date:20121024

HI,
May be this helps:
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
  trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
  group - gl(2,10,20, labels=c(Ctl,Trt))
  weight - c(ctl, trt)
  lm.D9 - lm(weight ~ group)



fun1-function(x){
 res-c(paste(as.character(summary(x)$call),collapse= ),
 x$coefficients[1],
 x$coefficients[2],
 length(x$model),
 summary(x)$coefficients[2,2],
 summary(x)$r.squared,
 summary(x)$adj.r.squared,
 summary(x)$fstatistic,
 
pf(summary(x)$fstatistic[1],summary(x)$fstatistic[2],summary(x)$fstatistic[3],lower.tail=FALSE))
 names(res)-c(call,intercept,slope,n,slope.SE,r.squared,Adj. 
r.squared,
 F-statistic,numdf,dendf,p.value)
 return(res)}  
res2-fun1(lm.D9)

write.csv(res2,newregsummary.csv)
 

 x 
call lm weight ~ group 
intercept 5.032 
slope -0.371 
n 2 
slope.SE 0.3114348514 
r.squared 0.073077599 
Adj. r.squared 0.02158191 
F-statistic 1.4191012974 
numdf 1 
dendf 18 
p.value 0.249023166 

A.K.




- Original Message -
From: fxen3k f.seha...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Tuesday, October 23, 2012 7:28 AM
Subject: [R] Export summary from regression output

Hi there,

I tried it many times but didn't get it worked.

I just want to export the summary of a OLS regression (lm() function) into a
csv-file including the call-formula, coefficients, r-squared, 
adjusted r-squared  and f statistic. 

I know I can export:
write.csv2(Regression_60d_ann$coefficients, Regression_60d_ann.csv)
But then I only get the coefficients, but not all the other output...

I tried creating a matrix and I wanted to put in
Regression_60d_ann$coefficients, Regression_60d_ann$adj.r.squared,
Regression_60d_ann$r.squared, etc. but it didn't work due to different
length of rows.


Can anyone help or has a better solution?

Thanks in advance
Felix



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Re: [R] incomplete final line found by readTableHeader

2012-10-24 Thread Shane2012

Thanks, JeffThis is from a course, but the course is just needed to take some 
slides for presentation. I just learn R by myself, and want this skill more 
practical. So I try to use the innovative way to perform a more professional 
presentation.hence, no worry about the homework support issue. However, I still 
got the problem, I think I did the right thing to make the excel, then change 
the extension to csv to make sure it has the appropriate comma in it.But the 
warning still be there. I will try other more. Thank you.
 Date: Tue, 23 Oct 2012 20:14:53 -0700
From: ml-node+s789695n4647264...@n4.nabble.com
To: wenbo...@live.cn
Subject: Re: incomplete final line found by readTableHeader

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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DCN:[hidden email]Basics: ##.#.   ##.#.  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.

Shane2012 [hidden email] 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] How to use tapply with more than one variables grouped



Hi,

Just a modification of David's method:

apply(dat1,2,function(x) names(which.max(x[!is.na(x)]))==Obama)
#   AL    AR    CA    NY 
#FALSE FALSE  TRUE  TRUE 
names(dat1)[apply(dat1,2,function(x) names(which.max(x[!is.na(x)]))==Obama)] 
#[1] CA NY
A.K.



- Original Message -
From: David Winsemius dwinsem...@comcast.net
To: noobmin pseudov...@hotmail.com
Cc: r-help@r-project.org
Sent: Tuesday, October 23, 2012 8:45 PM
Subject: Re: [R] How to use tapply with more than one variables grouped


On Oct 23, 2012, at 1:25 PM, noobmin wrote:

             AL  AR  CA  NY 
 Doug    250 250 250  NA 
 Jennifer  20 340 300 100 
 Michele  250 500 250  60 
 Obama    15  45 520 600 
 
 My English is not very good, I'll try again. I want to list ALL states in
 the country where Obama had greater contribution. The table above shows the
 total contribution received by each candidate in a given state. To AL state
 obama not received more than Doug. For the AR state he received no more than
 others candidates. For the CA state he received a total of $ 520, which is
 520300250=250 and should be selected. In NY also had the largest
 contribution, $ 600, 60010060 and should therefore be selected.
 
 I want to make it to the N presidency candidates and M states of the
 country. The table above is only an example.

Perhaps:

 apply(dat, 2, function(x) x[Obama] == max(x, na.rm=TRUE) )
   AL    AR    CA    NY 
FALSE FALSE  TRUE  TRUE 

Or perhaps:
 names(dat)[ apply(dat, 2, function(x) x[Obama] == max(x, na.rm=TRUE) ) ]
[1] CA NY

 

 Sorry again, for me it was clear. = ( 
 Thanks
 
 
 
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[R] Z score

2012-10-24 Thread Vedant Sharma

Hi,

I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it should be a simple thing, but I am unable to do it right
now !

The example re the structure of the spreadsheet is  -

# Example:

MyFile - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

And, I think this formula that can be used for z score is -

(x-mean(x))/sd(x)

And, apply() function for rows should work. But bottomline - I am unable to
do it correctly.

Could you show me - using apply () or some other alternative function.

Thank you.

Cheers,
Ved

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[R] R untar error

2012-10-24 Thread zhaohu44

Hi,
When I use untar command and I got an error.

 untar(/public03/ara_gse_raw_data/GSE26679_RAW.tar,exdir=/public03/data/arabidopsis_data/GSE26679_RAW)
sh: /tmp/RtmpzQGJvh/file5f09b232: No such file or directory

I want to know how to solve this problem.

Thanks very much for any tips.


2012-10-24



zhaohu44
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Re: [R] ROC with more than one predictor

2012-10-24 Thread Suzen, Mehmet

On Tue, Oct 23, 2012 at 11:52 PM, Luigi marongiu.lu...@gmail.com wrote:

 Any tip on how to proceed?

You may want to do check rocplus package. Its vignette is pretty good.
http://cran.r-project.org/web/packages/rocplus/vignettes/rocplus.pdf
In your case, I think pairwise comparison would be one approach


Best,
Mehmet

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Re: [R] Changing radii line type in radial plots

2012-10-24 Thread Peter Ehlers


On 2012-10-23 15:22, bwone wrote:

Hello all,

Is it possible to change the radii line type in radial plots? I wasn't able
to find anything online.


Do please be more specific: which package's radial plot function
are you using? Package plotrix has radial.plot() with an 'lty'
argument.

Peter Ehlers



Thanks,
Bern



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Re: [R] List of multidimensional arrays


On Oct 23, 2012, at 11:36 PM, Loukia Spineli wrote:

 Dear all,
 
 I am trying to create a list, where each list element is a vector of
 different length arrays that contain 2by2 matrices. To be more specific
 there are 11 treatments that are compared with placebo (we have 11
 comparisons) and each comparison is studied by a different number of trials
 and each trial has a different number of missing participants in both arms.
 The length of the list is equal to the number of comparisons. In each
 comparison the number of arrays is equal to the number of trials that study
 this comparison. For instance 4 trials compare PAR with placebo. So, for
 this comparison we have 4 arrays and each array has a length equal to the
 producy of the number of participants in each arm. These arrays contain 2x2
 matrices.
 
 I have attached a document with the data and the code.

Well, A for effort but somewhat lower grade for not reading the Posting Guide 
and attaching a MIME-text file.

 I cannot create the
 list results the way I have described above. It creates only the  matrices
 for the first array (that has length equal to 135) of the first comparison
 leaving the rest 10 comparisons NULL.
 Any suggestion would be  really helpful
 
 Thank you in advance,
 Loukia
 __
 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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David Winsemius, MD
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Re: [R] Results not dispalying in R console

2012-10-24 Thread Pascal Oettli


Hello,

And what do you expect from us, without example?

Regards,
Pascal

Le 12/10/24 15:44, arunkumar a écrit :

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] R untar error


On Oct 23, 2012, at 7:43 PM, zhaohu44 wrote:

 Hi,
 When I use untar command and I got an error.
 
 untar(/public03/ara_gse_raw_data/GSE26679_RAW.tar,exdir=/public03/data/arabidopsis_data/GSE26679_RAW)
 sh: /tmp/RtmpzQGJvh/file5f09b232: No such file or directory
 
 I want to know how to solve this problem.

This does not look like an R problem.
 
 
-- 

David Winsemius, MD
Alameda, CA, USA

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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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--- 
Sent from my phone. Please excuse my brevity.

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] List of multidimensional arrays

2012-10-24 Thread Loukia Spineli

You are absolutely right. I read the guidelines! Mistake to attach 'word'
file! The comfort is that at least  I got an A :).
I revise my question approapriately!

On Wed, Oct 24, 2012 at 10:36 AM, David Winsemius dwinsem...@comcast.netwrote:


 On Oct 23, 2012, at 11:36 PM, Loukia Spineli wrote:

  Dear all,
 
  I am trying to create a list, where each list element is a vector of
  different length arrays that contain 2by2 matrices. To be more specific
  there are 11 treatments that are compared with placebo (we have 11
  comparisons) and each comparison is studied by a different number of
 trials
  and each trial has a different number of missing participants in both
 arms.
  The length of the list is equal to the number of comparisons. In each
  comparison the number of arrays is equal to the number of trials that
 study
  this comparison. For instance 4 trials compare PAR with placebo. So, for
  this comparison we have 4 arrays and each array has a length equal to the
  producy of the number of participants in each arm. These arrays contain
 2x2
  matrices.
 
  I have attached a document with the data and the code.

 Well, A for effort but somewhat lower grade for not reading the Posting
 Guide and attaching a MIME-text file.

  I cannot create the
  list results the way I have described above. It creates only the
  matrices
  for the first array (that has length equal to 135) of the first
 comparison
  leaving the rest 10 comparisons NULL.
  Any suggestion would be  really helpful
 
  Thank you in advance,
  Loukia
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 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.

 David Winsemius, MD
 Alameda, CA, USA



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[R] List of multidimensional arrays

2012-10-24 Thread Loukia Spineli

Dear all,

I am trying to create a list, where each list element is a vector of
different length arrays that contain 2by2 matrices. To be more specific
there are 11 treatments that are compared with placebo (we have 11
comparisons) and each comparison is studied by a different number of trials
and each trial has a different number of missing participants in both arms.
The length of the list is equal to the number of comparisons. In each
comparison the number of arrays is equal to the number of trials that study
this comparison. For instance 4 trials compare PAR with placebo. So, for
this comparison we have 4 arrays and each array has a length equal to the
producy of the number of participants in each arm. These arrays contain 2x2
matrices.

I have attached a document with the data and the code. I cannot create the
list results the way I have described above. It creates only the  matrices
for the first array (that has length equal to 135) of the first comparison
leaving the rest 10 comparisons NULL.
Any suggestion would be  really helpful

Thank you in advance,
Loukia


List of multidimensional arrays.pdf
Description: Adobe PDF document
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Re: [R] incomplete final line found by readTableHeader

2012-10-24 Thread Milan Bouchet-Valat

Le mardi 23 octobre 2012 à 20:36 -0700, Shane2012 a écrit :
 Thanks, Jeff
 This is from a course, but the course is just needed to take some
 slides for presentation. I just learn R by myself, and want this skill
 more practical. So I try to use the innovative way to perform a more
 professional presentation.hence, no worry about the homework support
 issue. However, I still got the problem, I think I did the right thing
 to make the excel, then change the extension to csv to make sure it
 has the appropriate comma in it.But the warning still be there. I will
 try other more. Thank you.
The warning is slightly ambiguous in this context. What it means is
*not* the record/observation/individual is incomplete, i.e. needs one
more comma but the last line of the file is incomplete, i.e. does not
end with a line break.

This warning is completely innocuous in this case as David stated: you
can get rid of it by opening your CSV file with a text editor and adding
an empty line at the end of the file, but if you don't do this, things
will work as expected. To make it short: you don't have any problem. ;-)


Regards

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[R] R5: Lock a class field from within a method?

2012-10-24 Thread Johannes Graumann

Hello,

testclass - setRefClass(
  testclass,
  fields = list(testfield = logical),
  methods = list(validate=function(){testfield-TRUE}))

 test - testclass$new()
 test$testfield
logical(0)
 test$validate()
 test$testfield
[1] TRUE

Works just fine for me.

I would love to be able to do something like

testclass - setRefClass(
  testclass,
  fields = list(testfield = logical),
  methods = list(validate=function(){
testfield-TRUE
.self$lock(testfield)
  }))

but am unabel to achieve that. Can anyone point out how to go about 
rendering a field immutable after execution of a specific method?

Sincerely, Joh

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Re: [R] plotting multiple variables in 1 bar graph

2012-10-24 Thread Jim Lemon


On 10/23/2012 07:22 PM, Macy Anonuevo wrote:

I'd greatly appreciate your help in making a bar graph with multiple
variables plotted on it. All the help sites I've seen so far only plot 1
variable on the y-axis

Data set:
I have 6 sites, each measured 5 times over the past year. During each
sampling time, I counted the occurrences of different benthic components
(coral, dead coral, sand, etc.) over 5 transects in each site

site time transect coral deadcoral sand rubble .
S1   time1   trans110   15  10 4
S1   time1   trans2 5 410
6
S1  time1   trans3 10   2 5
7
.
.
.
S5  time5trans5 63   1
6

I used aggregate to get the means of the individual variables (coral, dead
coral, etc.) using the site and time as grouping factors.

aggregate.plot(deadcoral, by=list(SITE=site, TIME=time), FUN=c(mean),
error=c(sd), legend.site=topright, bar.col=rainbow(6))

What I need now is to plot all the variables in 1 site as they change over
time.

What Excel produced:
http://r.789695.n4.nabble.com/file/n4647099/abdeens_benthic_cover.jpg
(The image has mean %cover as the y-value instead of mean count but the
example still applies)

I've spent several hours looking for code to do this but didn't find
anything. I'd use the Excel graph except that it doesn't have the sd or se
bars.



Hi Macy,
This isn't exactly the same as the Excel plot, but it might help you 
out. If your data is in a data frame named abs:


library(plotrix)
barpos-barp(abs,names.arg=names(abs),
 do.first=grid(nx=NA,lty=1,ny=9),col=rainbow(5),
 ylim=c(0,100),staxx=TRUE,ylab=Percent,
 main=Abdeen Benthic Cover Through Time,
 height.at=seq(0,100,by=10))
dispersion(barpos$x,barpos$y,ulim=barpos$y/10)
legend(5,80,paste(T,1:5,sep=),fill=rainbow(5),
 bg=white)

Jim

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Re: [R] SEM multigroup modeling

2012-10-24 Thread John Fox

Dear Maya,

You and I corresponded about this problem yesterday, so I'll respond on the 
r-help list only briefly:

On Tue, 23 Oct 2012 23:12:23 +0300
 Maya Abou Zeid ma...@aub.edu.lb wrote:
 Hello,
 
 I am using the SEM package in R to fit a multigroup latent variable model and 
 ran into some difficulties. I have 2 questions:
 
 
 1.   First, I am getting the following error message and wondering what 
 to do to fix it:
 
 Error in solve.default((N[g] - 1) * robustVcov(mod.g, adj.obj = 
 adj.objects[[g]])) :
 
   system is computationally singular: reciprocal condition number = 
 4.52055e-23

It's not really possible to try to locate the source of the problem without the 
data. You could try several things to get a proper solution, as mentioned in 
?sem, and you might try suppressing the computation of the robust coefficient 
covariance matrix, which is where the problem surfaces.

 
 
 
 2.   Second, I want to specify different models for two groups (Drivers 
 vs. Nondrivers), but I want only 1 parameter to be different, while 
 everything else remains the same. Should I just create two files with the 
 model equations? But then how do I use these 2 files as an input to the 
 multigroupModel or sem functions? I tried the following but got the error 
 message {Error in model[, 2] : incorrect number of dimensions}. Here is a 
 sample of the code:
 
 
 mod.QT.1 - specifyEquations(...) # model equations for category 1
 
 mod.QT.2 - specifyEquations(...) # model equations for category 2
 
 Lst.model - list(mod.QT.1, mod.QT.2)
 
 
 
 mod.mg - multigroupModel(Lst.model, groups=c(Driver, Nondriver))
 
 
 
 sem.QT - sem(mod.mg, data=QTrav, group=Driver_Class1, formula = ~ ..., 
 fixed.x=c(...))

multigroupModel() doesn't take a list of models as an argument but rather 
arbitrarily named iniital arguments, one for each group. See ?multigroupModel 
for (some) more information.

 
 Are there any sample codes for multigroup modeling using the SEM package that 
 I can look at?

There are a couple of examples in ?sem, but I know you've already seen them. 
These are given in a little more detail in 
http://socserv.socsci.mcmaster.ca/jfox/Books/Companion/appendix/Appendix-SEMs.pdf.

Best,
 John


John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

 
 Thanks in advance,
 Maya
 
 
 
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Re: [R] Z score


Hello,

Try the following.

apply(MyFile, 1, scale)

Hope this helps,

Rui Barradas
Em 24-10-2012 07:17, Vedant Sharma escreveu:

Hi,

I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it should be a simple thing, but I am unable to do it right
now !

The example re the structure of the spreadsheet is  -

# Example:

MyFile - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

And, I think this formula that can be used for z score is -

(x-mean(x))/sd(x)

And, apply() function for rows should work. But bottomline - I am unable to
do it correctly.

Could you show me - using apply () or some other alternative function.

Thank you.

Cheers,
Ved

[[alternative HTML version deleted]]

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Re: [R] barplot

2012-10-24 Thread Jim Lemon


On 10/24/2012 03:21 AM, Thiho Jules wrote:

Hi,

I want to make a barplot with the following datasets:

I have a file as following:

name  chr position A1 A2 pop1 pop1 pop2 pop2

I have calculated a measure using all the values in columns pops, the
values are saved in a vector.

Now I want to make a barplot using the values in this vector as the y axis
and the values in the column position in the x-axis.


Hi Thiho,
You can do this with the barp function in the plotrix package. See the 
x argument for placing the bars at arbitrary positions.


Jim

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Re: [R] Z score

Hi,
Try this:
 res-do.call(rbind,lapply(lapply(apply(MyFile,1,function(x) 
x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x) 
x[c(Sample_1,Sample_2,Sample_3)]))
 res
# Sample_1   Sample_2   Sample_3
#Gene_1  0.4931970 -1.1507929  0.6575959
#Gene_2  1.1421818 -0.7179429 -0.4242390
#Gene_3 -0.5773503  1.1547005 -0.5773503
#Gene_4 -0.7071068 NA  0.7071068
A.K.





- Original Message -
From: Vedant Sharma vedantg...@gmail.com
To: R help r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 2:17 AM
Subject: [R] Z score

Hi,

I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it should be a simple thing, but I am unable to do it right
now !

The example re the structure of the spreadsheet is  -

# Example:

MyFile - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

And, I think this formula that can be used for z score is -

(x-mean(x))/sd(x)

And, apply() function for rows should work. But bottomline - I am unable to
do it correctly.

Could you show me - using apply () or some other alternative function.

Thank you.

Cheers,
Ved

    [[alternative HTML version deleted]]

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Re: [R] Z score

Hi,

In cases, with more sample columns, you could also use this:
 res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x) 
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] ))
res2
 #    Sample_1   Sample_2   Sample_3
#Gene_1  0.4931970 -1.1507929  0.6575959
#Gene_2  1.1421818 -0.7179429 -0.4242390
#Gene_3 -0.5773503  1.1547005 -0.5773503
#Gene_4 -0.7071068 NA  0.7071068
A.K.



- Original Message -
From: Vedant Sharma vedantg...@gmail.com
To: R help r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 2:17 AM
Subject: [R] Z score

Hi,

I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it should be a simple thing, but I am unable to do it right
now !

The example re the structure of the spreadsheet is  -

# Example:

MyFile - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

And, I think this formula that can be used for z score is -

(x-mean(x))/sd(x)

And, apply() function for rows should work. But bottomline - I am unable to
do it correctly.

Could you show me - using apply () or some other alternative function.

Thank you.

Cheers,
Ved

    [[alternative HTML version deleted]]

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Re: [R] factor or character


Hello,

Sorry but, what doesn't work? What is the error message or result?
As for stringsAsFactors not keeping character strings as character 
strings maybe this is from calling data.frame from within the 
environment used by with(), but you can use other ways of converting to 
character strings, such as


mod5.sig$snps - as.character(mod5.sig$snps)

before subsetting. And you can also check the result of

colnames(todos) %in% mod5.sig$snp

What does this give you?

Hope this helps,

Rui Barradas
Em 24-10-2012 11:58, Silvano Cesar da Costa escreveu:

Hi,

still doesn't work.



Hello,

I've just seen the error, you are _not_ searching for colnames in
mod5.sig$snps. Corrected:

Selec = todos[ , colnames(todos) %in% mod5.sig$snps]

Hope this helps,

Rui Barradas
Em 23-10-2012 21:17, Silvano Cesar da Costa escreveu:

Hi Rui,


it doesn't work:


(mod5.sig = with(mod5, data.frame(snps = SNP[pvalor5e-8],

stringsAsFactors=FALSE)))

str(mod5.sig)

'data.frame':   76 obs. of  1 variable:
   $ snps: Factor w/ 220 levels rs10058955_A,..: 89 59 88 73 40 35 97
55
87 204 ...

Selec = todos[ , colnames(todos) %in% mod5.sig]
head(Selec)

data frame com 0 colunas e 6  linhas


the problem is the same.

Thanks,





Hello,

When creating the data.frame of snps use the option stringsAsFactors =
FALSE
I believe your error comes from the fact that you are trying to find
colnames in a variable coded as integers (a factor, like str shows).

Hope this helps,

Rui Barradas
Em 23-10-2012 19:02, Silvano Cesar da Costa escreveu:

Hi,

The program below work very well.

(snps = c('rs621782_G', 'rs8087639_G', 'rs8094221_T', 'rs7227515_A',
'rs537202_C'))
Selec = todos[ , colnames(todos) %in% snps]
head(Selec)


But, I have a data set with 1.000 columns and I need extract 70 to use
(like snps in command above).

This 70 snps are in a file. So I create a file to extract them with

(mod5.sig = with(mod5, data.frame(snps = SNP[pvalor  5e-8])))
str(mod5.sig)
(snps = (mod5.sig))

The structure is:
'data.frame':   76 obs. of  1 variable:
$ snps: Factor w/ 220 levels rs10058955_A,..: 89 59 88 73 40 35
97
55
87 204 ...

But it doesn't work. The output is:

Selec = todos[ , colnames(todos) %in% snps]
head(Selec)

data frame with 0 columns and 6 rows

What's is wrong?

Thanks a lot,


-
Silvano Cesar da Costa

Universidade Estadual de Londrina
Centro de Ciências Exatas
Departamento de Estatística

Fone: (43) 3371-4346

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-
Silvano Cesar da Costa

Universidade Estadual de Londrina
Centro de Ciências Exatas
Departamento de Estatística

Fone: (43) 3371-4346
-






-
Silvano Cesar da Costa

Universidade Estadual de Londrina
Centro de Ciências Exatas
Departamento de Estatística

Fone: (43) 3371-4346
-



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[R] Five cases of the Multivariate VECM

2012-10-24 Thread 敷田治誠　クラウジオ

Hello,

I was studying several packages related to time series analysis (urca,
vars, tseries). I understand that we can estimate a VECM and also test
restrictions on alphas and betas. However, I couldn't find a function that
allows me to specify the five cases of VECM (restricted constant,
unrestricted constant, restricted and unrestricted trend and no constant).
Is there any function that fulfills this aspect of VECM analysis in any
package?

Thanks a lot for your time and attention.

Claudio D. Shikida
http://www.shikida.net  and http://works.bepress.com/claudio_shikida/

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[R] recursive function on a structured list of lists (dendrogram)

2012-10-24 Thread Ivan Alves

Dear all,

I have been trying the following without avail and would be very grateful for 
any help.  From a dendrogram (recursive list of lists with some structure), I 
would like to obtain some information of the component lists and of the 
enclosing list at the same time.  In dendrogram-speech I basically would like 
the label of the leaf and the height of the enclosing branch.

A dendrogram example (from the help file of stats::dendrogram), and some 
functions showing how it is structured:

hc - hclust(dist(USArrests), ave)
dend1 - as.dendrogram(hc)
plot(dend1)
str(dend1)
Similarly to dendrapply(), I tried o recursively obtain from the tree a list 
including, for each member (leaf) the height of the list containing it. 
However, I fail to fully grasp how the 'recursiveness' is made within the 
function saving both elements at the leaf and branch levels.  For reference the 
dendrapply function is as follows:

function (X, FUN, ...) 
{
FUN - match.fun(FUN)
if (!inherits(X, dendrogram)) 
stop('X' is not a dendrogram)
Napply - function(d) {
r - FUN(d, ...)
if (!is.leaf(d)) {
if (!is.list(r)) 
r - as.list(r)
if (length(r)  (n - length(d))) 
r[seq_len(n)] - vector(list, n)
r[] - lapply(d, Napply)
}
r
}
Napply(X)
}

I essentially don't manage to 'save' the height of a branch (a list of lists) 
so that it can be used at the next iterations for adding to the leafs there. 
Many thanks for any guidance on how to recursively implement a function.

Kind regards,
Ivan

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Re: [R] After sorting a dataframe by date

2012-10-24 Thread martiny

Ai...I see, i didn't know I can sort rowname.
more study to do :)

cheers,
martin



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Re: [R] Results not dispalying in R console

2012-10-24 Thread Rolf Turner


On 24/10/12 19:44, arunkumar wrote:

Hi

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


I *suspect* that you are displaying the results inside some sort of 
function

(e.g. a for loop, or source() ) whence the results are invisible unless
explicitly print()-ed.   However, given the award-winning vagueness of your
question, it's hard to tell.

Advice:

(1) Learn something about R; don't just hammer and hope.  Read the
introductory manuals and scan the FAQ.

(2) Learn to ask questions in such a manner that they can actually
be answered.  As fortune(182) has it, ``R is lacking a mind_read()
function!''

cheers,

Rolf Turner

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[R] equation solver

2012-10-24 Thread pina

Hi, 

I'm Pina and I'm a student in geology. I'm working with spectral profile of
sand and I have to find the similarity between one spectral profile selected
by hyperspectral image anche one that I created to mix different percentage
of 4 mineral component. I have to find the best mix of percentage of this 4
mineral in order to have the best likeness with the spectral profile chose
by hyperspectral image. Untill now I worked with excel using the function
equation solver to find the best fit but I 'd like to know if is possible do
this operation working in R. 
Thanks for the attention. 
I'm looking forward to hearing from you. 

Regards. 

Pina.



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Re: [R] repeat takes along time

2012-10-24 Thread JPM

Hi Petr

Thanks for reply.



On 24 October 2012 07:26, PIKAL Petr [via R] 
ml-node+s789695n464726...@n4.nabble.com wrote:

 Hi

 It is a piece of my code. At this, i create a matrix(deltab) with each
column is a binomial, but in there are several NAs, and i must restart
the cycle the number of Nas is zero. So, the problem this process takes a
very long time.

 
  Hi
 
  I want help for repeat because it takes a long time
 
  repeat{
  for (i in 1:n){
  probb[i]=sum(Wc[z[,j]=yb[i]])
  }
  deltab[,b]=rbinom(n,1,probb)
  if(length(which(is.na(deltab[,b])==T))==0){break}
  }
 
  This code works but takes much computation time.

 What is this code supposed to do? It gives an error (see below)

  repeat{
 + for (i in 1:n){
 + probb[i]=sum(Wc[z[,j]=yb[i]])
 + }
 + deltab[,b]=rbinom(n,1,probb)
 + if(length(which(is.na(deltab[,b])==T))==0){break}
 + }
 Error: object 'n' not found
 

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

 and some data to test the code.

 Anyway, this kind of code shall be implemented probably in other language
 like C+.

 Regards
 Petr


 
  Anyone could help me to find a solution to take a shor time.
 
  Thanks
 
  Jorge
 
 
 
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[R] Recode function car package erases previous values

2012-10-24 Thread Pancho Mulongeni

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, 
Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
 No Yes 
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
 No Yes no resp 
716   7   0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
  0 no resp 
723   6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table 
pharm
 No Yes no resp 
716   7   6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia
 
Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

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Re: [R] factor or character

2012-10-24 Thread Silvano Cesar da Costa

Hi,

still doesn't work.


 Hello,

 I've just seen the error, you are _not_ searching for colnames in
 mod5.sig$snps. Corrected:

 Selec = todos[ , colnames(todos) %in% mod5.sig$snps]

 Hope this helps,

 Rui Barradas
 Em 23-10-2012 21:17, Silvano Cesar da Costa escreveu:
 Hi Rui,


 it doesn't work:

 (mod5.sig = with(mod5, data.frame(snps = SNP[pvalor5e-8],
 stringsAsFactors=FALSE)))
 str(mod5.sig)
 'data.frame':76 obs. of  1 variable:
   $ snps: Factor w/ 220 levels rs10058955_A,..: 89 59 88 73 40 35 97
 55
 87 204 ...
 Selec = todos[ , colnames(todos) %in% mod5.sig]
 head(Selec)
 data frame com 0 colunas e 6  linhas


 the problem is the same.

 Thanks,




 Hello,

 When creating the data.frame of snps use the option stringsAsFactors =
 FALSE
 I believe your error comes from the fact that you are trying to find
 colnames in a variable coded as integers (a factor, like str shows).

 Hope this helps,

 Rui Barradas
 Em 23-10-2012 19:02, Silvano Cesar da Costa escreveu:
 Hi,

 The program below work very well.

 (snps = c('rs621782_G', 'rs8087639_G', 'rs8094221_T', 'rs7227515_A',
 'rs537202_C'))
 Selec = todos[ , colnames(todos) %in% snps]
 head(Selec)


 But, I have a data set with 1.000 columns and I need extract 70 to use
 (like snps in command above).

 This 70 snps are in a file. So I create a file to extract them with

 (mod5.sig = with(mod5, data.frame(snps = SNP[pvalor  5e-8])))
 str(mod5.sig)
 (snps = (mod5.sig))

 The structure is:
 'data.frame':  76 obs. of  1 variable:
$ snps: Factor w/ 220 levels rs10058955_A,..: 89 59 88 73 40 35
 97
 55
 87 204 ...

 But it doesn't work. The output is:
 Selec = todos[ , colnames(todos) %in% snps]
 head(Selec)
 data frame with 0 columns and 6 rows

 What's is wrong?

 Thanks a lot,


 -
 Silvano Cesar da Costa

 Universidade Estadual de Londrina
 Centro de Ciências Exatas
 Departamento de Estatística

 Fone: (43) 3371-4346

 __
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 -
 Silvano Cesar da Costa

 Universidade Estadual de Londrina
 Centro de Ciências Exatas
 Departamento de Estatística

 Fone: (43) 3371-4346
 -






-
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] bugs and misfeatures in polr(MASS).... fixed!

2012-10-24 Thread tjb

Sorry, I have been away. I'll have a look at this and get back to you
tomorrow.

Tim

On Thu, Oct 18, 2012 at 10:35 PM, ahs [via R] 
ml-node+s789695n4646600...@n4.nabble.com wrote:

 Hello,
 I am trying to use this fix for the convergence problem in polr, but I
 don't seem to get the change of code right. I redefine the function polr by
 the lines

 tjb wrote
 if(missing(start)) {
   # try something that should always work -tjb
   u - as.integer(table(y))
   u - (cumsum(u)/sum(u))[1:q]
   zetas -
  switch(method,
 logistic= qlogis(u),
 probit=   qnorm(u),
 cauchit=  qcauchy(u),
 cloglog=  -log(-log(u)) )
   s0 - c(rep(0,pc),zetas[1],log(diff(zetas)))

 but - which part should I erase? The whole old if(missing(start)) {...} or
 something else or nothing? I have tried all ways I could think of.

 I also can't see where the new object s0 is used in the old code?

 Any input on my problem would be very much appreciated!


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Re: [R] equation solver

2012-10-24 Thread Berend Hasselman


On 24-10-2012, at 10:04, pina pina@hotmail.it wrote:

 Hi, 
 
 I'm Pina and I'm a student in geology. I'm working with spectral profile of
 sand and I have to find the similarity between one spectral profile selected
 by hyperspectral image anche one that I created to mix different percentage
 of 4 mineral component. I have to find the best mix of percentage of this 4
 mineral in order to have the best likeness with the spectral profile chose
 by hyperspectral image. Untill now I worked with excel using the function
 equation solver to find the best fit but I 'd like to know if is possible do
 this operation working in R. 

It looks like you need something that can do optimization.
R provides optim() and nlminb().

A similar question was posted several days ago.
See  
http://r.789695.n4.nabble.com/Optimization-in-R-similar-to-MS-Excel-Solver-td4646759.html

You might have some use for package limSolve and function lsei.

Also see the CRAN Optimization Task View 
(http://cran.r-project.org/web/views/Optimization.html)

Berend

 Thanks for the attention. 
 I'm looking forward to hearing from you. 
 
 Regards. 
 
 Pina.
 
 
 
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Re: [R] multi-panel figure: overall title for each row

2012-10-24 Thread Jean V Adams

Hannes,

A bit inelegant, but it works.  Try this:
mtext(Overall Title Row 2, outer=TRUE, line=-17) 
mtext(Overall Title Row 3, outer=TRUE, line=-34) 

Jean


capy_bara hettl...@few.vu.nl wrote on 10/23/2012 09:08:39 AM:
 
 Dear all,
 I have a 3x2 plot and in addition to the title of the individual plots I
 would like to have an overall title
 for each row.  I managed to get an overall title for the whole plot 
matrix
 with mtext:
 
 par(mfrow=(c(3,2)), mar=c(6.4,4.5,4.2, 1.8), oma=c(0,0,3,0))
 for (i in 1:6)
   barplot(sample(1:10,5), main=paste(Plot #,i))
 mtext(Overall Title Row 1, outer=TRUE)
 
 but I  cannot put an overall title in the middle above the titles of the
 plots in the second or third row. 
 If I use mtext within the loop, I can only add another title above an
 individual plot.
 
 I hope someone can help me, many thanks in advance
 
 Hannes

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Re: [R] Understanding lattice barchart() display

2012-10-24 Thread Rich Shepard


On Tue, 23 Oct 2012, Peter Ehlers wrote:


As Bert correctly points out, you have more than one quant value per
tclass/stream combination. You need to summarize your 1987-case data.frame
to a 56-case data.frame (8 levels of tclass and 7 levels of stream). The
easy way to do that is:

benthos2 - aggregate(quant ~ tclass + stream, data = benthos,
  FUN = sum)

Then just use benthos2 as the data argument to barchart().


Peter,

  Thank you. I'll learn the aggregate function and do as you suggest.

Regards,

Rich

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Re: [R] Recode function car package erases previous values

Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
#  B20_C1 B20_C2 nrB20C
#1 NA NA  1
#2  0  0 NA
#3  0  1 NA
#4  0  0 NA
#5  1  1 NA
#6 NA  1 NA
 dat1$pharm-ifelse(apply(dat1,1,function(x) all(is.na(x[1:2]))),no 
response,ifelse(dat1[,1]==1,Yes,No))
dat2-within(dat1,{pharm-factor(pharm)})
 levels(dat2$pharm)
#[1] No  no response Yes    
 dat2
#  B20_C1 B20_C2 nrB20C   pharm
#1 NA NA  1 no response
#2  0  0 NA  No
#3  0  1 NA  No
#4  0  0 NA  No
#5  1  1 NA Yes
#6 NA  1 NA    NA

table(dat2$pharm)

# No no response Yes 
 # 3   1   1 

A.K.




- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 5:17 AM
Subject: [R] Recode function car package erases previous values

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, 
Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
No Yes 
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
     No     Yes no resp 
    716       7       0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
      0 no resp 
    723       6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table 
pharm
     No     Yes no resp 
    716       7       6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia
 
Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

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Re: [R] Results not dispalying in R console

2012-10-24 Thread Bert Gunter

I **think** he is saying that he is not getting anything at all back
on the console. If so, he needs to *follow the posting guide* and
tell us what OS and setup he is using and **exactly** what is meant by
the R console (GUI, command line, ...?) . Your last suggestion to
post elsewhere may also be apropos.

-- Bert

On Wed, Oct 24, 2012 at 12:40 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
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.
---
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.

arunkumar akpbond...@gmail.com wrote:

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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Bert Gunter
Genentech Nonclinical Biostatistics

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

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Re: [R] Recode function car package erases previous values

2012-10-24 Thread Pancho Mulongeni

-Original Message-
From: arun [mailto:smartpink...@yahoo.com] 
Sent: Wednesday, October 24, 2012 3:31 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
#  B20_C1 B20_C2 nrB20C
#1 NA NA  1
#2  0  0 NA
#3  0  1 NA
#4  0  0 NA
#5  1  1 NA
#6 NA  1 NA
 dat1$pharm-ifelse(apply(dat1,1,function(x) all(is.na(x[1:2]))),no 
response,ifelse(dat1[,1]==1,Yes,No))
dat2-within(dat1,{pharm-factor(pharm)})
 levels(dat2$pharm)
#[1] No  no response Yes
 dat2
#  B20_C1 B20_C2 nrB20C   pharm
#1 NA NA  1 no response
#2  0  0 NA  No
#3  0  1 NA  No
#4  0  0 NA  No
#5  1  1 NA Yes
#6 NA  1 NA    NA

table(dat2$pharm)

# No no response Yes
 # 3   1   1 

A.K.

Thank you, once I understand what you did I will use it, for now, I use a 
workaround
1. First I use the recode function
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
2. Now I just subset
pharm[nr.B20C==1]-'no resp'
and this gives me the desired output
 No Yes no resp
716   7   6 

- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 5:17 AM
Subject: [R] Recode function car package erases previous values

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
No Yes
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
     No     Yes no resp
    716       7       0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
      0 no resp
    723       6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table pharm
     No     Yes no resp
    716       7       6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia

Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

__
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[R] for ( i in 23:0 ) { V - vector [ 1 : i ] }

2012-10-24 Thread colaiutachambers

I'm used to run that on Matlab 
 for indice=1:23,
V=vector(1:indice)
end  
that give me 23 vectors as output.

Why the same command on R   for ( i in 23:0 ) { V - vector [ 1 : i ] } 
returns one only vector?
How can I obtain the 23 vectors that I need ? 

Thank you.





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Re: [R] bugs and misfeatures in polr(MASS).... fixed!

2012-10-24 Thread ahs

Great!
You can skip my question about s0 though, I found where it is being used, 
but I still struggle with the code and convergence problem.

I also found something here
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/CharlesDupontStuff/newPolr.R
that seems like someone tried to fix it, but with this code I get mu
starting values out of range, nor do I get it right when I try to force
starting values.

If you got your code to work fine, you might have the answer to my problem.
If you want, maybe you could post your whole *fixed-polr.R*? :-)



I also can't see where the new object s0 is used in the old code?



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Re: [R] Interper output from cajorls and VECM

2012-10-24 Thread Laura Catalina Echeverri

Hello Rmillan,

I was looking for the same issue and I found that the variables ECT1,
ECT2... are the loading matrix, as you can check in the example of the paper
found at  http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf
http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf  

The example is on pages 24-25

Best,
Laura



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Re: [R] multi-panel figure: overall title for each row

2012-10-24 Thread Bert Gunter

A perhaps more elegant way to do it would be to use the ?layout
function rather than the ancient mfrow() to create space for both
titles and graphs. You could then e.g. draw your titles in the space
you created for titles.

-- Bert

On Wed, Oct 24, 2012 at 5:33 AM, Jean V Adams jvad...@usgs.gov wrote:
 Hannes,

 A bit inelegant, but it works.  Try this:
 mtext(Overall Title Row 2, outer=TRUE, line=-17)
 mtext(Overall Title Row 3, outer=TRUE, line=-34)

 Jean


 capy_bara hettl...@few.vu.nl wrote on 10/23/2012 09:08:39 AM:

 Dear all,
 I have a 3x2 plot and in addition to the title of the individual plots I
 would like to have an overall title
 for each row.  I managed to get an overall title for the whole plot
 matrix
 with mtext:

 par(mfrow=(c(3,2)), mar=c(6.4,4.5,4.2, 1.8), oma=c(0,0,3,0))
 for (i in 1:6)
   barplot(sample(1:10,5), main=paste(Plot #,i))
 mtext(Overall Title Row 1, outer=TRUE)

 but I  cannot put an overall title in the middle above the titles of the
 plots in the second or third row.
 If I use mtext within the loop, I can only add another title above an
 individual plot.

 I hope someone can help me, many thanks in advance

 Hannes

 [[alternative HTML version deleted]]

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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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

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Re: [R] for ( i in 23:0 ) { V - vector [ 1 : i ] }

2012-10-24 Thread Bert Gunter

Quoting Rolf Turner:

(1) Learn something about R; don't just hammer and hope.  Read the
introductory manuals and scan the FAQ.

-- Bert

On Wed, Oct 24, 2012 at 5:25 AM, colaiutachambers
gabriele.carrar...@gmail.com wrote:
 I'm used to run that on Matlab
 for indice=1:23,
 V=vector(1:indice)
 end
 that give me 23 vectors as output.

 Why the same command on R   for ( i in 23:0 )  { V - vector [ 1 : i ] } 
 returns one only vector?
 How can I obtain the 23 vectors that I need ?

 Thank you.





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 View this message in context: 
 http://r.789695.n4.nabble.com/for-i-in-23-0-V-vector-1-i-tp4647306.html
 Sent from the R help mailing list archive at Nabble.com.

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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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

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Re: [R] List of multidimensional arrays

2012-10-24 Thread Richard M. Heiberger

I am looking at your pdf file.

it doesn't match your description.

There are no missing values in the vectors
x, y, t.
Each vector has length 55 which is not a multiple of 4,
so we don't know where the 2x2 matrices come from.

The code doesn't run.  There are are too many }.
PDF files are formatted and are not ascii.

Please send your code in ascii text in the body of the email.
Please pick it up from your email and paste it into
a fresh R session to be sure that it works.

Perhaps consrtruct manually an example of what you want the answer to look
like.

Rich



On Wed, Oct 24, 2012 at 3:51 AM, Loukia Spineli
spinelilouki...@gmail.comwrote:

 Dear all,

 I am trying to create a list, where each list element is a vector of
 different length arrays that contain 2by2 matrices. To be more specific
 there are 11 treatments that are compared with placebo (we have 11
 comparisons) and each comparison is studied by a different number of trials
 and each trial has a different number of missing participants in both arms.
 The length of the list is equal to the number of comparisons. In each
 comparison the number of arrays is equal to the number of trials that study
 this comparison. For instance 4 trials compare PAR with placebo. So, for
 this comparison we have 4 arrays and each array has a length equal to the
 producy of the number of participants in each arm. These arrays contain 2x2
 matrices.

 I have attached a document with the data and the code. I cannot create the
 list results the way I have described above. It creates only the  matrices
 for the first array (that has length equal to 135) of the first comparison
 leaving the rest 10 comparisons NULL.
 Any suggestion would be  really helpful

 Thank you in advance,
 Loukia

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Re: [R] Recode function car package erases previous values

Hi Pancho,

I tried ur method:
pharm-as.factor(recode(dat1$B20_C1,1='Yes';0='No'))
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes


pharm[dat1$nrB20C==1]-'no resp'
#Warning message:
#In `[-.factor`(`*tmp*`, dat1$nr.B20C == 1, value = no resp) :
 # invalid factor level, NAs generated
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes


Not sure how you got the result.
A.K.

 



- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Wednesday, October 24, 2012 9:51 AM
Subject: RE: [R] Recode function car package erases previous values



-Original Message-
From: arun [mailto:smartpink...@yahoo.com] 
Sent: Wednesday, October 24, 2012 3:31 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
#  B20_C1 B20_C2 nrB20C
#1 NA NA  1
#2  0  0 NA
#3  0  1 NA
#4  0  0 NA
#5  1  1 NA
#6 NA  1 NA
 dat1$pharm-ifelse(apply(dat1,1,function(x) all(is.na(x[1:2]))),no 
response,ifelse(dat1[,1]==1,Yes,No))
dat2-within(dat1,{pharm-factor(pharm)})
 levels(dat2$pharm)
#[1] No  no response Yes
 dat2
#  B20_C1 B20_C2 nrB20C   pharm
#1 NA NA  1 no response
#2  0  0 NA  No
#3  0  1 NA  No
#4  0  0 NA  No
#5  1  1 NA Yes
#6 NA  1 NA    NA

table(dat2$pharm)

# No no response Yes
 # 3   1   1 

A.K.

Thank you, once I understand what you did I will use it, for now, I use a 
workaround
1. First I use the recode function
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
2. Now I just subset
pharm[nr.B20C==1]-'no resp'
and this gives me the desired output
     No     Yes no resp
    716       7       6 


- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 5:17 AM
Subject: [R] Recode function car package erases previous values

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
No Yes
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
     No     Yes no resp
    716       7       0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
      0 no resp
    723       6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table pharm
     No     Yes no resp
    716       7       6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia
 
Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

__
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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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Re: [R] find similarity between two spectral profile

2012-10-24 Thread Bart Joosen

If you provide some data we can work with (a reproducible example), then we
can help you.
Now we can only guess?



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Re: [R] Recode function car package erases previous values

2012-10-24 Thread Pancho Mulongeni

Hi Arun,
I also used to get that error, but what class is your nr.B20C, mine is not a 
factor, it is numeric and perhaps that's why it works

-Original Message-
From: arun [mailto:smartpink...@yahoo.com] 
Sent: Wednesday, October 24, 2012 4:35 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi Pancho,

I tried ur method:
pharm-as.factor(recode(dat1$B20_C1,1='Yes';0='No'))
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes

pharm[dat1$nrB20C==1]-'no resp'
#Warning message:
#In `[-.factor`(`*tmp*`, dat1$nr.B20C == 1, value = no resp) :
 # invalid factor level, NAs generated
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes

Not sure how you got the result.
A.K.

- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Wednesday, October 24, 2012 9:51 AM
Subject: RE: [R] Recode function car package erases previous values

-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Wednesday, October 24, 2012 3:31 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
#  B20_C1 B20_C2 nrB20C
#1 NA NA  1
#2  0  0 NA
#3  0  1 NA
#4  0  0 NA
#5  1  1 NA
#6 NA  1 NA
 dat1$pharm-ifelse(apply(dat1,1,function(x) all(is.na(x[1:2]))),no 
response,ifelse(dat1[,1]==1,Yes,No))
dat2-within(dat1,{pharm-factor(pharm)})
 levels(dat2$pharm)
#[1] No  no response Yes
 dat2
#  B20_C1 B20_C2 nrB20C   pharm
#1 NA NA  1 no response
#2  0  0 NA  No
#3  0  1 NA  No
#4  0  0 NA  No
#5  1  1 NA Yes
#6 NA  1 NA    NA

table(dat2$pharm)

# No no response Yes
 # 3   1   1 

A.K.

Thank you, once I understand what you did I will use it, for now, I use a 
workaround 1. First I use the recode function
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
2. Now I just subset
pharm[nr.B20C==1]-'no resp'
and this gives me the desired output
     No     Yes no resp
    716       7       6 

- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 5:17 AM
Subject: [R] Recode function car package erases previous values

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
No Yes
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
     No     Yes no resp
    716       7       0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
      0 no resp
    723       6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table pharm
     No     Yes no resp
    716       7       6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia

Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

__
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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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Re: [R] R5: Lock a class field from within a method?



On Oct 24, 2012, at 2:14 AM, Johannes Graumann wrote:


Hello,

testclass - setRefClass(
 testclass,
 fields = list(testfield = logical),
 methods = list(validate=function(){testfield-TRUE}))


test - testclass$new()
test$testfield

logical(0)

test$validate()
test$testfield

[1] TRUE

Works just fine for me.

I would love to be able to do something like

testclass - setRefClass(
 testclass,
 fields = list(testfield = logical),
 methods = list(validate=function(){
testfield-TRUE
.self$lock(testfield)
 }))

but am unabel to achieve that. Can anyone point out how to go about
rendering a field immutable after execution of a specific method?



The fact that you used only lock in your code and I am unable to  
find such a function makes me wonder whether that was an implicit  
psuedo-code effort and that you do not know about:


?lockBinding

--

David Winsemius, MD
Alameda, CA, USA

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Re: [R] After sorting a dataframe by date



On Oct 24, 2012, at 2:27 AM, martiny wrote:


Ai...I see, i didn't know I can sort rowname.
more study to do :)


You deleted the context of this thread, but while you are looking at  
the rownames documentation you might also keep your eyes open for the  
zoo and xts packages which could convert to a matrix with the dates or  
date-times as rownames.


And you should learn to post in context. Nabble is NOT the way most  
people correspond on Rhelp. You will find that if you continue to  
flout local convention that people will begin to ignore your postings.



View this message in context: 
http://r.789695.n4.nabble.com/After-sorting-a-dataframe-by-date-tp4647173p4647291.html
Sent from the R help mailing list archive at Nabble.com.

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David Winsemius, MD
Alameda, CA, USA

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Re: [R] plotting multiple variables in 1 bar graph

2012-10-24 Thread S Ellison


 I'd greatly appreciate your help in making a bar graph with multiple
 variables plotted on it. All the help sites I've seen so far only plot 1
 variable on the y-axis
 ...

 I've spent several hours looking for code to do this but didn't find
 anything. I'd use the Excel graph except that it doesn't have the sd or se
 bars.

Perhaps something like lattice or ggplot would serve better?

Here's something using ggplot (which has prettier colours than lattice)


#Something like your data - with considerable licence on unreadable names!
abst -structure(list(Count = c(17.03, 22.94, 28.38, 29.72, 28.37, 14.45, 
1.51, 0.54, 0.62, 1.52, 62.3, 70.6, 68.82, 64.75, 63.77, 3.17, 
2.78, 2.22, 2.03, 1.94, 0.61, 0.33, 0.74, 0.74, 0.58, 0.44, 0.12, 
0.37, 0.08, 0.41, 0.04, 0, 0.08, 0, 0.08, 1.96, 1.68, 2.84, 2.06, 
3.32, 0, 0, 0, 0, 0), Benthic = structure(c(2L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 9L, 9L, 9L, 9L, 9L, 4L, 
4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 8L, 8L, 
8L, 8L, 8L, 7L, 7L, 7L, 7L, 7L), .Label = c(Algae, Coral, 
Deadcoral, Ind, other, softcoral, something, sponges, 
xBiotic), class = factor), Time = structure(c(1L, 2L, 3L, 
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 
5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 
1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), .Label = c(T 1, T 2, 
T 3, T 4, T 5), class = factor)), .Names = c(Count, 
Benthic, Time), row.names = c(NA, -45L), class = data.frame)

head(abst)

#Add an arbitrary 'std error'

abst$stderr - 0.05*abst$Count

 library(ggplot2)
 b - ggplot(subset(abst, Benthic!=something), aes(x = Time, y = Count, 
fill=Time)) 
 bptot-b + geom_bar(stat = identity, position = dodge)
 bptot + facet_grid(. ~ Benthic , scales=free_y, margins=T)
 
#or, for unequal scale heights - much easier to see individual trends
bwrap - bptot + facet_wrap( ~ Benthic , scales=free, nrow=2)
bwrap

#Now add error bars
bwrap+geom_errorbar(aes(ymin=Count-stderr, ymax=Count+stderr), width=.3)

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Re: [R] rJava install - %1 is not a valid Win32 application.

2012-10-24 Thread David Stevens

Hello all.

Steven's approach didn't work for me. I'm running 64 bit R

  R.Version()
$platform
[1] x86_64-pc-mingw32

$arch
[1] x86_64

$os
[1] mingw32

$system
[1] x86_64, mingw32

and downloaded and installed the 64 bit version of Java directly from 
Oracle. I then tried to load the rJava library and got

Error : .onLoad failed in loadNamespace() for 'rJava', details:
   call: inDL(x, as.logical(local), as.logical(now), ...)
   error: unable to load shared object 'D:/Users/David 
Stevens/Documents/R/win-library/2.15/rJava/libs/x64/rJava.dll':
   LoadLibrary failure:  %1 is not a valid Win32 application.

Error: package/namespace load failed for 'rJava'

I searched my computer for rJava.dll and found several ( 10) copies 
scattered around.

d:\Oneida\CalWaiparadir /s \rjava.dll
...
  Directory of d:\Program Files\R\R-2.13.1\library\rJava\libs\i386
06/24/2011  10:38 AM80,384 rJava.dll
1 File(s) 80,384 bytes
  Directory of d:\Program Files\R\R-2.13.1\library\rJava\libs\x64
06/24/2011  10:39 AM55,296 rJava.dll
1 File(s) 55,296 bytes
  Directory of d:\Program Files\R\R-2.14.1\library\rJava\libs\i386
12/29/2011  11:40 PM81,408 rJava.dll
1 File(s) 81,408 bytes
  Directory of d:\Program Files\R\R-2.14.1\library\rJava\libs\x64
12/29/2011  11:40 PM55,296 rJava.dll
1 File(s) 55,296 bytes
  Directory of d:\Program Files\R\R-2.15.0\library\rJava\libs\i386
12/29/2011  11:40 PM81,408 rJava.dll
1 File(s) 81,408 bytes
  Directory of d:\Program Files\R\R-2.15.0\library\rJava\libs\x64
12/29/2011  11:40 PM55,296 rJava.dll
1 File(s) 55,296 bytes
  Directory of d:\Program Files\R\R-2.15.1\library\rJava\libs\i386
08/29/2012  01:12 PM57,856 rJava.dll
1 File(s) 57,856 bytes
  Directory of d:\Program Files\R\R-2.15.1\library\rJava\libs\x64
08/29/2012  01:12 PM58,368 rJava.dll
1 File(s) 58,368 bytes
  Directory of d:\Users\David 
Stevens\Documents\R\win-library\2.13\rJava\libs\i386
07/06/2011  10:06 AM80,384 rJava.dll
1 File(s) 80,384 bytes
  Directory of d:\Users\David 
Stevens\Documents\R\win-library\2.13\rJava\libs\x64
07/06/2011  10:06 AM55,296 rJava.dll
1 File(s) 55,296 bytes
  Directory of d:\Users\David 
Stevens\Documents\R\win-library\2.15\rJava\libs\i386
10/23/2012  12:30 PM57,856 rJava.dll
1 File(s) 57,856 bytes
  Directory of d:\Users\David 
Stevens\Documents\R\win-library\2.15\rJava\libs\x64
10/23/2012  12:30 PM58,368 rJava.dll


Is there something else I can try?  Short of erasing and reinstalling 
everything?

Regards

David Stevens

On 10/16/2012 9:19 AM, Steven Ranney wrote:
 Thanks for the information, Simon.

 I had both the 64-bit version of R and the 64-bit Windows 7, but not the
 64-bit version of Java.

 Now that I've downloaded the proper version of Java, the problem has been
 solved.

 Thanks -

 SR
 Steven H. Ranney



 On Mon, Oct 15, 2012 at 5:12 PM, Simon Knapp sleepingw...@gmail.com wrote:

 My guess would be that your running the 32 bit version of R - and
 rJava is looking for the 64 bit dll. I'd suggest starting the 64 bit
 version of R explicitly (e.g. the 64 bit version of Rgui lives at
 R_HOME/bin/x64/Rgui.exe, whereas the 32 bit version lives at
 R_HOME/bin/i386/Rgui.exe).

 On Tue, Oct 16, 2012 at 2:37 AM, Steven Ranney steven.ran...@gmail.com
 wrote:
 All --



 I'm having a problem with the rJava package.  I can download it to my
 machine (Win 7 64-bit) but when I try to load the package into R (2.15.1,
 64-bit version), I get the following error:


 require(rJava)
 Loading required package: rJava

 Error : .onLoad failed in loadNamespace() for 'rJava', details:

call: inDL(x, as.logical(local), as.logical(now), ...)

error: unable to load shared object
 'C:/Users/sranney/Documents/R/win-library/2.15/rJava/libs/x64/rJava.dll':

LoadLibrary failure:  %1 is not a valid Win32 application.


 I have verified that the file R is looking for is in the appropriate
 place,
 but I continue to get the error.  I have tried to download rJava from
 another source, but still get the error.


 I have not been able to find another user with this same issue.


 Thanks for your help --


 Steven Ranney

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Re: [R] Summary of variables with NA, empty

2012-10-24 Thread Lopez, Dan

The examples I gave--Null, Empty string, white space, etc where just examples 
based on SPSS Modeler's Data Audit node. 

I just want something that both identifies the columns having missing values-- 
regardless of what they technically are stored as(NA or a field with space bar 
hit a couple of times,etc) -- and tabulates based on what type of missing 
value. This is a basic data exploration step that I thought just maybe comes 
standard in R and that I just don't know of yet.

Hmisc::describe is good and may have to suffice. Missing for the example 
below using Hmisc::describe was 0 although there was a . And I know that's 
because of the technical difference. 

#EXAMPLE data - just one column in this case
 dput(sample(mydata$COMMUTE_BIN,100))
structure(c(2L, 5L, 3L, 2L, 6L, 3L, 2L, 3L, 4L, 2L, 2L, 4L, 3L, 
4L, 3L, 3L, 3L, 6L, 2L, 2L, 2L, 4L, 6L, 4L, 2L, 3L, 2L, 2L, 6L, 
3L, 2L, 6L, 3L, 2L, 3L, 4L, 4L, 4L, 5L, 7L, 3L, 5L, 2L, 3L, 2L, 
2L, 6L, 7L, 7L, 4L, 3L, 3L, 2L, 2L, 2L, 5L, 2L, 2L, 2L, 2L, 2L, 
2L, 5L, 2L, 3L, 3L, 6L, 4L, 6L, 2L, 7L, 4L, 6L, 2L, 3L, 2L, 2L, 
2L, 3L, 2L, 3L, 4L, 3L, 5L, 3L, 4L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 
4L, 5L, 3L, 2L, 2L, 3L, 1L), .Label = c(, 15, 15 - 24, 
25 - 34, 35 - 44, 45 - 54, 55+), class = factor)

As David mentioned maybe I will have to create my own function. Maybe something 
similar to what I got here for identifying a factor columns, column labels and 
number of levels.
#EXAMPLE of formula I will probably need to create for identifying and listing 
column names and counts of NA and  and other missing in a dataframe or 
table. In this case however I am listing factor columns and excluding columns 
w/ 32 levels
set.seed(1)
dat1- 
data.frame(col1=factor(sample(1:25,10,replace=TRUE)),col2=sample(letters[1:10],10,replace=TRUE),col3=factor(rep(1:5,each=2)))
PrintLvls2 - function(x) {print(data.frame(Lvls=sapply(x[sapply(x,function(x) 
is.factor(x)length(levels(x))=32)],nlevels), 
  
Names=sapply(x[sapply(x, function(x) is.factor(x)length(levels(x))=32)], 
 
function(y) paste0(levels(y), collapse=, ))), right=FALSE)}
 PrintLvls2(dat1)
 Lvls Names  
col1 92, 6, 7, 10, 15, 16, 17, 23, 24
col2 7b, c, d, e, g, h, j
col3 51, 2, 3, 4, 5 

Thanks.
Dan

-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com] 
Sent: Tuesday, October 23, 2012 3:15 PM
To: David Winsemius
Cc: Lopez, Dan; R help (r-help@r-project.org)
Subject: Re: [R] Summary of variables with NA, empty

To highlight:

Basically all Null values is a meaningless phrase in R. ?Null ?NA ?NaN have 
**very specific meanings** in R and have nothing to do with the various sorts 
of whitespace characters that David mentions (spaces, tabs...). If you wish to 
use R, you **must** understand the distinctions (the Intro to R tutorial 
discusses some of this -- have you read it?).

There is functionality to test for these sorts of things (is.na, is.null, etc). 
You need to put in the effort to learn about this if you mean to use R in any 
serious way, as these will occur in either data I/O (NA's) or data manipulation 
(e.g. 0/0)

-- Bert

On Tue, Oct 23, 2012 at 2:44 PM, David Winsemius dwinsem...@comcast.net wrote:

 On Oct 23, 2012, at 11:17 AM, Lopez, Dan wrote:

 Hi,

 Is there a function I can use on my dataframe to give me a concise summary 
 of variables that are NA,blank,etc? Basically all Null values, Empty 
 strings, white space, blank values. Ideally it would look something like the 
 below:

 # it should only includes the fields with NAs, blanks, etc. Added bonus 
 would be to include column Index.
 #Valid Records = records that are not NA, blank,etc #ColIndex - what 
 place is column in the original dataframe...1,2,3, ...xth

Valid Records  Null (NA?)Empty String  White 
 Space   Blank ValueColIndex

 Would a Valid Record be defined by grep([^ ], column)? ... i.e. has 
 a non-space character in it What is a ColIndex?
 How is an Empty String different than White Space or a Blank Value



 Var1   528   
  2
 Var2   40   20   
 10   10  
  3
 Var3   58
2 
  20
 ..


 I generally use describe from package:Hmisc. There are other versions of 
 describe in other packages. It's not going to classify items composed 
 entirely of a varying number of spaces and other non-character items like 
 tabs as a single group. And it's unclear what you will use as an operational 
 definition to separate blanks and

Re: [R] Recode function car package erases previous values

2012-10-24 Thread John Fox

Dear Pancho,

I'm not going to respond to the subsequent messages in this thread, since you 
appear to have solved your problem, just explain that what you did originally 
was to recode the variable B20_C1, creating the new variable pharm. Then you 
recoded another variable, nr.B20C, replacing the original version of pharm. I'm 
not sure why you expected this to give you what you want -- it really doesn't 
make sense.

Best,
 John


John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

On Wed, 24 Oct 2012 09:17:25 +
 Pancho Mulongeni p.mulong...@namibia.pharmaccess.org wrote:
 Hi all,
 I am attempting to create a new variable based on values of other variables. 
 The variable is called pharm. It basically takes the numeric code of 1 as yes 
 and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
 would also like to have a level for non-respondents and these are captured in 
 the variable nr.B20C, which is a 1 when there is a non-response on the whole 
 group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
 have three levels, 
 Yes No and no resp.
 See below what happens
 
  pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
  table(pharm)
 pharm
  No Yes 
 716   7 
  levels(pharm)-c('No','Yes','no resp')
  table(pharm)
 pharm
  No Yes no resp 
 716   7   0 
  pharm-as.factor(recode(nr.B20C,1='no resp'))
  table(pharm)
 pharm
   0 no resp 
 723   6 
 
 The recode variable just cannot seem to 'remember' I had just recoded 7 
 values to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' 
 which comes from nr.B20C (it has values 0 or 1). 
 This inconvenient as I would like to have ultimately the following table 
 pharm
  No Yes no resp 
 716   7   6 (FROM nr.B20C where row has value 1).
 
 Background. The variable pharm assess where you used the pharmacy to get your 
 contraception.
 
 Pancho Mulongeni
 Research Assistant
 PharmAccess Foundation
 1 Fouché Street
 Windhoek West
 Windhoek
 Namibia
  
 Tel:   +264 61 419 000
 Fax:  +264 61 419 001/2
 Mob: +264 81 4456 286
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] [r] How to pick colums from a ragged array?

HI Stuart,

Just a doubt:
When I run Rui's function:
I got results as:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]
# ID DATE DG
#22  841 20050421  1
#23  841 20050421  2
#38 1019 19870508  2
#39 1019 19870508  1

But, when I look at the id.d
16  814 20020814  2
17  814 20021125  2
18  814 20040429  2
19  814 20040429  2
20  814 20071205  2
21  814 20071205  2
-

   323 2005  3
8   323 20060111  2
9   323 20071119  3
10  323 20080107  2
11  323 20080407  1
12  323 20080521  2
13  323 20080521  3

According to the  function I wrote earlier, I get 
res1
  flag
58   FALSE
167   TRUE
323   TRUE
547  FALSE
794  FALSE
814   TRUE
841   TRUE
910  FALSE
999  FALSE
1019  TRUE

If I understand correctly, you wanted to eliminate 814 and 167 because the DGs 
are the same.  That can be done.  I also want to make it clear that why 323 is 
not included.  It is duplicated in the last visit.
Now, to eliminate the ones with duplicate DGs:
res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)|tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2-id.d[id.d[,1]%in%names(res1[res1$flag==TRUE,])(duplicated(id.d[,1:2])|duplicated(id.d[,1:2],fromLast=TRUE)),]
res2[!res2$ID%in% res2[duplicated(res2)|duplicated(res2,fromLast=TRUE),]$ID,]
# ID DATE DG
#12  323 20080521  2
#13  323 20080521  3
#22  841 20050421  1
#23  841 20050421  2
#38 1019 19870508  2
#39 1019 19870508  1
A.K.

 
 






- Original Message -
From: Stuart Leask stuart.le...@nottingham.ac.uk
To: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
petr.pi...@precheza.cz; Rui Barradas (ruipbarra...@sapo.pt) 
ruipbarra...@sapo.pt
Cc: 
Sent: Wednesday, October 24, 2012 11:41 AM
Subject: RE: [r] How to pick colums from a ragged array?

(And, considering  the real application, the functions ideally should probably 
output a variable INCLUDE, the same length as the original data, with TRUE and 
FALSE for whether or not that row should be included...)

-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you have written.

You'll recall I wanted to remove these first (or last) duplicates, because they 
represented instances where two different diagnoses (in this case, variable DG, 
value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which 
was 'first' (or 'last').

Your functions have revealed something I wasn't expecting: In some cases, the 
diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two different 
departments in a clinic, and both departments submit data. I have to say that 
crazy things like this are often a feature of real data, which I'm sure you've 
come across yourselves.

Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.

You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
wonder if any of the functions you've written could do this with a little more
Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first (or last) 
DATE duplicate, the DGs for these two dates are different.

Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)    # defaults to first = TRUE getRepeat(id.d, first = 
FALSE)  to get the last ones
g.rr-do.call(rbind, g.r) # put the data into a matrix

# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

I'm not sure how to add this to your suggestions, Arun  Petr...


Stuart


-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?

Hi

I assumed that id.d is data frame

id.d - data.frame (ID,DATE )

and

fff(id.d)

works for me

Petr


 -Original Message-
 From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
 Sent: Tuesday, October 23, 2012 3:13 PM
 To: PIKAL Petr
 Subject: RE: [r] How to pick

Re: [R] help using optim function

2012-10-24 Thread Ben Bolker

tmuman mumantariq at gmail.com writes:

  Hi, am very new to R and I've written an optim function, but can't
 get it to work
 
 least.squares.fitter-function(start.params,gr,
  low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
  {
 result-optim(par=start.params,
  method=c('Nelder-Mead'),
   fn=least.squares.fit,
lower=low.constraints,
   upper=high.constraints,data=data,scale=scale,ploton=ploton)
return(result)  
}

 least.squares.fitter(c(2,2),c(0,0),c(Inf,Inf),ricker.one.step,dat,'linear')
 
 where least.squares.fit is the function returning the difference between
 model data points and predicted points (residuals, residuals^2 and SSR),
 which i am trying to minimise
 
 and ricker.one.step is the function that generated the predictions
 
 I've been trying to figure out whats wrong for hours but can't.
 

  You haven't given us much to work with here -- have you
read the posting guide (URL at the bottom of every message)?
It's not reproducible ( http://tinyurl.com/reproducible-000 ),
and you haven't told us what's going wrong.  Are you getting
error messages?  What are they?  Warnings?  No errors but
you have reason to believe the answer is wrong?

A few comments:
*   'scale' and 'ploton' are not arguments to optim(),
so they'd better be accepted by your least.squares.fit() function
* the Nelder-Mead algorithm doesn't accept constraints (lower/upper)
  (see the nloptr package for a Nelder-Mead implementation that does)
* except for the constraints, you might be better off with ?nls
if all you want to do is least-squares fitting; nlsLM in the
minpack.lm package does least-squares fitting with constraints

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

Re: [R] for ( i in 23:0 ) { V - vector [ 1 : i ] }


Hello,

Try the following.

lapply(1:23, function(i) 1:i)

As for why yuor code doesn't work, you're assigning 24 times a value to 
V, on loop exit only the last one is there. A correct way using loops 
would be



V - list()
for(i in 1:23)
V[[i]] - 1:i


Beware that the last vector goes from 1 _down_ to 0.

And do read An Introduction to R, file R-intro.pdf in your R/doc directory.

Hope this helps,
Em 24-10-2012 13:25, colaiutachambers escreveu:

I'm used to run that on Matlab

for indice=1:23,

V=vector(1:indice)
end
that give me 23 vectors as output.

Why the same command on R   for ( i in 23:0 ) { V - vector [ 1 : i ] } 
returns one only vector?
How can I obtain the 23 vectors that I need ?

Thank you.





--
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http://r.789695.n4.nabble.com/for-i-in-23-0-V-vector-1-i-tp4647306.html
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Re: [R] [r] How to pick colums from a ragged array?

Hi,

With the dataset that you provided, and considering that my solution works (ID  
323 included), the below code outputs a variable INCLUDE.
res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)|tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2-id.d[id.d[,1]%in%names(res1[res1$flag==TRUE,])(duplicated(id.d[,1:2])|duplicated(id.d[,1:2],fromLast=TRUE)),]
res3-res2[!res2$ID%in% 
res2[duplicated(res2)|duplicated(res2,fromLast=TRUE),]$ID,]
id.d1-id.d
bad-id.d1[id.d1$ID%in%res3$ID,]
id.d1$id.d_INCLUDE-TRUE
bad$INCLUDE-FALSE
res4-merge(id.d1,bad,all=TRUE)
res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE
res5-res4[,-4]
 head(res5)
#   ID DATE DG INCLUDE
#1  58 20060821  1    TRUE
#2  58 20061207  2    TRUE
#3  58 20080102  1    TRUE
#4  58 20090904  1    TRUE
#5 167 20040205  4    TRUE
#6 167 20040205  4    TRUE
 tail(res5)
# ID DATE DG INCLUDE
#35  910 20080521  4    TRUE
#36  910 20091224  2    TRUE
#37  999 20050503  2    TRUE
#38 1019 19870508  1   FALSE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1   FALSE
A.K.





- Original Message -
From: Stuart Leask stuart.le...@nottingham.ac.uk
To: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
petr.pi...@precheza.cz; Rui Barradas (ruipbarra...@sapo.pt) 
ruipbarra...@sapo.pt
Cc: 
Sent: Wednesday, October 24, 2012 11:41 AM
Subject: RE: [r] How to pick colums from a ragged array?

(And, considering  the real application, the functions ideally should probably 
output a variable INCLUDE, the same length as the original data, with TRUE and 
FALSE for whether or not that row should be included...)

-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you have written.

You'll recall I wanted to remove these first (or last) duplicates, because they 
represented instances where two different diagnoses (in this case, variable DG, 
value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which 
was 'first' (or 'last').

Your functions have revealed something I wasn't expecting: In some cases, the 
diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two different 
departments in a clinic, and both departments submit data. I have to say that 
crazy things like this are often a feature of real data, which I'm sure you've 
come across yourselves.

Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.

You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
wonder if any of the functions you've written could do this with a little more
Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first (or last) 
DATE duplicate, the DGs for these two dates are different.

Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)    # defaults to first = TRUE getRepeat(id.d, first = 
FALSE)  to get the last ones
g.rr-do.call(rbind, g.r) # put the data into a matrix

# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

I'm not sure how to add this to your suggestions, Arun  Petr...


Stuart


-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?

Hi

I assumed that id.d is data frame

id.d - data.frame (ID,DATE )

and

fff(id.d)

works for me

Petr


 -Original Message-
 From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
 Sent: Tuesday, October 23, 2012 3:13 PM
 To: PIKAL Petr
 Subject: RE: [r] How to pick colums from a ragged array?

 Hi Petr.
 I see what you mean it should do, but when I run it I get an error
 (see below).
 Stuart


  ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
 + ,547,794,814,814,814,814,814,814,841,841,841,841,841
 + ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
 + ,1019)
 
  DATE -
 +  c(20060821,20061207,20080102,20090904,20040205,20040205,2005
 +

Re: [R] parsing a structured object

2012-10-24 Thread BenM

That's simple and appears to work.  Thanks for the prompt response.

Ben



--
View this message in context: 
http://r.789695.n4.nabble.com/parsing-a-structured-object-tp4647246p4647334.html
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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] ifelse reformulation




Hi,
Try this:
test1-read.table(text= 
   id x1 x2 x3 x4 x5 x6 x7 
1   1 36 26 21 32 31 27 31 
2   2 45 21 46 50 22 36 29 
3   3 49 47 35 44 33 31 46 
4   4 42 32 38 28 39 45 32 
5   5 29 42 39 48 25 35 34 
6   6 39 31 30 37 46 43 44 
7   7 41 40 25 23 42 40 24 
8   8 27 29 47 34 26 38 28 
9   9 25 35 29 36 43 34 23 
10 10 24 44 37 26 27 46 22 
11 11 38 50 32 49 37 24 40 
12 12 20 34 48 25 30 41 36 
13 13 26 46 20 40 29 20 43 
14 14 33 37 49 31 47 30 30 
15 15 43 39 27 35 48 47 27 
,sep=,header=TRUE) 


group-c(40,50,60,70)
as.vector(apply(test1,1,function(x) ifelse(any(x[4:6]%in%group),0,1)))
# [1] 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1

#Comparing the results individually

ifelse(test1$x4==40|test1$x4==50|test1$x4==60|test1$x4==70|test1$x5==40|test1$x5==50|test1$x5==60|test1$x5==70|test1$x6==40|test1$x6==50|test1$x6==60|test1$x7==70,0,1)
# [1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1

 as.vector(apply(test1,1,function(x) 
ifelse(any(x[4:6]==40|x[4:6]==50|x[4:6]==60|x[4:6]==70),0,1)))
 #[1] 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1
A.K.

From: Bruno Marques brunos...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Wednesday, October 24, 2012 12:47 PM
Subject: Re: [R] ifelse reformulation


Hi Arun, let me ask you just one more thing, please.

If i want to compare, not one but a group of values, can i do this:

group-c(40,50,60,70)

as.vector(apply(test1,1,function(x) ifelse(any(x[4:6]==group),0,1)))

What i want is: IF it finds any of the values in group, then 0.
I mean, it has to work like a OR... but i think this code will work like an 
AND... am i wrong?

Sorry for my bad english!

Best regards,

Bruno


2012/10/12 arun smartpink...@yahoo.com

HI,
Try this:
count40- ifelse(test1$x3==40|test1$x4==40|test1$x5==40,0,1)
count40

# [1] 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1
as.vector(apply(test1,1,function(x) ifelse(any(x[4:6]==40),0,1)))
# [1] 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1
A.K.




- Original Message -
From: brunosm brunos...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, October 12, 2012 10:18 AM
Subject: Re: [R] ifelse reformulation

That was exacly what i was looking for! Thanks a lot!
Cheers!



--
View this message in context: 
http://r.789695.n4.nabble.com/ifelse-reformulation-tp4645981p4645990.html

Sent from the R help mailing list archive at Nabble.com.

__
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[R] how to get underlying values being plotted in garchfit predict plot

2012-10-24 Thread Kulkarni, Miheer

Hello,

I have a time series of 500 returns. I am fitting a garch(1,1) model to this 
series using

fit - garchFit(~garch(1,2), data=y)

Then I run:  predict(fit,n.ahead=100,plot=TRUE)

To get the plot.

My question, how can I access the data that is being plotted with the black 
line. Basically I want the list of predictions for the next 100 times, the 
function returns the meanForecast(constant), meanError, standardDeviation, and 
lower and upper intervals based on mean +/- 1.96*MSE but not the values 
actually being plotted (for the first 100 times, beyond that it shows the mean 
and conf. interval).

Thanks!



[[alternative HTML version deleted]]

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Re: [R] [r] How to pick colums from a ragged array?


Hello,

Using one of Arun's ideas, some post ago, this new function returns a 
logical index into id.d of the rows that should be _removed_, hence rm1 
and rm2. I think




getRepLogical - function(x, first = TRUE){
fun - if(first) head else tail
dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2)))
len - tapply(x[,2], x[,1], FUN = length)
lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))

lst - if(first) lst else lapply(lst, rev)
i1 - unlist(lst)
dg - tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x, 2)))
lst - lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))

lst - if(first) lst else lapply(lst, rev)
i2 - unlist(lst)
i1  i2
}

rm1 - getRepLogical(id.d)
rm2 - getRepLogical(id.d, first = FALSE)

id.d[rm1, ]
id.d[rm2, ]

id.d$INCLUDE - !(rm1 | rm2)


Hope this helps,

Rui Barradas
Em 24-10-2012 16:41, Stuart Leask escreveu:

(And, considering  the real application, the functions ideally should probably 
output a variable INCLUDE, the same length as the original data, with TRUE and 
FALSE for whether or not that row should be included...)

-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you have written.

You'll recall I wanted to remove these first (or last) duplicates, because they 
represented instances where two different diagnoses (in this case, variable DG, 
value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which 
was 'first' (or 'last').

Your functions have revealed something I wasn't expecting: In some cases, the 
diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two different 
departments in a clinic, and both departments submit data. I have to say that 
crazy things like this are often a feature of real data, which I'm sure you've 
come across yourselves.

Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.

You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
wonder if any of the functions you've written could do this with a little more
Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first (or last) 
DATE duplicate, the DGs for these two dates are different.

Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
  c(20060821,20061207,20080102,20090904,20040205,20040205,2005
  ,20060111,20071119,20080107,20080407,20080521,20080521,20041005
  ,20070905,20020814,20021125,20040429,20040429,20071205,20071205
  ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
  ,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
  ,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)# defaults to first = TRUE getRepeat(id.d, first = 
FALSE)  to get the last ones
g.rr-do.call(rbind, g.r) # put the data into a matrix

# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

I'm not sure how to add this to your suggestions, Arun  Petr...


Stuart


-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?

Hi

I assumed that id.d is data frame

id.d - data.frame (ID,DATE )

and

fff(id.d)

works for me

Petr



-Original Message-
From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
Sent: Tuesday, October 23, 2012 3:13 PM
To: PIKAL Petr
Subject: RE: [r] How to pick colums from a ragged array?

Hi Petr.
I see what you mean it should do, but when I run it I get an error
(see below).
Stuart



ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323

+ ,547,794,814,814,814,814,814,814,841,841,841,841,841
+ ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
+ ,1019)

DATE -

+  c(20060821,20061207,20080102,20090904,20040205,20040205,2005
+  ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
+  ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
+  ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
+  ,20070112,20070514, 19870508,20040205,20040205, 20091120,20091210
+  ,20091224,20050503,19870508,19870508,19880330)

  id.d - cbind (ID,DATE )
fff-function(data, first=TRUE, remove=FALSE) {

+
+ testfirst - function(x) x[1,2]==x[2,2] testlast -

Re: [R] daylight

2012-10-24 Thread Greg Snow

Do you care about local topography/terrain?  I think most of the
calculators/tables that are commonly used assume that you are at a
fairly flat place on the earth's surface, but that is not always true.
 The area where my wife grew up had its longest day closer to the
equinox than the summer solstice and if I wanted to calculate the
number of hours that direct sunlight would hit solar panels (which I
don't have yet) on my house or the solar oven (which I do have) in my
back yard, then assuming sunset and sunrise occurred at 90 degrees
from vertical would definitely overestimate the time.

On Tue, Oct 23, 2012 at 1:45 PM, 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



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/daylight-tp4647213.html
 Sent from the R help mailing list archive at Nabble.com.

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



-- 
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com

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Re: [R] [r] How to pick colums from a ragged array?

Hi Rui,

I think now our results are matching except in the INCLUDE column 

id.d[c(11:13,22:24,38:40),]
# ID DATE DG INCLUDE
#11  323 20080407  1    TRUE
#12  323 20080521  2   FALSE
#13  323 20080521  3    TRUE
#22  841 20050421  1    TRUE
#23  841 20050421  2   FALSE
#24  841 20060428  1    TRUE
#38 1019 19870508  2    TRUE
#39 1019 19870508  1   FALSE
#40 1019 19880330  1    TRUE


I thought all the rows with the above IDS would be FALSE (from my solution):

res4[c(11:13,22:24,38:40),]
 ID DATE DG INCLUDE
#11  323 20080407  1   FALSE
#12  323 20080521  2   FALSE
#13  323 20080521  3   FALSE
#22  841 20050421  1   FALSE
#23  841 20050421  2   FALSE
#24  841 20060428  1   FALSE
#38 1019 19870508  1   FALSE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1   FALSE

A.K.




- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: Stuart Leask stuart.le...@nottingham.ac.uk
Cc: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
petr.pi...@precheza.cz; r-help r-help@r-project.org
Sent: Wednesday, October 24, 2012 1:41 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Using one of Arun's ideas, some post ago, this new function returns a 
logical index into id.d of the rows that should be _removed_, hence rm1 
and rm2. I think



getRepLogical - function(x, first = TRUE){
     fun - if(first) head else tail
     dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2)))
     len - tapply(x[,2], x[,1], FUN = length)
     lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))
     lst - if(first) lst else lapply(lst, rev)
     i1 - unlist(lst)
     dg - tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x, 2)))
     lst - lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))
     lst - if(first) lst else lapply(lst, rev)
     i2 - unlist(lst)
     i1  i2
}

rm1 - getRepLogical(id.d)
rm2 - getRepLogical(id.d, first = FALSE)

id.d[rm1, ]
id.d[rm2, ]

id.d$INCLUDE - !(rm1 | rm2)


Hope this helps,

Rui Barradas
Em 24-10-2012 16:41, Stuart Leask escreveu:
 (And, considering  the real application, the functions ideally should 
 probably output a variable INCLUDE, the same length as the original data, 
 with TRUE and FALSE for whether or not that row should be included...)

 -Original Message-
 From: Leask Stuart
 Sent: 24 October 2012 16:25
 To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
 (ruipbarra...@sapo.pt)
 Subject: RE: [r] How to pick colums from a ragged array?

 Arun, Petr, Rui, many thanks for your help, and the functions you have 
 written.

 You'll recall I wanted to remove these first (or last) duplicates, because 
 they represented instances where two different diagnoses (in this case, 
 variable DG, value 1, 2, 3, 4 or 5) had been recorded on the same day - so I 
 can't say which was 'first' (or 'last').

 Your functions have revealed something I wasn't expecting: In some cases, the 
 diagnoses recorded on the duplicated DATEs are the same!
 This is a surprise to me, but probably reflects someone going to two 
 different departments in a clinic, and both departments submit data. I have 
 to say that crazy things like this are often a feature of real data, which 
 I'm sure you've come across yourselves.

 Of course, I don't want to remove records in which I can determine an 
 unambiguous 'first diagnosis'.

 You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
 wonder if any of the functions you've written could do this with a little more
 Indexing and the 'duplicate' function
 So the function should only exclude an ID, having identified a first (or 
 last) DATE duplicate, the DGs for these two dates are different.

 Test dataset:

 ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
 ,547,794,814,814,814,814,814,814,841,841,841,841,841
 ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
 ,1019)

 DATE -
   c(20060821,20061207,20080102,20090904,20040205,20040205,2005
   ,20060111,20071119,20080107,20080407,20080521,20080521,20041005
   ,20070905,20020814,20021125,20040429,20040429,20071205,20071205
   ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
   ,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
   ,20091224,20050503,19870508,19870508,19880330)

 DG-
 c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

 id.d-data.frame(ID,DATE,DG)
 id.d

 # Considering Ruis  getRepeat function:

 g.r-getRepeat(id.d)    # defaults to first = TRUE getRepeat(id.d, first = 
 FALSE)  to get the last ones
 g.rr-do.call(rbind, g.r) # put the data into a matrix

 # I can remove the date duplicates with:
 g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

 I'm not sure how to add this to your suggestions, Arun  Petr...


 Stuart


 -Original Message-
 From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
 Sent: 23 October

Re: [R] Summary of variables with NA, empty


On Oct 24, 2012, at 8:32 AM, Lopez, Dan wrote:

 The examples I gave--Null, Empty string, white space, etc where just examples 
 based on SPSS Modeler's Data Audit node. 
 
 I just want something that both identifies the columns having missing 
 values-- regardless of what they technically are stored as(NA or a field with 
 space bar hit a couple of times,etc) -- and tabulates based on what type of 
 missing value. This is a basic data exploration step that I thought just 
 maybe comes standard in R and that I just don't know of yet.

In none of your examples below do you create factor columns with any of the 
features you say that you are hoping to identify. No NA's, no white-space 
levels, no 999 values.  I do not (and never have) used SPSS Modeler's Data 
Audit, so definition by analogy is not going to work for me.
 
 
 Hmisc::describe is good and may have to suffice. Missing for the example 
 below using Hmisc::describe was 0 although there was a . And I know that's 
 because of the technical difference. 

Right. Hmisc::describe counts the number of NA's. A value of  or   is not 
the same as the R NA missing.

  is.na(factor())
[1] FALSE
 is.na(factor(NA))
[1] TRUE
 is.na(factor( ))
[1] FALSE

If you want to test for an empty string use nchar(vec) == 0

I offered an earlier suggestion for a grepl test for all spaces.


 #EXAMPLE data - just one column in this case
 dput(sample(mydata$COMMUTE_BIN,100))
 structure(c(2L, 5L, 3L, 2L, 6L, 3L, 2L, 3L, 4L, 2L, 2L, 4L, 3L, 
 4L, 3L, 3L, 3L, 6L, 2L, 2L, 2L, 4L, 6L, 4L, 2L, 3L, 2L, 2L, 6L, 
 3L, 2L, 6L, 3L, 2L, 3L, 4L, 4L, 4L, 5L, 7L, 3L, 5L, 2L, 3L, 2L, 
 2L, 6L, 7L, 7L, 4L, 3L, 3L, 2L, 2L, 2L, 5L, 2L, 2L, 2L, 2L, 2L, 
 2L, 5L, 2L, 3L, 3L, 6L, 4L, 6L, 2L, 7L, 4L, 6L, 2L, 3L, 2L, 2L, 
 2L, 3L, 2L, 3L, 4L, 3L, 5L, 3L, 4L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 
 4L, 5L, 3L, 2L, 2L, 3L, 1L), .Label = c(, 15, 15 - 24, 
 25 - 34, 35 - 44, 45 - 54, 55+), class = factor)
 
 As David mentioned maybe I will have to create my own function. Maybe 
 something similar to what I got here for identifying a factor columns, column 
 labels and number of levels.
 #EXAMPLE of formula I will probably need to create for identifying and 
 listing column names and counts of NA and  and other missing in a 
 dataframe or table. In this case however I am listing factor columns and 
 excluding columns w/ 32 levels
 set.seed(1)
 dat1- 
 data.frame(col1=factor(sample(1:25,10,replace=TRUE)),col2=sample(letters[1:10],10,replace=TRUE),col3=factor(rep(1:5,each=2)))

 PrintLvls2 - function(x) 
 {print(data.frame(Lvls=sapply(x[sapply(x,function(x) 
 is.factor(x)length(levels(x))=32)],nlevels), 
 
 Names=sapply(x[sapply(x, function(x) is.factor(x)length(levels(x))=32)], 

 function(y) paste0(levels(y), collapse=, ))), right=FALSE)}
 PrintLvls2(dat1)
 Lvls Names  
 col1 92, 6, 7, 10, 15, 16, 17, 23, 24
 col2 7b, c, d, e, g, h, j
 col3 51, 2, 3, 4, 5 

I find it good to put in counter-examples such as a column that is non-factor. 
I thought that a non-factor column would probably break your code, but happily 
it survived. You might think about writing two functions: one to pick the 
columns to be assessed and the other to return a structured object from the 
candidates.

-- 
david.

 
 Thanks.
 Dan
 
 -Original Message-
 From: Bert Gunter [mailto:gunter.ber...@gene.com] 
 Sent: Tuesday, October 23, 2012 3:15 PM
 To: David Winsemius
 Cc: Lopez, Dan; R help (r-help@r-project.org)
 Subject: Re: [R] Summary of variables with NA, empty
 
 To highlight:
 
 Basically all Null values is a meaningless phrase in R. ?Null ?NA ?NaN have 
 **very specific meanings** in R and have nothing to do with the various sorts 
 of whitespace characters that David mentions (spaces, tabs...). If you wish 
 to use R, you **must** understand the distinctions (the Intro to R tutorial 
 discusses some of this -- have you read it?).
 
 There is functionality to test for these sorts of things (is.na, is.null, 
 etc). You need to put in the effort to learn about this if you mean to use R 
 in any serious way, as these will occur in either data I/O (NA's) or data 
 manipulation (e.g. 0/0)
 
 -- Bert
 
 On Tue, Oct 23, 2012 at 2:44 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
 
 On Oct 23, 2012, at 11:17 AM, Lopez, Dan wrote:
 
 Hi,
 
 Is there a function I can use on my dataframe to give me a concise summary 
 of variables that are NA,blank,etc? Basically all Null values, Empty 
 strings, white space, blank values. Ideally it would look something like 
 the below:
 
 # it should only includes the fields with NAs, blanks, etc. Added bonus 
 would be to include column Index.
 #Valid Records = records that are not NA, blank,etc #ColIndex - what 
 place is column in the original dataframe...1,2,3, ...xth
 
   Valid Records  Null

Re: [R] [r] How to pick colums from a ragged array?


Hello,

I just realized that function getRepLogical marks the second, not the 
first (eventually from last) to be removed. The first tapply should be


dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2), 
fromLast = TRUE))


in order to remove the first (or last).

Rui Barradas
Em 24-10-2012 18:41, Rui Barradas escreveu:

Hello,

Using one of Arun's ideas, some post ago, this new function returns a 
logical index into id.d of the rows that should be _removed_, hence 
rm1 and rm2. I think




getRepLogical - function(x, first = TRUE){
fun - if(first) head else tail
dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2)))
len - tapply(x[,2], x[,1], FUN = length)
lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))

lst - if(first) lst else lapply(lst, rev)
i1 - unlist(lst)
dg - tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x, 2)))
lst - lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE, 
if(len[[i]]  2) len[[i]] - 2 else 0)))

lst - if(first) lst else lapply(lst, rev)
i2 - unlist(lst)
i1  i2
}

rm1 - getRepLogical(id.d)
rm2 - getRepLogical(id.d, first = FALSE)

id.d[rm1, ]
id.d[rm2, ]

id.d$INCLUDE - !(rm1 | rm2)


Hope this helps,

Rui Barradas
Em 24-10-2012 16:41, Stuart Leask escreveu:
(And, considering  the real application, the functions ideally should 
probably output a variable INCLUDE, the same length as the original 
data, with TRUE and FALSE for whether or not that row should be 
included...)


-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)

Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you 
have written.


You'll recall I wanted to remove these first (or last) duplicates, 
because they represented instances where two different diagnoses (in 
this case, variable DG, value 1, 2, 3, 4 or 5) had been recorded on 
the same day - so I can't say which was 'first' (or 'last').


Your functions have revealed something I wasn't expecting: In some 
cases, the diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two 
different departments in a clinic, and both departments submit data. 
I have to say that crazy things like this are often a feature of real 
data, which I'm sure you've come across yourselves.


Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.


You have all put in so much effort on my behalf, I'm ashamed to ask, 
but I wonder if any of the functions you've written could do this 
with a little more

Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first 
(or last) DATE duplicate, the DGs for these two dates are different.


Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
  ,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
  ,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1) 



id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)# defaults to first = TRUE getRepeat(id.d, 
first = FALSE)  to get the last ones

g.rr-do.call(rbind, g.r) # put the data into a matrix

# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

I'm not sure how to add this to your suggestions, Arun  Petr...


Stuart


-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?

Hi

I assumed that id.d is data frame

id.d - data.frame (ID,DATE )

and

fff(id.d)

works for me

Petr



-Original Message-
From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
Sent: Tuesday, October 23, 2012 3:13 PM
To: PIKAL Petr
Subject: RE: [r] How to pick colums from a ragged array?

Hi Petr.
I see what you mean it should do, but when I run it I get an error
(see below).
Stuart



ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323

+ ,547,794,814,814,814,814,814,814,841,841,841,841,841
+ ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
+ ,1019)

DATE -

+ c(20060821,20061207,20080102,20090904,20040205,20040205,2005
+ ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
+

Re: [R] [r] How to pick colums from a ragged array?


Hello,

Inline.
Em 24-10-2012 19:05, arun escreveu:

Hi Rui,

I think now our results are matching except in the INCLUDE column

id.d[c(11:13,22:24,38:40),]
# ID DATE DG INCLUDE
#11  323 20080407  1TRUE
#12  323 20080521  2   FALSE
#13  323 20080521  3TRUE
#22  841 20050421  1TRUE
#23  841 20050421  2   FALSE
#24  841 20060428  1TRUE
#38 1019 19870508  2TRUE
#39 1019 19870508  1   FALSE
#40 1019 19880330  1TRUE


I thought all the rows with the above IDS would be FALSE


Why? Look at the last ID, 1019. The last of all must be included, the 
date doesn't repeat. And one of the first must also be included, if not 
we would be completely excluding that date. Or at least this is how I'm 
understanding the problem.


Rui Barradas

  (from my solution):

res4[c(11:13,22:24,38:40),]
  ID DATE DG INCLUDE
#11  323 20080407  1   FALSE
#12  323 20080521  2   FALSE
#13  323 20080521  3   FALSE
#22  841 20050421  1   FALSE
#23  841 20050421  2   FALSE
#24  841 20060428  1   FALSE
#38 1019 19870508  1   FALSE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1   FALSE

A.K.




- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: Stuart Leask stuart.le...@nottingham.ac.uk
Cc: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
petr.pi...@precheza.cz; r-help r-help@r-project.org
Sent: Wednesday, October 24, 2012 1:41 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Using one of Arun's ideas, some post ago, this new function returns a
logical index into id.d of the rows that should be _removed_, hence rm1
and rm2. I think



getRepLogical - function(x, first = TRUE){
  fun - if(first) head else tail
  dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2)))
  len - tapply(x[,2], x[,1], FUN = length)
  lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE,
if(len[[i]]  2) len[[i]] - 2 else 0)))
  lst - if(first) lst else lapply(lst, rev)
  i1 - unlist(lst)
  dg - tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x, 2)))
  lst - lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE,
if(len[[i]]  2) len[[i]] - 2 else 0)))
  lst - if(first) lst else lapply(lst, rev)
  i2 - unlist(lst)
  i1  i2
}

rm1 - getRepLogical(id.d)
rm2 - getRepLogical(id.d, first = FALSE)

id.d[rm1, ]
id.d[rm2, ]

id.d$INCLUDE - !(rm1 | rm2)


Hope this helps,

Rui Barradas
Em 24-10-2012 16:41, Stuart Leask escreveu:

(And, considering  the real application, the functions ideally should probably 
output a variable INCLUDE, the same length as the original data, with TRUE and 
FALSE for whether or not that row should be included...)

-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you have written.

You'll recall I wanted to remove these first (or last) duplicates, because they 
represented instances where two different diagnoses (in this case, variable DG, 
value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which 
was 'first' (or 'last').

Your functions have revealed something I wasn't expecting: In some cases, the 
diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two different 
departments in a clinic, and both departments submit data. I have to say that 
crazy things like this are often a feature of real data, which I'm sure you've 
come across yourselves.

Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.

You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
wonder if any of the functions you've written could do this with a little more
Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first (or last) 
DATE duplicate, the DGs for these two dates are different.

Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)# defaults to first = TRUE getRepeat(id.d, first = 
FALSE)  to get the last ones
g.rr-do.call(rbind, g.r) # put the data into

Re: [R] Changing radii line type in radial plots

2012-10-24 Thread bwone

I am using the package plotrix radial.plot(). Yes, radial.plot() has a line
type argument, lty, but that is for the polygons or the radial lines, not
the radii or axes of the radial plot.unless I am doing something wrong.
Thanks!



--
View this message in context: 
http://r.789695.n4.nabble.com/Changing-radii-line-type-in-radial-plots-tp4647238p4647339.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] [r] How to pick colums from a ragged array?

HI Stuart,

Just a small comment:

id.d1$id.d_INCLUDE-TRUE #is not needed.


res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)|tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2-id.d[id.d[,1]%in%names(res1[res1$flag==TRUE,])(duplicated(id.d[,1:2])|duplicated(id.d[,1:2],fromLast=TRUE)),]
res3-res2[!res2$ID%in% 
res2[duplicated(res2)|duplicated(res2,fromLast=TRUE),]$ID,]
id.d1-id.d
bad-id.d1[id.d1$ID%in%res3$ID,]
bad$INCLUDE-FALSE
res4-merge(id.d1,bad,all=TRUE)
res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE
tail(res4)
#   ID DATE DG INCLUDE
#35  910 20080521  4    TRUE
#36  910 20091224  2    TRUE
#37  999 20050503  2    TRUE
#38 1019 19870508  1   FALSE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1   FALSE
A.K.




- Original Message -
From: Stuart Leask stuart.le...@nottingham.ac.uk
To: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
petr.pi...@precheza.cz; Rui Barradas (ruipbarra...@sapo.pt) 
ruipbarra...@sapo.pt
Cc: 
Sent: Wednesday, October 24, 2012 11:41 AM
Subject: RE: [r] How to pick colums from a ragged array?

(And, considering  the real application, the functions ideally should probably 
output a variable INCLUDE, the same length as the original data, with TRUE and 
FALSE for whether or not that row should be included...)

-Original Message-
From: Leask Stuart
Sent: 24 October 2012 16:25
To: arun (smartpink...@yahoo.com); 'PIKAL Petr'; Rui Barradas 
(ruipbarra...@sapo.pt)
Subject: RE: [r] How to pick colums from a ragged array?

Arun, Petr, Rui, many thanks for your help, and the functions you have written.

You'll recall I wanted to remove these first (or last) duplicates, because they 
represented instances where two different diagnoses (in this case, variable DG, 
value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which 
was 'first' (or 'last').

Your functions have revealed something I wasn't expecting: In some cases, the 
diagnoses recorded on the duplicated DATEs are the same!
This is a surprise to me, but probably reflects someone going to two different 
departments in a clinic, and both departments submit data. I have to say that 
crazy things like this are often a feature of real data, which I'm sure you've 
come across yourselves.

Of course, I don't want to remove records in which I can determine an 
unambiguous 'first diagnosis'.

You have all put in so much effort on my behalf, I'm ashamed to ask, but I 
wonder if any of the functions you've written could do this with a little more
Indexing and the 'duplicate' function
So the function should only exclude an ID, having identified a first (or last) 
DATE duplicate, the DGs for these two dates are different.

Test dataset:

ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
,547,794,814,814,814,814,814,814,841,841,841,841,841
,841,841,841,841,910,910,910,910,910,910,999,1019,1019
,1019)

DATE -
c(20060821,20061207,20080102,20090904,20040205,20040205,2005
,20060111,20071119,20080107,20080407,20080521,20080521,20041005
,20070905,20020814,20021125,20040429,20040429,20071205,20071205
,20050421,20050421,20060428,20060602,20060816,20061025,20061129
,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
,20091224,20050503,19870508,19870508,19880330)

DG-
c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4,3,2,2,2,1,1)

id.d-data.frame(ID,DATE,DG)
id.d

# Considering Ruis  getRepeat function:

g.r-getRepeat(id.d)    # defaults to first = TRUE getRepeat(id.d, first = 
FALSE)  to get the last ones
g.rr-do.call(rbind, g.r) # put the data into a matrix

# I can remove the date duplicates with:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]

I'm not sure how to add this to your suggestions, Arun  Petr...


Stuart


-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: 23 October 2012 15:24
To: Stuart Leask
Subject: RE: [r] How to pick colums from a ragged array?

Hi

I assumed that id.d is data frame

id.d - data.frame (ID,DATE )

and

fff(id.d)

works for me

Petr


 -Original Message-
 From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
 Sent: Tuesday, October 23, 2012 3:13 PM
 To: PIKAL Petr
 Subject: RE: [r] How to pick colums from a ragged array?

 Hi Petr.
 I see what you mean it should do, but when I run it I get an error
 (see below).
 Stuart


  ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
 + ,547,794,814,814,814,814,814,814,841,841,841,841,841
 + ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
 + ,1019)
 
  DATE -
 +  c(20060821,20061207,20080102,20090904,20040205,20040205,2005
 +  ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
 +  ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
 +  ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
 +  ,20070112,20070514, 19870508,20040205,20040205, 20091120,20091210
 +  ,20091224,20050503,19870508,19870508,19880330)

[R] randomly select another observation with same grouping factor and year value, do for every record in dataframe

2012-10-24 Thread Meredith, Christy S -FS

Hello,
I am trying to create a function that will move through each record of  a data 
frame, find the value in the HUC column, then
randomly select another observation from the dataframe with the same value in 
HUC column, as well as the same value in Yr column as the first 
observation. I want the function to produce a list of the RchID of the first 
observation, the RchID of the second randomly chosen observation, and several 
other characteristics of the randomly chosen observation.  Below is the code I 
have written, but it doesn't work.

Thanks for any help.
Christy


test
roads=read.csv(streamland23.csv)

for (i in 1:nrow (roads)){
Sitetype= roads$Sitetype
yr=roads$REACH_Yr
initRchid=roads$RchID
huc1=roads$HUC

sample.df - function(df, n) df[sample(nrow(df), n), , drop = FALSE]

selected=sample.df(roads[roads$HUC == huc1 roads$REACH_Yr ==yr , ], 1)


output=cbind (initRchid,selected$RchID,selected$Sitetype,selected$REACH_Yr)

}




Christy Meredith
USDA Forest Service
Rocky Mountain Research Station
PIBO Monitoring
Data Analyst
Voice: 435-755-3573
Fax: 435-755-3563





This electronic message contains information generated by the USDA solely for 
the intended recipients. Any unauthorized interception of this message or the 
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subject the violator to civil or criminal penalties. If you believe you have 
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[[alternative HTML version deleted]]

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

Re: [R] [r] How to pick colums from a ragged array?

Hi,

 According to the OP So the function should only exclude an ID, having 
identified a first (or last) DATE duplicate, the DGs for these two dates are 
different.
Rui:
By running your modified function (using dte - tapply(x[,2], x[,1], FUN = 
function(x) duplicated(fun(x, 2),fromLast = TRUE))), 

 id.d$INCLUDE - !(rm1 | rm2)
 head(id.d)
# ID DATE DG INCLUDE
#1    58 20060821  1    TRUE
#2    58 20061207  2    TRUE
#3    58 20080102  1    TRUE
#4    58 20090904  1    TRUE
#5   167 20040205  4   FALSE
#6   167 20040205  4   FALSE

For #167, DGs are same.  Not sure whether to exclude it or not.


My modified solution is similar but I am excluding 167 and 814.


fun1-function(dat){
res1first- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
 res1last- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2first-dat[dat[,1]%in%names(res1first[res1first$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res2last-dat[dat[,1]%in%names(res1last[res1last$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res3first-res2first[!res2first$ID%in% 
res2first[duplicated(res2first)|duplicated(res2first,fromLast=TRUE),]$ID,]
res3last-res2last[!res2last$ID%in% 
res2last[duplicated(res2last)|duplicated(res2last,fromLast=TRUE),]$ID,]
res3firstsubset-do.call(rbind,lapply(split(res3first,res3first$ID),head,1))
res3firstsubset$INCLUDE-FALSE
res3lastsubset-do.call(rbind,lapply(split(res3last,res3last$ID),tail,1))
res3lastsubset$INCLUDE-FALSE
 res4-merge(dat,merge(res3first,merge(res3firstsubset,merge(res3lastsubset,res3last,all=TRUE),all=TRUE),all=TRUE),all=TRUE)
 res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE
res4
}

tail(fun1(id.d))
# ID DATE DG INCLUDE
#35  910 20080521  4    TRUE
#36  910 20091224  2    TRUE
#37  999 20050503  2    TRUE
#38 1019 19870508  1    TRUE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1    TRUE

A.K.












- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Stuart Leask stuart.le...@nottingham.ac.uk
Sent: Wednesday, October 24, 2012 2:50 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Inline.
Em 24-10-2012 19:05, arun escreveu:
 Hi Rui,

 I think now our results are matching except in the INCLUDE column

 id.d[c(11:13,22:24,38:40),]
 #     ID     DATE DG INCLUDE
 #11  323 20080407  1    TRUE
 #12  323 20080521  2   FALSE
 #13  323 20080521  3    TRUE
 #22  841 20050421  1    TRUE
 #23  841 20050421  2   FALSE
 #24  841 20060428  1    TRUE
 #38 1019 19870508  2    TRUE
 #39 1019 19870508  1   FALSE
 #40 1019 19880330  1    TRUE


 I thought all the rows with the above IDS would be FALSE

Why? Look at the last ID, 1019. The last of all must be included, the 
date doesn't repeat. And one of the first must also be included, if not 
we would be completely excluding that date. Or at least this is how I'm 
understanding the problem.

Rui Barradas
   (from my solution):

 res4[c(11:13,22:24,38:40),]
       ID     DATE DG INCLUDE
 #11  323 20080407  1   FALSE
 #12  323 20080521  2   FALSE
 #13  323 20080521  3   FALSE
 #22  841 20050421  1   FALSE
 #23  841 20050421  2   FALSE
 #24  841 20060428  1   FALSE
 #38 1019 19870508  1   FALSE
 #39 1019 19870508  2   FALSE
 #40 1019 19880330  1   FALSE

 A.K.




 - Original Message -
 From: Rui Barradas ruipbarra...@sapo.pt
 To: Stuart Leask stuart.le...@nottingham.ac.uk
 Cc: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL Petr 
 petr.pi...@precheza.cz; r-help r-help@r-project.org
 Sent: Wednesday, October 24, 2012 1:41 PM
 Subject: Re: [r] How to pick colums from a ragged array?

 Hello,

 Using one of Arun's ideas, some post ago, this new function returns a
 logical index into id.d of the rows that should be _removed_, hence rm1
 and rm2. I think



 getRepLogical - function(x, first = TRUE){
       fun - if(first) head else tail
       dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2)))
       len - tapply(x[,2], x[,1], FUN = length)
       lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE,
 if(len[[i]]  2) len[[i]] - 2 else 0)))
       lst - if(first) lst else lapply(lst, rev)
       i1 - unlist(lst)
       dg - tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x, 2)))
       lst - lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE,
 if(len[[i]]  2) len[[i]] - 2 else 0)))
       lst - if(first) lst else lapply(lst, rev)
       i2 - unlist(lst)
       i1  i2
 }

 rm1 - getRepLogical(id.d)
 rm2 - getRepLogical(id.d, first = FALSE)

 id.d[rm1, ]
 id.d[rm2, ]

 id.d$INCLUDE - !(rm1 | rm2)


 Hope this helps,

 Rui Barradas
 Em 24-10-2012 16:41, Stuart Leask escreveu:
 (And, considering  the real application, the functions ideally should 
 probably output a variable INCLUDE, the same length as the original data, 
 with TRUE and

[R] Testing proportional odds assumption in R

2012-10-24 Thread Thomas Yee


Hi,

M1 and M2 are extreme in that all or none of the variables have
parallel lines on the logit scale.  One can try fitting a partial
POM, which remains fraught (but not as much as M2) because if
the lines intersect for a particular variable where the data lie
then there will be numerical problems.

For example, try something like

M3 - vglm(y ~ x1 + x2 + x3, data = data,
  family = cumulative(parallel = FALSE ~ x3 - 1))

which will only allow nonparallelism wrt the x3 variable.

Another idea is to try transforming each x variable if
it crashes... that might help.

BTW, can use lrtest(), e.g., lrtest(M3, M2)

cheers

Thomas

I want to test whether the proportional odds assumption for an ordered
regression is met.   
 
The UCLA website points out that there is no mathematical way to test the
proportional odds assumption 
(http://www.ats.ucla.edu/stat//R/dae/ologit.htm),
and use graphical inspection (We were unable to locate a facility in 
R to   
perform any of the tests commonly used to test the parallel 
slopes   
assumption.).

  

However, I found a pdf by Agresti suggesting a method using the 
vglm 
function, the pdf is called Examples of Using R for Modeling 
Ordinal
Data.


 M1 - vglm(y ~ x1 + x2 + x3, data=data, 
family=cumulative(parallel=TRUE))

 M2 - vglm(y ~ x1 + x2 + x3, data=data, family=cumulative, maxit=100)

pchisq(deviance(Mo.1)-deviance(Mo.2), 
df=df.residual(Mo.1)-df.residual(Mo.2),

lower.tail=FALSE)

If the test is significant, the proportional odds assumption is (might be)
violated.

However, running this procedure in my dataset with 5 predictors (3
dichotomous, 2 z-standardized metric) and an ordinal dependent variable
(0,1,2,3) in a sample of N=2500 leads to various problems, maybe you can
help me out how to solve these.

Warning messages:
(1) Error in dotC(name = tapplymat1, mat = as.double(mat),
as.integer(nr),  :
  NA/NaN/Inf in foreign function call (arg 1
(2) In matrix.power(wz, M = M, power = 0.5, fast = TRUE) :
  Some weight matrices have negative eigenvalues. They
will be assigned NAs
(3)
In Deviance.categorical.data.vgam(mu = mu, y = y, w = w, residuals =
residuals,  :
  fitted values close to 0 or 1
(4) In log(prob) : NaNs produced

The last two don't seem to be as critical and the first two, seeing that
models with that errors at least provide a p-value (the first to errors
lead to a p--value of 1).

Thank you

   [[alternative HTML version deleted]]


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Re: [R] Recode function car package erases previous values

Hi,

I think the as.factor in ur code created trouble.
library(car)
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))

  

pharm-recode(dat1$B20_C1,1='Yes';0='No')
 pharm[dat1$nrB20C==1]-no resp
pharm
#[1] no resp No  No  Yes Yes NA   
 table(pharm)
#pharm
  #   No no resp Yes 
    #  2   1   2 
A.K.




- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Wednesday, October 24, 2012 10:41 AM
Subject: RE: [R] Recode function car package erases previous values

Hi Arun,
I also used to get that error, but what class is your nr.B20C, mine is not a 
factor, it is numeric and perhaps that's why it works


-Original Message-
From: arun [mailto:smartpink...@yahoo.com] 
Sent: Wednesday, October 24, 2012 4:35 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi Pancho,

I tried ur method:
pharm-as.factor(recode(dat1$B20_C1,1='Yes';0='No'))
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes


pharm[dat1$nrB20C==1]-'no resp'
#Warning message:
#In `[-.factor`(`*tmp*`, dat1$nr.B20C == 1, value = no resp) :
 # invalid factor level, NAs generated
 pharm
#[1] NA No   No   Yes  Yes  NA
#Levels: No Yes


Not sure how you got the result.
A.K.

 



- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Wednesday, October 24, 2012 9:51 AM
Subject: RE: [R] Recode function car package erases previous values



-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Wednesday, October 24, 2012 3:31 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values

Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
#  B20_C1 B20_C2 nrB20C
#1 NA NA  1
#2  0  0 NA
#3  0  1 NA
#4  0  0 NA
#5  1  1 NA
#6 NA  1 NA
 dat1$pharm-ifelse(apply(dat1,1,function(x) all(is.na(x[1:2]))),no 
response,ifelse(dat1[,1]==1,Yes,No))
dat2-within(dat1,{pharm-factor(pharm)})
 levels(dat2$pharm)
#[1] No  no response Yes
 dat2
#  B20_C1 B20_C2 nrB20C   pharm
#1 NA NA  1 no response
#2  0  0 NA  No
#3  0  1 NA  No
#4  0  0 NA  No
#5  1  1 NA Yes
#6 NA  1 NA    NA

table(dat2$pharm)

# No no response Yes
 # 3   1   1 

A.K.

Thank you, once I understand what you did I will use it, for now, I use a 
workaround 1. First I use the recode function
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
2. Now I just subset
pharm[nr.B20C==1]-'no resp'
and this gives me the desired output
     No     Yes no resp
    716       7       6 


- Original Message -
From: Pancho Mulongeni p.mulong...@namibia.pharmaccess.org
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 5:17 AM
Subject: [R] Recode function car package erases previous values

Hi all,
I am attempting to create a new variable based on values of other variables. 
The variable is called pharm. It basically takes the numeric code of 1 as yes 
and 0 to be No from the variable B20_C1 (a question on a survey). However, I 
would also like to have a level for non-respondents and these are captured in 
the variable nr.B20C, which is a 1 when there is a non-response on the whole 
group of variable B20_C1 to B20_C5. So ultimately the variable pharmacy will 
have three levels, Yes No and no resp.
See below what happens

 pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
 table(pharm)
pharm
No Yes
716   7 
 levels(pharm)-c('No','Yes','no resp')
 table(pharm)
pharm
     No     Yes no resp
    716       7       0 
 pharm-as.factor(recode(nr.B20C,1='no resp'))
 table(pharm)
pharm
      0 no resp
    723       6 

The recode variable just cannot seem to 'remember' I had just recoded 7 values 
to 'Yes' and 716 to be 'No'  and instead it assigns the level '0' which comes 
from nr.B20C (it has values 0 or 1). 
This inconvenient as I would like to have ultimately the following table pharm
     No     Yes no resp
    716       7       6 (FROM nr.B20C where row has value 1).

Background. The variable pharm assess where you used the pharmacy to get your 
contraception.

Pancho Mulongeni
Research Assistant
PharmAccess Foundation
1 Fouché Street
Windhoek West
Windhoek
Namibia
 
Tel:   +264 61 419 000
Fax:  +264 61 419 001/2
Mob: +264 81 4456 286

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Re: [R] Assigning values to several consecutives rows in a sequence while leaving some empty

2012-10-24 Thread nymphita


Thank you very much Michael and Berend for your ideas and feddback. 

I apologize for my mistakes. It's true that I still have much to learn (and
I sometimes forget what I read). I'll surely use the dput() command next
time I post something and I'll try to be more clear with my questions too. 

Berend's suggestion worked perfectly! 
Thanks a lot.





--
View this message in context: 
http://r.789695.n4.nabble.com/Assigning-values-to-several-consecutives-rows-in-a-sequence-while-leaving-some-empty-tp4646956p4647352.html
Sent from the R help mailing list archive at Nabble.com.

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

[R] Kaplan Meier Post Hoc?

2012-10-24 Thread Charles Determan Jr

This is more of a general question without data.  After doing 'survdiff',
from the 'survival' package, on strata including four groups (so 4 curves
on a Kaplan Meier curve) you get a chi squared p-value whether to reject
the null hypothesis or not.  Is there a method to followup with pairwise
testing on the respective groups?  I have searched the library but have
come up with nothing.  Perhaps I am mistaken in something here.

Regards,
Charles

[[alternative HTML version deleted]]

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

Re: [R] [r] How to pick colums from a ragged array?

2012-10-24 Thread Stuart Leask

I mis-typed, missing an if. I think you've got it, but let me try again:

The function should:
-  put FALSE in a column for every instance of an ID
IF ( that ID has a first (or last) DATE duplicated )
AND
IF (the DGs for the duplicated dates are different).

So for the earliest/first date function, INCLUDE should be TRUE, apart from 
FALSE for _all_ the instances of IDs 167, 841 and 1019
For the latest/last date function, INCLUDE should be TRUE, apart from FALSE for 
all the instances of ID  323.

 Stuart

-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: 24 October 2012 21:30
To: Rui Barradas
Cc: R help; Stuart Leask
Subject: Re: [r] How to pick colums from a ragged array?

Hi,

 According to the OP So the function should only exclude an ID, having 
identified a first (or last) DATE duplicate, the DGs for these two dates are 
different.
Rui:
By running your modified function (using dte - tapply(x[,2], x[,1], FUN = 
function(x) duplicated(fun(x, 2),fromLast = TRUE))),

 id.d$INCLUDE - !(rm1 | rm2)
 head(id.d)
# ID DATE DG INCLUDE
#158 20060821  1TRUE
#258 20061207  2TRUE
#358 20080102  1TRUE
#458 20090904  1TRUE
#5   167 20040205  4   FALSE
#6   167 20040205  4   FALSE

For #167, DGs are same.  Not sure whether to exclude it or not.


My modified solution is similar but I am excluding 167 and 814.


fun1-function(dat){
res1first- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
 res1last- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2first-dat[dat[,1]%in%names(res1first[res1first$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res2last-dat[dat[,1]%in%names(res1last[res1last$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res3first-res2first[!res2first$ID%in% 
res2first[duplicated(res2first)|duplicated(res2first,fromLast=TRUE),]$ID,]
res3last-res2last[!res2last$ID%in% 
res2last[duplicated(res2last)|duplicated(res2last,fromLast=TRUE),]$ID,]
res3firstsubset-do.call(rbind,lapply(split(res3first,res3first$ID),head,1))
res3firstsubset$INCLUDE-FALSE
res3lastsubset-do.call(rbind,lapply(split(res3last,res3last$ID),tail,1))
res3lastsubset$INCLUDE-FALSE
 
res4-merge(dat,merge(res3first,merge(res3firstsubset,merge(res3lastsubset,res3last,all=TRUE),all=TRUE),all=TRUE),all=TRUE)
 res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE
res4
}

tail(fun1(id.d))
# ID DATE DG INCLUDE
#35  910 20080521  4TRUE
#36  910 20091224  2TRUE
#37  999 20050503  2TRUE
#38 1019 19870508  1TRUE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1TRUE

A.K.












- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Stuart Leask stuart.le...@nottingham.ac.uk
Sent: Wednesday, October 24, 2012 2:50 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Inline.
Em 24-10-2012 19:05, arun escreveu:
 Hi Rui,

 I think now our results are matching except in the INCLUDE column

 id.d[c(11:13,22:24,38:40),]
 # ID DATE DG INCLUDE
 #11  323 20080407  1TRUE
 #12  323 20080521  2   FALSE
 #13  323 20080521  3TRUE
 #22  841 20050421  1TRUE
 #23  841 20050421  2   FALSE
 #24  841 20060428  1TRUE
 #38 1019 19870508  2TRUE
 #39 1019 19870508  1   FALSE
 #40 1019 19880330  1TRUE


 I thought all the rows with the above IDS would be FALSE

Why? Look at the last ID, 1019. The last of all must be included, the date 
doesn't repeat. And one of the first must also be included, if not we would be 
completely excluding that date. Or at least this is how I'm understanding the 
problem.

Rui Barradas
   (from my solution):

 res4[c(11:13,22:24,38:40),]
   ID DATE DG INCLUDE
 #11  323 20080407  1   FALSE
 #12  323 20080521  2   FALSE
 #13  323 20080521  3   FALSE
 #22  841 20050421  1   FALSE
 #23  841 20050421  2   FALSE
 #24  841 20060428  1   FALSE
 #38 1019 19870508  1   FALSE
 #39 1019 19870508  2   FALSE
 #40 1019 19880330  1   FALSE

 A.K.




 - Original Message -
 From: Rui Barradas ruipbarra...@sapo.pt
 To: Stuart Leask stuart.le...@nottingham.ac.uk
 Cc: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL
 Petr petr.pi...@precheza.cz; r-help r-help@r-project.org
 Sent: Wednesday, October 24, 2012 1:41 PM
 Subject: Re: [r] How to pick colums from a ragged array?

 Hello,

 Using one of Arun's ideas, some post ago, this new function returns a
 logical index into id.d of the rows that should be _removed_, hence
 rm1 and rm2. I think



 getRepLogical - function(x, first = TRUE){
   fun - if(first) head else tail
   dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x,
2)))
   len - tapply(x[,2], x[,1], FUN = length)
   lst - lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE,
if(len[[i]]  2) len[[i]] - 2 else 0)))
   lst

Re: [R] [r] How to pick colums from a ragged array?


Hello,

Inline.
Em 24-10-2012 22:40, Stuart Leask escreveu:

I mis-typed, missing an if. I think you've got it, but let me try again:

The function should:
-  put FALSE in a column for every instance of an ID
IF ( that ID has a first (or last) DATE duplicated )
AND
IF (the DGs for the duplicated dates are different).

So for the earliest/first date function, INCLUDE should be TRUE, apart from 
FALSE for _all_ the instances of IDs 167, 841 and 1019
For the latest/last date function, INCLUDE should be TRUE, apart from FALSE for 
all the instances of ID  323.


In this case forget my last post and use getRepLogical as I posted it 
originaly


Rui Barradas


  Stuart

-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: 24 October 2012 21:30
To: Rui Barradas
Cc: R help; Stuart Leask
Subject: Re: [r] How to pick colums from a ragged array?

Hi,

  According to the OP So the function should only exclude an ID, having identified 
a first (or last) DATE duplicate, the DGs for these two dates are different.
Rui:
By running your modified function (using dte - tapply(x[,2], x[,1], FUN = 
function(x) duplicated(fun(x, 2),fromLast = TRUE))),

  id.d$INCLUDE - !(rm1 | rm2)
  head(id.d)
# ID DATE DG INCLUDE
#158 20060821  1TRUE
#258 20061207  2TRUE
#358 20080102  1TRUE
#458 20090904  1TRUE
#5   167 20040205  4   FALSE
#6   167 20040205  4   FALSE

For #167, DGs are same.  Not sure whether to exclude it or not.


My modified solution is similar but I am excluding 167 and 814.


fun1-function(dat){
res1first- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
  res1last- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) 
tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2first-dat[dat[,1]%in%names(res1first[res1first$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res2last-dat[dat[,1]%in%names(res1last[res1last$flag==TRUE,])(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res3first-res2first[!res2first$ID%in% 
res2first[duplicated(res2first)|duplicated(res2first,fromLast=TRUE),]$ID,]
res3last-res2last[!res2last$ID%in% 
res2last[duplicated(res2last)|duplicated(res2last,fromLast=TRUE),]$ID,]
res3firstsubset-do.call(rbind,lapply(split(res3first,res3first$ID),head,1))
res3firstsubset$INCLUDE-FALSE
res3lastsubset-do.call(rbind,lapply(split(res3last,res3last$ID),tail,1))
res3lastsubset$INCLUDE-FALSE
  
res4-merge(dat,merge(res3first,merge(res3firstsubset,merge(res3lastsubset,res3last,all=TRUE),all=TRUE),all=TRUE),all=TRUE)
  res4$INCLUDE[is.na(res4$INCLUDE)]-TRUE
res4
}

tail(fun1(id.d))
# ID DATE DG INCLUDE
#35  910 20080521  4TRUE
#36  910 20091224  2TRUE
#37  999 20050503  2TRUE
#38 1019 19870508  1TRUE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1TRUE

A.K.












- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Stuart Leask stuart.le...@nottingham.ac.uk
Sent: Wednesday, October 24, 2012 2:50 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Inline.
Em 24-10-2012 19:05, arun escreveu:

Hi Rui,

I think now our results are matching except in the INCLUDE column

id.d[c(11:13,22:24,38:40),]
# ID DATE DG INCLUDE
#11  323 20080407  1TRUE
#12  323 20080521  2   FALSE
#13  323 20080521  3TRUE
#22  841 20050421  1TRUE
#23  841 20050421  2   FALSE
#24  841 20060428  1TRUE
#38 1019 19870508  2TRUE
#39 1019 19870508  1   FALSE
#40 1019 19880330  1TRUE


I thought all the rows with the above IDS would be FALSE

Why? Look at the last ID, 1019. The last of all must be included, the date 
doesn't repeat. And one of the first must also be included, if not we would be 
completely excluding that date. Or at least this is how I'm understanding the 
problem.

Rui Barradas

   (from my solution):

res4[c(11:13,22:24,38:40),]
   ID DATE DG INCLUDE
#11  323 20080407  1   FALSE
#12  323 20080521  2   FALSE
#13  323 20080521  3   FALSE
#22  841 20050421  1   FALSE
#23  841 20050421  2   FALSE
#24  841 20060428  1   FALSE
#38 1019 19870508  1   FALSE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1   FALSE

A.K.




- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: Stuart Leask stuart.le...@nottingham.ac.uk
Cc: arun (smartpink...@yahoo.com) smartpink...@yahoo.com; PIKAL
Petr petr.pi...@precheza.cz; r-help r-help@r-project.org
Sent: Wednesday, October 24, 2012 1:41 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Using one of Arun's ideas, some post ago, this new function returns a
logical index into id.d of the rows that should be _removed_, hence
rm1 and rm2. I think



getRepLogical - function(x, first = TRUE){
   fun - if(first) head else tail
   dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x,
2)))
   len - tapply(x[,2], x[,1], FUN

Re: [R] Kaplan Meier Post Hoc?

2012-10-24 Thread Marc Schwartz

On Oct 24, 2012, at 4:33 PM, Charles Determan Jr deter...@umn.edu wrote:

 This is more of a general question without data.  After doing 'survdiff',
 from the 'survival' package, on strata including four groups (so 4 curves
 on a Kaplan Meier curve) you get a chi squared p-value whether to reject
 the null hypothesis or not.  Is there a method to followup with pairwise
 testing on the respective groups?  I have searched the library but have
 come up with nothing.  Perhaps I am mistaken in something here.
 
 Regards,
 Charles



Take a look at ?p.adjust, which provides a generic framework for multiple 
pairwise comparison adjustment methods.

The most conservative, but not always the best approach, would of course be 
Bonferroni.

Regards,

Marc Schwartz

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

[R] Defining categories

2012-10-24 Thread bibek sharma

Hello R user,

Data below represent year in decimal. I would like to catagorize it
in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to
catagory 2 and so on..
Any suggestion how it can be done with if else statement or any other way?

2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
4.116667
1.63
2.147222

Thank you for  your help.
Bibek

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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] Defining categories

Hi,
May be this:
dat1-read.table(text=
2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
4.116667
1.63
2.147222
,sep=,header=FALSE)
dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11  
dat1$V1=2,2,ifelse(dat1$V12dat1$V1=3,3,ifelse(dat1$V13dat1$V1=4,4,ifelse(dat1$V14dat1$V1=5,5,6)


 head(dat1)
#    V1 category
#1 2.880556    3
#2 0.616667    1
#3 5.08    6
#4 0.858333    1
#5 0.47    1
#6 2.936111    3
A.K.



- Original Message -
From: bibek sharma mbhpat...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, October 24, 2012 6:52 PM
Subject: [R] Defining categories

Hello R user,

Data below represent year in decimal. I would like to catagorize it
in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to
catagory 2 and so on..
Any suggestion how it can be done with if else statement or any other way?

2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
4.116667
1.63
2.147222

Thank you for  your help.
Bibek

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

Re: [R] Defining categories

2012-10-24 Thread Jorge I Velez

See ?cut for a simpler way of doing this.
HTH,
Jorge.-


On Thu, Oct 25, 2012 at 10:02 AM, arun  wrote:

 Hi,
 May be this:
 dat1-read.table(text=
 2.880556
 0.616667
 5.08
 0.858333
 0.47
 2.936111
 4.258333
 0.258333
 2.03
 2.58
 1.09
 0.447222
 1.87
 0.080556
 4.03
 4.116667
 1.63
 2.147222
 ,sep=,header=FALSE)
 dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11 
 dat1$V1=2,2,ifelse(dat1$V12dat1$V1=3,3,ifelse(dat1$V13dat1$V1=4,4,ifelse(dat1$V14dat1$V1=5,5,6)


  head(dat1)
 #V1 category
 #1 2.8805563
 #2 0.6166671
 #3 5.086
 #4 0.8583331
 #5 0.471
 #6 2.9361113
 A.K.



 - Original Message -
 From: bibek sharma 
 To: r-help@r-project.org
 Cc:
 Sent: Wednesday, October 24, 2012 6:52 PM
 Subject: [R] Defining categories

 Hello R user,

 Data below represent year in decimal. I would like to catagorize it
 in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to
 catagory 2 and so on..
 Any suggestion how it can be done with if else statement or any other way?

 2.880556
 0.616667
 5.08
 0.858333
 0.47
 2.936111
 4.258333
 0.258333
 2.03
 2.58
 1.09
 0.447222
 1.87
 0.080556
 4.03
 4.116667
 1.63
 2.147222

 Thank you for  your help.
 Bibek

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


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

[R] extracting variance co variance matrix from system package

2012-10-24 Thread Dereje Bacha

I run system of equations. The system package gives variance
covariance matrix for each equation. I want to extract the variance covariance 
matrix
(of the coefficients) for each equation. How do we extract it?. 
Y1~X1+X2
Y2~X1+X2
fit.ols-system (system, ols,
data)â¦â¦. To extract it I wrote as follows:
fit.ols$coefCov [[eq1]].
Â it did not work.Â 

I
want it to apply to delta method. I need help please. Â Thank you!. . 
Dereje
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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] Z score



Hi Ved,

Sorry, I didn't test it well enough at that time.  

In your example file,
 #there were NAs
MyFile1 - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )


#Here, the apply() function outputs a list when I remove the NA from the last 
row.
 apply(MyFile1,1,function(x) x[!is.na(x)]) #outputs a list
#$Gene_1
#Sample_1 Sample_2 Sample_3 
 # 87   77   88 

#$Gene_2
#Sample_1 Sample_2 Sample_3 
 # 98   22   34 

#$Gene_3
#Sample_1 Sample_2 Sample_3 
 # 33   43   33 

#$Gene_4
#Sample_1 Sample_3 
 # 78   81 

# Without NAs
MyFile2 - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,48,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

apply(dat3,1,function(x) x[!is.na(x)]) # the output is a matrix
# Gene_1 Gene_2 Gene_3 Gene_4
#Sample_1 87 98 33 78
#Sample_2 77 22 43 48
#Sample_3 88 34 33 81
is.matrix(apply(dat3,1,function(x) x[!is.na(x)]) )
#[1] TRUE

#Consider another case
MyFile3 - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )

t(sapply(lapply(apply(MyFile3,1,function(x) x[!is.na(x)]),function(x) 
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile3)] )) #works because the 
apply() output is a list
#    Sample_1   Sample_2   Sample_3
#Gene_1  0.4931970 -1.1507929  0.6575959
#Gene_2 NA -0.7071068  0.7071068
#Gene_3 -0.5773503  1.1547005 -0.5773503
#Gene_4 -0.7071068 NA  0.7071068


#Yet another case:
MyFile4 - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77
Gene_2,,22,34
Gene_3,33,,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )
 apply(MyFile4,1,function(x) x[!is.na(x)]) #output is a matrix because equal 
number of NAs were present in each row
# Gene_1 Gene_2 Gene_3 Gene_4
#[1,] 87 22 33 78
#[2,] 77 34 33 81
t(sapply(lapply(apply(MyFile4,1,function(x) x[!is.na(x)]),function(x) 
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile4)] )) #doesn't work



#In your dataset, there were no NAs
dat1-read.csv(Bcl2_With_expressions.csv,sep=\t,row.names=1)
MyFile-dat1[,-1]

 str(apply(MyFile,1,function(x) x[!is.na(x)])) # a matrix 
# num [1:29, 1:18] 10.48 10.96 9.28 11.1 10.95 ...
 #- attr(*, dimnames)=List of 2
 # ..$ : chr [1:29] ALL2 MLL8 ALL42 MLL5 ...
 # ..$ : chr [1:18] BAX BCL2L15 BCL2 BMF ...

#In this case, 
either
 res2-apply(MyFile,1,function(x) (x-mean(x))/sd(x))

#or

 res1-apply(apply(MyFile,1,function(x) x[!is.na(x)]),2,function(x) 
(x-mean(x))/sd(x)) #works

 
 identical(res1,res2)
#[1] TRUE

 head(res1,2)
 #  BAX   BCL2L15 BCL2    BMF    BAD  MCL1 BCL2L1
#ALL2 0.1216373 -0.215256 1.040758 -0.4078606 -0.2427741 0.6967070 -0.1054749
#MLL8 0.6565878 -1.446252 1.052566 -0.1825442 -0.2312166 0.9882503 -0.9687260
  #  BOK BCL2A1    BCL2L14   BAK1  BBC3    BCL2L11
#ALL2 -0.1465807  0.5353133 -0.1772439 -0.3751981 0.6341806 -1.2432273
#MLL8  0.2918296 -0.8466821  0.3088331 -1.4025846 0.7056799  0.9944288
  #  BID NOXA1    BIK  HRK    BCL2L2
#ALL2 -2.2961643 0.2105960 -0.9195998 -0.001731806 1.6691590
#MLL8 -0.5103087 0.3433778  1.2352986 -0.568548518 0.3674839


Hope it helps
A.K.








From: Vedant Sharma vedantg...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Wednesday, October 24, 2012 7:56 PM
Subject: Re: [R] Z score


Hello Arun,

Thank you. I could manage to get the answer. 

However, this particular code, however, doesn't seem to work when I try to read 
from a .csv file (as attached). And, I am inquisitive to find out the reason ! 

MyFile - read.csv (file.choose(), header=T, row.names=1)
MyFile - MyFile [,-1]
res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x) 
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] ))

Thanks again !! 

Cheers,
Ved

=


On Wed, Oct 24, 2012 at 9:53 PM, arun smartpink...@yahoo.com wrote:

Hi,

In cases, with more sample columns, you could also use this:
 res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x) 
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] ))
res2

 #    Sample_1   Sample_2   Sample_3
#Gene_1  0.4931970 -1.1507929  0.6575959
#Gene_2  1.1421818 -0.7179429 -0.4242390
#Gene_3 -0.5773503  1.1547005 -0.5773503
#Gene_4 -0.7071068 NA  0.7071068
A.K.



- Original Message -
From: Vedant Sharma vedantg...@gmail.com
To: R help r-help@r-project.org
Cc:
Sent: Wednesday, October 24, 2012 2:17 AM
Subject: [R] Z score


Hi,

I need to find the z-score of the data present in a speardsheet. The values
needs to be

Re: [R] Defining categories