date:20120524

The combination of column means in cbind is what I am trying to do.  I just
don't know how to do it for every component (2 in this case) of the list. 
I've been able to work with R to apply a function across all components when
there is just one list but I am having trouble working with multiple lists
at the same time.  

--
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Re: [R] Question on if i am allowed to do something

2012-05-24 Thread Duncan Murdoch


On 12-05-24 8:23 AM, Giannis Mamalikidis wrote:

Your answers were very helpful, thanks :)

But, you see, the program I'm writing will be a team effort - and while I
have not problem on making this program an open source - we don't really see
eye to eye with the team on this.
So no source code will be available - just freeware.

R will not be at all altered! the only thing that R will do is be used to
show a couple of plots, and make some t.test and ks.test.

Does that qualify as a derivative work of R? Or would derivative work be if
I altered R's Source code?


I would say that makes your program a derivative work, so you should not 
distribute R as part of it.


But I am not a lawyer; in fact, I'm an interested party:  a copyright 
holder on parts of R.  So I don't know if the law is on my side, but you 
can be assured that you are violating the wishes of at least one R 
copyright holder if you do distribute it in a closed source package.


Duncan Murdoch



I've read the license but English is not my Native language, hence some
terminology is unclear to me.

And I'm just an undergraduate with no income in the middle of an economy
crisis. I don't really see me paying a lawyer to ask the question.

---
Giannis Mamalikidis
e-mail: giannis_mamaliki...@msn.com
Member of the STAINS Research Group
(STAtistics  INformation Systems Group) of the
Aristotle University of Thessaloniki
Site: http://stains.csd.auth.gr
B.Sc Student in Electrical and Mechanical Engineering
---
-Original Message-
From: Marc Schwartz
Sent: Thursday, May 24, 2012 3:58 AM
To: Duncan Murdoch
Cc: Giannis Mamalikidis ; r-help@r-project.org
Subject: Re: [R] Question on if i am allowed to do something


On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote:


On 12-05-23 1:31 PM, Giannis Mamalikidis wrote:

Hello all.

I would like to know: provided that I absolutely state that R is not mine
and I also include the R’s License which will be shown so people know R
and
R’s license,
(provided the above) am I allowed to include R’s folder (the folder that
has
its binaries) on my freeware program or not?


R is licensed under the GPL v 2 or 3.  The recommended packages that come
with it are separately licensed, but are also (mainly) GPL.  So just
follow the terms of that license, which is included with R.

Duncan Murdoch



Just to add on to Duncan's reply, you might want to review the GPL FAQ (and
possibly consult a lawyer):

   http://www.gnu.org/licenses/gpl-faq.html

The use and distribution of R within your 'freeware' can place certain
obligations on you, including possibly requiring that you license your
application with a GPL compatible license and make the source code of your
application available in situations where your application is 'linked' to R.
If you do not intend to make your own source code available, you should be
very clear on what the GPL requires of you before proceeding.

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.

Re: [R] Using NA as a break point for indicator variable?


Hello,

Assuming that 'd' is your original data.frame and that you've set entire 
rows to NA, try this



d$leak_num - NA
ix - !is.na(d[, 1])  # any column will do, entire row is NA
## alternative, if other rows may have NAs, due to something else
#ix - apply(d, 1, function(x) all(!is.na(x)))
r - rle(ix)
v - cumsum(r$values)
d$leak_num[ix] - rep(v[r$values], r$lengths[r$values])
d


Hope this helps,

Rui Barradas

Em 24-05-2012 11:00, Max Brondfield mbro...@post.harvard.edu escreveu:

Date: Wed, 23 May 2012 16:42:02 -0400
From: Max Brondfieldmbro...@post.harvard.edu
To:r-help@r-project.org
Subject: [R] Using NA as a break point for indicator variable?
Message-ID:
cadu+jdpcjuhztxxrsxyqvjaemw_n0ilbl6zjjhzc-rsbcmn...@mail.gmail.com
Content-Type: text/plain

Hi all,
I am working with a spatial data set for which I am only interested in high
concentration values (leaks). The low values (  90th percentile) have
already been turned into NA's, leaving me with a matrix like this:

  CH4_leak

   lonlatCH4
1  -71.11954 42.35068 2.595834
2  -71.11954 42.35068 2.595688
3   NA   NA   NA
4   NA   NA   NA
5   NA   NA   NA
6  -71.11948 42.35068 2.435762
7  -71.11948 42.35068 2.491003
8  NANA   NA
9  -71.11930 42.35068 2.464475
10 -71.11932 42.35068 2.470865

Every time an NA comes up, it means the leak is gone, and the next valid
value would represent a different leak (at a different location). My goal
is to tag all of the remaining values with an indicator variable to
spatially distinguish the leaks. I am envisioning a simple numeric
indicator such as:

  lonlatCH4leak_num
1  -71.11954 42.35068 2.595834   1
2  -71.11954 42.35068 2.595688   1
3   NA   NA   NA NA
4   NA   NA   NA NA
5   NA   NA   NA NA
6  -71.11948 42.35068 2.435762   2
7  -71.11948 42.35068 2.491003   2
8  NANA   NA NA
9  -71.11930 42.35068 2.064475   3
10 -71.11932 42.35068 2.070865  3

Does anyone have any thoughts on how to code this, perhaps using the NA
values as a break point? The data set is far too large to do this
manually, and I must admit I'm completely at a loss. Any help would be much
appreciated! Best,

Max

[[alternative HTML version deleted]]




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Re: [R] Question on if i am allowed to do something

2012-05-24 Thread Marc Schwartz

Giannis,

While your economic situation is appreciated, that you plan to give away your 
application and not generate any income from it, is irrelevant to the question.

The nuances as to how you are interfacing with R and that you plan to include R 
(importantly, possibly only a part of it) as a component of your application 
are. Not altering R's source code is similarly only one part of the issue, 
though if you were, you would be clearly crossing the line to a derivative 
work. However, even in that scenario there are other considerations as to 
whether or not those changes impact your application versus just the modified R 
code.

You are asking a legal question on a list that is focused on technical 
considerations and expecting the list participants to provide free legal 
advice, based upon which you may be making life altering decisions. While you 
may not be able to afford to pay a lawyer now, you and you partners may find 
yourselves needing to pay a lawyer in the future to defend yourselves against a 
lawsuit for copyright and licensing infringement. A situation, by the way, that 
won't be at all helpful to your future career.

You may wish to see if there are any lawyers locally or at your institution 
with specific experience in this domain who may be will to provide you with pro 
bono advice. Lacking expert legal advice, the only thing we can tell you is 
don't do it because you may be violating the GPL license and therefore may 
expose yourself and your partners to future legal risk.

Regards,

Marc

On May 24, 2012, at 7:23 AM, Giannis Mamalikidis wrote:

 Your answers were very helpful, thanks :)
 
 But, you see, the program I'm writing will be a team effort - and while I 
 have not problem on making this program an open source - we don't really see 
 eye to eye with the team on this.
 So no source code will be available - just freeware.
 
 R will not be at all altered! the only thing that R will do is be used to 
 show a couple of plots, and make some t.test and ks.test.
 
 Does that qualify as a derivative work of R? Or would derivative work be if I 
 altered R's Source code?
 I've read the license but English is not my Native language, hence some 
 terminology is unclear to me.
 
 And I'm just an undergraduate with no income in the middle of an economy 
 crisis. I don't really see me paying a lawyer to ask the question.
 
 ---
 Giannis Mamalikidis
 e-mail: giannis_mamaliki...@msn.com
 Member of the STAINS Research Group
 (STAtistics  INformation Systems Group) of the
 Aristotle University of Thessaloniki
 Site: http://stains.csd.auth.gr
 B.Sc Student in Electrical and Mechanical Engineering
 ---
 -Original Message- From: Marc Schwartz
 Sent: Thursday, May 24, 2012 3:58 AM
 To: Duncan Murdoch
 Cc: Giannis Mamalikidis ; r-help@r-project.org
 Subject: Re: [R] Question on if i am allowed to do something
 
 
 On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote:
 
 On 12-05-23 1:31 PM, Giannis Mamalikidis wrote:
 Hello all.
 
 I would like to know: provided that I absolutely state that R is not mine
 and I also include the R’s License which will be shown so people know R and
 R’s license,
 (provided the above) am I allowed to include R’s folder (the folder that has
 its binaries) on my freeware program or not?
 
 R is licensed under the GPL v 2 or 3.  The recommended packages that come 
 with it are separately licensed, but are also (mainly) GPL.  So just follow 
 the terms of that license, which is included with R.
 
 Duncan Murdoch
 
 
 Just to add on to Duncan's reply, you might want to review the GPL FAQ (and 
 possibly consult a lawyer):
 
 http://www.gnu.org/licenses/gpl-faq.html
 
 The use and distribution of R within your 'freeware' can place certain 
 obligations on you, including possibly requiring that you license your 
 application with a GPL compatible license and make the source code of your 
 application available in situations where your application is 'linked' to R. 
 If you do not intend to make your own source code available, you should be 
 very clear on what the GPL requires of you before proceeding.
 
 Regards,
 
 Marc Schwartz
 

__
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] building transformation macro



On May 24, 2012, at 4:43 AM, davv.vikas wrote:


Hi All,

I am writing a macro which perform following functionality
1) take a file say excel or csv
2)read all the labels
3) dynamically genrate square,square-root,log  of each column with
appropiate column name and write it to a new file,

i am stuck at dynamically genrateing the column names like  
Revenue_square

etc.

Exmaple  i have a excel with 3 columns Revenue, cost  profit
Now my Macro should be able to  read the values and for each coulmn  
and
perform square,square root and log of   Revenue, cost  profit and  
write to
excel with column names as Revenue_square,Revenue_squareroot,  
Revenue_log

cost_square etc...

Below is my code

test$Rev_square = test[c(1)]^2

I would use test[[1]]^2. Generally one wants to pull the vector out of  
the list container.


data=read.delim2(//ARLMSAN01/CTRX_Data/vikasK.sharma/Desktop/ 
balancesheet_example.csv,header=T,sep=,)


headings = names(data)
show (headings)
HEADINGS = toupper(headings)
# Perhaps replacing your for-loop body with this

for ( i in 1:length(HEADINGS))

{


assign(paste(HEADINGS[i],  Rev_Sqr, sep=), data[[i]]^2)


}


The generalization to additional transformations seems obvious.

--

David Winsemius, MD
West Hartford, CT

__
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] edgeR: design matrix setup

2012-05-24 Thread Martin Morgan


On 05/23/2012 07:24 AM, rmje wrote:

I have a data frame like this:

T0h T0.25h T0.5h  T1h  T2h  T3h  T6h T12h T24h T48h
C0h C0.25h C0.5h  C1h  C2h  C3h  C6h C12h C24h C48h
NM_001001130   68 9556   43   66   62   68   90   63   89   65
8558   49 81   49   76   73   48   77
NM_001001152   7912952   50   24   45  130  154  112  147   56
8567   33 52   31   93   77   65  109
...
...
...


It has about 29000 rows and 20 columns (incl col.name).
How would you set this up in edgeR if you want to compare the 10 first with
the 10 last?


Please ask questions about Bioconductor packages on the Bioconductor 
mailing list


  http://bioconductor.org/help/mailing-list/

There have been recent time-course questions on the mailing list, so 
it's good to browse recents posts. Remember to include sessionInfo(). 
Have you looked at the extensive vignette for this package


  http://bioconductor.org/packages/2.10/bioc/html/edgeR.html

and limma, and taken a stab at answering your own question? It helps to 
see where you're coming from...; for this particular question and if 
you're statistically naive it might help to have a conversation with 
someone local who is a little versed in statistics.


Martin




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



--
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Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109

Location: M1-B861
Telephone: 206 667-2793

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

Re: [R] Question on if i am allowed to do something

2012-05-24 Thread Giannis Mamalikidis


Your answers were very helpful, thanks :)

But, you see, the program I'm writing will be a team effort - and while I 
have not problem on making this program an open source - we don't really see 
eye to eye with the team on this.

So no source code will be available - just freeware.

R will not be at all altered! the only thing that R will do is be used to 
show a couple of plots, and make some t.test and ks.test.


Does that qualify as a derivative work of R? Or would derivative work be if 
I altered R's Source code?
I've read the license but English is not my Native language, hence some 
terminology is unclear to me.


And I'm just an undergraduate with no income in the middle of an economy 
crisis. I don't really see me paying a lawyer to ask the question.


---
Giannis Mamalikidis
e-mail: giannis_mamaliki...@msn.com
Member of the STAINS Research Group
(STAtistics  INformation Systems Group) of the
Aristotle University of Thessaloniki
Site: http://stains.csd.auth.gr
B.Sc Student in Electrical and Mechanical Engineering
---
-Original Message- 
From: Marc Schwartz

Sent: Thursday, May 24, 2012 3:58 AM
To: Duncan Murdoch
Cc: Giannis Mamalikidis ; r-help@r-project.org
Subject: Re: [R] Question on if i am allowed to do something


On May 23, 2012, at 6:51 PM, Duncan Murdoch wrote:


On 12-05-23 1:31 PM, Giannis Mamalikidis wrote:

Hello all.

I would like to know: provided that I absolutely state that R is not mine
and I also include the R’s License which will be shown so people know R 
and

R’s license,
(provided the above) am I allowed to include R’s folder (the folder that 
has

its binaries) on my freeware program or not?


R is licensed under the GPL v 2 or 3.  The recommended packages that come 
with it are separately licensed, but are also (mainly) GPL.  So just 
follow the terms of that license, which is included with R.


Duncan Murdoch



Just to add on to Duncan's reply, you might want to review the GPL FAQ (and 
possibly consult a lawyer):


 http://www.gnu.org/licenses/gpl-faq.html

The use and distribution of R within your 'freeware' can place certain 
obligations on you, including possibly requiring that you license your 
application with a GPL compatible license and make the source code of your 
application available in situations where your application is 'linked' to R. 
If you do not intend to make your own source code available, you should be 
very clear on what the GPL requires of you before proceeding.


Regards,

Marc Schwartz

__
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] How to retrieve value form dataframe by Key?


Hello,

Try

?subset
subset(dtfr, Col1 == xxx)
?[
dtfr[Col1 == xxy, ]

Hope this helps,

Rui Barradas

Em 24-05-2012 11:00, Manish Gupta mandecent.gu...@gmail.com escreveu:

Date: Thu, 24 May 2012 01:09:58 -0700 (PDT)
From: Manish Guptamandecent.gu...@gmail.com
To:r-help@r-project.org
Subject: [R] How to retrieve value form dataframe by Key?
Message-ID:1337846998835-4631173.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii

Hi,

I have one datagrame

Col1  Col2Col3
Col3
xxx erThis is third record of 1st line.
This is 4th record of first line.
xxy erThis is third record of 1st line.
This is 4th record of second line
xyx erThis is third record of 1st line.
This is 4th record of third line

I want to get records corresponding to key (here Col1). How can i implement
it?

e.g.
if i key in xxx i should get
xxx erThis is third record of 1st line.
This is 4th record of first line.

if i key in xxy i should get
xxy erThis is third record of 1st line.
This is 4th record of second line

Regards


--
View this message in 
context:http://r.789695.n4.nabble.com/How-to-retrieve-value-form-dataframe-by-Key-tp4631173.html
Sent from the R help mailing list archive at Nabble.com.


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

Re: [R] How to avoid numbering while printing list element?

2012-05-24 Thread jim holtman

Will this do it for you:

 x - list(1,2,3)
 x[[1]]
[1] 1
 cat(x[[1]], '\n')
1



On Thu, May 24, 2012 at 5:50 AM, Manish Gupta mandecent.gu...@gmail.com wrote:
 HI,

 I am printing list element.

 print(desc[[2]])
 [1] XXXxx

 I want to remove [1]
 Output should be
 XXXxx

 How can i implement it?

 Thanks


 --
 View this message in context: 
 http://r.789695.n4.nabble.com/How-to-avoid-numbering-while-printing-list-element-tp4631180.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
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-- 
Jim Holtman
Data Munger Guru

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

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Re: [R] building transformation macro



On May 24, 2012, at 9:06 AM, David Winsemius wrote:



On May 24, 2012, at 4:43 AM, davv.vikas wrote:


Hi All,

I am writing a macro which perform following functionality
1) take a file say excel or csv
2)read all the labels
3) dynamically genrate square,square-root,log  of each column with
appropiate column name and write it to a new file,

i am stuck at dynamically genrateing the column names like  
Revenue_square

etc.

Exmaple  i have a excel with 3 columns Revenue, cost  profit
Now my Macro should be able to  read the values and for each coulmn  
and
perform square,square root and log of   Revenue, cost  profit and  
write to
excel with column names as Revenue_square,Revenue_squareroot,  
Revenue_log

cost_square etc...

Below is my code

test$Rev_square = test[c(1)]^2

I would use test[[1]]^2. Generally one wants to pull the vector out  
of the list container.


data=read.delim2(//ARLMSAN01/CTRX_Data/vikasK.sharma/Desktop/ 
balancesheet_example.csv,header=T,sep=,)


headings = names(data)
show (headings)
HEADINGS = toupper(headings)
# Perhaps replacing your for-loop body with this

for ( i in 1:length(HEADINGS))

{


assign(paste(HEADINGS[i],  Rev_Sqr, sep=), data[[i]]^2)


}


The generalization to additional transformations seems obvious.


The solution offered in StackOverflow by Greg Snow is better.

AND please stop cross-posting. It is not specifically deprecated on  
SO, but it is on Rhelp.


--

David Winsemius, MD
West Hartford, CT

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

Re: [R] Question on if i am allowed to do something

2012-05-24 Thread Giannis Mamalikidis

I see. It is decided then. As it might be possible that the license might be 
violated, I certainly will not include R in my programme.


Thank you all for your help, you've been very helpful.

---
Giannis Mamalikidis
e-mail: giannis_mamaliki...@msn.com
Member of the STAINS Research Group
(STAtistics  INformation Systems Group) of the
Aristotle University of Thessaloniki
Site: http://stains.csd.auth.gr
B.Sc Student in Electrical and Mechanical Engineering

__
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[R] package metafor: specify weights?

2012-05-24 Thread Anke Stein


Dear R-experts,
Dear Wolfgang,

Weighted model fitting in metafor uses the inverse of the study specific 
variances as weights.
I am wondering if it is possible to specify different weights.

In my meta-analysis, there are two types of studies with (intrinsic) 
differences in their range of sample sizes (which are used to calculate the 
variances of Fisher's z).
I would like to try normalizing the sample sizes within each set of the two 
study types and use these normalized sample sizes as weights.
Would that be possible with rma()? So far, I only found the option weighted = 
TRUE/FALSE, but no possibility to specify which weights should be used.

Many thanks in advance,
Anke






--
__
Anke Stein (Dipl.-Biol.)

Biodiversity, Macroecology  Conservation Biogeography
Head Prof. Dr. Holger Kreft
Georg-August University of Göttingen
Büsgenweg 2 | 37077 Göttingen | Germany

phone +49(0)551-39-13761
fax +49(0)551-39-3618
ast...@uni-goettingen.de
http://www.uni-goettingen.de/biodiversity

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[R] Upper an lower bound in linear programing

2012-05-24 Thread Haio

Hello everyone!
I´m trying to solve a linear optimization problem with r using the LPsolve
function and i want to set upper and lower bounds for the obejctive
function.  Any idea´s on how to do this??


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Re: [R] transform 1 col to 2 col

2012-05-24 Thread MacQueen, Don

See insertion below

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 5/24/12 1:43 AM, Jim Lemon j...@bitwrit.com.au wrote:

On 05/24/2012 05:10 PM, Soheila Khodakarim wrote:
 Dear All

 How can I transform 1 column to 2 columns in R?

 211217_s_at

 GO:0005249

 211217_s_at

 GO:0005251

 211217_s_at

 GO:0005515

 211217_s_at

 GO:0015271

211217_s_at

GO:0030955





 211217_s_at

 GO:0005249

 211217_s_at

 GO:0005251

211217_s_at

 GO:0005515

 211217_s_at

 GO:0015271

211217_s_at

 GO:0030955

Hi Soheila,
Let us begin with the assumption that your col is roughly what you
posted, and that text resides in a file named inscrutable.dat. Let us
first read that text into a data frame in R:

inscrutatble.df-read.table(inscrutable.dat)
  inscrutable.df
 V1 

1 211217_s_at
2  GO:0005249
3  211217_s_at
4   GO:0005251
5  211217_s_at
6   GO:0005515
7  211217_s_at
8   GO:0015271
9  211217_s_at
10  GO:0030955
11 211217_s_at
12  GO:0005249
13 211217_s_at
14  GO:0005251
15 211217_s_at
16  GO:0005515
17 211217_s_at
18  GO:0015271
19 211217_s_at
20  GO:0030955

Now we will further assume that you want to place every odd element in
one col and every even element in the other col.


Or perhaps:

  matrix(inscrutable.df$V1, ncol=2, byrow=TRUE)

or
  data.frame(matrix(inscrutable.df$V1, ncol=2, byrow=TRUE))

to get a data frame.

-Don


inscrutable2.df-data.frame(
  col1=inscrutable.df$V1[seq(1,length(inscrutable.df$V1),by=2)],
  col2=inscrutable.df$V1[seq(2,length(inscrutable.df$V1),by=2)])
inscrutable2.df
   col1   col2
1  211217_s_at GO:0005249
2  211217_s_at GO:0005251
3  211217_s_at GO:0005515
4  211217_s_at GO:0015271
5  211217_s_at GO:0030955
6  211217_s_at GO:0005249
7  211217_s_at GO:0005251
8  211217_s_at GO:0005515
9  211217_s_at GO:0015271
10 211217_s_at GO:0030955

There we are - easy as pie.

Jim

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Re: [R] Using NA as a break point for indicator variable?

2012-05-24 Thread William Dunlap

 Does anyone have any thoughts on how to code this, perhaps using the NA
 values as a break point?

You can count the cumulative number of NA breakpoints in a vector
with cumsum(is.na(vector)), as in

   cbind(d, LeakNo=with(d, cumsum(is.na(lon)|is.na(lat)|is.na(CH4
   lon  lat  CH4 LeakNo
  1  -71.11954 42.35068 2.595834  0
  2  -71.11954 42.35068 2.595688  0
  3 NA   NA   NA  1
  4 NA   NA   NA  2
  5 NA   NA   NA  3
  6  -71.11948 42.35068 2.435762  3
  7  -71.11948 42.35068 2.491003  3
  8 NA   NA   NA  4
  9  -71.11930 42.35068 2.464475  4
  10 -71.11932 42.35068 2.470865  4

Add 1 if you want to start with 1.  If you only want to increase the count
after each sequence of NA's then you could use rle() or
   na - with(d, is.na(lon)|is.na(lat)|is.na(CH4))
   cbind(d, LeakNo=cumsum(c(TRUE, na[-1]  na[-length(na)])))
   lon  lat  CH4 LeakNo
  1  -71.11954 42.35068 2.595834  1
  2  -71.11954 42.35068 2.595688  1
  3 NA   NA   NA  1
  4 NA   NA   NA  1
  5 NA   NA   NA  1
  6  -71.11948 42.35068 2.435762  2
  7  -71.11948 42.35068 2.491003  2
  8 NA   NA   NA  2
  9  -71.11930 42.35068 2.464475  3
  10 -71.11932 42.35068 2.470865  3

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Max Brondfield
 Sent: Wednesday, May 23, 2012 1:42 PM
 To: r-help@r-project.org
 Subject: [R] Using NA as a break point for indicator variable?
 
 Hi all,
 I am working with a spatial data set for which I am only interested in high
 concentration values (leaks). The low values ( 90th percentile) have
 already been turned into NA's, leaving me with a matrix like this:
 
  CH4_leak
 
   lonlatCH4
 1  -71.11954 42.35068 2.595834
 2  -71.11954 42.35068 2.595688
 3   NA   NA   NA
 4   NA   NA   NA
 5   NA   NA   NA
 6  -71.11948 42.35068 2.435762
 7  -71.11948 42.35068 2.491003
 8  NANA   NA
 9  -71.11930 42.35068 2.464475
 10 -71.11932 42.35068 2.470865
 
 Every time an NA comes up, it means the leak is gone, and the next valid
 value would represent a different leak (at a different location). My goal
 is to tag all of the remaining values with an indicator variable to
 spatially distinguish the leaks. I am envisioning a simple numeric
 indicator such as:
 
  lonlatCH4leak_num
 1  -71.11954 42.35068 2.595834   1
 2  -71.11954 42.35068 2.595688   1
 3   NA   NA   NA NA
 4   NA   NA   NA NA
 5   NA   NA   NA NA
 6  -71.11948 42.35068 2.435762   2
 7  -71.11948 42.35068 2.491003   2
 8  NANA   NA NA
 9  -71.11930 42.35068 2.064475   3
 10 -71.11932 42.35068 2.070865  3
 
 Does anyone have any thoughts on how to code this, perhaps using the NA
 values as a break point? The data set is far too large to do this
 manually, and I must admit I'm completely at a loss. Any help would be much
 appreciated! Best,
 
 Max
 
   [[alternative HTML version deleted]]
 
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Re: [R] set tkscale by tkentry

2012-05-24 Thread Greg Snow

I believe that what is happening is that when you try to edit the
entry widget any intermediate values get sent to the slider widget
which then checks to see if they are in the allowable range and if it
is not then it sets the value to either the minimum and maximum and
sends that back to the entry widget while you are still trying to edit
it.  Even if you highlight a single digit and try to replace it with a
different digit it first deletes the highlighted digit resulting in a
number smaller than the minimum of the slider which then updates the
entry widget to the minimum before the new digit can go in, then
adding the new digit makes it larger than the slider maximum.

There may be a way to slow down the updates or validations so that you
can edit the entry before the slider tries to validate and change the
value.  Or you can just include 0 as the minimum of the slider so that
the intermediate values in the entry widget are in the allowable
range.

On Thu, May 24, 2012 at 2:20 AM, Alexander juschitz_alexan...@yahoo.de wrote:
 Hi, I am working under Windows and I am using R2.11
 I want to use tkscale in my GUI. As the interval is quite big, I can't set
 the scale to a certain specific value. Therefore I want to add tkentry to
 allow the user to set tkscale to a certain value.

 Here is the code

 library(tcltk)

 tt-tktoplevel()
 tkpack(m1-tkscale(tt,from=306870.00, to=3026741, label=alpha,
 variable=varalpha,showvalue=TRUE, resolution=1,
 orient=horiz),side=bottom)
 tkpack(tkentry(tt,width=10,textvariable=varalpha,validate=key,validatecommand=string
 is double %P),side=bottom)

 As you can see, varalpha is a global variable (I think in tcl, not in R),
 and if you change the parameter by the scale, the value in tkentry is
 updated. If I want to set now the value of varalpha by tkentry, it bugs.
 Does anyone have an idea, why? I already tried the interval from 0 to any
 number, which works fine...

 library(tcltk)

 tt-tktoplevel()
 tkpack(m1-tkscale(tt,from=0.00, to=1000, label=alpha,
 variable=varalpha,showvalue=TRUE, resolution=1,
 orient=horiz),side=bottom)
 tkpack(tkentry(tt,width=10,textvariable=varalpha,validate=key,validatecommand=string
 is double %P),side=bottom)

 Thanks in advance,
 Alexander

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/set-tkscale-by-tkentry-tp4631174.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] package metafor: specify weights?

2012-05-24 Thread Viechtbauer Wolfgang (STAT)

At the moment, there is no possibility of specifying the weights with the rma() 
function. While the main model fitting part could be easily adapted to 
incorporate user-specified weights, the problem comes in with all the 
additional statistics that can be computed based on a fitted model. How should 
the predict() function now work? What would be the definition of I^2 now? How 
would one generalize the influence and outlier statistics to that case? Just to 
give some examples.

Of course, I could leave out such things when the user has specified the 
weights, but then things also get confusing for the user. For example, it is 
already less than ideal that you can only use the trim and fill method with 
models that do not incorporate moderators. Nobody has (as of yet) generalized 
the trim and fill method to that case. But when there are too many special 
cases, the package becomes unusable.

I did consider user-specified weights at one point, but it opened up so many 
cans of worms that I preferred to quickly put the lid pack on those cans. That 
item is written down in my to-do list, but to be honest, it is somewhere at the 
very end of that list.

If you are hesitant to combine the results from those two types of studies, 
what about simply using a moderator to distinguish the two groups?

Best,

Wolfgang

--   
Wolfgang Viechtbauer, Ph.D., Statistician   
Department of Psychiatry and Psychology   
School for Mental Health and Neuroscience   
Faculty of Health, Medicine, and Life Sciences   
Maastricht University, P.O. Box 616 (VIJV1)   
6200 MD Maastricht, The Netherlands   
+31 (43) 388-4170 | http://www.wvbauer.com   


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of Anke Stein
 Sent: Thursday, May 24, 2012 15:59
 To: r-help@r-project.org
 Subject: [R] package metafor: specify weights?
 
 Dear R-experts,
 Dear Wolfgang,
 
 Weighted model fitting in metafor uses the inverse of the study specific
 variances as weights.
 I am wondering if it is possible to specify different weights.
 
 In my meta-analysis, there are two types of studies with (intrinsic)
 differences in their range of sample sizes (which are used to calculate
 the variances of Fisher's z).
 I would like to try normalizing the sample sizes within each set of the
 two study types and use these normalized sample sizes as weights.
 Would that be possible with rma()? So far, I only found the option
 weighted = TRUE/FALSE, but no possibility to specify which weights
 should be used.
 
 Many thanks in advance,
 Anke
 
 
 
 
 
 
 --
 __
 Anke Stein (Dipl.-Biol.)
 
 Biodiversity, Macroecology  Conservation Biogeography Head Prof. Dr.
 Holger Kreft Georg-August University of Göttingen Büsgenweg 2 | 37077
 Göttingen | Germany
 
 phone +49(0)551-39-13761
 fax +49(0)551-39-3618
 ast...@uni-goettingen.de
 http://www.uni-goettingen.de/biodiversity
 
 __
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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] applying cbind (or any function) across all components in a list

2012-05-24 Thread Greg Snow

The do.call function may be what you want, but it is not completely
clear.  If that does not solve your problem then try giving us an
example of what your list looks like (the dput function can help) and
what you want your end result to look like.

On Thu, May 24, 2012 at 5:38 AM, Hans Thompson hans.thomps...@gmail.com wrote:
 The combination of column means in cbind is what I am trying to do.  I just
 don't know how to do it for every component (2 in this case) of the list.
 I've been able to work with R to apply a function across all components when
 there is just one list but I am having trouble working with multiple lists
 at the same time.

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/applying-cbind-or-any-function-across-all-components-in-a-list-tp4631128p4631187.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] Using NA as a break point for indicator variable?

2012-05-24 Thread Max Brondfield

This method worked perfectly! The rle() function was key and I was
completely unfamiliar with it. Thanks so much,
Max

On Thu, May 24, 2012 at 8:52 AM, Rui Barradas ruipbarra...@sapo.pt wrote:

 Hello,

 Assuming that 'd' is your original data.frame and that you've set entire
 rows to NA, try this


 d$leak_num - NA
 ix - !is.na(d[, 1])  # any column will do, entire row is NA
 ## alternative, if other rows may have NAs, due to something else
 #ix - apply(d, 1, function(x) all(!is.na(x)))
 r - rle(ix)
 v - cumsum(r$values)
 d$leak_num[ix] - rep(v[r$values], r$lengths[r$values])
 d


 Hope this helps,

 Rui Barradas

 Em 24-05-2012 11:00, Max Brondfield mbro...@post.harvard.edu escreveu:

 Date: Wed, 23 May 2012 16:42:02 -0400
 From: Max Brondfieldmbrondf@post.**harvard.edumbro...@post.harvard.edu
 
 To:r-help@r-project.org
 Subject: [R] Using NA as a break point for indicator variable?
 Message-ID:
CADu+jDpcJUHZTXxrsxyQvjaEmw_**N0iLbL6ZJjHZC-rSBCMneiw@mail.**
 gmail.comcadu%2bjdpcjuhztxxrsxyqvjaemw_n0ilbl6zjjhzc-rsbcmn...@mail.gmail.com
 
 Content-Type: text/plain


 Hi all,
 I am working with a spatial data set for which I am only interested in
 high
 concentration values (leaks). The low values (  90th percentile) have
 already been turned into NA's, leaving me with a matrix like this:

   CH4_leak

   lonlatCH4
 1  -71.11954 42.35068 2.595834
 2  -71.11954 42.35068 2.595688
 3   NA   NA   NA
 4   NA   NA   NA
 5   NA   NA   NA
 6  -71.11948 42.35068 2.435762
 7  -71.11948 42.35068 2.491003
 8  NANA   NA
 9  -71.11930 42.35068 2.464475
 10 -71.11932 42.35068 2.470865

 Every time an NA comes up, it means the leak is gone, and the next valid
 value would represent a different leak (at a different location). My goal
 is to tag all of the remaining values with an indicator variable to
 spatially distinguish the leaks. I am envisioning a simple numeric
 indicator such as:

  lonlatCH4leak_num
 1  -71.11954 42.35068 2.595834   1
 2  -71.11954 42.35068 2.595688   1
 3   NA   NA   NA NA
 4   NA   NA   NA NA
 5   NA   NA   NA NA
 6  -71.11948 42.35068 2.435762   2
 7  -71.11948 42.35068 2.491003   2
 8  NANA   NA NA
 9  -71.11930 42.35068 2.064475   3
 10 -71.11932 42.35068 2.070865  3

 Does anyone have any thoughts on how to code this, perhaps using the NA
 values as a break point? The data set is far too large to do this
 manually, and I must admit I'm completely at a loss. Any help would be
 much
 appreciated! Best,

 Max

[[alternative HTML version deleted]]





-- 
Max Brondfield
Research Assistant
Department of Geography  Environment
Boston University

[[alternative HTML version deleted]]

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

[R] Memory allocation error

2012-05-24 Thread swaraj basu

Dear All,

I am running R in a system with the following configuration

*Processor: Intel(R) Xeon(R) CPU   X5650  @ 2.67GHz
OS: Ubuntu X86_64 10.10
RAM: 24 GB*

The R session info is
*
R version 2.14.1 (2011-12-22)
Platform: x86_64-pc-linux-gnu (64-bit)

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


I have a matrix of dimensions 12 rows X  29318 columns. The matrix contains
numeric as well as NA values. I am using the* rcorr *function from the *
Hmisc* package to get correlation information from the matrix (*
rcorr(matrix)*). During the calculation I get the error *cannot allocate
vector of size 6.7 GB*. When I check the memory allocation of my R session
I get the following information

*gc()
  used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
Ncells  249638 13.4 467875 25.0 NA   407500 21.8
Vcells 1499217 11.52335949 17.9   7000  1970005 15.1

*Can someone please help me in finding a workaround to the problem.

-Regards
-- 
Swaraj Basu
PhD Student (Bioinformatics - Functional Genomics)
Animal Physiology and Evolution
Stazione Zoologica Anton Dohrn
Naples

[[alternative HTML version deleted]]

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[R] use list as function arguments

2012-05-24 Thread Alexander Shenkin

Hello Folks,

Is there any way to pass a list into a function such that the function
will use the list as its arguments?  I haven't been able to figure that out.

The background: I'm trying to build a function that will apply another
function multiple times, each time with a different set of specified
arguments.  I'm trying to figure out how to pass in a list of argument
lists, then loop through the top-level list, passing in the lower-level
list as function arguments.

pseudocode:

b = list( list( arg1 = 1, arg2 = 2 ),
  list( arg1 = 3, arg2 = 4 )
)

a - apply_function(arglist) {
for (i in length(arglist)) {
b(arglist[i])
}
}


Specifically, the actual use I'm trying to implement is a function to
format columns of data frames and matrices independently.  What I have
so far is below, but it's not working.  Perhaps I'm going about this the
wrong way?


format_cols - function(x, format_list = list()) {
# usage: length(format_list) must equal ncol(x)
#format list should be a list of lists of key=value pairs
corresponding to format settings for each column

if (is.data.frame(x)) {
newout = data.frame()
} else if (is.matrix(x)) {
newout = matrix()
}

for (i in 1:ncol(x)){
newout = cbind(newout, format(x,format_list[[i]]))
x[,i] = format(x,format_list[[i]])
}
return(newout)
}

Thanks,
Allie

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[R] inner_perc_table?

Hello,

Does anyone on this list know what inner_perc_table is or where it is
typically found?  I am trying to modify some source code and it is used
with the .C() function.  When I try and run it, it states that
'inner_perc_table is not found'.  It is only called in such a way and isn't
defined at any point in the previous code in which the function works.
Should this be in some other directory of the source code files?  I have
been scouring through them and have continue to come up empty.

Many thanks for any assistance,
Charles

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[R] R does not recognise columns and rows as they are supposed to be

2012-05-24 Thread Jonsson

Dear All,
The code given bellow is to extract  values of one region and write that to
a text file(there are 365 binary files in the directory).
The problem which I am facing is that all my files are binary with size of
360 rows and 720 columns.
I specified that in this line:file2-matrix(data=file,ncol=720,nrow=360)
but I got an error : Error in mean(file2[X, Y], na.rm = TRUE) : subscript
out of bounds.
and then I rewrote the above line as
:file2-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720
which is not right.But that worked and I didn't get any error.however,the
results were not correct.
Any help please

dir1- list.files(C:\\Users\\aalyaari\\Desktop\\New folder
(10)\\Climate_Rad_f_GAMMA_%d.img, full.names = TRUE)
listfile-dir()
for (i in c(1:365)) {
  conne - file(listfile[i], rb)
  file- readBin(conne, double(), size=4,  n=720*360, signed=T)
 file2-matrix(data=file,ncol=720,nrow=360)  
extract[i]-mean(file2[X,Y],na.rm=TRUE)
  close(conne)
write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
(10)\\sam.txt)}


--
View this message in context: 
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Re: [R] R does not recognise columns and rows as they are supposed to be



On May 24, 2012, at 11:51 AM, Jonsson wrote:


Dear All,
The code given bellow is to extract  values of one region and write  
that to

a text file(there are 365 binary files in the directory).
The problem which I am facing is that all my files are binary with  
size of

360 rows and 720 columns.
I specified that in this line:file2- 
matrix(data=file,ncol=720,nrow=360)
but I got an error : Error in mean(file2[X, Y], na.rm = TRUE) :  
subscript

out of bounds.
and then I rewrote the above line as
:file2-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows  
=720
which is not right.But that worked and I didn't get any  
error.however,the

results were not correct.
Any help please

dir1- list.files(C:\\Users\\aalyaari\\Desktop\\New folder
(10)\\Climate_Rad_f_GAMMA_%d.img, full.names = TRUE)
listfile-dir()
for (i in c(1:365)) {
 conne - file(listfile[i], rb)
 file- readBin(conne, double(), size=4,  n=720*360, signed=T)
file2-matrix(data=file,ncol=720,nrow=360)
extract[i]-mean(file2[X,Y],na.rm=TRUE)
 close(conne)
write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
(10)\\sam.txt)}


Please do not cross post to SO and Rhelp. (See Posting Guide.)

(But if you do, rhelp generally expects reproducible examples which  
this is not.)


--
David Winsemius, MD
West Hartford, CT

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Re: [R] use list as function arguments

2012-05-24 Thread Greg Snow

?do.call

On Thu, May 24, 2012 at 9:32 AM, Alexander Shenkin ashen...@ufl.edu wrote:
 Hello Folks,

 Is there any way to pass a list into a function such that the function
 will use the list as its arguments?  I haven't been able to figure that out.

 The background: I'm trying to build a function that will apply another
 function multiple times, each time with a different set of specified
 arguments.  I'm trying to figure out how to pass in a list of argument
 lists, then loop through the top-level list, passing in the lower-level
 list as function arguments.

 pseudocode:

 b = list( list( arg1 = 1, arg2 = 2 ),
          list( arg1 = 3, arg2 = 4 )
        )

 a - apply_function(arglist) {
        for (i in length(arglist)) {
                b(arglist[i])
        }
 }


 Specifically, the actual use I'm trying to implement is a function to
 format columns of data frames and matrices independently.  What I have
 so far is below, but it's not working.  Perhaps I'm going about this the
 wrong way?


 format_cols - function(x, format_list = list()) {
    # usage: length(format_list) must equal ncol(x)
    #        format list should be a list of lists of key=value pairs
 corresponding to format settings for each column

    if (is.data.frame(x)) {
        newout = data.frame()
    } else if (is.matrix(x)) {
        newout = matrix()
    }

    for (i in 1:ncol(x)){
        newout = cbind(newout, format(x,format_list[[i]]))
        x[,i] = format(x,format_list[[i]])
    }
    return(newout)
 }

 Thanks,
 Allie

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

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[R] Manually modifying an hclust dendrogram to remove singletons

2012-05-24 Thread r-help . 20 . trevva

Dear R-Help,

I have a clustering problem with hclust that I hope someone can help
me with. Consider the classic hclust example:

 hc - hclust(dist(USArrests), ave)
 plot(hc)

I would like to cut the tree up in such a way so as to avoid small
clusters, so that we get a minimum number of items in each cluster,
and therefore avoid singletons. e.g. in this example, you can see that
Hawaii is split off onto its own at quite a high level. I would like
to avoid having a single item clustered on its own like this. How can
I achieve this?

I have tried manually modifying the tree using dendrapply but have not
been able to produce a valid solution thus far..

Suggestions are welcome.

Best wishes,

Mark

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Re: [R] How to evaluate R things from Visual Studio?

2012-05-24 Thread Giannis Mamalikidis

I'll answer line by line so:

You ask for help from people who have used R.NET, give them time to
answer.
I have.. I'm not sure why you seem so certain I didn't. But it is beside the
point anyway

Also, what about folks who have not used it? Does not work
is not enough information. What did you try? Give enough detail so
that someone might follow your footsteps, including VS versions, R
location and version, assumptions made, and links to resources used on
the way.

Folks who haven't used it won't be able to help me anyway.
People who did get it working will be able to help me, and those are the
people I'm counting on.
The sole reason I did not disclose any R.NET related info is to avoid making
an already large message, larger.
When I find someone, if I find someone who does have R.NET and can help me,
I will gladly give any and all needed information

FWIW, I got the R.NET example to work on Windows 7 (64-bit OS) with
Visual Studio 10 Express (32-bit only) with their example program
here:

http://rdotnet.codeplex.com/

I downloaded this file, and set the VS reference to it:

http://rdotnet.codeplex.com/downloads/get/211446

I used R-2.12.0 as per their example

REngine.SetDllDirectory(@C:\Program Files\R\R-2.12.0\bin\i386);

It's worth much, believe me. I really needed another person who got R.NET
working.

It does work on R 2.12.0, I agree. but i need it to work any latest R.
I can't make my users use old and obsolete versions of R. And people have
made it work on 2.15.0 (currently latest version) as well.

I set R_HOME and R_LIBS to paths for the R-2.12.0 installation.
Can you please tell me exactly what string you put inside R_HOME and R_LIBS
environment variables? (i.e. C:\Progs\R\R-2.15.0\bin?
C:\Progs\R\R-2.15.0? what exactly?)
Also, it was YOU who created those Environmental Variables, right? Because
they do not exist in my computer and i didn't delete them.
Safe to assume that we're talking about System - Advanced - Environment
Variables - System Variables. correct?

I read this page:

http://rdotnet.codeplex.com/documentation?referringTitle=Home

I had a quick stab at using a recent R build, but dropped it to try
the stated version in their example. There are error messages printed
from the Windows application when it fails and they are worth reading
(and reporting when asking for help).

I've already read every page on rdotnet.codeplex.com

Of course there are. I need not repeat myself. I explained above why no
direct R.NET info were given.

Now that I have found one guy who has R.NET working, I shall :)

My R version is: 2.15.0 and is installed on C:\Progs\R\R-2.15.0 instead of
C:\Program Files\R\R-2.15.0 (but it doesnt work on that either, tried it)

The code I try to run on R.NET version 1.4.1 (changeset 6d2c3f161801) is the
example they have on the official page:

Private Sub Form1_Load(sender As System.Object, e As System.EventArgs)
Handles MyBase.Load
Dim Rdo As REngine = REngine.CreateInstance(RDotNet, -q) '
quiet mode

' .NET Framework array to R vector.
Dim group1 As NumericVector = Rdo.CreateNumericVector(New
Double() {30.02, 29.99, 30.11, 29.97, 30.01, 29.99})

Rdo.SetSymbol(group1, group1)

' Direct parsing from R script.
Dim group2 As NumericVector = Rdo.Evaluate(group2 - c(29.89,
29.93, 29.72, 29.98, 30.02, 29.98)).AsNumeric()

' Test difference of mean and get the P-value.
Dim testResult As GenericVector = Rdo.Evaluate(t.test(group1,
group2)).AsList()

Dim p As Double = testResult(p.value).AsNumeric().First()
End Sub

The Error i get is this:
System.DllNotFoundException: Dll was not found.
at RDotNet.NativeLibrary.UnmanagedDll..ctor(String dllName) in
C:\Users\N1h1l1sT\Downloads\rdotnet_6d2c3f161801\RDotNet.NativeLibrary\UnmanagedDll.cs:line
42

at RDotNet.REngine..ctor(String id, String dll)
at RDotNet.REngine.CreateInstance(String id, String dll)
at RAttemptFour.Form1.Form1_Load(Object sender, EventArgs e) in
C:\Users\N1h1l1sT\Dropbox\Visual Basic 2010\Test
Projects\RAttemptFour\RAttemptFour\Form1.vb:line 35

No matter what i put inside the HOME or R_Home environment variables on my
windows, it just won't work...
the line 35 refers to this line: Dim Rdo As REngine =
REngine.CreateInstance(RDotNet, -q) ' quiet mode

---
Giannis Mamalikidis
e-mail: giannis_mamaliki...@msn.com
Member of the STAINS Research Group
(STAtistics INformation Systems Group) of the
Aristotle University of Thessaloniki
Site: http://stains.csd.auth.gr
B.Sc Student in Electrical and Mechanical Engineering

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[R] svychisq using two frames

2012-05-24 Thread lasciel

Hello,

I’m hoping you have a few minutes to help out someone very new to R.  I’ve
done some searching, but cannot find this particular issue.

I have survey data from two different time periods (years).  Both years are
stratified samples and have the same variables (and variable names), but are
different people in the community answering in the different years. 
Everything loads into the survey design fine and I can calculate the means
and CIs I need.

I want to calculate whether there was a change in a single variable between
years.  The variable of interest has binary yes/no answers.  I was planning
to use the ‘svychisq’ function, but cannot figure out how to do this across
what is currently two data frames.  I haven’t actually tried to run a
svychisq with the two frames yet as I am waiting on the data for the second
frame.  I’m trying to plan out the logic ahead of time.

So, let’s say I have survey design frame Year1 with variables Wt1 and MyVar,
and survey design frame Year2 with variables Wt2 and MyVar – remember the
variable name is the same in both frames.

My first inclination is that I would use a statement like:
svychisq(~MyVar+MyVar, Year1+Year2, statistic=”Chisq”)

However, it seems from reading the help files that I can only use the
‘svychisq’ function if I declare a single data frame in the design
statement, rather than referencing a frame for Year1 and a different frame
for Year2.  

So, I could use ‘rbind’ to stack the two years of data together with their
appropriate weights into a single data frame.  
svychisq(~MyVar+MyVar, BothYears, statistic=”Chisq”)

But now I have a problem that the variable name is the same across years, so
how do I differentiate the different time periods in the syntax for the
formula?  Do I need to also create two new variables per:
MyVar.1 [year = 1] - MyVar
MyVar.2 [year = 2] - MyVar
svychisq(~MyVar.1+MyVar.2, BothYears, statistic=”Chisq”, na.rm= TRUE) 


(I feel like I may be overthinking this and the answer is much simpler)


--
View this message in context: 
http://r.789695.n4.nabble.com/svychisq-using-two-frames-tp4631220.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Upper an lower bound in linear programing

2012-05-24 Thread Petr Savicky

On Thu, May 24, 2012 at 07:16:20AM -0700, Haio wrote:
 Hello everyone!
 I´m trying to solve a linear optimization problem with r using the LPsolve
 function and i want to set upper and lower bounds for the obejctive
 function.  Any idea´s on how to do this??

Hello:

The formulation of an LP problem allows to use the objective
function also as a row in the constraint matrix. So, it is
possible to set a constraint on it. For example,
maximizing x + y
under the constraints
  x/4 = y = 2 + x/3
  x + y = 9  (constraint on the objective function)
may be done as follows

  library(lpSolve)
 
  crit - c(1, 1)
  mat - rbind(
  c(-1/4, 1),
  c(-1/3, 1),
  c( 1,   1))
  dir - c(=, =, =)
  rhs - c(0, 2, 9)
 
  out - lp(max, objective.in=crit, const.mat=mat, const.dir=dir, 
const.rhs=rhs)
  out

  Success: the objective function is 9

  out$solution

  [1] 7.2 1.8

Hope this helps.

Petr Savicky.

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[R] transform 1 col to 2 col

2012-05-24 Thread arun

Hi,

Just an addendum.

Suppose if you want to add more columns and do the split, you can try,

inscrutable.df2-data.frame(inscrutable.df1,V2=(c(rnorm(1),GTDF,SQ:1234,DFFD,DFDSE)[rep(c(1,2,3,4,5),times=4)]))
inscrutable3.df3-data.frame(matrix(inscrutable.df1$V1,ncol=2,byrow=TRUE),matrix(inscrutable.df2$V2,ncol=2,byrow=TRUE))  

A.K.



- Original Message -
From: MacQueen, Don macque...@llnl.gov
To: Soheila Khodakarim lkhodaka...@gmail.com
Cc: r-help@r-project.org r-help@r-project.org
Sent: Thursday, May 24, 2012 10:19 AM
Subject: Re: [R] transform 1 col to 2 col

See insertion below

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 5/24/12 1:43 AM, Jim Lemon j...@bitwrit.com.au wrote:

On 05/24/2012 05:10 PM, Soheila Khodakarim wrote:
 Dear All

 How can I transform 1 column to 2 columns in R?

 211217_s_at

 GO:0005249

 211217_s_at

 GO:0005251

 211217_s_at

     GO:0005515

 211217_s_at

 GO:0015271

    211217_s_at

    GO:0030955





 211217_s_at

 GO:0005249

 211217_s_at

 GO:0005251

    211217_s_at

     GO:0005515

 211217_s_at

 GO:0015271

    211217_s_at

 GO:0030955

Hi Soheila,
Let us begin with the assumption that your col is roughly what you
posted, and that text resides in a file named inscrutable.dat. Let us
first read that text into a data frame in R:

inscrutatble.df-read.table(inscrutable.dat)
  inscrutable.df
             V1 

1 211217_s_at
2  GO:0005249
3  211217_s_at
4   GO:0005251
5  211217_s_at
6   GO:0005515
7  211217_s_at
8   GO:0015271
9  211217_s_at
10  GO:0030955
11 211217_s_at
12  GO:0005249
13 211217_s_at
14  GO:0005251
15 211217_s_at
16  GO:0005515
17 211217_s_at
18  GO:0015271
19 211217_s_at
20  GO:0030955

Now we will further assume that you want to place every odd element in
one col and every even element in the other col.


Or perhaps:

  matrix(inscrutable.df$V1, ncol=2, byrow=TRUE)

or
  data.frame(matrix(inscrutable.df$V1, ncol=2, byrow=TRUE))

to get a data frame.

-Don


inscrutable2.df-data.frame(
  col1=inscrutable.df$V1[seq(1,length(inscrutable.df$V1),by=2)],
  col2=inscrutable.df$V1[seq(2,length(inscrutable.df$V1),by=2)])
inscrutable2.df
           col1       col2
1  211217_s_at GO:0005249
2  211217_s_at GO:0005251
3  211217_s_at GO:0005515
4  211217_s_at GO:0015271
5  211217_s_at GO:0030955
6  211217_s_at GO:0005249
7  211217_s_at GO:0005251
8  211217_s_at GO:0005515
9  211217_s_at GO:0015271
10 211217_s_at GO:0030955

There we are - easy as pie.

Jim

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Re: [R] inner_perc_table?

2012-05-24 Thread Duncan Murdoch


On 24/05/2012 11:35 AM, Charles Determan Jr wrote:

Hello,

Does anyone on this list know what inner_perc_table is or where it is
typically found?  I am trying to modify some source code and it is used
with the .C() function.  When I try and run it, it states that
'inner_perc_table is not found'.  It is only called in such a way and isn't
defined at any point in the previous code in which the function works.
Should this be in some other directory of the source code files?  I have
been scouring through them and have continue to come up empty.


Look for the C function R_registerRoutines and/or the R function 
getNativeSymbolInfo.


Duncan Murdoch

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Re: [R] inner_perc_table?

Thank you Duncan,

I used the R function getNativeSymbolInfo(inner_perc_table) and the object
is not found:

Error in is.vector(X) : object 'inner_perc_table' not found
Done.

I also tried the defining the directory to the nlme.dll file and it had the
same error.
Is there something I missed while doing this?

Thanks,
Charles


On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch
murdoch.dun...@gmail.comwrote:

 On 24/05/2012 11:35 AM, Charles Determan Jr wrote:

 Hello,

 Does anyone on this list know what inner_perc_table is or where it is
 typically found?  I am trying to modify some source code and it is used
 with the .C() function.  When I try and run it, it states that
 'inner_perc_table is not found'.  It is only called in such a way and
 isn't
 defined at any point in the previous code in which the function works.
 Should this be in some other directory of the source code files?  I have
 been scouring through them and have continue to come up empty.


 Look for the C function R_registerRoutines and/or the R function
 getNativeSymbolInfo.

 Duncan Murdoch


[[alternative HTML version deleted]]

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Re: [R] Exclude when sd=0

2012-05-24 Thread David L Carlson

Assuming you mean that the column is constant so that the sd is 0

test - matrix(rnorm(50), nrow=10, ncol=5)
test[,2] - 1
test[,4] - 2
# test[,which(apply(test, 2, sd)  0)] could fail on rounding errors
Test2 - test[, which(apply(test, 2, sd)  1e-10)] 


--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Chris Burns
 Sent: Wednesday, May 23, 2012 7:45 PM
 To: r-help@r-project.org
 Subject: [R] Exclude when sd=0
 
 How do I trim a matrix to exclude columns that have no standard
 deviation?
 
 Thanks,
 
 Chris
 
   [[alternative HTML version deleted]]
 
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 guide.html
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Re: [R] set tkscale by tkentry

2012-05-24 Thread j verzani

Greg Snow 538280 at gmail.com writes:

 
 I believe that what is happening is that when you try to edit the
 entry widget any intermediate values get sent to the slider widget
 which then checks to see if they are in the allowable range and if it
 is not then it sets the value to either the minimum and maximum and
 sends that back to the entry widget while you are still trying to edit
 it.  Even if you highlight a single digit and try to replace it with a
 different digit it first deletes the highlighted digit resulting in a
 number smaller than the minimum of the slider which then updates the
 entry widget to the minimum before the new digit can go in, then
 adding the new digit makes it larger than the slider maximum.
 

Greg is right. You might try validating on focusout, rather than the key, 
but this is easy enough to do in R code, rather than let tcl do that work:

a - 306870; b - 3026741

tt-tktoplevel()
varalpha - tclVar(a)
charalpha - tclVar(as.character(a))


scale - tkscale(tt, from=a, to=b, resolution=1, label=alpha,
  variable=varalpha,
  showvalue=TRUE, orient=horiz)
ed - tkentry(tt, textvariable=charalpha)

tkpack(ed)
tkpack(scale)

## connect
tkconfigure(scale, command=function(...) {
  tclvalue(charalpha) - as.character(tclvalue(varalpha))
})

valid_input - function(...) {
  val - as.numeric(tclvalue(charalpha))
  if(a = val  val = b) {
message(set to , val)
tclvalue(varalpha) - val
  } else {
message(not valid...)
tkfocus(ed)
  }
}

tkbind(ed, Return, valid_input)
tkbind(ed, FocusOut, valid_input)

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Re: [R] inner_perc_table?

My apologies, feel a little foolish.  I forgot the  for the object.  This
worked, however, once I have this information, I am not sure what I should
be using to fix the issue.  I can determine the location in  the nlme.dll
file but how can I use this information to have the other function work?

Regards,
Charles

On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch
murdoch.dun...@gmail.comwrote:

 On 24/05/2012 11:35 AM, Charles Determan Jr wrote:

 Hello,

 Does anyone on this list know what inner_perc_table is or where it is
 typically found?  I am trying to modify some source code and it is used
 with the .C() function.  When I try and run it, it states that
 'inner_perc_table is not found'.  It is only called in such a way and
 isn't
 defined at any point in the previous code in which the function works.
 Should this be in some other directory of the source code files?  I have
 been scouring through them and have continue to come up empty.


 Look for the C function R_registerRoutines and/or the R function
 getNativeSymbolInfo.

 Duncan Murdoch


[[alternative HTML version deleted]]

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[R] Fwd: help needed

2012-05-24 Thread QAMAR MUHAMMAD UZAIR





 --- segue il messaggio inoltrato (the forwarded 
message follows) ---
---BeginMessage---

Dear R users,
i am very new to R. i am working on the following data.

01.01.1967  0.87
02.01.1967  0.87
03.01.1968  0.87
04.01.1968  0.87
05.01.1969  0.87
06.01.1969  0.87
07.01.1970  0.87
08.01.1970  0.87
09.01.1971  0.87
10.01.1971  0.87
11.01.1972  0.87
12.01.1972  0.87
13.01.1973  0.69
14.01.1973  0.70
15.01.1974  0.71
16.01.1974  0.72




I want to reshape it in the following FORMAT

19671968196919701971197219731974
1   0.870.870.870.870.71
2   0.870.870.870.870.72

OBVIOUSLY, I HAVE A LARGE AMOUNT OF DATA TO WORK WITH.i 
would also like to take into account the effect of leap 
year. For example if 1969 in a leap year then the column 
under it, has to have 1 extra reading.




Thanks in advance

---End Message---
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Re: [R] applying cbind (or any function) across all components in a list

I'm confused why I haven't made clear what I am asking for help with.  

I have two different lists with two (or many) components, [[1]] and [[2]]. 
One of the list has components with dim=c(2,3) and the other has dim=c(2,2). 
I want to create a new list with components dim=c(2,4) by binding together
the averages of the columns using 

 cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2,
 (l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2)

which should put out:

 [,1] [,2] [,3] [,4]
[1,]1234
[2,]2345

my problem is finding out how I can apply this function to all the
components within the list in the same function.

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Re: [R] R does not recognise columns and rows as they are supposed to be

Hello,

There are several things with your code, most of which are not errors.

1. X-c(364:369) ; Y-c(82:92 and later c(1:365)
Expressions of the form m:n are integer sequences, the c() is not needed.

2. extract-vector()
First you create the vector, then keep extending it throughout the loop.
This is much slower.
If you know the final length (and type), create it like this:
extract - numeric(365)   # or double(365)

3. 'dir1' is not used.

4. Now the readBin instruction:
4.1. 'file' is the name of a function, use, say, 'file1' instead.
4.2. Option 'signed' defaults to TRUE but it's only valid for integer types.

5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get rid
of option 'size', like the help page says,

If size is specified and not the natural size of the object, each element
of the vector is coerced to an appropriate type before being written or as
it is read.

You are NOT reading doubles.

As for the rest, it seems ok to me. (Keep option 'n' in readBin.)

Hope this helps,

Rui Barradas

Jonsson wrote
 
 Dear All,
 The code given bellow is to extract  values of one region and write that
 to a text file(there are 365 binary files in the directory).
 The problem which I am facing is that all my files are binary with size of
 360 rows and 720 columns.
 I specified that in this line:   
 file2-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in
 mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds.
 and then I rewrote the above line as
 :file2-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720
 which is not right.But that worked and I didn't get any error.however,the
 results were not correct.
 Any help please
  X-c(364:369) ; Y-c(82:92) #   for sellected region
 extract-vector()
 dir1- list.files(C:\\Users\\aalyaari\\Desktop\\New folder
 (10)\\Climate_Rad_f_GAMMA_%d.img, full.names = TRUE)
 listfile-dir()
 for (i in c(1:365)) {
   conne - file(listfile[i], rb)
   file- readBin(conne, double(), size=4,  n=720*360, signed=T)
  file2-matrix(data=file,ncol=720,nrow=360)  
   extract[i]-mean(file2[X,Y],na.rm=TRUE)
   close(conne)
 write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
 (10)\\sam.txt)}
 


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

Re: [R] R does not recognise columns and rows as they are supposed to be

Sorry, forgot the last line in the op code.

write.table(extract, ...) is called every time through the loop, not just
one time after it. Put it after closing '}'.

Rui Barradas



Rui Barradas wrote
 
 Hello,
 
 There are several things with your code, most of which are not errors.
 
 1. X-c(364:369) ; Y-c(82:92 and later c(1:365)
 Expressions of the form m:n are integer sequences, the c() is not needed.
 
 2. extract-vector()
 First you create the vector, then keep extending it throughout the loop.
 This is much slower.
 If you know the final length (and type), create it like this:
 extract - numeric(365)   # or double(365)
 
 3. 'dir1' is not used.
 
 4. Now the readBin instruction:
 4.1. 'file' is the name of a function, use, say, 'file1' instead.
 4.2. Option 'signed' defaults to TRUE but it's only valid for integer
 types.
 
 5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get
 rid of option 'size', like the help page says,
 
 If size is specified and not the natural size of the object, each element
 of the vector is coerced to an appropriate type before being written or as
 it is read.
 
 You are NOT reading doubles.
 
 As for the rest, it seems ok to me. (Keep option 'n' in readBin.)
 
 Hope this helps,
 
 Rui Barradas
 
 Jonsson wrote
 
 Dear All,
 The code given bellow is to extract  values of one region and write that
 to a text file(there are 365 binary files in the directory).
 The problem which I am facing is that all my files are binary with size
 of 360 rows and 720 columns.
 I specified that in this line:   
 file2-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in
 mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds.
 and then I rewrote the above line as
 :file2-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720
 which is not right.But that worked and I didn't get any error.however,the
 results were not correct.
 Any help please
  X-c(364:369) ; Y-c(82:92) #   for sellected region
 extract-vector()
 dir1- list.files(C:\\Users\\aalyaari\\Desktop\\New folder
 (10)\\Climate_Rad_f_GAMMA_%d.img, full.names = TRUE)
 listfile-dir()
 for (i in c(1:365)) {
   conne - file(listfile[i], rb)
   file- readBin(conne, double(), size=4,  n=720*360, signed=T)
  file2-matrix(data=file,ncol=720,nrow=360)  
  extract[i]-mean(file2[X,Y],na.rm=TRUE)
   close(conne)
 write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
 (10)\\sam.txt)}
 
 


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Re: [R] inner_perc_table?

2012-05-24 Thread William Dunlap

You need to supply a character string, e.g., inner_perc_table, that names
the C entry point.  If you supply a name, e.g., entry_point,  or a more general 
expression,
e.g, entry_points[j], then it will be evaluated with the usual R evaluation
rules and the result must be a character string.

   getNativeSymbolInfo(inner_perc_table)
  $name
  [1] inner_perc_table 

  $address
  pointer: 0x6b1cb650
  attr(,class)
  [1] NativeSymbol

  $package
  DLL name: nlme
  Filename: c:/Program Files/R/R-2.15.0/library/nlme/libs/x64/nlme.dll
  Dynamic lookup: FALSE

  $numParameters
  [1] 6

  attr(,class)
  [1] CRoutine NativeSymbolInfo

or
   entry_point - inner_perc_table
   entry_points - c(outer_perc_table, inner_perc_table)
   identical(getNativeSymbolInfo(entry_point), 
getNativeSymbolInfo(inner_perc_table))
  [1] TRUE
   identical(getNativeSymbolInfo(entry_points[2]), 
getNativeSymbolInfo(inner_perc_table))
  [1] TRUE

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Charles Determan Jr
 Sent: Thursday, May 24, 2012 10:02 AM
 To: Duncan Murdoch
 Cc: r-help@r-project.org
 Subject: Re: [R] inner_perc_table?
 
 Thank you Duncan,
 
 I used the R function getNativeSymbolInfo(inner_perc_table) and the object
 is not found:
 
 Error in is.vector(X) : object 'inner_perc_table' not found
 Done.
 
 I also tried the defining the directory to the nlme.dll file and it had the
 same error.
 Is there something I missed while doing this?
 
 Thanks,
 Charles
 
 
 On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch
 murdoch.dun...@gmail.comwrote:
 
  On 24/05/2012 11:35 AM, Charles Determan Jr wrote:
 
  Hello,
 
  Does anyone on this list know what inner_perc_table is or where it is
  typically found?  I am trying to modify some source code and it is used
  with the .C() function.  When I try and run it, it states that
  'inner_perc_table is not found'.  It is only called in such a way and
  isn't
  defined at any point in the previous code in which the function works.
  Should this be in some other directory of the source code files?  I have
  been scouring through them and have continue to come up empty.
 
 
  Look for the C function R_registerRoutines and/or the R function
  getNativeSymbolInfo.
 
  Duncan Murdoch
 
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

__
R-help@r-project.org mailing list
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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] inner_perc_table?

Thank you William,

Does this mean I should use  when inner_perc_table is called in the code
then or am I supposed to define a full path?

Regards,
Charles

On Thu, May 24, 2012 at 12:31 PM, William Dunlap wdun...@tibco.com wrote:

 You need to supply a character string, e.g., inner_perc_table, that names
 the C entry point.  If you supply a name, e.g., entry_point,  or a more
 general expression,
 e.g, entry_points[j], then it will be evaluated with the usual R evaluation
 rules and the result must be a character string.

   getNativeSymbolInfo(inner_perc_table)
  $name
  [1] inner_perc_table

  $address
  pointer: 0x6b1cb650
  attr(,class)
  [1] NativeSymbol

  $package
  DLL name: nlme
  Filename: c:/Program Files/R/R-2.15.0/library/nlme/libs/x64/nlme.dll
  Dynamic lookup: FALSE

  $numParameters
  [1] 6

  attr(,class)
  [1] CRoutine NativeSymbolInfo

 or
   entry_point - inner_perc_table
   entry_points - c(outer_perc_table, inner_perc_table)
   identical(getNativeSymbolInfo(entry_point),
 getNativeSymbolInfo(inner_perc_table))
  [1] TRUE
   identical(getNativeSymbolInfo(entry_points[2]),
 getNativeSymbolInfo(inner_perc_table))
  [1] TRUE

 Bill Dunlap
 Spotfire, TIBCO Software
 wdunlap tibco.com


  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf
  Of Charles Determan Jr
  Sent: Thursday, May 24, 2012 10:02 AM
  To: Duncan Murdoch
  Cc: r-help@r-project.org
  Subject: Re: [R] inner_perc_table?
 
  Thank you Duncan,
 
  I used the R function getNativeSymbolInfo(inner_perc_table) and the
 object
  is not found:
 
  Error in is.vector(X) : object 'inner_perc_table' not found
  Done.
 
  I also tried the defining the directory to the nlme.dll file and it had
 the
  same error.
  Is there something I missed while doing this?
 
  Thanks,
  Charles
 
 
  On Thu, May 24, 2012 at 11:51 AM, Duncan Murdoch
  murdoch.dun...@gmail.comwrote:
 
   On 24/05/2012 11:35 AM, Charles Determan Jr wrote:
  
   Hello,
  
   Does anyone on this list know what inner_perc_table is or where it is
   typically found?  I am trying to modify some source code and it is
 used
   with the .C() function.  When I try and run it, it states that
   'inner_perc_table is not found'.  It is only called in such a way and
   isn't
   defined at any point in the previous code in which the function works.
   Should this be in some other directory of the source code files?  I
 have
   been scouring through them and have continue to come up empty.
  
  
   Look for the C function R_registerRoutines and/or the R function
   getNativeSymbolInfo.
  
   Duncan Murdoch
  
 
[[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.


[[alternative HTML version deleted]]

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

Re: [R] R does not recognise columns and rows as they are supposed to be

2012-05-24 Thread Jonsson

Yes this helped a lot .
I exactly followed what you suggested:
X-(82:92) ; Y-(364:369) #   for sellected region
 extract - double(365) 
 setwd(C:\\Users\\aalyaari\\Desktop\\New folder (10)\\)
 listfile-dir()
 for (i in 1:365) {
+   conne - file(listfile[i], rb)
+   file1- readBin(conne, double(),  n=360*720)
+  file2-matrix(data=file1,ncol=720,nrow=360)
+  extract[i]-mean(file2[X,Y],na.rm=TRUE)
+   close(conne)
+ write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
(10)\\samregion1.txt)}

But I wonder why  I got all values like

-3.75E+306
-1.30E+54
-1.22E+58
and the right ones should be like:
22.25
22.76
33.25



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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] applying cbind (or any function) across all components in a list

2012-05-24 Thread David L Carlson

You don't make it clear if you want the two lists processed in parallel, but
if so, this may be a step toward the function you want:

Combine - function(l1, l2) {
  pattern - cbind(c(1, 2, 2, 3), c(1, 1, 2, 2))
  for (i in 1:length(l1)) {
print((l1[[i]][,pattern[,1]]+l2[[i]][,pattern[,2]])/2)
  }
}
Combine(l1, l2)

This simply prints the results, but you didn't indicate how you wanted the
output organized.

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Greg Snow
 Sent: Thursday, May 24, 2012 10:17 AM
 To: Hans Thompson
 Cc: r-help@r-project.org
 Subject: Re: [R] applying cbind (or any function) across all components
 in a list
 
 The do.call function may be what you want, but it is not completely
 clear.  If that does not solve your problem then try giving us an
 example of what your list looks like (the dput function can help) and
 what you want your end result to look like.
 
 On Thu, May 24, 2012 at 5:38 AM, Hans Thompson
 hans.thomps...@gmail.com wrote:
  The combination of column means in cbind is what I am trying to do.
  I just
  don't know how to do it for every component (2 in this case) of the
 list.
  I've been able to work with R to apply a function across all
 components when
  there is just one list but I am having trouble working with multiple
 lists
  at the same time.
 
  --
  View this message in context: http://r.789695.n4.nabble.com/applying-
 cbind-or-any-function-across-all-components-in-a-list-
 tp4631128p4631187.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
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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

Re: [R] plot(summary) quantreg - Not all outputs needed

2012-05-24 Thread Peter Ehlers


On 2012-05-24 03:26, stefan23 wrote:

Hi Folks,
I am currently trying to present some results I obtained by using the
quantreg package developed by Roger Koenker. After calculating
fit-summary(rq(Y~X1+X2, tau=2:98/100) ) the function plot(fit) presents a
really nice the results by showing the values for all regressors in the
given interval tau. But in my case, I only need the output of a single
variable, say X1 and I am not interested in plotting the others. Is there a
way to hide the other graphics?


Isn't this accomplished by specifying the argument 'parm'?

  plot(fit, parm = 2)

etc.

Peter Ehlers


Thank you very much for your help,

Cheers
Stefan

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] applying cbind (or any function) across all components in a list



On May 24, 2012, at 12:49 PM, Hans Thompson wrote:


I'm confused why I haven't made clear what I am asking for help with.

I have two different lists with two (or many) components, [[1]] and  
[[2]].
One of the list has components with dim=c(2,3) and the other has  
dim=c(2,2).
I want to create a new list with components dim=c(2,4) by binding  
together

the averages of the columns using


cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2,
(l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 )


which should put out:

[,1] [,2] [,3] [,4]
[1,]1234
[2,]2345

my problem is finding out how I can apply this function to all the
components within the list in the same function.


Well, it's not clear what function you want applied to what. In  
fact, it's even less clear in this posting than it was in your first  
one (since you assume, incorrectly, that we are looking at this on  
Nabble. Readers of this list expect you to include context.)  I could  
not figure out in your first posting how we should be considering the  
fourth expression


  {(l1[[1]][,3]+l2[[1]][,2])/2 }

... to be an average of anything since its indices do not match?  
Perhaps equivalently, why should a 2 x 3 structure merged somehow to  
a 2 x 2 structure yield a 2 x 4 structure?


--
David Winsemius, MD
West Hartford, CT

__
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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 does not recognise columns and rows as they are supposed to be

2012-05-24 Thread William Dunlap

X-c(364:369) ; Y-c(82:92) #   for sellected region
...
   file2-matrix(data=file,ncol=720,nrow=360)
  extract[i]-mean(file2[X,Y],na.rm=TRUE)

Note that 2-dimensional subscripts are given as
mat[ROWS, COLUMNS]
and you are using the reverse order, hence you get the
error message that you are asking for rows 364:369 of
a 360-row matrix.

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Rui Barradas
 Sent: Thursday, May 24, 2012 9:48 AM
 To: r-help@r-project.org
 Subject: Re: [R] R does not recognise columns and rows as they are supposed 
 to be
 
 Hello,
 
 There are several things with your code, most of which are not errors.
 
 1. X-c(364:369) ; Y-c(82:92 and later c(1:365)
 Expressions of the form m:n are integer sequences, the c() is not needed.
 
 2. extract-vector()
 First you create the vector, then keep extending it throughout the loop.
 This is much slower.
 If you know the final length (and type), create it like this:
 extract - numeric(365)   # or double(365)
 
 3. 'dir1' is not used.
 
 4. Now the readBin instruction:
 4.1. 'file' is the name of a function, use, say, 'file1' instead.
 4.2. Option 'signed' defaults to TRUE but it's only valid for integer types.
 
 5. And, maybe this is the problem, in R, doubles are 8 bytes, not 4. Get rid
 of option 'size', like the help page says,
 
 If size is specified and not the natural size of the object, each element
 of the vector is coerced to an appropriate type before being written or as
 it is read.
 
 You are NOT reading doubles.

The file may well contain single precision numbers and using
  what=double, size=4
will read them and convert them to doubles so other functions
in R can use them.  (Likewise, readBin can read files containing
1- or 2-byte signed or unsigned integers and store them as the
4-byte signed integers that the rest of R can deal with.)

If the file was created on another sort of machine, then you may
have worry about the byte order - try adding endian=big or
endian=little.

 
 As for the rest, it seems ok to me. (Keep option 'n' in readBin.)
 
 Hope this helps,
 
 Rui Barradas
 
 Jonsson wrote
 
  Dear All,
  The code given bellow is to extract  values of one region and write that
  to a text file(there are 365 binary files in the directory).
  The problem which I am facing is that all my files are binary with size of
  360 rows and 720 columns.
  I specified that in this line:
  file2-matrix(data=file,ncol=720,nrow=360) but I got an error : Error in
  mean(file2[X, Y], na.rm = TRUE) : subscript out of bounds.
  and then I rewrote the above line as
  :file2-matrix(data=file,ncol=360,nrow=720.I put ncol=360 and nrows =720
  which is not right.But that worked and I didn't get any error.however,the
  results were not correct.
  Any help please
   X-c(364:369) ; Y-c(82:92) #   for sellected region
  extract-vector()
  dir1- list.files(C:\\Users\\aalyaari\\Desktop\\New folder
  (10)\\Climate_Rad_f_GAMMA_%d.img, full.names = TRUE)
  listfile-dir()
  for (i in c(1:365)) {
conne - file(listfile[i], rb)
file- readBin(conne, double(), size=4,  n=720*360, signed=T)
   file2-matrix(data=file,ncol=720,nrow=360)
  extract[i]-mean(file2[X,Y],na.rm=TRUE)
close(conne)
  write.table(extract,C:\\Users\\aalyaari\\Desktop\\New folder
  (10)\\sam.txt)}
 
 
 
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/R-does-not-recognise-
 columns-and-rows-as-they-are-supposed-to-be-tp4631217p4631227.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.

__
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] R Error: System is computationally singular

Greetings,

I am trying to fit a zero-inflated Poisson model using zeroinfl() from the
pscl library. I have 5 covariates (4 continuous, 1 categorical); the
categorical variable has 7 levels.  I have had success fitting models that
contain only the continuous covariates; however, when I add the categorical
variable to any of the models (or if I run it by itself) I get the following
error: 

Error in solve.default(as.matrix(fit$hessian)) : 

  system is computationally singular: reciprocal condition number =
3.46934e-20

 The code I am using is:

library(pscl)
f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
D_GRASS)
ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)
 
There is no correlation between my covariates. Also, I tried reducing my
categorical covariate to 3 levels and still receive the same error. Can
anyone suggest why I may be getting this error when I add the categorical
covariate?
 
I appreciate your time and input. Thank you,
 
Nate
 
Nathan Svoboda
Graduate Research Assistant
Carnivore Ecology Lab
Mississippi State University

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Re: [R] R Error: System is computationally singular



On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:


Greetings,

I am trying to fit a zero-inflated Poisson model using zeroinfl()  
from the

pscl library. I have 5 covariates (4 continuous, 1 categorical); the
categorical variable has 7 levels.  I have had success fitting  
models that
contain only the continuous covariates; however, when I add the  
categorical
variable to any of the models (or if I run it by itself) I get the  
following

error:

Error in solve.default(as.matrix(fit$hessian)) :

 system is computationally singular: reciprocal condition number =
3.46934e-20

The code I am using is:

library(pscl)
f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
D_GRASS)
ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)

There is no correlation between my covariates. Also, I tried  
reducing my
categorical covariate to 3 levels and still receive the same error.  
Can
anyone suggest why I may be getting this error when I add the  
categorical

covariate?



What does this show:

with( FAWNS, table(LOCS, LCOVER) )

--
David Winsemius, MD
West Hartford, CT

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Re: [R] R Error: System is computationally singular

Thank you for your quick reply! 
 
When I run the code you provide I get this output:
 
LCOVER
LOCS  1  2  3  4  5  6  7  9
   0 214507  79939  69803 778359  22932  32391  99630   8082
   1 15  7  1 32  0  0  0  0
   2  2  1  0  0  0  0  0  0
   3  0  0  0  1  0  0  0  0
 
Nate
 
 
 



From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thu 5/24/2012 1:54 PM
To: Nathan Svoboda
Cc: r-help@r-project.org
Subject: Re: [R] R Error: System is computationally singular




On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:

 Greetings,

 I am trying to fit a zero-inflated Poisson model using zeroinfl() 
 from the
 pscl library. I have 5 covariates (4 continuous, 1 categorical); the
 categorical variable has 7 levels.  I have had success fitting 
 models that
 contain only the continuous covariates; however, when I add the 
 categorical
 variable to any of the models (or if I run it by itself) I get the 
 following
 error:

 Error in solve.default(as.matrix(fit$hessian)) :

  system is computationally singular: reciprocal condition number =
 3.46934e-20

 The code I am using is:

 library(pscl)
 f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
 D_GRASS)
 ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)

 There is no correlation between my covariates. Also, I tried 
 reducing my
 categorical covariate to 3 levels and still receive the same error. 
 Can
 anyone suggest why I may be getting this error when I add the 
 categorical
 covariate?


What does this show:

with( FAWNS, table(LOCS, LCOVER) )

--
David Winsemius, MD
West Hartford, CT




[[alternative HTML version deleted]]

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Re: [R] R Error: System is computationally singular

Thank you for your quick reply, 

When I run the code you provide I get this output:

LCOVER
LOCS  1  2  3  4  5  6  7  9
   0 214507  79939  69803 778359  22932  32391  99630   8082
   1 15  7  1 32  0  0  0  0
   2  2  1  0  0  0  0  0  0
   3  0  0  0  1  0  0  0  0

Nate



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Re: [R] R Error: System is computationally singular

Hi David,
 
My apologies, I am not sure if this makes a big difference in your assessment 
of the problem, but the results I just sent were only from a portion (1/15) of 
the data. The dataset is rather large and the computer I am currently using to 
set up the models is limited in its capabilities to analyze large datasets. 
When I run the code you provided on a larger portion of the data (1/2) this is 
the output I receive:
 
 LCOVER
LOCS   1   2   3   4   5   6   7   9
   0 1692196  630659  550623 6140352  180896  255512  785929   63756
   1 141  30  48 279   9  14  36   1
   2  17   4   5  14   3   3   4   1
   3   0   0   0   3   0   0   1   0
   5   2   0   0   0   0   0   0   0
 
Thanks again for your time and assistance,
 
Nate
 
Nathan Svoboda
Graduate Research Assistant
Mississippi State University
 
 


From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thu 5/24/2012 1:54 PM
To: Nathan Svoboda
Cc: r-help@r-project.org
Subject: Re: [R] R Error: System is computationally singular




On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:

 Greetings,

 I am trying to fit a zero-inflated Poisson model using zeroinfl() 
 from the
 pscl library. I have 5 covariates (4 continuous, 1 categorical); the
 categorical variable has 7 levels.  I have had success fitting 
 models that
 contain only the continuous covariates; however, when I add the 
 categorical
 variable to any of the models (or if I run it by itself) I get the 
 following
 error:

 Error in solve.default(as.matrix(fit$hessian)) :

  system is computationally singular: reciprocal condition number =
 3.46934e-20

 The code I am using is:

 library(pscl)
 f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
 D_GRASS)
 ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)

 There is no correlation between my covariates. Also, I tried 
 reducing my
 categorical covariate to 3 levels and still receive the same error. 
 Can
 anyone suggest why I may be getting this error when I add the 
 categorical
 covariate?


What does this show:

with( FAWNS, table(LOCS, LCOVER) )

--
David Winsemius, MD
West Hartford, CT




[[alternative HTML version deleted]]

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

Re: [R] R Error: System is computationally singular

2012-05-24 Thread Marc Schwartz

Nathan,

This does help, as in the first cut you provided, there was no variability in 
LOCS for LCOVER = 5 and you have very few values of LOCS  0 (you still do, 
relative to the scale of the total).

Have you tried using a zero inflated negative binomial model (dist = negbin) 
rather than poisson? I am not sure that the assumption of a zero inflated 
poisson distribution is reasonable with your data. Also, at least in this cut 
of the data, you have no 4's in LOCS and no 8's in LCOVER (same as before).

If my math is correct only 0.006% of your LOCS values are  0. I am also not 
convinced that you have enough data to differentiate between 1 and =1 of 
whatever it is you are counting in LOCS.

If that is the case, you might want to consider using logistic regression with 
a dichotomous response variable of LOCS == 0 versus LOCS = 1. You seem to be 
in the general realm of very rare events given the distribution of LOCS in your 
data. 

Regards,

Marc Schwartz


On May 24, 2012, at 2:41 PM, Nathan Svoboda wrote:

 Hi David,
 
 My apologies, I am not sure if this makes a big difference in your assessment 
 of the problem, but the results I just sent were only from a portion (1/15) 
 of the data. The dataset is rather large and the computer I am currently 
 using to set up the models is limited in its capabilities to analyze large 
 datasets. When I run the code you provided on a larger portion of the data 
 (1/2) this is the output I receive:
 
 LCOVER
 LOCS   1   2   3   4   5   6   7   9
   0 1692196  630659  550623 6140352  180896  255512  785929   63756
   1 141  30  48 279   9  14  36   1
   2  17   4   5  14   3   3   4   1
   3   0   0   0   3   0   0   1   0
   5   2   0   0   0   0   0   0   0
 
 Thanks again for your time and assistance,
 
 Nate
 
 Nathan Svoboda
 Graduate Research Assistant
 Mississippi State University
 
 
 On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:
 
 Greetings,
 
 I am trying to fit a zero-inflated Poisson model using zeroinfl() 
 from the
 pscl library. I have 5 covariates (4 continuous, 1 categorical); the
 categorical variable has 7 levels.  I have had success fitting 
 models that
 contain only the continuous covariates; however, when I add the 
 categorical
 variable to any of the models (or if I run it by itself) I get the 
 following
 error:
 
 Error in solve.default(as.matrix(fit$hessian)) :
 
 system is computationally singular: reciprocal condition number =
 3.46934e-20
 
 The code I am using is:
 
 library(pscl)
 f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
 D_GRASS)
 ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)
 
 There is no correlation between my covariates. Also, I tried 
 reducing my
 categorical covariate to 3 levels and still receive the same error. 
 Can
 anyone suggest why I may be getting this error when I add the 
 categorical
 covariate?
 
 
 What does this show:
 
 with( FAWNS, table(LOCS, LCOVER) )
 
 --
 David Winsemius, MD
 West Hartford, CT

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

Re: [R] svychisq using two frames

2012-05-24 Thread Thomas Lumley

On Fri, May 25, 2012 at 4:10 AM, lasciel corte...@msu.edu wrote:


 So, I could use ‘rbind’ to stack the two years of data together with their
 appropriate weights into a single data frame.
 svychisq(~MyVar+MyVar, BothYears, statistic=”Chisq”)

 But now I have a problem that the variable name is the same across years, so
 how do I differentiate the different time periods in the syntax for the
 formula?  Do I need to also create two new variables per:
 MyVar.1 [year = 1] - MyVar
 MyVar.2 [year = 2] - MyVar
 svychisq(~MyVar.1+MyVar.2, BothYears, statistic=”Chisq”, na.rm= TRUE)


 (I feel like I may be overthinking this and the answer is much simpler)




You don't need to create two new variables; you just need a year variable

 svychisq(~MyVar+Year, BothYears, statistic=”Chisq”, na.rm= TRUE)

tests whether MyVar is independent of Year.

   -thomas

-- 
Thomas Lumley
Professor of Biostatistics
University of Auckland

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[R] R base package grid does not output raster image

2012-05-24 Thread Patrick Nicholson

I am running 64-bit R 2.15.0 on a Windows Server 2008 R2 Amazon EC2 instance. 
grid does not produce output. For example, the following code should print the 
R logo to the window() device:


library(grid)
library(png)

img.path - system.file(img, Rlogo.png, package=png)
bg - readPNG(img.path)
background - rasterGrob(unclass(bg))

grid.draw(background)


I, however, see nothing. Is there a flag or option I should be using on Windows 
Server 2008 R2 to enable grid output?

For clarity, the following is how I am actually using images in R plots. This 
code works on my laptop (R 2.15.0, ggplot2 0.8.9):


library(ggplot2)
library(grid)
library(maps)
library(mapproj)
library(png)
library(RgoogleMaps)

counties - map_data(county, region=virginia)
states - map_data(state)

tmp - tempfile(fileext=.png)
bg - GetMap.bbox(range(counties$long), range(counties$lat), destfile=tmp, 
maptype=satellite, format=png32)
background - readPNG(tmp)
background - rasterGrob(unclass(background))

ggplot(counties, aes(long, lat)) + 
  coord_map(xlim=c(bg$BBOX$ll[2], bg$BBOX$ur[2]), ylim=c(bg$BBOX$ll[1], 
bg$BBOX$ur[1])) +
  geom_path(aes(group=group), color=darkgrey) +
  geom_path(data=states, aes(group=group), color=white, size=1) +
  opts(axis.line=theme_blank(), 
   axis.text.x=theme_blank(), 
   axis.text.y=theme_blank(),
   axis.ticks=theme_blank(), 
   axis.title.x=theme_blank(), 
   axis.title.y=theme_blank(),
   axis.ticks.length=unit(0, lines), 
   axis.ticks.margin=unit(0, lines),
   panel.border=theme_blank(), 
   panel.background=function(...)background,
   panel.grid.major=theme_blank(), 
   panel.grid.minor=theme_blank(),
   panel.margin=unit(0, lines), 
   legend.position=none, 
   legend.title=theme_blank(),
   legend.background=theme_blank(),
   plot.margin=unit(0*c(-1.5, -1.5, -1.5, -1.5), lines))


Thank you,
Patrick

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Re: [R] applying cbind (or any function) across all components in a list

Yes. This gives me:

 [,1] [,2] [,3] [,4]
[1,]1234
[2,]2345
 [,1] [,2] [,3] [,4]
[1,]6789
[2,]789   10

BUT, how can I have it still within components like

[[1]]
 [,1] [,2] [,3] [,4]
[1,]1234
[2,]2345

[[2]]
 [,1] [,2] [,3] [,4]
[1,]6789
[2,]789   10

How should I have phrased my question to be specific to this result?

Thanks.

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Re: [R] applying cbind (or any function) across all components in a list

The function I am giving for context is cbind.  Are you asking how I would
like to apply the answer to my question?

I am trying to take the results of a Fluidigm SNP microarray, organized by
assay into a list (each component is the results of one assay), find
coordinate midpoints ([1,] and [2,] of my XX, XY, and YY clusters (these are
genotypes) and is represented by l1.  l2 is the midpoint between XX/XY and
XY/YY although I did not give this in my example for simplicity, and I am
now trying to find the midpoint between these new midpoints and their
closest genotype clusters. This is represented as

cbind((l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2,
(l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2)

but only works for one assay in the list of 96.  I want to apply this to the
entire list.  My entire code so far is:

## OPEN .CSV and ORGANIZE BY ASSAY


file=
{
rawdata - read.csv(file, skip = 15)
OrgAssay - split(rawdata, rawdata$Assay)


## RETURN MIDPOINTS FOR EACH CLUSTER WITHOUT NO CALLS

#for loop
ClustMidPts -list()

for(locus in 1:length(names(OrgAssay))){
ClustMidPts[[locus]]-t(cbind(tapply(OrgAssay[[locus]][,Allele.X.1],
OrgAssay[[locus]][,Final], mean,na.rm=T),
   tapply(OrgAssay[[locus]][,Allele.Y.1],
OrgAssay[[locus]][,Final], mean,na.rm=T)))}

names(ClustMidPts)=names(OrgAssay)


## CREATE CLUSTER-CLUSTER MIDPOINT

#for loop
ClustClustMidPts - list()

for(locus in 1:length(names(ClustMidPts))){
ClustClustMidPts[[locus]] -
cbind(XXYX=(ClustMidPts[[locus]][,XX]+ClustMidPts[[locus]][,YX])/2,
YXYY=(ClustMidPts[[locus]][,YX]+ClustMidPts[[locus]][,YY])/2)
}

names(ClustClustMidPts)=names(ClustMidPts)


Please also let me know how I messed up the formatting because it shows up
fine in gmail even when I post on Nabble.  How did I assume you were using
Nabble?  Is this topic included in the posting guide?

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[R] How to open a file with a name changed?

2012-05-24 Thread Minerva Mora

Hi, I apologize for my english.

I´m trying to read a file, but the name of this file changes every day, for
example: today is May 24, 2012

bonos- read.table(C:/Bonos/*20120524*.csv, header=TRUE, sep=\t)

So, tomorrow I want to read the file again, but i don´t want to put the date
by myself, i want this automatically. I know that if a put day() this
instruccion gives me the date.

My problem is, I don´t know how to concatenate these:
-C:/Bonos/
-the date
-.csv
 
I tried this:
bonos- read.table(cat(cit,dia,extension), header=TRUE, sep=\t)
where
--cit is *\C:/Bonos/*
--dia is *dia-format(today, %Y%m%d)*
--extension is *extension-.csv\*

but i have this problem:
C:/Bonos/ 20120524 .csv*Error in read.table(cat(cit, dia, extension),
header = TRUE, sep = \t) : * * 'file' must be a character string or
connection*

What can i do?

Thanks!

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Re: [R] R Error: System is computationally singular



On May 24, 2012, at 3:41 PM, Nathan Svoboda wrote:


Re: [R] R Error: System is computationally singular
Hi David,

My apologies, I am not sure if this makes a big difference in your  
assessment of the problem, but the results I just sent were only  
from a portion (1/15) of the data. The dataset is rather large and  
the computer I am currently using to set up the models is limited in  
its capabilities to analyze large datasets. When I run the code you  
provided on a larger portion of the data (1/2) this is the output I  
receive:


 LCOVER
LOCS   1   2   3   4   5   6   7   9
   0 1692196  630659  550623 6140352  180896  255512  785929   63756
   1 141  30  48 279   9  14  36   1
   2  17   4   5  14   3   3   4   1
   3   0   0   0   3   0   0   1   0
   5   2   0   0   0   0   0   0   0


I do not see linear dependence (aka computational singularity) in that  
data, but if there are no LOCS values of 4, an missing levels has been  
reported as a show-stopper with zinf models with pscl in the past.  
There could also easily emerge linear dependence if you tabulated the  
entire data set. If level 4 had 3 at level 4 of LCOVER or 1 at level 7  
then there would be linear dependence.


Marc Schwartz, a smarter guy than I,  has already suggested to you  
that your Poisson error structure might not be a good description of  
the data.





Thanks again for your time and assistance,

Nate

Nathan Svoboda
Graduate Research Assistant
Mississippi State University


From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thu 5/24/2012 1:54 PM
To: Nathan Svoboda
Cc: r-help@r-project.org
Subject: Re: [R] R Error: System is computationally singular


On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:

 Greetings,

 I am trying to fit a zero-inflated Poisson model using zeroinfl()
 from the
 pscl library. I have 5 covariates (4 continuous, 1 categorical); the
 categorical variable has 7 levels.  I have had success fitting
 models that
 contain only the continuous covariates; however, when I add the
 categorical
 variable to any of the models (or if I run it by itself) I get the
 following
 error:

 Error in solve.default(as.matrix(fit$hessian)) :

  system is computationally singular: reciprocal condition number =
 3.46934e-20

 The code I am using is:

 library(pscl)
 f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER +  
D_EDGE +

 D_GRASS)
 ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)

 There is no correlation between my covariates. Also, I tried
 reducing my
 categorical covariate to 3 levels and still receive the same error.
 Can
 anyone suggest why I may be getting this error when I add the
 categorical
 covariate?


What does this show:

with( FAWNS, table(LOCS, LCOVER) )

--
David Winsemius, MD
West Hartford, CT




David Winsemius, MD
West Hartford, CT

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Re: [R] applying cbind (or any function) across all components in a list



On May 24, 2012, at 6:37 PM, Hans Thompson wrote:

The function I am giving for context is cbind.  Are you asking how I  
would

like to apply the answer to my question?

I am trying to take the results of a Fluidigm SNP microarray,  
organized by

assay into a list (each component is the results of one assay), find
coordinate midpoints ([1,] and [2,] of my XX, XY, and YY clusters  
(these are
genotypes) and is represented by l1.  l2 is the midpoint between XX/ 
XY and
XY/YY although I did not give this in my example for simplicity, and  
I am

now trying to find the midpoint between these new midpoints and their
closest genotype clusters. This is represented as

cbind( l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][,2]+l2[[1]][,1])/2,
(l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 )


cbind would not be much different than c() in this case.



but only works for one assay in the list of 96.


It's not really a function, but rather an expression that will be  
evaluated. If you had a function it might look like :


bindfn - function(l1, l1) { c( l1[[1]][,1]+l2[[1]][,1])/2, (l1[[1]][, 
2]+l2[[1]][,1])/2,

(l1[[1]][,2]+l2[[1]][,2])/2, (l1[[1]][,3]+l2[[1]][,2])/2 ) }

And you could perhaps (depending on whether l1 and l2 were  
constructed) properly execute:


do.call(bindfn, list(l1,l2))

perhaps:

mapply(bindfn, l1, l2)

You posted an example higher up in the thread but my wife is calling  
me to dinner. Will return later.



 I want to apply this to the
entire list.  My entire code so far is:

## OPEN .CSV and ORGANIZE BY ASSAY


file=
   {
   rawdata - read.csv(file, skip = 15)
   OrgAssay - split(rawdata, rawdata$Assay)


   ## RETURN MIDPOINTS FOR EACH CLUSTER WITHOUT NO CALLS

   #for loop
   ClustMidPts -list()

   for(locus in 1:length(names(OrgAssay))){
   ClustMidPts[[locus]]-t(cbind(tapply(OrgAssay[[locus]][,Allele.X. 
1],

OrgAssay[[locus]][,Final], mean,na.rm=T),
  tapply(OrgAssay[[locus]][,Allele.Y.1],
OrgAssay[[locus]][,Final], mean,na.rm=T)))}

   names(ClustMidPts)=names(OrgAssay)


## CREATE CLUSTER-CLUSTER MIDPOINT

   #for loop
   ClustClustMidPts - list()

   for(locus in 1:length(names(ClustMidPts))){
   ClustClustMidPts[[locus]] -
cbind(XXYX=(ClustMidPts[[locus]][,XX]+ClustMidPts[[locus]][,YX])/ 
2,

YXYY=(ClustMidPts[[locus]][,YX]+ClustMidPts[[locus]][,YY])/2)
}

   names(ClustClustMidPts)=names(ClustMidPts)


Please also let me know how I messed up the formatting because it  
shows up
fine in gmail even when I post on Nabble.  How did I assume you were  
using

Nabble?


Because you included no context. We cannot see your earlier postings.  
Expecting us to page out to Nabble or refer back to other item s in  
the thread is considered presumptuous on your part.



 Is this topic included in the posting guide?

YES. It is. And you were asked to read the Posting Guide about 10  
times by now.



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David Winsemius, MD
West Hartford, CT

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Re: [R] How to open a file with a name changed?

2012-05-24 Thread Steve Taylor

You probably want to use paste or paste0 instead of cat.
paste0(cit,dia,extension)

Also, this might work: 
tail(dir(C:/Bonos/,patt=csv$),1)


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Minerva Mora
Sent: Friday, 25 May 2012 11:45a
To: r-help@r-project.org
Subject: [R] How to open a file with a name changed?

Hi, I apologize for my english.

I´m trying to read a file, but the name of this file changes every day, for
example: today is May 24, 2012

bonos- read.table(C:/Bonos/*20120524*.csv, header=TRUE, sep=\t)

So, tomorrow I want to read the file again, but i don´t want to put the date
by myself, i want this automatically. I know that if a put day() this
instruccion gives me the date.

My problem is, I don´t know how to concatenate these:
-C:/Bonos/
-the date
-.csv
 
I tried this:
bonos- read.table(cat(cit,dia,extension), header=TRUE, sep=\t)
where
--cit is *\C:/Bonos/*
--dia is *dia-format(today, %Y%m%d)*
--extension is *extension-.csv\*

but i have this problem:
C:/Bonos/ 20120524 .csv*Error in read.table(cat(cit, dia, extension),
header = TRUE, sep = \t) : * * 'file' must be a character string or
connection*

What can i do?

Thanks!

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Re: [R] R Error: System is computationally singular

Thank you both, 
 
I will try using a zero inflated negative binomial as suggested. I had success 
with negative binomial on previous runs but only when I had fewer covariates 
and only ran a portion (10%) of the data. 
 
I may also try to reduce the number of covariates in the model (i.e., combine 
some of my landcover [LCOVER] classifications). 
 
I have considered using logistic regression and may end up trying that.   
 
I appreciate both of your input and will let you know (i.e., post) the results 
of these suggestions and what ends up working for the benefit of you and others.
 
Thank you for your time and quick responses!!
 
Nate
 
 
Nathan Svoboda
Graduate Research Assistant
Carnivore Ecology Lab
Mississippi State University
 



From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Thu 5/24/2012 3:09 PM
To: Nathan Svoboda
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] R Error: System is computationally singular



Nathan,

This does help, as in the first cut you provided, there was no variability in 
LOCS for LCOVER = 5 and you have very few values of LOCS  0 (you still do, 
relative to the scale of the total).

Have you tried using a zero inflated negative binomial model (dist = negbin) 
rather than poisson? I am not sure that the assumption of a zero inflated 
poisson distribution is reasonable with your data. Also, at least in this cut 
of the data, you have no 4's in LOCS and no 8's in LCOVER (same as before).

If my math is correct only 0.006% of your LOCS values are  0. I am also not 
convinced that you have enough data to differentiate between 1 and =1 of 
whatever it is you are counting in LOCS.

If that is the case, you might want to consider using logistic regression with 
a dichotomous response variable of LOCS == 0 versus LOCS = 1. You seem to be 
in the general realm of very rare events given the distribution of LOCS in your 
data.

Regards,

Marc Schwartz


On May 24, 2012, at 2:41 PM, Nathan Svoboda wrote:

 Hi David,

 My apologies, I am not sure if this makes a big difference in your assessment 
 of the problem, but the results I just sent were only from a portion (1/15) 
 of the data. The dataset is rather large and the computer I am currently 
 using to set up the models is limited in its capabilities to analyze large 
 datasets. When I run the code you provided on a larger portion of the data 
 (1/2) this is the output I receive:

 LCOVER
 LOCS   1   2   3   4   5   6   7   9
   0 1692196  630659  550623 6140352  180896  255512  785929   63756
   1 141  30  48 279   9  14  36   1
   2  17   4   5  14   3   3   4   1
   3   0   0   0   3   0   0   1   0
   5   2   0   0   0   0   0   0   0

 Thanks again for your time and assistance,

 Nate

 Nathan Svoboda
 Graduate Research Assistant
 Mississippi State University


 On May 24, 2012, at 1:57 PM, Nathan Svoboda wrote:

 Greetings,

 I am trying to fit a zero-inflated Poisson model using zeroinfl()
 from the
 pscl library. I have 5 covariates (4 continuous, 1 categorical); the
 categorical variable has 7 levels.  I have had success fitting
 models that
 contain only the continuous covariates; however, when I add the
 categorical
 variable to any of the models (or if I run it by itself) I get the
 following
 error:

 Error in solve.default(as.matrix(fit$hessian)) :

 system is computationally singular: reciprocal condition number =
 3.46934e-20

 The code I am using is:

 library(pscl)
 f1 - formula(LOCS ~ as.factor(LCOVER) + D_ROADS + D_WATER + D_EDGE +
 D_GRASS)
 ZIP1 - zeroinfl(f1, dist=poisson, link = logit, data = FAWNS)

 There is no correlation between my covariates. Also, I tried
 reducing my
 categorical covariate to 3 levels and still receive the same error.
 Can
 anyone suggest why I may be getting this error when I add the
 categorical
 covariate?


 What does this show:

 with( FAWNS, table(LOCS, LCOVER) )

 --
 David Winsemius, MD
 West Hartford, CT




[[alternative HTML version deleted]]

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Re: [R] How to open a file with a name changed?



On May 24, 2012, at 7:45 PM, Minerva Mora wrote:


Hi, I apologize for my english.

I´m trying to read a file, but the name of this file changes every  
day, for

example: today is May 24, 2012

bonos- read.table(*20120524*.csv, header=TRUE, sep=\t)



yr = 2012
mo = 01
dt = 25
filnam =paste(C:/Bonos/ , yr, mo, dt,sep=)

bonos - read.table(filnam, header=TRUE, sep=\t)


So, tomorrow I want to read the file again, but i don´t want to put  
the date

by myself, i want this automatically. I know that if a put day() this
instruccion gives me the date.

My problem is, I don´t know how to concatenate these:
-C:/Bonos/
-the date
-.csv

I tried this:
bonos- read.table(cat(cit,dia,extension), header=TRUE, sep=\t)
where
--cit is *\C:/Bonos/*
--dia is *dia-format(today, %Y%m%d)*
--extension is *extension-.csv\*

but i have this problem:
C:/Bonos/ 20120524 .csv*Error in read.table(cat(cit, dia,  
extension),

header = TRUE, sep = \t) : * * 'file' must be a character string or
connection*

What can i do?

Thanks!

--
View this message in context: 
http://r.789695.n4.nabble.com/How-to-open-a-file-with-a-name-changed-tp4631261.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.


David Winsemius, MD
West Hartford, CT

__
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
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Re: [R] How to open a file with a name changed?