* r-help@r-project.org
 *Subject:* Re: [R] how to calculate a variance and covariance matrix for
 a vector

 ** **

 Hi,

 ** **

 Thanks for your response. 

 ** **

 I tried to use cov() before, but got NA all over the place.

 ** **

 a and b1 are scalars in the example. So that is the problem. I will start
 over from the formula. 

 ** **

 Thanks again.

 ** **

 Sincerely,

 Jialin Huang

 ** **

 On Mon, Mar 12, 2012 at 10:47 AM, Doran, Harold hdo...@air.org wrote:***
 *

 Can you provide sample data? This suggests that a and b1, for example, are
 scalars. If that is what you actually have then you cannot produce a
 covariance between two scalars.


  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On
  Behalf Of huang jialin
  Sent: Monday, March 12, 2012 11:43 AM
  To: r-help@r-project.org
  Subject: [R] how to calculate a variance and covariance matrix for a
 vector
 
  Hello,
 
  I have a vector {a, b1, b2, b3, b4}. How can I calculate the following
  matrix:
 
  var(a)  cov(a, b1)  cov(a, b2)  cov(a, b3)  cov(a, b4)
  cov(a, b1) var(b1)  cov(a, b2)  cov(a, b3)  cov(a, b4)
  ...
  ...
  cov(a, b1)  cov(a, b2)  cov(a, b3)  cov(a, b4) var(b4)
 
  I would very appreciate your inputs. Thank you very much.
 
  Sincerely,
  Jialin Huang
 

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 ** **


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Re: [R] Running Total

2012-03-05 Thread Mark Lamias 
Modifying the method others have already proposed:
 
x - c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
cbind(x, cumsum(x==0))

 
--Mark Lamias



From: Edward Patzelt patze...@umn.edu
To: jim holtman jholt...@gmail.com 
Cc: r-help@r-project.org 
Sent: Monday, March 5, 2012 1:59 PM
Subject: Re: [R] Running Total

Actually in looking at this I need it to only add if a 0 occurs instead of
a 1.

On Mon, Mar 5, 2012 at 12:57 PM, jim holtman jholt...@gmail.com wrote:

 cumsum

 is probably what you want:

  x - c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
 + 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
 + 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
 + 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
 + 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
 + 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
 + 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
 + 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
  cbind(x, cumsum(x))
      x
  [1,] 1  1
  [2,] 1  2
  [3,] 0  2
  [4,] 1  3
  [5,] 1  4
  [6,] 1  5
  [7,] 1  6
  [8,] 1  7
  [9,] 0  7
  [10,] 1  8
  [11,] 0  8
  [12,] 1  9
  [13,] 1  10
  [14,] 1  11
  [15,] 1  12
  [16,] 1  13
  [17,] 0  13


 On Mon, Mar 5, 2012 at 1:51 PM, Edward Patzelt patze...@umn.edu wrote:
  I'm am trying to create a vector that has a running total that adds each
  time a 1 occurs.  here's the code and data
 
  c(1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 1L, 1L, 1L,
  1L, 0L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L,
  0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
  1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L,
  1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L,
  0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L,
  1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L,
  0L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L)
 
 
  total - {};
 
  for(i in 1:length(dat$Valid)){
  total[i] - ifelse(dat$Valid[i-1]==1, total[i] + 1, total[i])
  }
 
 
  Cheers,
 
 
 
  --
  Edward H. Patzelt
  Research Assistant – TRiCAM Lab
  University of Minnesota – Psychology/Psychiatry
  VA Medical Center
  S355 Elliot Hall: 612-626-0072
  www.psych.umn.edu/research/tricam
 
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  and provide commented, minimal, self-contained, reproducible code.
 



 --
 Jim Holtman
 Data Munger Guru

 What is the problem that you are trying to solve?
 Tell me what you want to do, not how you want to do it.




-- 
Edward H. Patzelt
Research Assistant – TRiCAM Lab
University of Minnesota – Psychology/Psychiatry
VA Medical Center
S355 Elliot Hall: 612-626-0072
www.psych.umn.edu/research/tricam

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Re: [R] Is there an easier way to iterate over multiple data frames in R?

2011-11-23 Thread Mark Lamias 
Depending upon what you are trying to do with them, how about storing the 
dataframes as a multidimensional array of size 4x3x3 in the case below.  After 
that  you can use the sapply function with the 
appropriate MARGIN argument specified:

For example


mydfs - as.array(list(df1, df2, df3))

sapply(X=mydfs, MARGIN=c(1,2), FUN=mean)


 [,1]    [,2]   [,3]
V1  0.8206239 -0.51637848 -0.9025252
V2 -0.2828919  0.41837384 -0.4042646
V3 -0.8996891 -0.06284681 -0.2459261
--Mark Lamias



 From: Kaiyin Zhong kindlych...@gmail.com
To: r-help r-help@r-project.org 
Sent: Wednesday, November 23, 2011 10:34 AM
Subject: [R] Is there an easier way to iterate over multiple data frames in R?

 for (d in paste('df', 1:3, sep='')) {
+  assign(d, as.data.frame(replicate(3, rnorm(4
+ }
 dats = list(df1,df2,df3)
 for (i in 1:length(dats)) {
+  names(dats[[i]]) = c('w', 'l', 'h')
+ }
 dats
[[1]]
            w           l           h
1  1.24319239 -0.05543649  0.05409178
2  0.05124331 -1.89346950  0.33896273
3 -1.69686777 -0.35963008 -0.91720034
4  1.30786112 -0.23953238  0.94139356

[[2]]
          w           l          h
1 -1.238519 -0.12352187 -1.2577607
2  1.180469  2.38836107  2.9139199
3  1.494369 -0.07088712  0.2372746
4  1.942535  1.47911615  1.1431675

[[3]]
           w          l          h
1  1.0198692 -1.4222194  1.9486072
2  0.3057461  1.7630326 -0.6501801
3 -0.5576854 -1.1637263 -0.1782680
4  0.6625268  0.6913202  0.9588915
 i = 1
 for (n in paste('df', 1:3, sep='')) {
+  assign(n, dats[[i]])
+  i = i+1
+ }
 df1
            w           l           h
1  1.24319239 -0.05543649  0.05409178
2  0.05124331 -1.89346950  0.33896273
3 -1.69686777 -0.35963008 -0.91720034
4  1.30786112 -0.23953238  0.94139356

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Re: [R] R-latex syntax highlighting?

2011-11-23 Thread Mark Lamias 
I use pgfSweave so that I can embed R code and output in LaTeX;
pgfSweave depends on the formatR package to format the code, and the highlight
package to highlight R code; graphics are based on tikzDevice, which has better 
quality than R's PDF or postscript output, in my opinion.

Check out the developer's page here:  
http://yihui.name/en/2011/05/sweave-and-pgfsweave-in-lyx-2-0-x-experimental/ abd

http://www.r-bloggers.com/how-to-start-using-pgfsweave-in-lyx-in-one-minute/





 From: Giovanni Azua brave...@gmail.com
To: r-help@r-project.org 
Sent: Wednesday, November 23, 2011 9:33 AM
Subject: [R] R-latex syntax highlighting?

Hello,

Can anyone provide or point me to a good setup for the listings latex package 
that would produce nice R-syntax highlighting?

I am using an example I found in internet for setting up listings like this:
\lstset{
language=R,
basicstyle=\scriptsize\ttfamily,
commentstyle=\ttfamily\color{gray},
numbers=left,
numberstyle=\ttfamily\color{red}\footnotesize,
stepnumber=1,
numbersep=5pt,
backgroundcolor=\color{white},
showspaces=false,
showstringspaces=false,
showtabs=false,
frame=single,
tabsize=2,
captionpos=b,
breaklines=true,
breakatwhitespace=false,
title=\lstname,
escapeinside={},
keywordstyle={},
morekeywords={}
}

But I know that using the color latex package it can look a lot nicer and I 
guess some people might have configured that before and are willing to share it?

I'm also interested in the a similar problem best way to put R code with syntax 
highlighting into a presentation. I use Apple's Keynote. 

Many thanks in advance,
Best regards,
Giovanni
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Re: [R] one sample Wilcoxon test using 'coin'

2011-10-31 Thread Mark Lamias 
The wilcoxon test in coin are meant only for Testing the equality of the 
distributions of a numeric response in *two or more* independent groups against 
shift alternatives whereas the wilcoxon test in the base pacakge Performs 
*one-* and two-sample Wilcoxon tests on vectors of data; the latter is also 
known as ‘Mann-Whitney’ test.   So, wilcoxon_test and wilcoxonsign_test 
in coin will not work for one-sample.  Why not just use wilcox.test?
 
Mark J. Lamias



From: Holger Taschenberger holger.taschenber...@mpi-bpc.mpg.de
To: r-help@r-project.org
Sent: Monday, October 31, 2011 11:12 AM
Subject: [R] one sample Wilcoxon test using 'coin'

Hi,

R allows me to run a one sample Wilcoxon test like this:
wilcox.test(c(1,3.5,2.1,4,1.5,5), mu=2, exact=TRUE)

The function 'wilcoxsign_test' from the package 'coin' should (I
suppose) be able to calculate exact p values even if there are ties in
the ranks. However, I couldn't find information on how to run a one
sample test using 'wilcoxsign_test' like in the example above.

Can anybody help?

        Thanks,
                Holger

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Re: [R] MLE estimation

2011-02-27 Thread Mark Lamias 
Check out Casella and Berger's Statistical Inference.  ISBN 978-81-315-0394-2 
or http://en.wikipedia.org/wiki/Maximum_likelihood as an online reference.

--Mark J. Lamias


From: Ning Cheng wakinchaue...@gmail.com
To: r-help@r-project.org
Sent: Sunday, February 27, 2011 3:19 PM
Subject: [R] MLE estimation

Dear List,
This problem is more a statistic one than a R one.

Any one can recommend me some references website or online paper on
maximum likelihood estimation?I'm now working on that,while still
doubt how to prove that the estimated parameters are normal
distributed.

Thanks for your time and help!

Best,
Ning

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[R] R version of SAS/IntrNet

2010-05-12 Thread Mark Lamias 
Does an R package exist that is similar to SAS/Intrnet 
(http://support.sas.com/documentation/onlinedoc/intrnet/index.html) that will 
allow users to take parameters passed to it by a web request, produce results 
and return them to the web-browser in HTML format?  I tried looking through the 
package descriptions, but didnt' see anything like this.  Any direction or 
suggestions you could provide would be greatly appreciated.

Thanks.

Sincerely yours,

Mark J. Lamias
Statistical Consultant  Survey Methodologist



  
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[R] Improving loop performance

2010-05-11 Thread Mark Lamias 
R-users,

I have the following piece of code which I am trying to run on a dataframe 
(aga2) with about a half million records.  While the code works, it is 
extremely slow.  I've read some of the help archives indicating that I should 
allocate space to the p1 and ags1 vectors, which I have done, but this doesn't 
seem to improve speed much.  Would anyone be able to provide me with advice on 
how I might be able to speed this up?


p1 - character(dim(aga2)[1])
ags - character(dim(aga2)[1])
for (i in 1:dim(aga2)[1])
{
 if (aga2$first.exon[i]==TRUE)
 {
  p1[i]-as.character(aga2[i, AP])
  ags[i]-as.character(aga2[i, AS])
  
 }
 else 
 {
  p1[i]-paste(p1[i-1], aga2[i, AP], sep=,)
  ags[i]-paste(ags[i-1], aga2[i, AS], sep=,)
 }
}

Thanks.

--Mark Lamias


  
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[R] Fw: Improving loop performance

2010-05-11 Thread Mark Lamias 
I will clarify my problem as others has asked for more detail:

I have a dataframe, aga2, that looks like this:

   Row.ID AgilentProbe GeneSymbol GeneID Exons AgilentStart first.geneid 
first.exon last.geneid last.exon
8    1348 A_23_P116898    A2M  2    34  9112685 TRUE   
TRUE    TRUE  TRUE
62  19410  A_23_P95594   NAT1  9 4 18124656 TRUE   
TRUE    TRUE  TRUE
39  10323  A_23_P31798   NAT2 10 2 18302422 TRUE   
TRUE    TRUE  TRUE
21   5353 A_23_P162918   SERPINA3 12 5 94150936 TRUE   
TRUE   FALSE FALSE
22    A_23_P162913   SERPINA3 12 5 94150800    FALSE  
FALSE   FALSE FALSE
98  29990 A_32_P151937   SERPINA3 12 5 94150720    FALSE  
FALSE   FALSE  TRUE
33   9516   A_23_P2920   SERPINA3 12 7 94158435    FALSE   
TRUE   FALSE  TRUE
96  29595 A_32_P124727   SERPINA3 12 8 94160018    FALSE   
TRUE    TRUE  TRUE
57  18176  A_23_P80570  AADAC 13 5    153028473 TRUE   
TRUE    TRUE  TRUE
46  16139  A_23_P56529   AAMP 14 9    218838396 TRUE   
TRUE    TRUE  TRUE

For the above example, I would like to end up with a vector, probe1, like this, 
based upon the AgilentProbe values:

A_23_P116898
A_23_P95594  
A_23_P31798 
A_23_P162918   
A_23_P162918,A_23_P162913,
A_23_P162918,A_23_P162913,A_32_P151937 
A_23_P2920  
A_32_P124727
A_23_P80570
A_23_P56529

I build up each element of the vector based upon the value of last.exon.  If 
the value of last.exon is FALSE, I'd like to obtain the previous value of 
AgilentProbe and concatenate it with the current value, and then move on to the 
next element. 

As stated previously, this code works, but it is very slow with larger datasets:

probe1 - character(dim(aga2)[1])
agstart - character(dim(aga2)[1])

for (i in 1:dim(aga2)[1])
{
 if (aga2$first.exon[i]==TRUE)
 {
  probe1[i]-as.character(aga2[i, AgilentProbe])
  agstart[i]-as.character(aga2[i, AgilentStart])
  
 }
 else 
 {
  probe1[i]-paste(probe1[i-1], aga2[i, AgilentProbe], sep=,)
  agstart[i]-paste(agstart[i-1], aga2[i, AgilentStart], sep=,)
 }
}


I tried a few of the previous suggestions (and tried modifying them), but they 
didn't seem to quite do the trick.  Any assistance would be greatly appreciated.

Thanks a million.

--Mark Lamias






 


- Forwarded Message 
From: jim holtman jholt...@gmail.com
To: Mark Lamias mlam...@yahoo.com
Cc: r-help@r-project.org
Sent: Tue, May 11, 2010 12:46:26 PM
Subject: Re: [R] Improving loop performance

It was supposed to be  'head(p1, -1)'  instead of  'tail(p1, -1)'


On Tue, May 11, 2010 at 12:17 PM, Mark Lamias mlam...@yahoo.com wrote:

R-users,

I have the following piece of code which I am trying to run on a dataframe 
(aga2) with about a half million records.  While the code works, it is 
extremely slow.  I've read some of the help archives indicating that I should 
allocate space to the p1 and ags1 vectors, which I have done, but this doesn't 
seem to improve speed much.  Would anyone be able to provide me with advice on 
how I might be able to speed this up?


p1 - character(dim(aga2)[1])
ags - character(dim(aga2)[1])
for (i in 1:dim(aga2)[1])
{
 if (aga2$first.exon[i]==TRUE)
 {
  p1[i]-as.character(aga2[i, AP])
  ags[i]-as.character(aga2[i, AS])
  
 }
 else
 {
  p1[i]-paste(p1[i-1], aga2[i, AP], sep=,)
  ags[i]-paste(ags[i-1], aga2[i, AS], sep=,)
 }
}

Thanks.

--Mark Lamias



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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?



  
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