Re: [R] What is the most cost effective hardware for R?

[Hugh Morgan]

Thank you all for the help.  We have decided against using for example 
Amazon cloud for basicly paperwork issues.  We have money available now 
for buying kit, this may not be available for buying services, and may 
not be available next year, or the next.  We shall certainly consider it 
as a fall back at times of high load.


We are looking at the Dell poweredge M915.  It has 64 cores and we are 
getting it with 256 GB memory, and it really not that expensive.  I am 
surprised what power you can get these days for not very much money.


Thanks again.

__
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] Dotchart showing mean and median by group

[maxbre]

hi all

I have another question reated to the dotchart: is it possible by means of
par() to set a logaritmic scale? 
If yes, how ? and if not, any alternative solution?

thanks

--
View this message in context: 
http://r.789695.n4.nabble.com/Dotchart-showing-mean-and-median-by-group-tp4619597p4622618.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.

Re: [R] reception of (Vegan) envfit analysis by manuscript reviewers

[Gavin Simpson]

On Wed, 2012-05-09 at 15:51 -0600, Matt Bakker wrote:
 I'm getting lots of grief from reviewers about figures generated with
 the envfit function in the Vegan package. Has anyone else struggled to
 effectively explain this analysis? If so, can you share any helpful
 tips?
 
 The most recent comment I've gotten back: What this shows is which
 NMDS axis separates the communities, not the relationship between the
 edaphic factor and the Bray-Curtis distance.

Without further context for that quote and your manuscript to see how
you are using the method it is difficult to say whether you are doing
something silly or the reviewer is bone-headed.

I've had similar comments from reviewers about my use of the ordisurf()
function. In each case it was the reviewers' failure to understand the
methods applied that was the cause of the confusion.

As you provide little or no context I'll explain what envfit() does etc.

The idea goes back a long way (!) and is in my 1995 edition of Jongman
et al Data Analysis in Community and Landscape Ecology (Cambridge
University Press) though most likely was in 1987 version too. See
Section 5.4 of the Ordination chapter by Ter Braak in that book.

The idea is to find the direction (in the k-dimensional ordination
space) that has maximal correlation with an external variable.
Essentially, we have:

E(z_j) = b_0 + b_1x_1 + b_2x_2

where E(z_j) is the expectation (or mean, or fitted values) of the jth
external (environmental) variable, x_1 and x_2 are the axis scores in
ordination dimensions 1 and 2, and b_y are unknown regression
coefficients. This generalises to more than 2 dimensions or axes.

The biplot arrow drawn goes from (0,0) to (b_1, b_2).

You can see that the aim is to model or predict the values of the jth
environmental variable (z_j) as a linear combination of the axis or
site scores of the samples in the ordination space. Exactly the same
idea underlies the ordisurf() function except that we use a GAM and for
the right hand side of the equation multivariate splines are used which
allow a non-linear surface instead of a plane.

When applied to nMDS, if the nMDS provides a reasonable approximation to
the original dissimilarities, then envfit() will estimate and show the
strengths of the correlation and direction of maximal correlation
between the nMDS configuration and the jth enviromental variable. This
technique can be used to indicate if one or more environmental variables
are associated with differences between sites/samples as represented in
the nMDS ordination.

The big caveat is the implication that the correlation or relationship
between z_j and the ordination space is linear. ordisurf() allows you to
relax this assumption as we fit a potentially non-linear surface to the
ordination space instead of the plane that envfit() effectively produces
(though we show only the direction of change with the arrow).

So without seeing your manuscript or more context (and I'm not promising
to read it or comment more if you provide it) I would suggest that, *if*
you have applied nMDS and used envfit() correctly the combined analysis
*does* reflect the *linear* relationship between the edaphic factor and
the Bray-Curtis distance, assuming of course that the nMDS has low
stress (i.e fits the original dissimilarities well).

In future, you should consider posting similar questions
(ecological/environmental) to the R-SIG-Ecology list instead of the main
R-Help list. I know Jari (lead developer of vegan and author of
envfit() ) has stopped regularly reading the main R-Help list and you
will get far more eyes familiar with these techniques on the
R-SIG-Ecology list.

I have taken the liberty of cc'ing this to the R-Sig-Ecology list so
others can comment.

HTH

G

 Thanks for any suggestions!
 
 
 Matt
 
 __
 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.
 

-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Dr. Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
 UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

__
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] Fwd: Finding local maxima on a loess surface

[David Winsemius]

On May 10, 2012, at 12:10 AM, Diego Rojas wrote:


-- Forwarded message --
From: Diego Rojas diroj...@gmail.com
Date: Wed, May 9, 2012 at 3:05 PM
Subject: Re: [R] Finding local maxima on a loess surface
To: David Winsemius dwinsem...@comcast.net


Thanks again, would you please try to ilustrate further your point  
with
this example code. How would you find the coordinates for the maxima  
on the
surface. Consider that in the surface I'm dealing with there are at  
least

35 maxima.

require(MASS)
topo.lo - loess(z ~ x * y, topo, degree = 1, span = 0.25,
normalize = FALSE)
topo.mar - list(x = seq(0, 6.5, 0.1), y = seq(0, 6.5, 0.1))
new.dat - expand.grid(topo.mar)
topo.pred - predict(topo.lo, new.dat)
## draw the contour map based on loess predictions

library(rgl)

persp3d(topo.mar$x, topo.mar$y, topo.pred, shade=0.5, col=blue)


hasmax - function(mtx, x, y)  if(  (mtx[x,y]  mtx[x,y-1]) 
 (mtx[x,y]  mtx[x,y+1]) 
 (mtx[x,y]  mtx[x-1,y]) 
 (mtx[x,y]  mtx[x+1,y]) ) {return(TRUE ) } else {return(FALSE)}

for(x in 3:(dim(topo.pred)[1] -4)) {
 for(y in 3:(dim(topo.pred)[2]-4) )  {
 if( hasmax(topo.pred, x , y) ){print(c(x,y))}  }}
#
[1] 40  7

Note: that topo.pred has a border of two and three row/columns of NA's  
that made this very annoying to debug. A proper function would  
probably need to pre-qualify the index ranges.



I tried a sign change approach but generalizing to 2d created  
conceptual difficulties I could not resolve, so I just checked in both  
directions for the local point being greater than its neighbors.  You  
obviously could do something other than printing coordinates at a  
maximum


--
David.



Thanks fot your help

On Sat, May 5, 2012 at 9:34 AM, David Winsemius dwinsem...@comcast.net 
wrote:




On May 4, 2012, at 3:00 PM, Diego Rojas wrote:

Thanks, I know about it but i wat to find several local maxima, so in
other words I need a way to identify the places in the surface  
where both

slopes are equal to 0 and the second derivative is negative.



There is no way that I know that will produce a mathematical  
function that

would support symbolic manipulations of that sort for the results
obtainable from a loess-object. I was expecting that you would be
approaching this numerically and doing evaluations on a grid.  
Testing for
equality to 0 is not a good practice if following that route. Sign  
reversal

would be a more sensible criterion. ( And you _would_ be using
predict.loess(). )

Still no data example or code offered, so not pursuing further  
efforts at

illustration.



On Fri, May 4, 2012 at 9:28 AM, David Winsemius dwinsem...@comcast.net 


wrote:

On May 3, 2012, at 6:09 PM, Diego Rojas wrote:

If a run a LOESS model and then produce a smoothed surface: Is  
there any

way to determine the coordinates of the local maxima on the surface?

?predict# it has a loess method.





David Winsemius, MD
West Hartford, CT

__
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] Dotchart showing mean and median by group

[David Winsemius]

On May 10, 2012, at 2:24 AM, maxbre wrote:


hi all

I have another question reated to the dotchart: is it possible by  
means of

par() to set a logaritmic scale?
If yes, how ? and if not, any alternative solution?


Looking at the dotchart code it appears to me that the log parameter  
to plot.window is hard-coded at , i.e both scales are linear.  
Testing with the xlog parameter to par does fail.  You can always  
define a new dochart2 on the basis of that code.


--
David Winsemius, MD
West Hartford, CT

__
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] Dotchart showing mean and median by group

[David Winsemius]

On May 10, 2012, at 5:03 AM, David Winsemius wrote:



On May 10, 2012, at 2:24 AM, maxbre wrote:


hi all

I have another question reated to the dotchart: is it possible by  
means of

par() to set a logaritmic scale?
If yes, how ? and if not, any alternative solution?


Looking at the dotchart code it appears to me that the log parameter  
to plot.window is hard-coded at , i.e both scales are linear.  
Testing with the xlog parameter to par does fail.  You can always  
define a new dochart2 on the basis of that code.


Another alternative would be lattice (a simple mod to one of its  
examples shows it works):


dotplot(variety ~ yield | site, data = barley, groups = year,
key = simpleKey(levels(barley$year), space = right),  
scales=list(x=list(log=TRUE)),

xlab = Barley Yield (bushels/acre) ,
aspect=0.5, layout = c(1,6), ylab=NULL)




--

David Winsemius, MD
West Hartford, CT

__
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] There must be a better way to do this

[Jim Lemon]

On 05/10/2012 02:17 AM, David Perlman wrote:

Thanks, that is very helpful.  I agree that my example plot was a bit 
cluttered, but this is what I actually wanted:
http://brainimaging.waisman.wisc.edu/~perlman/data/MNPT1T2_h_unp_raw.pdf
I just needed to get example code out quickly.  You get better help when you 
have a self-contained demo of the question.  :)

I have replaced my old horrible code with the nice concise segments code.  
Thanks!

That's a rather nice way of illustrating what happened. I can see the 
trends in the bumpcharts on the bottom.


Jim

__
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] bigmemory

[ya]

Hi all,

I have a question about using bigmemory package.

Here is my code:

 
x=read.big.matrix(acc3.dat,backingfile=acc3.bin,descriptorfile=acc3.desc,type=double)

Error in filebacked.big.matrix(nrow = nrow, ncol = ncol, type = type,  :
  A big.matrix must have at least one row and one column

And here is the example code:

 x - read.big.matrix(airline.csv, header=TRUE,
+ backingfile=airline.bin,
+ descriptorfile=airline.desc,
+ type=integer)

So, what was wrong? Any suggestions please?

Thank you very much.

ya

__
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] [R-sig-eco] reception of (Vegan) envfit analysis by manuscript reviewers

[Jari Oksanen]

On 10/05/2012, at 11:45 AM, Gavin Simpson wrote:

 On Wed, 2012-05-09 at 15:51 -0600, Matt Bakker wrote:
 I'm getting lots of grief from reviewers about figures generated with
 the envfit function in the Vegan package. Has anyone else struggled to
 effectively explain this analysis? If so, can you share any helpful
 tips?
 
 The most recent comment I've gotten back: What this shows is which
 NMDS axis separates the communities, not the relationship between the
 edaphic factor and the Bray-Curtis distance.
 
 Without further context for that quote and your manuscript to see how
 you are using the method it is difficult to say whether you are doing
 something silly or the reviewer is bone-headed.
 
 I've had similar comments from reviewers about my use of the ordisurf()
 function. In each case it was the reviewers' failure to understand the
 methods applied that was the cause of the confusion.
 
 As you provide little or no context I'll explain what envfit() does etc.
 
 The idea goes back a long way (!) and is in my 1995 edition of Jongman
 et al Data Analysis in Community and Landscape Ecology (Cambridge
 University Press) though most likely was in 1987 version too. See
 Section 5.4 of the Ordination chapter by Ter Braak in that book.
 
 The idea is to find the direction (in the k-dimensional ordination
 space) that has maximal correlation with an external variable.


Hello,

The method was indeed in the first edition of ter Braak's book. However, the 
idea is much older. The vegan implementation was based on an unpublished report 
from the Bell Labs from 1970s (or earlier). In this Bell Labs memorandum the 
method was specifically suggested for NMDS. Vegan uses different algorithm, but 
the method is the same. The early history in vegan can be traced in  ORDNEWS 
correspondence from 2001 or so, but it is so old that I cannot find that 
message via this computer any longer.

Then about Bray-Curtis. The referee may be correct when writing that the fitted 
vectors are not directly related to Bray-Curtis. You fit the vectors to the 
NMDS ordination, and that is a non-linear mapping from Bray-Curtis to the 
metric ordination space.  There are two points here: non-linearity and stress. 
Because of these, it is not strictly about B-C. Of course, the referee is wrong 
when writing about NMDS axes: the fitted vector has nothing to do with axes 
(unless you rotate your axis parallel to the fitted vector which you can do). 
The NMDS is based on Bray-Curtis, but it is not the same, and the vector 
fitting is based on NMDS. So why not write that is about NMDS? Why to insist on 
Bray-Curtis which is only in the background?

Cheers, Jari Oksanen

-- 
Jari Oksanen, Dept Biology, Univ Oulu, 90014 Finland

__
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] bigmemory

[ya]

Hi all,

In addition, I am planning to do a multiple imputation with MICE package 
using the data read by bigmemory package.


So usually, the multiple imputation code is like this:

 imp=mice(data.frame,m=50,seed=1234,print=F)

the data.frame is required. How can I change the big.matrix class 
generated by bigmemory package to a data.frame?


Thank you very much.

ya

__
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] stop calculation in a function

[jeff6868]

Hi dear R-users,

I have a question about a function I'm trying to improve.
How can I stop the function calculation at the last numeric value of my
data?
The problem is that the end of my data contains missing values (NAs). And
the aim of my function is to compare the first numeric value with the next
one (till the end). For the moment, It works well when my data doesn't
contains any NAs at the end of my file. I think that the problem is, as I
have NAs at the end of my data, R tries to compare my last numeric value
with the next numeric value wich doesn't exists, and so tries to modify the
length of my data (the error message is that the output has not the same
length as the input).
Could somebody tell me what I should modify or add in my function in order
to fix this problem?
Here's the function. Thanks for your advises! 

out2NA - function(x,seuil){
st1 = NULL
# Temporal variable memorysing the last correct numeric value#
temp - st1[1] - x[1]
ind_temp - 1
# Max time gap between two comparisons #
ecart_temps - 10
tps - time(x)

for (i in 2:length(x)){
if((!is.na(x[i]))){
if((tps[i]-tps[ind_temp]  ecart_temps)  (abs(x[i]-temp)  seuil)){
#(abs(x[i+1]-x[i])1)){
st1[i] - NA
}
else {
temp - st1[i] - x[i]
ind_temp - i
}
}
}
return(st1)
}

dat1 - myts[,2]
myts[,2] - apply(dat1,2,function(x) out2NA(x,2))

--
View this message in context: 
http://r.789695.n4.nabble.com/stop-calculation-in-a-function-tp4622964.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.

Re: [R] Interweaving of two datasets

[lunarossa]

it doesn't work.
Find attached what I need explained in xls.
Thank you very very much!
http://r.789695.n4.nabble.com/file/n4622912/interweaving_of_2_datasets.xls
interweaving_of_2_datasets.xls 

--
View this message in context: 
http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4622912.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] Classification of Cluster-Correlated data

[Yohann R]

Dear R-Help,

I'm dealing with a supervized binary classification issue. My dataset is
composed of 1500 individuals, living in 600 households. I have
approximately 4000 variables to classify my subjects as
infected/uninfected.

I was wondering how would it be possible to account for the hierarchical
nature of my data in a data mining classification method, such as CART,
MARS or other methods, as it is done for instance in mixed-effects models ?
I suppose that the hierarchical structure of the data cannot be ignored,
because the risk of a individual to be infected is higher is there is
already an infected individual in his household.

Thank you

Yohann Mansiaux

[[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.

Re: [R] outlier identify in qqplot

[agent dunham]

Sorry, but I need the same and i don't understand your help. 

So, after fitting my lm model, how can i identify my data? I was trying the
following, but it doesn't work.

/identify(qnorm(c(0.25,
0.75)),quantile(rstandard(mymodel)[!is.na(rstandard(mymodel))], c(0.25,
0.75)),row.names(mydata))
warning: no point within 0.25 inches/

/In identify.default(qnorm(c(0.25, 0.75)),
quantile(rstandard(lmt2t)[!is.na(rstandard(lmt2t))],  :
  more 'labels'  than points/

Thanks in advance, show u...@host.com

--
View this message in context: 
http://r.789695.n4.nabble.com/outlier-identify-in-qqplot-tp4076587p4623088.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.

Re: [R] stop calculation in a function

[Sarah Goslee]

Well, if i goes from 2 to length(x) and you try to access x[i+1], of
course odd things will happen. Why not construct the loop to
(length(x)-1)  instead, so that x[i+1] is defined.

Sarah


On Thu, May 10, 2012 at 5:14 AM, jeff6868
geoffrey_kl...@etu.u-bourgogne.fr wrote:
 Hi dear R-users,

 I have a question about a function I'm trying to improve.
 How can I stop the function calculation at the last numeric value of my
 data?
 The problem is that the end of my data contains missing values (NAs). And
 the aim of my function is to compare the first numeric value with the next
 one (till the end). For the moment, It works well when my data doesn't
 contains any NAs at the end of my file. I think that the problem is, as I
 have NAs at the end of my data, R tries to compare my last numeric value
 with the next numeric value wich doesn't exists, and so tries to modify the
 length of my data (the error message is that the output has not the same
 length as the input).
 Could somebody tell me what I should modify or add in my function in order
 to fix this problem?
 Here's the function. Thanks for your advises!

 out2NA - function(x,seuil){
    st1 = NULL
    # Temporal variable memorysing the last correct numeric value#
    temp - st1[1] - x[1]
    ind_temp - 1
    # Max time gap between two comparisons #
    ecart_temps - 10
    tps - time(x)

    for (i in 2:length(x)){
    if((!is.na(x[i]))){
    if((tps[i]-tps[ind_temp]  ecart_temps)  (abs(x[i]-temp)  seuil)){
    #(abs(x[i+1]-x[i])1)){
    st1[i] - NA
    }
    else {
    temp - st1[i] - x[i]
    ind_temp - i
    }
    }
    }
    return(st1)
    }

    dat1 - myts[,2]
    myts[,2] - apply(dat1,2,function(x) out2NA(x,2))

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

__
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] outlier identify in qqplot

[R. Michael Weylandt]

Reproducible example? https://github.com/hadley/devtools/wiki/Reproducibility

Michael

On Thu, May 10, 2012 at 6:31 AM, agent dunham crossp...@hotmail.com wrote:
 Sorry, but I need the same and i don't understand your help.

 So, after fitting my lm model, how can i identify my data? I was trying the
 following, but it doesn't work.

 /identify(qnorm(c(0.25,
 0.75)),quantile(rstandard(mymodel)[!is.na(rstandard(mymodel))], c(0.25,
 0.75)),row.names(mydata))
 warning: no point within 0.25 inches/

 /In identify.default(qnorm(c(0.25, 0.75)),
 quantile(rstandard(lmt2t)[!is.na(rstandard(lmt2t))],  :
  more 'labels'  than points/

 Thanks in advance, show u...@host.com

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/outlier-identify-in-qqplot-tp4076587p4623088.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
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] Modifying R package code

[Brian Smith]

Hi,

I was trying to change some code in an existing package. I downloaded the
source package (say 'package_xx') from CRAN, and changed the R code
provided in the /package_xx/R/xx.R. I then saved the changes and did the R
CMD INSTALL -l /path to modified package/.

Do I need to do something else before the changes will be effective?

thanks!

[[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.

Re: [R] Axes value format

[Vihan Pandey]

On 08/05/2012, David Winsemius dwinsem...@comcast.net wrote:

 On May 8, 2012, at 2:23 PM, Vihan Pandey wrote:

 On 8 May 2012 19:47, John Kane jrkrid...@inbox.com wrote:
 Quite likely, but we need to know what you are doing and what
 graphics package you are using.

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

 Frightfully sorry about that. I'm using R on a Mac, and its a simple
 plot using plot() which is taking values from a CSV file, let me
 illustrate for one dataset :

 See if this is helpful. Untested,  since you still do not have a
 reproducible example for us to work with.

 x - c(12345, 1234567, 123)
 paste( round( x/c(1, 1000, 100)[findInterval(x,
 c(1,1000,100)) ], 2),
  c(,K,M)[findInterval(x, c(1,1000,100)) ]  )

 [1] 12.35 K 1.23 M 123 

Thanks, xaxt, yaxt in plot() did the job. Sorry for not posting a
reproducible example I just needed a quick fix which I was sure would
be a simple option in plot() or elsewhere.

Thanks and Cheers!

- vihan


 --
 David.


 ==
 #!/usr/bin/Rscript

 out_file = foobar.pdf
 pdf(out_file, height=8.5, width=11)

 my_values - read.csv(foo.csv,head=TRUE,sep=,)

 plot(my_values$num_sims,
  my_values$exec_time,
  xlab=Number of Simulations,
  ylab=Execution Time(in milliseconds),
  col=red,
  main=Execution Time for Simulations)

 lines(my_values$num_sims,my_values$exec_time,col=red)

 my_values2 - read.csv(bar.csv,head=TRUE,sep=,)

 lines(my_values2$num_sims,my_values2$exec_time,col=blue)
 points(my_values2$num_sims,my_values2$exec_time,col=blue)


 legend(topright,
  lty=c(1,1),
c(foo,bar),
col=c(red,blue)
);

 dev.off()

 print(paste(Plot was saved in:, getwd()))
 ==

 foo.csv and bar.csv have values like:

 num_sims,exec_time
 100,44556
 200,89112
 300,133668

 etc.

 Please let me know if you require any additional information.

 Cheers!

 - vihan


 John Kane
 Kingston ON Canada


 -Original Message-
 From: vihanpan...@gmail.com
 Sent: Tue, 8 May 2012 19:29:45 +0200
 To: r-help@r-project.org
 Subject: [R] Axes value format

 Hi all,

 I have some graphs where the values on the X and Y axes are by
 default
 in exponent form like 2e+05 or 1.0e+07. Is it possible to make
 them in
 a more readable form like 10M for 1.0e+07 or 200K for 2e+05?

 Thanks and Regards,

 - vihan

 __
 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.

 
 FREE ONLINE PHOTOSHARING - Share your photos online with your
 friends and family!
 Visit http://www.inbox.com/photosharing to find out more!



 __
 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.

 David Winsemius, MD
 West Hartford, CT



__
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] Problem with SQLDF - Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table:

[Shivam]

Thanks Gabor, Jim, POSIXct is working fine :)

Regards,
Shivam

On Thu, May 10, 2012 at 5:22 AM, Gabor Grothendieck ggrothendi...@gmail.com
 wrote:

 On Wed, May 9, 2012 at 7:24 PM, Shivam shivamsi...@gmail.com wrote:
  Ohh ... Thanks Gabor. I have a few related queries then, kindly have a
 look:
 
  1. Does it only hinder the sqldf package or are there other issues with
  using POSIXlt in a dataframe? Am asking because I have a few dataframes
 with
  columns of class(POSIXlt ).
 
  2. I have columns containing 'date+timestamp', something like '2011-01-03
  09:07:07' which are of class POSIXlt. I need to perform some arithmetic
  operations on these columns. Which class would be most appropriate for
 such
  kind of data?
 

 Its not just sqldf.  You will have other problems too if you put
 POSIXlt objects in data frames too.

 See R News 4/1.

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




-- 
*Victoria Concordia Crescit*

[[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] additional axis, different scale

[pannigh]

Dear list,
I am looking for a possibility to present results in a more graphical way by
adding an axis. But I have trouble relating my data to the added axis.
Imagine the following example:

a  - c(10, 20, 30, 40)
b  - c(50, 250, 500, 600)
ba - b/a

par(las=1, mar=c(5,5,.5,5))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
  axis(4, at=c(seq(0,600,length=6)), lab=c(seq(0,25,length=6)), col.axis=2 )
  lines(a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)

I want the red line to relate its values to the x-axis (a) and axis 4 (on
the right) and not as usual to the x-axis (a) and the y-axis (b).
This would show the tendency of the red line much clearer which now can't be
seen because of the very different scaling.
E.g. I want R to know that I am trying to plot the first point of the red
line P1(50/5) using the x-axis and the right axis, not the y-axis on the
left ect.

I would like to solve this without using a factor solution like:
  bb -600/25 * ba
  lines(a,bb, type=b, pch=21, cex=2, col=3, lwd=2, lty=1)


For any kind of help I would be grateful !

--
View this message in context: 
http://r.789695.n4.nabble.com/additional-axis-different-scale-tp4623210.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.

Re: [R] Axes value format

[Vihan Pandey]

On 08/05/2012, John Kane jrkrid...@inbox.com wrote:
 Actually I meant a working example and some data (See ?dput for a handy way
 to supply data)

 It is also a good idea to include the information from sessionInfo()


  I think David W has a good approach.

 Otherwise you might just want to write the axis yourself.

 =
 x  - c(100, 200, 300)
 y  - c( 44556, 89112, 133668)

 nms  - c(1M, 2M, 3M)

 plot(x,y, xaxt=n)
 axis(1, x, labels=nms)
 

Sorry for not posting a working example and some data, I just wanted a
quick fix and didn't know people would work so much for my query few
mailing lists do so :-) Thanks though, your xaxt option did the trick.

I will learn from this and read up on dput and sessioninfo()

Thanks and Cheers!

- vihan




 John Kane
 Kingston ON Canada


 -Original Message-
 From: vihanpan...@gmail.com
 Sent: Tue, 8 May 2012 20:23:21 +0200
 To: jrkrid...@inbox.com
 Subject: Re: [R] Axes value format

 On 8 May 2012 19:47, John Kane jrkrid...@inbox.com wrote:
 Quite likely, but we need to know what you are doing and what graphics
 package you are using.

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

 Frightfully sorry about that. I'm using R on a Mac, and its a simple
 plot using plot() which is taking values from a CSV file, let me
 illustrate for one dataset :

 ==
 #!/usr/bin/Rscript

 out_file = foobar.pdf
 pdf(out_file, height=8.5, width=11)

 my_values - read.csv(foo.csv,head=TRUE,sep=,)

 plot(my_values$num_sims,
  my_values$exec_time,
  xlab=Number of Simulations,
  ylab=Execution Time(in milliseconds),
  col=red,
  main=Execution Time for Simulations)

 lines(my_values$num_sims,my_values$exec_time,col=red)

 my_values2 - read.csv(bar.csv,head=TRUE,sep=,)

 lines(my_values2$num_sims,my_values2$exec_time,col=blue)
 points(my_values2$num_sims,my_values2$exec_time,col=blue)


 legend(topright,
  lty=c(1,1),
c(foo,bar),
col=c(red,blue)
);

 dev.off()

 print(paste(Plot was saved in:, getwd()))
 ==

 foo.csv and bar.csv have values like:

 num_sims,exec_time
 100,44556
 200,89112
 300,133668

 etc.

 Please let me know if you require any additional information.

 Cheers!

 - vihan


 John Kane
 Kingston ON Canada


 -Original Message-
 From: vihanpan...@gmail.com
 Sent: Tue, 8 May 2012 19:29:45 +0200
 To: r-help@r-project.org
 Subject: [R] Axes value format

 Hi all,

 I have some graphs where the values on the X and Y axes are by default
 in exponent form like 2e+05 or 1.0e+07. Is it possible to make them in
 a more readable form like 10M for 1.0e+07 or 200K for 2e+05?

 Thanks and Regards,

 - vihan

 __
 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.

 
 FREE ONLINE PHOTOSHARING - Share your photos online with your friends
 and family!
 Visit http://www.inbox.com/photosharing to find out more!



 
 GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at
 http://www.inbox.com/smileys
 Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google Talk™ and
 most webmails




__
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] Axes value format

[Vihan Pandey]

On 08/05/2012, arun smartpink...@yahoo.com wrote:
 Hi Vihan,

 The link below might be helpful.


 (http://stackoverflow.com/questions/3415097/controlling-number-formatting-at-axis-of-r-plots)

Thanks, the xaxt parameter in plot() solved the problem.

Thanks and Cheers!

- vihan


 A.K.







 - Original Message -
 From: Vihan Pandey vihanpan...@gmail.com
 To: r-help r-help@r-project.org
 Cc:
 Sent: Tuesday, May 8, 2012 1:29 PM
 Subject: [R] Axes value format

 Hi all,

 I have some graphs where the values on the X and Y axes are by default
 in exponent form like 2e+05 or 1.0e+07. Is it possible to make them in
 a more readable form like 10M for 1.0e+07 or 200K for 2e+05?

 Thanks and Regards,

 - vihan

 __
 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
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] stop calculation in a function

[Berend Hasselman]

On 10-05-2012, at 12:59, Sarah Goslee wrote:

 Well, if i goes from 2 to length(x) and you try to access x[i+1], of
 course odd things will happen. Why not construct the loop to
 (length(x)-1)  instead, so that x[i+1] is defined.
 

The reference to x[i+1] in a commented line so I don't think this  is the cause.

The reason for the behaviour can be seen when indenting the code for the for 
loop part properly:

for (i in 2:length(x)){
   if((!is.na(x[i]))){
   if((tps[i]-tps[ind_temp]  ecart_temps)  (abs(x[i]-temp)  
seuil)){
   #(abs(x[i+1]-x[i])1))
   st1[i] - NA
   }
   else {
   temp - st1[i] - x[i]
   ind_temp - i
   }
   }
}
return(st1)

When is.na(x[i]) == TRUE then the st1 entry for that value of i is skipped.
So when x has trailing NA's st1 will be shorter.

It's better to declare st1 initially of the correct length

st1 - numeric(length(x))

which could also speed things up when x is very long (because st1 has to be 
increased every time i increases).

Berend


 Sarah
 
 
 On Thu, May 10, 2012 at 5:14 AM, jeff6868
 geoffrey_kl...@etu.u-bourgogne.fr wrote:
 Hi dear R-users,
 
 I have a question about a function I'm trying to improve.
 How can I stop the function calculation at the last numeric value of my
 data?
 The problem is that the end of my data contains missing values (NAs). And
 the aim of my function is to compare the first numeric value with the next
 one (till the end). For the moment, It works well when my data doesn't
 contains any NAs at the end of my file. I think that the problem is, as I
 have NAs at the end of my data, R tries to compare my last numeric value
 with the next numeric value wich doesn't exists, and so tries to modify the
 length of my data (the error message is that the output has not the same
 length as the input).
 Could somebody tell me what I should modify or add in my function in order
 to fix this problem?
 Here's the function. Thanks for your advises!
 
 out2NA - function(x,seuil){
st1 = NULL
# Temporal variable memorysing the last correct numeric value#
temp - st1[1] - x[1]
ind_temp - 1
# Max time gap between two comparisons #
ecart_temps - 10
tps - time(x)
 
for (i in 2:length(x)){
if((!is.na(x[i]))){
if((tps[i]-tps[ind_temp]  ecart_temps)  (abs(x[i]-temp)  seuil)){
#(abs(x[i+1]-x[i])1)){
st1[i] - NA
}
else {
temp - st1[i] - x[i]
ind_temp - i
}
}
}
return(st1)
}
 
dat1 - myts[,2]
myts[,2] - apply(dat1,2,function(x) out2NA(x,2))
 
 --
 -- 
 Sarah Goslee
 http://www.functionaldiversity.org
 
 __
 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
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] stop calculation in a function

[jeff6868]

Thank you for your reply sarah.
Well actually I don't try to access x[i+1]. The line where you saw it starts
with #. It was just try I wanted to keep (sorry I should have removed it
before posting).

But I ask him to access to the next value if conditions in the loop are not
verified (restart the comparison from the next value). It works well as long
as I have numeric values in my data. But if my data ends with NAs, I have
this problem.
That's why I'm trying to ask him to stop the calculation in the loop at the
last numeric value to avoid this error (don't know if it's the best way to
solve it, but it's the main idea I think).
Have you got any other idea about this?
Thanks a lot!




--
View this message in context: 
http://r.789695.n4.nabble.com/stop-calculation-in-a-function-tp4622964p4623259.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.

Re: [R] additional axis, different scale

[Jim Lemon]

On 05/10/2012 09:37 PM, pannigh wrote:

Dear list,
I am looking for a possibility to present results in a more graphical way by
adding an axis. But I have trouble relating my data to the added axis.
Imagine the following example:

a- c(10, 20, 30, 40)
b- c(50, 250, 500, 600)
ba- b/a

par(las=1, mar=c(5,5,.5,5))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
   axis(4, at=c(seq(0,600,length=6)), lab=c(seq(0,25,length=6)), col.axis=2 )
   lines(a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)

I want the red line to relate its values to the x-axis (a) and axis 4 (on
the right) and not as usual to the x-axis (a) and the y-axis (b).
This would show the tendency of the red line much clearer which now can't be
seen because of the very different scaling.
E.g. I want R to know that I am trying to plot the first point of the red
line P1(50/5) using the x-axis and the right axis, not the y-axis on the
left ect.

I would like to solve this without using a factor solution like:
   bb-600/25 * ba
   lines(a,bb, type=b, pch=21, cex=2, col=3, lwd=2, lty=1)



Hi pannigh,
Have a look at twoord.plot (plotrix)
Jim

__
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] additional axis, different scale

[John Kane]

I don't think there is any other way.  On the other hand, most gurus suggest 
that a dual scales on a graph are not a good thing.

What about using a two panel graph?

Quick rejigging of your code :
=
a  - c(10, 20, 30, 40)
b  - c(50, 250, 500, 600)
ba - b/a

op  -  par(las=1, mar=c(5,5,.5,5), mfrow=c(2, 1))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
 plot (a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)
par(op)
==



John Kane
Kingston ON Canada


 -Original Message-
 From: pann...@gwdg.de
 Sent: Thu, 10 May 2012 04:37:37 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] additional axis, different scale
 
 Dear list,
 I am looking for a possibility to present results in a more graphical way
 by
 adding an axis. But I have trouble relating my data to the added axis.
 Imagine the following example:
 
 a  - c(10, 20, 30, 40)
 b  - c(50, 250, 500, 600)
 ba - b/a
 
 par(las=1, mar=c(5,5,.5,5))
 plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
 xlim=c(0,45))
   axis(4, at=c(seq(0,600,length=6)), lab=c(seq(0,25,length=6)),
 col.axis=2 )
   lines(a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)
 
 I want the red line to relate its values to the x-axis (a) and axis 4 (on
 the right) and not as usual to the x-axis (a) and the y-axis (b).
 This would show the tendency of the red line much clearer which now can't
 be
 seen because of the very different scaling.
 E.g. I want R to know that I am trying to plot the first point of the red
 line P1(50/5) using the x-axis and the right axis, not the y-axis on the
 left ect.
 
 I would like to solve this without using a factor solution like:
   bb -600/25 * ba
   lines(a,bb, type=b, pch=21, cex=2, col=3, lwd=2, lty=1)
 
 
 For any kind of help I would be grateful !
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/additional-axis-different-scale-tp4623210.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.


GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at 
http://www.inbox.com/smileys
Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google Talk™ and most 
webmails

__
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] Dotchart showing mean and median by group

[Gabor Grothendieck]

On Thu, May 10, 2012 at 2:24 AM, maxbre mbres...@arpa.veneto.it wrote:
 hi all

 I have another question reated to the dotchart: is it possible by means of
 par() to set a logaritmic scale?
 If yes, how ? and if not, any alternative solution?


1. This is getting increasingly complicated as new requirements are
added but anyways here it is.  As before, for the first dotchart call
we substitute in our own dotchart (which is the same as R's dotchart
except its environment is reset to p so that it picks up anything in p
prior to similarly named functions elsewhere in R).  This time we also
add our own plot.window to p overriding log=.  The line marked ## is
optional and suppresses writing the axis annotations a second time.
As before, this code depends on the internals of dotchart so its not
ideal and you might wish to turn to lattice or ggplot 2 but it does
give the desired effect while sticking to classic graphics.

library(proto)

p - proto(dotchart = dotchart,
plot.window = function(..., log) graphics::plot.window(..., log = x))
with(p, dotchart(VADeaths, gdata = mean.values))

par(new = TRUE)

p[[axis]] - p[[mtext]] - list ##
with(p, dotchart(VADeaths, gdata = median.values, gpch = 20))

2. A variation is to use dotchart2 in Hmisc.   It has a version of
dotchart that directly supports adding to the plot.  Omit the
suppressWarnings call below if you don't mind a few spurious warnings.

library(Hmisc)

groups - col(VADeaths, as.factor = TRUE)
labels - rownames(VADeaths)[row(VADeaths)]

suppressWarnings({
  dotchart2(VADeaths, labels = labels, groups = groups,
gdata = mean.values, log = x)

  dotchart2(VADeaths, labels = labels, groups = groups,
gdata = median.values, log = x, pch = 1, add = TRUE)
})


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

[R] plotting legend-grob with grid.layout

[Johannes Radinger]

Hi,

I am using ggplot2 and arrange differnet plots into
one viewport by dividing it into rows and columns:


pushViewport(viewport(layout = grid.layout(nrow=2,ncol=2,widths = unit(c(50, 
50), mm),heights = unit(c(50, 50), mm

with following function I can extract the legend of a previously defined
plot (ggplot2-plot) as a grob:

legend - function(plot){
tmp - ggplot_gtable(ggplot_build(plot))
leg - which(sapply(tmp$grobs, function(x) x$name) == guide-box)
legend - tmp$grobs[[leg]]
}

The plot themselves are sent to the viewport with:

vplayout - function(x, y)
viewport(layout.pos.row = x, layout.pos.col = y)
print(p1, vp = vplayout(1, 1))


But, how can I put the legend into the viewport (e.g. colum=1,row=2)...

I tried with draw.grob() which works, but then I can define the posiiton...
This grid-arraning-viewport stuff is very confusing to me, so maybe someone can 
help out here...

/johannes

-- 

Jetzt informieren: http://mobile.1und1.de/?ac=OM.PW.PW003K20328T7073a

__
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] additional axis, different scale

[David Winsemius]

On May 10, 2012, at 8:07 AM, John Kane wrote:


I don't think there is any other way.


There is:

a  - c(10, 20, 30, 40)
b  - c(50, 250, 500, 600)
ba - b/a

par(las=1, mar=c(5,5,.5,5))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
par(new=TRUE)
plot(a,ba,type=b, pch=21, cex=2, col=2, lwd=2, lty=1, xlim=c(0,45),  
yaxt=n)
 axis(4, at=c(seq(0,25,length=6)), lab=c(seq(0,25,length=6)),  
col.axis=2 )


# could also specify ylim of c(0,25) to the second plot call.

# probably want to have ylab= in one or both of those plot calls, too.

But I do agree this can be considered deceptive plotting practice.

--
David.

On the other hand, most gurus suggest that a dual scales on a graph  
are not a good thing.


What about using a two panel graph?

Quick rejigging of your code :
=
a  - c(10, 20, 30, 40)
b  - c(50, 250, 500, 600)
ba - b/a

op  -  par(las=1, mar=c(5,5,.5,5), mfrow=c(2, 1))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
plot (a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)
par(op)
==



John Kane
Kingston ON Canada



-Original Message-
From: pann...@gwdg.de
Sent: Thu, 10 May 2012 04:37:37 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] additional axis, different scale

Dear list,
I am looking for a possibility to present results in a more  
graphical way

by
adding an axis. But I have trouble relating my data to the added  
axis.

Imagine the following example:

a  - c(10, 20, 30, 40)
b  - c(50, 250, 500, 600)
ba - b/a

par(las=1, mar=c(5,5,.5,5))
plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
xlim=c(0,45))
 axis(4, at=c(seq(0,600,length=6)), lab=c(seq(0,25,length=6)),
col.axis=2 )
 lines(a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)

I want the red line to relate its values to the x-axis (a) and axis  
4 (on

the right) and not as usual to the x-axis (a) and the y-axis (b).
This would show the tendency of the red line much clearer which now  
can't

be
seen because of the very different scaling.
E.g. I want R to know that I am trying to plot the first point of  
the red
line P1(50/5) using the x-axis and the right axis, not the y-axis  
on the

left ect.

I would like to solve this without using a factor solution like:
 bb -600/25 * ba
 lines(a,bb, type=b, pch=21, cex=2, col=3, lwd=2, lty=1)


For any kind of help I would be grateful !

--
View this message in context:
http://r.789695.n4.nabble.com/additional-axis-different-scale-tp4623210.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.



GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at 
http://www.inbox.com/smileys
Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google  
Talk™ and most webmails


__
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.


David Winsemius, MD
West Hartford, CT

__
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] Modifying R package code

[Duncan Murdoch]

On 12-05-10 7:38 AM, Brian Smith wrote:

Hi,

I was trying to change some code in an existing package. I downloaded the
source package (say 'package_xx') from CRAN, and changed the R code
provided in the /package_xx/R/xx.R. I then saved the changes and did the R
CMD INSTALL -l /path to modified package/.


You don't want -l in that statement.  It says where to install it, not 
what to install.


Simply using

R CMD INSTALL /path to modified package/

should be sufficient, assuming you have write permission on the default 
library.  If you don't, then you do need to use -l, but it needs to 
specify some other location in which to install your package, and after 
you start R, you need to tell it (using .libPaths()) to look there 
before looking in the standard place.


One other small bit of advice:  record the fact that you made a change 
in the Version field in the DESCRIPTION file.  Later, when you are 
tracking down bugs, it will be useful to remind you that nobody else can 
duplicate your problems.


Duncan Murdoch



Do I need to do something else before the changes will be effective?

thanks!

[[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
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] stop calculation in a function

[Sarah Goslee]

On Thu, May 10, 2012 at 8:02 AM, jeff6868
geoffrey_kl...@etu.u-bourgogne.fr wrote:
 Thank you for your reply sarah.
 Well actually I don't try to access x[i+1]. The line where you saw it starts
 with #. It was just try I wanted to keep (sorry I should have removed it
 before posting).

Sorry, I should really have more coffee before replying to the list.

Berend makes a good point about st1. You don't provide sample data, so
I can't try it, but here's a modified version of your function that
incorporates his advice plus a calculation of the number of terminal
NA values to skip. I don't know whether x can have internal NA values
as well as terminal, so I wrote a more complex check. If NA values
only appear at the end, you can simply use sum(is.na(x)) instead.

You could also interrupt the loop, but it seems easier to me to
determine the right number of iterations beforehand since you already
have the information needed.


out2NA - function(x,seuil){
   st1 = NULL
   # Temporal variable memorysing the last correct numeric value#
   temp - st1[1] - x[1]
   ind_temp - 1
   # Max time gap between two comparisons #
   ecart_temps - 10
   tps - time(x)

st1 - numeric(length(x))
endNA - 0
if(is.na(x[length(x)])) {
endNA - rle(rev(is.na(x)))$lengths[1]
}

   for (i in 2:(length(x)-endNA)){
if((!is.na(x[i]))){
if((tps[i]-tps[ind_temp]  ecart_temps)  (abs(x[i]-temp)  
seuil)){
st1[i] - NA
}
else {
temp - st1[i] - x[i]
ind_temp - i
}
}
   }
   return(st1)
}

 But I ask him to access to the next value if conditions in the loop are not
 verified (restart the comparison from the next value). It works well as long
 as I have numeric values in my data. But if my data ends with NAs, I have
 this problem.
 That's why I'm trying to ask him to stop the calculation in the loop at the
 last numeric value to avoid this error (don't know if it's the best way to
 solve it, but it's the main idea I think).
 Have you got any other idea about this?
 Thanks a lot!



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

__
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] file path

[Upton, Stephen (Steve) (CIV)]

Why not just construct a valid file name and use that in cat? You can then use 
file.path to join paths together if you want to write to a specific location, 
as in your example.

steve

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Wincent
Sent: Wednesday, May 09, 2012 11:15 AM
To: Tal Galili
Cc: r help
Subject: Re: [R] file path

Hmm, I don't think it gives what I want.

For example, I assign a file name to f,
 f - a?b.txt
 file.path(e:,f)
[1] e:/a?b.txt

The resultant character is not accepted as a file name by Windows OS.

On 9 May 2012 20:32, Tal Galili tal.gal...@gmail.com wrote:
 Hi Wincent,
 Have a look at:
 ?file.path



 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845 Read me:
 www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)
 --
 




 On Wed, May 9, 2012 at 11:03 AM, Wincent ronggui.hu...@gmail.com wrote:

 Dear all, is there any function to assert whether a file path is
 legitimate, and to convert any potential file path to a legitimate
 file path?

 I automate a batch of files and write them to plain text files with
 cat(). The file argument of cat() is generated automatically which
 may contain characters such as ?  , unacceptable in Windows OS.
 What I do at this moment is to strip such characters off with gsub().
 Is there any direct way to make legitimate file path without detailed
 knowledge about the naming rule specific to a OS?

 Best

 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University PhD of City University of
 Hong Kong http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.





--
Wincent Ronggui HUANG
Sociology Department of Fudan University PhD of City University of Hong Kong 
http://homepage.fudan.edu.cn/rghuang/cv/

__
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
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] additional axis, different scale

[John Kane]

Oh,  yes of course.  Thanks.  

I really don't like dual plots. I ran into one a few days ago that looked to me 
to be very dubious but I don't have the raw data to replot it in another form.  
It's a side issue so I'm not going to bother about it but is annoyed me.



John Kane
Kingston ON Canada


 -Original Message-
 From: dwinsem...@comcast.net
 Sent: Thu, 10 May 2012 08:18:48 -0400
 To: jrkrid...@inbox.com
 Subject: Re: [R] additional axis, different scale
 
 
 On May 10, 2012, at 8:07 AM, John Kane wrote:
 
 I don't think there is any other way.
 
 There is:
 
 a  - c(10, 20, 30, 40)
 b  - c(50, 250, 500, 600)
 ba - b/a
 
 par(las=1, mar=c(5,5,.5,5))
 plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
 xlim=c(0,45))
 par(new=TRUE)
 plot(a,ba,type=b, pch=21, cex=2, col=2, lwd=2, lty=1, xlim=c(0,45),
 yaxt=n)
   axis(4, at=c(seq(0,25,length=6)), lab=c(seq(0,25,length=6)),
 col.axis=2 )
 
 # could also specify ylim of c(0,25) to the second plot call.
 
 # probably want to have ylab= in one or both of those plot calls, too.
 
 But I do agree this can be considered deceptive plotting practice.
 
 --
 David.
 
 On the other hand, most gurus suggest that a dual scales on a graph
 are not a good thing.
 
 What about using a two panel graph?
 
 Quick rejigging of your code :
 =
 a  - c(10, 20, 30, 40)
 b  - c(50, 250, 500, 600)
 ba - b/a
 
 op  -  par(las=1, mar=c(5,5,.5,5), mfrow=c(2, 1))
 plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
 xlim=c(0,45))
 plot (a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)
 par(op)
 ==
 
 
 
 John Kane
 Kingston ON Canada
 
 
 -Original Message-
 From: pann...@gwdg.de
 Sent: Thu, 10 May 2012 04:37:37 -0700 (PDT)
 To: r-help@r-project.org
 Subject: [R] additional axis, different scale
 
 Dear list,
 I am looking for a possibility to present results in a more
 graphical way
 by
 adding an axis. But I have trouble relating my data to the added
 axis.
 Imagine the following example:
 
 a  - c(10, 20, 30, 40)
 b  - c(50, 250, 500, 600)
 ba - b/a
 
 par(las=1, mar=c(5,5,.5,5))
 plot(a,b, type=b, pch=22, cex=2, col=4, lwd=2, ylim=c(0,650),
 xlim=c(0,45))
  axis(4, at=c(seq(0,600,length=6)), lab=c(seq(0,25,length=6)),
 col.axis=2 )
  lines(a,ba, type=b, pch=21, cex=2, col=2, lwd=2, lty=1)
 
 I want the red line to relate its values to the x-axis (a) and axis
 4 (on
 the right) and not as usual to the x-axis (a) and the y-axis (b).
 This would show the tendency of the red line much clearer which now
 can't
 be
 seen because of the very different scaling.
 E.g. I want R to know that I am trying to plot the first point of
 the red
 line P1(50/5) using the x-axis and the right axis, not the y-axis
 on the
 left ect.
 
 I would like to solve this without using a factor solution like:
  bb -600/25 * ba
  lines(a,bb, type=b, pch=21, cex=2, col=3, lwd=2, lty=1)
 
 
 For any kind of help I would be grateful !
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/additional-axis-different-scale-tp4623210.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.
 
 
 GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at
 http://www.inbox.com/smileys
 Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google
 Talk™ and most webmails
 
 __
 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.
 
 David Winsemius, MD
 West Hartford, CT


FREE ONLINE PHOTOSHARING - Share your photos online with your friends and 
family!
Visit http://www.inbox.com/photosharing to find out more!

__
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] file path

[Wincent]

As I said, the file name is derived automatically from text processing.
Thanks all the same.

On 10 May 2012 20:35, Upton, Stephen (Steve) (CIV) scup...@nps.edu wrote:
 Why not just construct a valid file name and use that in cat? You can then use
 file.path to join paths together if you want to write to a specific location,
 as in your example.

 steve

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
 Behalf Of Wincent
 Sent: Wednesday, May 09, 2012 11:15 AM
 To: Tal Galili
 Cc: r help
 Subject: Re: [R] file path

 Hmm, I don't think it gives what I want.

 For example, I assign a file name to f,
 f - a?b.txt
 file.path(e:,f)
 [1] e:/a?b.txt

 The resultant character is not accepted as a file name by Windows OS.

 On 9 May 2012 20:32, Tal Galili tal.gal...@gmail.com wrote:
 Hi Wincent,
 Have a look at:
 ?file.path



 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845 Read me:
 www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)
 --
 




 On Wed, May 9, 2012 at 11:03 AM, Wincent ronggui.hu...@gmail.com wrote:

 Dear all, is there any function to assert whether a file path is
 legitimate, and to convert any potential file path to a legitimate
 file path?

 I automate a batch of files and write them to plain text files with
 cat(). The file argument of cat() is generated automatically which
 may contain characters such as ?  , unacceptable in Windows OS.
 What I do at this moment is to strip such characters off with gsub().
 Is there any direct way to make legitimate file path without detailed
 knowledge about the naming rule specific to a OS?

 Best

 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University PhD of City University of
 Hong Kong http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.





 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University PhD of City University of Hong Kong
 http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.



-- 
Wincent Ronggui HUANG
Sociology Department of Fudan University
PhD of City University of Hong Kong
http://homepage.fudan.edu.cn/rghuang/cv/

__
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] stop calculation in a function

[jeff6868]

Thanks for your answer too Berend.
Yes you're right about x[i+1]. You answered juste before me.
Well your idea of declaring all in numeric is great. It avoids my problem.
But actually I also have small missing data gaps in the rest of my data (in
the middle of numeric values).
And one of the aim of my function is to avoid comparison between 2 numeric
values which are separated with a long period of time (with NA inside), in
order for example not to compare a value of the 1st january and the next
numeric value of the 1st april.
I'm trying to combine both. For the moment, it works only for data which
doesn't ends with NAs as you've understood. With numeric() for st1, the
problem of NAs at the end is solved but it creates a new problem with the
other NAs (which was OK before). Do you better understand what I'm trying to
do?
If you have an other idea, It'll be welcomed.
Thanks

--
View this message in context: 
http://r.789695.n4.nabble.com/stop-calculation-in-a-function-tp4622964p4623391.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.

Re: [R] stop calculation in a function

[Sarah Goslee]

You're moving the goal posts. If you need to deal with internal as
well as terminal NA values in a particular way, it would help to have
told us that up front.

Please give us some sample data that illustrates what you're really
trying to work with, and a full explanation of what you need to
accomplish.

Sarah

On Thu, May 10, 2012 at 8:41 AM, jeff6868
geoffrey_kl...@etu.u-bourgogne.fr wrote:
 Thanks for your answer too Berend.
 Yes you're right about x[i+1]. You answered juste before me.
 Well your idea of declaring all in numeric is great. It avoids my problem.
 But actually I also have small missing data gaps in the rest of my data (in
 the middle of numeric values).
 And one of the aim of my function is to avoid comparison between 2 numeric
 values which are separated with a long period of time (with NA inside), in
 order for example not to compare a value of the 1st january and the next
 numeric value of the 1st april.
 I'm trying to combine both. For the moment, it works only for data which
 doesn't ends with NAs as you've understood. With numeric() for st1, the
 problem of NAs at the end is solved but it creates a new problem with the
 other NAs (which was OK before). Do you better understand what I'm trying to
 do?
 If you have an other idea, It'll be welcomed.
 Thanks


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

__
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] file path

[jim holtman]

Has any mentioned

?make.names

On Thu, May 10, 2012 at 8:39 AM, Wincent ronggui.hu...@gmail.com wrote:
 As I said, the file name is derived automatically from text processing.
 Thanks all the same.

 On 10 May 2012 20:35, Upton, Stephen (Steve) (CIV) scup...@nps.edu wrote:
 Why not just construct a valid file name and use that in cat? You can then 
 use
 file.path to join paths together if you want to write to a specific location,
 as in your example.

 steve

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
 Behalf Of Wincent
 Sent: Wednesday, May 09, 2012 11:15 AM
 To: Tal Galili
 Cc: r help
 Subject: Re: [R] file path

 Hmm, I don't think it gives what I want.

 For example, I assign a file name to f,
 f - a?b.txt
 file.path(e:,f)
 [1] e:/a?b.txt

 The resultant character is not accepted as a file name by Windows OS.

 On 9 May 2012 20:32, Tal Galili tal.gal...@gmail.com wrote:
 Hi Wincent,
 Have a look at:
 ?file.path



 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845 Read me:
 www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)
 --
 




 On Wed, May 9, 2012 at 11:03 AM, Wincent ronggui.hu...@gmail.com wrote:

 Dear all, is there any function to assert whether a file path is
 legitimate, and to convert any potential file path to a legitimate
 file path?

 I automate a batch of files and write them to plain text files with
 cat(). The file argument of cat() is generated automatically which
 may contain characters such as ?  , unacceptable in Windows OS.
 What I do at this moment is to strip such characters off with gsub().
 Is there any direct way to make legitimate file path without detailed
 knowledge about the naming rule specific to a OS?

 Best

 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University PhD of City University of
 Hong Kong http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.





 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University PhD of City University of Hong Kong
 http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.



 --
 Wincent Ronggui HUANG
 Sociology Department of Fudan University
 PhD of City University of Hong Kong
 http://homepage.fudan.edu.cn/rghuang/cv/

 __
 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.



-- 
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.

__
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] Taking a lead in panel data

[Apoorva Gupta]

I have tried the diff and lag function.
I have given an example below. Diff gives me a$var2.shift.
I want a$var2.noshift, i.e. var^(t+2)-var^(t) should stay in the row with
year=t.
Diff shifts it to the t+2 row.

  a - data.frame(c(rep(2,5), rep(3,5)), c(2005:2009, 2004:2008),
c(NA,10,34,23,12, 23,45, NA, 45, NA))
 colnames(a) - c(firm,year,var)
 a - pdata.frame(a)
 a$var2.shift - diff(a$var, lag=2, difference=1)
 a$var2.noshift - c(NA, 13,-22,NA, NA,NA, 0, NA, NA, NA)
 a
   firm year var var2.shift var2.noshift
2-20052 2005  NA NA   NA
2-20062 2006  10 NA   13
2-20072 2007  34 NA  -22
2-20082 2008  23 13   NA
2-20092 2009  12-22   NA
3-20043 2004  23 NA   NA
3-20053 2005  45 NA0
3-20063 2006  NA NA   NA
3-20073 2007  45  0   NA
3-20083 2008  NA NA   NA

On Tue, May 8, 2012 at 6:21 PM, Liviu Andronic landronim...@gmail.comwrote:

 On Tue, May 8, 2012 at 12:14 PM, Apoorva Gupta apoorva.ni...@gmail.com
 wrote:
  I have checked that. It allows me to get the t-1, t-2 value but not the
 t+1
  value.
  Is there any other way of achieving this other than using the plm
 package?
 
 It would be easier to help if you provided a minimal reproducible
 example, as requested in the posting guide. Have you tried
 diff(x, lag = -1, ...)

 or
 lag(x, k = -1, ...)

 Perhaps this does what you want. Regards
 Liviu




-- 
Apoorva Gupta
Consultant
National Institute of Public Finance and Policy

[[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] Split the work for many cores

[Alaios]

Dear all,
I am using my code the vgram.matrix of packets fields. I have around 500 
matrices that I need to pass inside that function and then plot those results.

Even though my system has 16 cores is quite clear that I am only using one of 
those.
Would it be able to skip these 500 tasks to the 16 cores, with each processor 
having around 4 matrices to process?

What would you suggest me doing?

Regards
Alex

[[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.

Re: [R] additional axis, different scale

[pannigh]

Hi Jim, hi John,

thank you very much for your tips. The plotrix package solves the problem! 
However, thank you also for the advice that my way of plotting the data
might not be the best. I will think about it.

So once again thanx !

--
View this message in context: 
http://r.789695.n4.nabble.com/additional-axis-different-scale-tp4623210p4623313.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] envfit output (vegan package) - not sure what it means

[cernst]

I'm pretty new to R and would appreciate some help interpreting the output of
a function that was recommended to me.

I've used the *envfit *function in the vegan package to plot vectors of four
climate variables onto a species matrix ordination.  The output indicates
that only a single variable (mean_temp) is significant:

***VECTORS

NMDS1  NMDS2r2   Pr(r)   
precip0.41539 -0.90964  0.29480.112887   
wind  -0.61937  0.78510  0.0441 0.763237   
temp_mean  0.50571-0.86270  0.5839 0.004995 **
pressure-0.668910.74334  0.0688 0.622378   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
P values based on 1000 permutations.

I understand that the length of the plotted arrows is proportional to the
strength of the correlation between the variable and the ordination, but I'm
not sure how to interpret the direction of the arrows. I know that it is
the direction of the gradient, but *I don't really understand what
gradient that is referring to*.  So I'm not sure how to interpret what I'm
seeing in the plot, other than the significance of the relationship.

Here is the plot, if that helps. Each point is a sampling period; they are
numbered from week 1 - week 8. The different shapes represent two different
habitats. 

http://r.789695.n4.nabble.com/file/n4623315/species_abunance_with_environment.jpeg
 

--
View this message in context: 
http://r.789695.n4.nabble.com/envfit-output-vegan-package-not-sure-what-it-means-tp4623315.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.

Re: [R] registry vulnerabilities in R

[Zhou Fang]

What about using a Portable Apps style packaging of R? That might solve some
of the issues.

--
View this message in context: 
http://r.789695.n4.nabble.com/registry-vulnerabilities-in-R-tp4619217p4623388.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] Output result to a csv file

[eddie smith]

Hi guys,

I am applying linear regression model using lm, generate the residual
and then need to output the result to a csv file. Here is my script:
maxtempstation1.lm = lm( Year ~ station1, data=opencsv)
maxstation1.res = resid(maxstation1.lm); maxstation1.res
write.csv(maxtempstation1.res, file=maxtempstation1.csv)

I can do that without problem, but I have 300 stations!. The general
idea is to get all the residuals(for all years) into one csv file.
Here is a sample of my data.

Year   st1 st2 st3st4 st5st6 st7st8
1982 27.75 27.84 29.81 28.24 29.08 27.97 28.92 28.2
1983 28.74 28.71 29.32 29.34 29.48 28.99 29.45 29.43
1984 28.13 28.34 29.64 28.03 28.85 28.03 28.61 28.08

Anybody willing to help?

Thanks in advance guys.

Eddie

__
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] envfit output (vegan package) - not sure what it means

[Jari Oksanen]

cernst crystal.ernst at mail.mcgill.ca writes:

 
 I'm pretty new to R and would appreciate some help interpreting the output of
 a function that was recommended to me.
 
 I've used the *envfit *function in the vegan package to plot vectors of four
 climate variables onto a species matrix ordination.  The output indicates
 that only a single variable (mean_temp) is significant:
 
 ***VECTORS
 
 NMDS1  NMDS2r2   Pr(r)   
 precip0.41539 -0.90964  0.29480.112887   
 wind  -0.61937  0.78510  0.0441 0.763237   
 temp_mean  0.50571-0.86270  0.5839 0.004995 **
 pressure-0.668910.74334  0.0688 0.622378   
 ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
 P values based on 1000 permutations.
 
 I understand that the length of the plotted arrows is proportional to the
 strength of the correlation between the variable and the ordination, but I'm
 not sure how to interpret the direction of the arrows. I know that it is
 the direction of the gradient, but *I don't really understand what
 gradient that is referring to*.  So I'm not sure how to interpret what I'm
 seeing in the plot, other than the significance of the relationship.
 
Crystal, 

The arrows are labelled: the arrow with label temp_mean refers to the 
gradient of temp_mean. The arrow points to the direction to which the
linear change in temp_mean is the fastest.

To see what this means, you can add the corresponding linear trend surface,
or its isoclines to the graphs. These isoclines are regularly spaced
lines that are perpendicular to the arrow. So this is a linear trend
surface, and the arrow points the gradient down the plane.

I don't have your data, so I so how to do that in one of the vegan
data sets. Just change here your data sets, ordination results and
variable ('temp_mean' instead of 'A1' below):

require(vegan)
data(dune)
data(dune.env)
ord - rda(dune)
plot(ord, dis=sites)
ef - envfit(ord ~  A1, data=dune.env)
plot(ef)
ordisurf(ord ~ A1, data = dune.env, add = TRUE, knots = 1)

HTH, Jari Oksanen

PS. The density of vegan users is higher in r-sig-ecology than here: 
consider posting there for a more secured answer.

__
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] Random resampling of columns in species association matrices

[mariasve]

Hi David,

Thank you for your suggestions. I am quite the beginner at R and don’t
understand how to actually implement your suggestion and am hoping for some
further advice on that, if possible.

This is a subset of my data. Rows are host species, and columns parasite
species. Three of the parasites are generalists, but P4L is a strict
specialist on FORCOL (27 individuals have this parasite). 

  H17L  P25L  P41L  P4L
AUTINF39 0  0  0
GLYSPI16 2 15  0
FORCOL1  0 0   27
HYLPOE3  0 2   0
HYLNAE1  4 2   0
MYRMYO  25 2   0
THAARD0  8 0   0

This is a list of host trait values for each of the hosts:
  abundance weight  survival
AUTINF488   38  0.48
GLYSPI827   14.10.59
FORCOL156   44.30.55
HYLPOE322   17.50.54
HYLNAE309   14.50.73
MYRMYO  475 20.80.59
THAARD429   18.40.67

And this is an estimate of host specificity of the parasites, incorporating
prevalence and phylogeny:

Specificity
H17L2.08
P25L1.72
P41L2.19
P4L 0

I want to determine whether specificity of the parasites relates to any of
the host traits. For this, I would like to do a multiple regression. To
avoid psedureplication, I want to include a host species only once in the
matrix. So, for H17L, I could pick either of the hosts (except THAARD),
etc., but once a host is picked for one parasite, it cannot be picked for
another. For example, if I pick GLYSPI for H17L, GLYSPI has to be removed as
a choice for P25L and P41L. Thus, I also have to randomize which parasite
has its host picked first. In all cases, I want to lock FORCOL and P4L, so
FORCOL will not be an option for H17L anymore. This last part I’m still
uncertain about, I might just randomly pick hosts for all parasites and then
risk losing the strict host species specialists from some matrices. 

If I make 2 random selections I might end up with:
Random1 Random2 
H17LAUTINF  GLYSPI  
P25LGLYSPI  HYLNAE
P41LHYLPOE  MYRMYO   
P4L FORCOL  FORCOL  

For the first random table I would then do a multiple regression on the
dependent specificity variable and independent host trait values:
Specificity abundance   weight  survival
2.08   488  38  0.48
1.72   827  14.10.59
2.19   322  17.50.54
0 156   44.30.55

If I generate 1000 randomly selected host-parasite combinations, I would
have 1000 such tables, on which I would have to run 1000 independent
regressions. Since I’m using model selection and multimodel inference to
estimate parameter values, I will end up doing the model selection 1000
times. 

Your second suggestion makes most sense to me, but I don’t understand how to
implement it. Would you (or someone else) please give me some advise on
that? Also, once I have the 1000 random host-parasite matrices, how do I
link these to the tables of actual values (host traits and parasite
specificity)?

Thanks so much!
Maria


--
View this message in context: 
http://r.789695.n4.nabble.com/Random-resampling-of-columns-in-species-association-matrices-tp4620618p4623563.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.

Re: [R] stop calculation in a function

[jeff6868]

I tried your proposition Sarah (I was answering to Berend when you posted
your answer).
Well it seems to work! 
I just had to add afterwords a line to have my NAs again.
I converted values = 0 by NA (numeric() in the function did the contrary for
the calculation):

mydata[mydata==0] - NA 

At first it was working for such kind of data: NAs just in the middle of my
data
test -
data.frame(c(1,2,3,4,NA,NA,7,8,9,10),c(11,12,NA,14,15,16,17,NA,19,20))
colnames(test)- c(data1,data2)

but not for data with NAs at the beginning, in the middle and at the end:

test2 -
data.frame(c(NA,2,3,4,NA,NA,NA,NA,NA,NA),c(NA,12,13,NA,15,16,17,NA,NA,NA))
colnames(test2)- c(data3,data4)

But thanks to your proposition, it seems to work in both cases now!
Thanks a lot sarah!

--
View this message in context: 
http://r.789695.n4.nabble.com/stop-calculation-in-a-function-tp4622964p4623584.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] Fwd: Re: need help in R

[Leah Mathibela]

 i need help in my coding
 the first line and the second line works and the ggplot works
 when i have a code like this and i want to fetch a map and saves it to 
 a directory how can i put it
  because it gives me an error message
 Error in load(GGemz) : bad 'file' argument

 GGemz - ggooglemap(center=center,zoom=13,destfile=Emz.png)
 save(GGemz,file=GGemz.Rda)
 load(GGemz)
 and when i put the GGemz in brackets it gives me an error
 load(file=GGemz)
 Error in readChar(con, 5L, useBytes = TRUE) : cannot open the connection
 In addition:Warning message:
 In readChar(con, 5L, useBytes = TRUE) :

 i dont have a problem here
 ggplot()+
   geom_tile(aes(x = lon, y = lat, fill = fill), data = GGemz)+
   geom_jitter(aes(x=lon,y=lat, color=fieldworker),data=emz)+
   scale_fill_identity()+
   scale_x_continuous('Longitude', limits = lon_range) +
   scale_y_continuous('Latitude', limits = lat_range) +
   scale_colour_discrete('') +
   opts(title = 'Invitation map of Emzinini, 2012') +
   coord_equal()

 this code brings an error says
 Error in dev.copy2pdf(device = x11, file = paste(mapdir, EmzinoniGG.pdf,  
 :
no device to print from
 i am not sure what does the error mean
 dev.copy2pdf(device=x11,file=paste(mapdir,EmzinoniGG.pdf,sep=))
 dev.off()
 -- 
 Leah Mathibela
 mobile:078-3311-491
 work:087-7545-997
 www.nova.org.za

David Winsemius, MD
West Hartford, CT

No virus found in this message.
Checked by AVG - www.avg.com http://www.avg.com



-- 
Leah Mathibela
mobile:078-3311-491
work:087-7545-997
www.nova.org.za


[[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.

Re: [R] Interweaving of two datasets

[Rui Barradas]

Hello,

What doesn't work, exactly?
I can only see two things:

1. The order of the columns is different, first data.frame in merge
instruction comes first.
Solution: reverse the order of data.frames in merge.
2. The order of the rows is different, the merge function orders it's output
by the common col(s).
Solutions: leave as is or use 'order'.

# first is the first sheet in attached .xls file, second is the second and
result is the third.
# 1.
res2 - merge(second, first)

# 2.
result[order(result$name, result$request), ]
res2[order(res2$name, res2$request), ]

If this isn't it, please state what the problem is,  doesn't work is a bit
vague.

Hope this helps,

Rui Barradas



lunarossa wrote
 
 it doesn't work.
 Find attached what I need explained in xls.
 Thank you very very much!
 
 http://r.789695.n4.nabble.com/file/n4622912/interweaving_of_2_datasets.xls
 interweaving_of_2_datasets.xls 
 


--
View this message in context: 
http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4623624.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.

Re: [R] outlier identify in qqplot

[agent dunham]

Find the data attached, 

http://r.789695.n4.nabble.com/file/n4623493/mydata.txt mydata.txt 

The model would be /lmmodel - lm(log(vdep) ~ v1 + sqrt(v2) + v3 +v5 + v6 +
v7 + v8 + v9 + v10, data = mydata)/

Thanks again, 


u...@host.com

--
View this message in context: 
http://r.789695.n4.nabble.com/outlier-identify-in-qqplot-tp4076587p4623493.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] Using valgrind to debug R, extracting column of a mts object causes valgrind to crash

[Jouni Helske]

Dear all,

I'm trying to debug my R package with valgrind, but I cannot get past the
point where I load make the data, as valgrind crashes when trying to
extract a single time series object of a multivariate mts object. I'm using
R 2.15.0 with platform x86_64-redhat-linux-gnu (64-bit).

The valgrind.R contains code

data(Seatbelts)
y-Seatbelts[,VanKilled]

And I run valgrind with R -d valgrind --tool=memcheck --leak-check=yes
--no-save  valgrind.R

Results look like this:

 data(Seatbelts)
 y-Seatbelts[,VanKilled]
vex amd64-IR: unhandled instruction bytes: 0x66 0xF 0x3A 0xB 0xC0 0xC
==31160== valgrind: Unrecognised instruction at address 0x399fe26990.
==31160== Your program just tried to execute an instruction that Valgrind
==31160== did not recognise.  There are two possible reasons for this.
==31160== 1. Your program has a bug and erroneously jumped to a non-code
==31160==location.  If you are running Memcheck and you just saw a
==31160==warning about a bad jump, it's probably your program's fault.
==31160== 2. The instruction is legitimate but Valgrind doesn't handle it,
==31160==i.e. it's Valgrind's fault.  If you think this is the case or
==31160==you are not sure, please let us know and we'll try to fix it.
==31160== Either way, Valgrind will now raise a SIGILL signal which will
==31160== probably kill your program.

 *** caught illegal operation ***
address 0x399fe26990, cause 'illegal opcode'

Traceback:
 1: ts(y, start = start(x), frequency = frequency(x))
 2: `[.ts`(Seatbelts, , VanKilled)
 3: Seatbelts[, VanKilled]
aborting ...
==31160==
==31160== HEAP SUMMARY:
==31160== in use at exit: 29,782,377 bytes in 12,590 blocks
==31160==   total heap usage: 30,275 allocs, 17,685 frees, 57,159,124
bytes allocated
==31160==
==31160== LEAK SUMMARY:
==31160==definitely lost: 0 bytes in 0 blocks
==31160==indirectly lost: 0 bytes in 0 blocks
==31160==  possibly lost: 0 bytes in 0 blocks
==31160==still reachable: 29,782,377 bytes in 12,590 blocks
==31160== suppressed: 0 bytes in 0 blocks
==31160== Reachable blocks (those to which a pointer was found) are not shown.
==31160== To see them, rerun with: --leak-check=full --show-reachable=yes
==31160==
==31160== For counts of detected and suppressed errors, rerun with: -v
==31160== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 2 from 2)
Illegal instruction (core dumped)

It seems that the crash is caused by the fact that the rows of the
Seatbelts are not defined in assignment, but if I use command
y-Seatbelts[1:192,VanKilled] the resulting object is not a time series
object anymore. Is there some better way to do this?

best regards,

Jouni

[[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] Problem with factors in NMDS and Envfit

[rodrock]

Hi everybody, 
thanks in advance for your help!

I am working with genetic data performing an NMDS to visualize the genetic
differentiation of individuals from different species and populations in
different site conditions (site variables are all factors), and actually
genetic data also.
The data base is kind of big ( 1000 individuals x 551 variables)

Nmds so far has not worked with the genetic data input as factors it shows
me and error
 (maybe the distance is not the proper?)
as integer the NMDS works well

ord2-metaMDS(gen_all_nonas,k = 2, distance=euclid, autotransform=F,
noshare=T, na.rm=T)

When I use envfit to see the correlation of the site factors it does not
work until I remove all Nas. Once I do that it works almost good (I can
live with the Nas removing) but somehow It show me an error considering the
factor: Populations where the individuals come from (Pop)

exp_site-envfit(ord2~Pop+Sp+Location+Soil+Water+Flooding+Ca, strata=Sp,
na.rm=T)# with strata
Error en `colnames-`(`*tmp*`, value = Pop) : 
  attempt to set colnames on object with less than two dimensions

I do not know what I am doing wrong, there are more than 40 Populations in
the data set but I get this problem. Once I remove Pop from the command It
works ok e.g., 
attach(site_all_nonas)

exp_site2-envfit(ord2~Sp+Location+Soil+Water+Flooding+Ca, strata=Sp,
na.rm=T)# with strata Sp
exp_site2

Goodness of fit:
 r2 Pr(r)
Sp   0.5398  0.001 ***
Location 0.4174  0.001 ***
Soil 0.0090  0.511
Water0.0103  0.873
Flooding 0.0234  0.515
Ca   0.0334  0.901
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
P values based on 999 permutations, stratified within strata.



-
Rodrigo Vargas G.
-
Silviculture Institute
Freiburg University

--
View this message in context: 
http://r.789695.n4.nabble.com/Problem-with-factors-in-NMDS-and-Envfit-tp4623645.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.

Re: [R] Interweaving of two datasets

[lunarossa]

Hi Ruri, sorry for my vagueness. The problem is not the order.

I tried to merge the datasets as you wrote. The result is that when I input:

table(phone)

The output is: 
0

And this happens for all the variables.

--
View this message in context: 
http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4623672.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.

Re: [R] Automating R for Hypothesis Testing

[meredith]

Rui-
  Thanks this definitely helps, just one quick question. How would you code
the values of chi-fm and chi-fms to change based on the degrees of freedom
of each model H(i)?

Meredith


Rui Barradas wrote
 
 Hello,
 
 Yes, it does help. Now we can see your data and what you're doing.
 What follows is a suggestion on what you could do, not full solution.
 (You forgot to say what X1 is, but I don't think it's important to
 understand the suggestion.)
 (If I'm wrong, say something.)
 
 
 milwaukeephos - read.csv(milwaukeephos.csv, header=TRUE,
 stringsAsFactors=FALSE)
 # list of data.frames, one per month
 ls1 - split(milwaukeephos, milwaukeephos$month)
 
 #- if you want to keep the models, not needed if you don't.
 #  (yoy probably don't)
 modelH - vector(list, 12)
 modelHa - vector(list, 12)
 modelH2 - vector(list, 12)
 modelH2a - vector(list, 12)
 #- values to record, these are needed, create them beforehand.
 chi_fm - numeric(12)
 chi_fms - numeric(12)
 #
 seq_months - c(1:12, 1) # wrap months around.
 for(i in 1:12){
   month_this - seq_months[i]
   month_next - seq_months[i + 1]
 
   lload - c(ls1[[month_this]]$load_kg, ls1[[month_next]]$load_kg)
   lflow - c(ls1[[month_this]]$flow, ls1[[month_next]]$flow)
   modelH[[i]] - lm(lload ~ lflow)
   # If you don't want to keep the models, use modelH only
   # ( without [[i]] )
   # and do the same with X1
 
   # rest of your code for first test goes here
   chi_fm[i] - bfm %*% var_fm %*% (bunres_fm - bres_fm)
 
   # and the same for the second test
   chi_fms[i] - ...etc...
 }
 
 
 Hope this helps,
 
 Rui Barradas
 
 
 meredith wrote
 
 dput:  http://r.789695.n4.nabble.com/file/n4620188/milwaukeephos.csv
 milwaukeephos.csv 
 
 # Feb-march
 modelH_febmarch-lm(llfeb_march~lffeb_march)
modelHa_febmarch-lm(llfeb_march~X1feb_mar+lffeb_march)
 anova(modelHa_febmarch)
 coefficients(modelH_febmarch)
 (Intercept) lffeb_march 
   -2.4298901.172821 
 coefficients(modelHa_febmarch)
 (Intercept)   X1feb_mar lffeb_march 
  -2.8957776  -0.5272793   1.3016303 
 bres_fm-matrix(c(-2.429890,0,1.172821),nrow=3)
 bunres_fm-matrix(c(-2.8957776,-0.5272793,1.3016303),nrow=3)
bfm-t(bunres_fm-bres_fm)
 fmvect-seq(1,1,length=34)
 X1a_febmar-seq(0,0,length=9) # dummy variable step 1
 X1b_febmar-seq(1,1,length=25) # dummy variable step 2
 X1feb_mar-c(X1a_febmar,X1b_febmar) #dummy variable creation
 # Test Stat Equation for Chisq
 fmxx-cbind(fmvect,X1feb_mar,lffeb_march)
 tfmx-t(fmxx)
 xcom_fm-(tfmx %*% fmxx)
 xinv_fm-ginv(xcom_fm)
 var_fm-xinv_fm*0.307
 chi_fm-bfm %*% var_fm %*% (bunres_fm-bres_fm)
 chi_fm # chisq value for recording
 if less than CV move onto to slope modification
 modelH2_febmarch-lm(llfeb_march~X3feb_march)
 modelH2a_febmarch-lm(llfeb_march~X3feb_march+X4feb_march)
 anova(modelH2a_febmarch)
 coefficients(modelH2_febmarch) # get coefficients to make beta vectors
 for test
 (Intercept) X3feb_march 
5.3421301.172821 
 coefficients(modelH2a_febmarch)
 (Intercept) X3feb_march X4feb_march 
   5.2936263   1.0353752   0.2407557 
 # Test Stat
 bsres_fm-matrix(c(5.342130,1.172821,0),nrow=3)
 bsunres_fm-matrix(c(5.2936263,1.0353752,0.2407557),nrow=3)
 bsfm-t(bsunres_fm-bsres_fm)
 #X matrix
 fmxs-cbind(fmvect,X3feb_march,X4feb_march)
 tfmxs-t(fmxs)
 xcoms_fm-(tfmxs %*% fmxs)
 xinvs_fm-ginv(xcoms_fm)
 var_fms-xinvs_fm*0.341
 chi_fms-bsfm %*% var_fms %*% (bsunres_fm-bsres_fm)
 chi_fms
 # Record Chisq value
 
 Does this help?
 Here lffeb_march is the combination of Feb and March log flows
 and llfeb_march is the combination of Feb and March log loads
 X3: lffeb_march-mean(feb_march)
 X4: X1*X3
 
 Thanks
 
 Rui Barradas wrote
 
 Hello,
 
 I'm not at all sure if I understand your problem. Does this describe it?
 
 
 test first model for months 1 and 2
 if test statistic less than critical value{
 test second model for months 1 and 2
 print results of the first and second tests? just one of them?
 }
 move on to months 2 and 3
 etc, until months 12 and 1
 
 
 Please post example data using dput(dataset).
 Just copy it's output and paste it in your post.
 
 And example code, what you're already doing.
 (Possibly simplified)
 
 Rui Barradas
 
 
 meredith wrote
 
 R Users-
   I have been trying to automate a manual code that I have developed
 for calling in a .csv file, isolating certain rows and columns that
 correspond to specified months:
 something to the effect
 i=name.csv
 N=length(i$month)
 iphos1=0
 iphos2=0
 isphos3=0
 for i=1,N
  if month=1
 iphos1=iphos+1
 iphos1(iphos1)=i
 
 an so on to call out the months into there own arrays (unless there is
 a way I can wrap it into the next automation)
 
 Next: I would like to run a simple linear regression combining each of
 the months 1 by 1:
 for instance I want to run a regression on a combined model from months
 1 and 2 and a dummy model for 1 and 2, compare them using a Chi-sq
 distribution, if Chi-sq is less than the Critical 

Re: [R] Dotchart showing mean and median by group

[maxbre]

thank you  all for the high level contributions and the very helpful
feedback;

I think I have now enogh material to study for months: what a good lesson
learned!

cheers

max 

--
View this message in context: 
http://r.789695.n4.nabble.com/Dotchart-showing-mean-and-median-by-group-tp4619597p4623526.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] png support in R on linux

[Unger, Kristian, Dr.]

Hi there,

I am sure this question has been posted a couple times before but I could not 
find a good solution on the web.

Is there any way to get png() support in R on SUSE linux? We want to run 
GenePattern on that server and some workflows use R scripts plotting into pngs.

Best wishes

Kristian


R version 2.14.0 (2011-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)

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

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

other attached packages:
[1] png_0.1-4

loaded via a namespace (and not attached):
[1] tools_2.14.0

Linux version 2.6.32.12-0.7-default (geeko@buildhost) (gcc version 4.3.4 
[gcc-4_3-branch revision 152973] (SUSE Linux) ) #1 SMP 2010-05-20 11:14:20 +0200


Arbeitsgruppenleiter Integrative Biologie / Head of Integrative Biology Group
Abteilung für Strahlenzytogenetik / Research Unit of Radiation Cytogenetics

Tel.: +49-89-3187-3515



Helmholtz Zentrum München
Deutsches Forschungszentrum für Gesundheit und Umwelt (GmbH)
Ingolstädter Landstr. 1
85764 Neuherberg
www.helmholtz-muenchen.de
Aufsichtsratsvorsitzende: MinDir´in Bärbel Brumme-Bothe
Geschäftsführer: Prof. Dr. Günther Wess und Dr. Nikolaus Blum
Registergericht: Amtsgericht München HRB 6466
USt-IdNr: DE 129521671

[[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.

Re: [R] Interweaving of two datasets

[John Kane]

I don't quite follow what you are doing.  Here is what Rui was suggesting using 
your sample data
===
dat1 - structure(list(name = structure(c(4L, 3L, 5L, 2L, 1L, 6L), .Label = 
c(Andy, 
Bruce, Ella, John, Luna, Morgana), class = factor), 
address = structure(c(2L, 4L, 1L, 3L, 2L, 4L), .Label = c(Kiev street, 
london road, main avenue, Rome street), class = factor), 
phone = c(2123L, 2342L, 454L, 56775L, 287678L, 39876L)), .Names = c(name, 
address, phone), class = data.frame, row.names = c(NA, 
-6L))

dat2  -  structure(list(name = structure(c(4L, 4L, 3L, 3L, 3L, 5L, 5L, 
2L, 2L, 2L, 2L, 1L, 6L, 6L, 6L), .Label = c(Andy, Bruce, 
Ella, John, Luna, Morgana), class = factor), request = 
structure(c(1L, 
5L, 4L, 1L, 3L, 4L, 5L, 5L, 5L, 4L, 4L, 2L, 1L, 4L, 2L), .Label = c(book, 
cigarettes, drink, food, paper), class = factor)), .Names = c(name, 
request), class = data.frame, row.names = c(NA, -15L))

mydata  -  merge(dat1, dat2)

mydata
===
#If you need information about a variable in the merge try something like:

table(mydata$phone)

Is this of any help?

John Kane
Kingston ON Canada


 -Original Message-
 From: gloriaal...@yahoo.it
 Sent: Thu, 10 May 2012 07:33:45 -0700 (PDT)
 To: r-help@r-project.org
 Subject: Re: [R] Interweaving of two datasets
 
 Hi Ruri, sorry for my vagueness. The problem is not the order.
 
 I tried to merge the datasets as you wrote. The result is that when I
 input:
 
 table(phone)
 
 The output is:
 0
 
 And this happens for all the variables.
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4623672.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.


GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at 
http://www.inbox.com/smileys
Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google Talk™ and most 
webmails

__
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] converting raster image

[Mintewab Bezabih]

Dear R users, 

I was wondering how I can convert a raster image (that made R through 
interpolation)  into an ascii or csv format?

this is the last line of my command

p - interpolate(r, tpsfit)

So p is my raster file which I want to convert into ascii or csv

Many thanks
Regards
Mintewab
__
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] How to capture NA values

[Alaios]

Dear all,
I would like to capture the NA values as produced from my code
the str over the return values return

int 61
 int 65
 int 69
 int 73
 int 101
 int NA
 int NA

it looks like that I am getting returned some integer values that are NA and I 
want to have an if statement for catching those

which is the right operator for checking these int NA values in R?

Alex

[[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.

Re: [R] How to capture NA values

[Berend Hasselman]

On 10-05-2012, at 17:16, Alaios wrote:

 Dear all,
 I would like to capture the NA values as produced from my code
 the str over the return values return
 
 int 61
  int 65
  int 69
  int 73
  int 101
  int NA
  int NA
 
 it looks like that I am getting returned some integer values that are NA and 
 I want to have an if statement for catching those
 
 which is the right operator for checking these int NA values in R?


How about simply is.na?

Berend

__
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] disagreement in loglikelihood and deviace in GLM with weights leads to different models selected using step()

[Marian K Talbert]

In species distribution modeling where one uses a large sample of 
background points to capture background variation in 
presence\pseudo-absence or use\available models (0\1 response) it is 
frequently recommended that one weight the data so the sum of the absence 
weights is equal to the sum of presence weights so that the model isn?t 
swamped by an overwhelming and arbitrary number of background points.   

I?m trying to do this in R in the standard glm in the stats package and am 
a bit confused.  I understand the issue with noninteger weights in glm and 
specifying reasonable starting values to ensure convergence to something 
meaningful and for any individual model fit I can get mostly the same 
results.  For example I have a dataset with 75 presence and 75 absence if 
instead I had randomly sampled 3150 background points I could set the 
weights to 1 for presence and 75/3150 for the background points.  I 
simulate this example by repeating my 75 absences 42 times and setting the 
weights for the absence to 75/3150 and setting mustart equal to the 
predicted response from the 75/75 model and fitting glms to each data set. 
 The output agrees in the coefficient estimates null and residual deviance 
but NOT in the AIC or the value returned from logLik() it appears the AIC 
is calculated from the likelihood though as I understand it these should 
be directly related to deviance so I?m not sure why these don?t all agree. 



 
My ultimate goal is to fit these models and run automatic model selection 
using the step function.  This doesn?t seem to work for two reasons: 
1.it uses the logLik value of likelihood and I feel it should be 
using .5*resid deviance.  I'm not quite sure why these are different 
2.it doesn?t use the correct penalty because the degrees of 
freedom are 3225 rather than something reasonable (I adjust this by 
multiplying by what I assume is a reasonable penalty so if I were using 
75/75 I'd use the penalty k=2 but here I use 2*sum(weight)/length(weight) 

For an individual model fit for example we have 

For No Repeated observations: 

Call:  glm(formula = as.formula(paste(response, ~, 
paste(out$dat$used.covs, 
collapse = +))), family = binomial, data = test.data) 

Coefficients: 
  (Intercept) asp_2k_alb  beetle_yr  bio_1 bio_10  
  bio_11   
   -3.057e+01 -3.045e-03  4.474e-01 -1.577e-01 -9.599e-01  
   2.326e+00   
   bio_12 bio_13 bio_14 bio_15 bio_16  
  bio_17   
2.205e-02 NA  1.461e-01  1.227e-01 -4.705e-02  
  -8.138e-02   
   bio_18 bio_19  bio_4  bio_5  bio_6  
   bio_7   
   -9.938e-03  9.128e-03  2.713e-03 -5.724e+04  5.725e+04  
   5.725e+04   
bio_8  bio_9  damage_yr dem_2k_alb  dist_line_all  
dist_pts   
   -4.155e-01 -2.874e-01 -6.129e-01  4.910e-03 -5.575e-05  
   3.965e-06   
 dist_trail_3  dist_transgov  dist_trans_01  dist_util4568 dist_water  
eastness   
   -1.510e-04  1.158e-05 NA  8.621e-06  2.392e-04  
  -4.421e-01   
northness slp_2k_albbeetle_all1damage_all1   
3.066e-01  2.975e-02 -6.609e+00  1.894e+00   

Degrees of Freedom: 149 Total (i.e. Null);  118 Residual 
Null Deviance:  207.9 
Residual Deviance: 112.5AIC: 176.5 

 logLik(no.rep) 
'log Lik.' -56.23042 (df=32) 


For repeated observations (of course in a real example these wouldn't be 
repeated observations but oversampled background points): 

Call:  glm(formula = as.formula(paste(response, ~, 
paste(out$dat$used.covs, 
collapse = +))), family = binomial, data = psdAbsDat, 
weights =train$weight) 

Coefficients: 
  (Intercept) asp_2k_alb  beetle_yr  bio_1 bio_10  
  bio_11   
   -3.057e+01 -3.045e-03  4.474e-01 -1.577e-01 -9.599e-01  
   2.326e+00   
   bio_12 bio_13 bio_14 bio_15 bio_16  
  bio_17   
2.205e-02 NA  1.461e-01  1.227e-01 -4.705e-02  
  -8.138e-02   
   bio_18 bio_19  bio_4  bio_5  bio_6  
   bio_7   
   -9.938e-03  9.128e-03  2.713e-03 -5.724e+04  5.725e+04  
   5.725e+04   
bio_8  bio_9  damage_yr dem_2k_alb  dist_line_all  
dist_pts   
   -4.155e-01 -2.874e-01 -6.129e-01  4.910e-03 -5.575e-05  
   3.965e-06   
 dist_trail_3  dist_transgov  dist_trans_01  dist_util4568 dist_water  
eastness   
   -1.510e-04  1.158e-05 NA  8.621e-06  2.392e-04  
  -4.421e-01   
northness slp_2k_albbeetle_all1damage_all1   
3.066e-01  2.975e-02 -6.609e+00  1.894e+00   

Degrees of Freedom: 3224 Total (i.e. Null);  3193 Residual 
Null Deviance:  207.9 

Residual Deviance: 112.5AIC: 117 

logLik(weight) 
'log 

Re: [R] Automating R for Hypothesis Testing

[Rui Barradas]

Hello,

I'm glad it helped.

As for your second question, I don't know, but I'm not very comfortable with
the way you're doing things.
Why subtract the coefficients of model 1 from model 2?
And why the dummy? Why set model 1 to zero?

Isn't it better to use anova's F? After all, it's designed for it, for the
linear model...
And if you really want/need the dummy, wouldn't a nested anova do it? (F
statistic, once again.)

anova(model1, model2)

is simple and statistically speaking seems to me much better. (I specially
don't like the subtraction bit.)

Rui Barradas

meredith wrote
 
 Rui-
   Thanks this definitely helps, just one quick question. How would you
 code the values of chi-fm and chi-fms to change based on the degrees of
 freedom of each model H(i)?
 
 Meredith
 
 
 Rui Barradas wrote
 
 Hello,
 
 Yes, it does help. Now we can see your data and what you're doing.
 What follows is a suggestion on what you could do, not full solution.
 (You forgot to say what X1 is, but I don't think it's important to
 understand the suggestion.)
 (If I'm wrong, say something.)
 
 
 milwaukeephos - read.csv(milwaukeephos.csv, header=TRUE,
 stringsAsFactors=FALSE)
 # list of data.frames, one per month
 ls1 - split(milwaukeephos, milwaukeephos$month)
 
 #- if you want to keep the models, not needed if you don't.
 #  (yoy probably don't)
 modelH - vector(list, 12)
 modelHa - vector(list, 12)
 modelH2 - vector(list, 12)
 modelH2a - vector(list, 12)
 #- values to record, these are needed, create them beforehand.
 chi_fm - numeric(12)
 chi_fms - numeric(12)
 #
 seq_months - c(1:12, 1) # wrap months around.
 for(i in 1:12){
  month_this - seq_months[i]
  month_next - seq_months[i + 1]
 
  lload - c(ls1[[month_this]]$load_kg, ls1[[month_next]]$load_kg)
  lflow - c(ls1[[month_this]]$flow, ls1[[month_next]]$flow)
  modelH[[i]] - lm(lload ~ lflow)
  # If you don't want to keep the models, use modelH only
  # ( without [[i]] )
  # and do the same with X1
 
  # rest of your code for first test goes here
  chi_fm[i] - bfm %*% var_fm %*% (bunres_fm - bres_fm)
 
  # and the same for the second test
  chi_fms[i] - ...etc...
 }
 
 
 Hope this helps,
 
 Rui Barradas
 
 
 meredith wrote
 
 dput:  http://r.789695.n4.nabble.com/file/n4620188/milwaukeephos.csv
 milwaukeephos.csv 
 
 # Feb-march
 modelH_febmarch-lm(llfeb_march~lffeb_march)
modelHa_febmarch-lm(llfeb_march~X1feb_mar+lffeb_march)
 anova(modelHa_febmarch)
 coefficients(modelH_febmarch)
 (Intercept) lffeb_march 
   -2.4298901.172821 
 coefficients(modelHa_febmarch)
 (Intercept)   X1feb_mar lffeb_march 
  -2.8957776  -0.5272793   1.3016303 
 bres_fm-matrix(c(-2.429890,0,1.172821),nrow=3)
 bunres_fm-matrix(c(-2.8957776,-0.5272793,1.3016303),nrow=3)
bfm-t(bunres_fm-bres_fm)
 fmvect-seq(1,1,length=34)
 X1a_febmar-seq(0,0,length=9) # dummy variable step 1
 X1b_febmar-seq(1,1,length=25) # dummy variable step 2
 X1feb_mar-c(X1a_febmar,X1b_febmar) #dummy variable creation
 # Test Stat Equation for Chisq
 fmxx-cbind(fmvect,X1feb_mar,lffeb_march)
 tfmx-t(fmxx)
 xcom_fm-(tfmx %*% fmxx)
 xinv_fm-ginv(xcom_fm)
 var_fm-xinv_fm*0.307
 chi_fm-bfm %*% var_fm %*% (bunres_fm-bres_fm)
 chi_fm # chisq value for recording
 if less than CV move onto to slope modification
 modelH2_febmarch-lm(llfeb_march~X3feb_march)
 modelH2a_febmarch-lm(llfeb_march~X3feb_march+X4feb_march)
 anova(modelH2a_febmarch)
 coefficients(modelH2_febmarch) # get coefficients to make beta vectors
 for test
 (Intercept) X3feb_march 
5.3421301.172821 
 coefficients(modelH2a_febmarch)
 (Intercept) X3feb_march X4feb_march 
   5.2936263   1.0353752   0.2407557 
 # Test Stat
 bsres_fm-matrix(c(5.342130,1.172821,0),nrow=3)
 bsunres_fm-matrix(c(5.2936263,1.0353752,0.2407557),nrow=3)
 bsfm-t(bsunres_fm-bsres_fm)
 #X matrix
 fmxs-cbind(fmvect,X3feb_march,X4feb_march)
 tfmxs-t(fmxs)
 xcoms_fm-(tfmxs %*% fmxs)
 xinvs_fm-ginv(xcoms_fm)
 var_fms-xinvs_fm*0.341
 chi_fms-bsfm %*% var_fms %*% (bsunres_fm-bsres_fm)
 chi_fms
 # Record Chisq value
 
 Does this help?
 Here lffeb_march is the combination of Feb and March log flows
 and llfeb_march is the combination of Feb and March log loads
 X3: lffeb_march-mean(feb_march)
 X4: X1*X3
 
 Thanks
 
 Rui Barradas wrote
 
 Hello,
 
 I'm not at all sure if I understand your problem. Does this describe
 it?
 
 
 test first model for months 1 and 2
 if test statistic less than critical value{
test second model for months 1 and 2
print results of the first and second tests? just one of them?
 }
 move on to months 2 and 3
 etc, until months 12 and 1
 
 
 Please post example data using dput(dataset).
 Just copy it's output and paste it in your post.
 
 And example code, what you're already doing.
 (Possibly simplified)
 
 Rui Barradas
 
 
 meredith wrote
 
 R Users-
   I have been trying to automate a manual code that I have developed
 for calling in a .csv file, isolating certain rows and columns that
 correspond to 

Re: [R] Output result to a csv file

[Rui Barradas]

Hello,

Try

station - colnames(opencsv)[-1]
mat - matrix(0, nrow=length(station), ncol=nrow(opencsv))
dimnames(mat) - list(station, opencsv$Year)
for(st in station){
model - lm(Year~opencsv[, st], data=opencsv)
mat[st, ] - residuals(model)
}
write.csv(mat, opencsv.csv)


Hope this helps,

Rui Barradas


--
View this message in context: 
http://r.789695.n4.nabble.com/Output-result-to-a-csv-file-tp4623510p4623714.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.

Re: [R] Interweaving of two datasets

[Rui Barradas]

With the sample data and the merge as above,

table(res2$phone)

  454  2123  2342  9876 56775 87678 
  2 2 3 3   4  1 
with(res2, table(name, request))
 request
name  book cigarettes drink food paper
  Andy   0  1 00 0
  Bruce  0  0 02 2
  Ella   1  0 11 0
  John 1  0 00 1
  Luna0  0 01 1
  Morgana1  1 01 0

Rui Barradas


lunarossa wrote
 
 Hi Rui, sorry for my vagueness. The problem is not the order.
 
 I tried to merge the datasets as you wrote. The result is that when I
 input:
 
 table(phone)
 
 The output is: 
 0
 
 And this happens for all the variables.
 


--
View this message in context: 
http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4623724.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.

Re: [R] How to capture NA values

[David Winsemius]

On May 10, 2012, at 11:16 AM, Alaios wrote:


Dear all,
I would like to capture the NA values as produced from my code
the str over the return values return

int 61
 int 65
 int 69
 int 73
 int 101
 int NA
 int NA

it looks like that I am getting returned some integer values that  
are NA and I want to have an if statement for catching those


which is the right operator for checking these int NA values in R?


If you have a vector of mode 'integer' (or numeric) you can get their  
index locations with:


which(is.na(vec))

Logical tests for any NA's would be:

!length(which(is.na(vec)))

or

sum(is.na(vec))==0

or perhaps the most self-documenting method:

any(is.na(vec))

--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

__
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] averaging two tables (rows with columns)

[Kristi Glover]

Hi R user,I am struggling to figure out on how I can calculate the average from 
the two tables in R. Any one can help me? really your help  would be grateful- 
I am spending so much time to figure it out. It should not be so hard, I think. 
I have very big data but I have created a hypothetical data for simplification. 
for example 
I have : table 1




table 1: species occurance data
















speciesX


speciesY


speciesZ


speciesXX




Plot1


1


0


1


0




Plot2


0


1


1


0




Plot3


0


0


0


1




Plot4


1


0


1


0




Table 2



table 2. species tolerance data 












EnviA


EnviB


EnviC




speciesX


0.21


0.4


0.17




speciesY


0.1


0.15


0.18




speciesXX


0.14


0.16


0.19



You may noticed that table 2 does not have species Z which was in table 1.


Now I want to get the average value of species tolerance in each plot based on 
each environmental value (EnviA or EnviB etc)The example of the out come (final 
table I was looking for it) Results table 1a: average species tolerance in each 
plot based on EnviA




Result Table 3. Average species tolerance in each plot based on EnviA








speciesX


speciesY


speciesZ


speciesXX


Average




Plot1


0.21


NA


Nodata


0.14


0.175




Plot2


NA


0.1


Nodata


NA


0.1




Plot3


NA


NA


Nodata


0.14


0.14




Plot4


0.21


NA


Nodata


NA


0.21




Result table 1b: average species tolerance in plot based on EnviB



Table 4. Average species tolerance in each plot based on EnviB








speciesX


speciesY


speciesZ


speciesXX


Average




Plot1


0.4


NA


Nodata


0.16


0.28




Plot2


NA


0.15


Nodata


NA


0.15




Plot3


NA


NA


Nodata


0.16


0.16




Plot4


0.4


NA


Nodata


NA


0.4




Would any one help me how I can calculate these?Thanks
Kristi Golver==
  
[[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.

Re: [R] averaging two tables (rows with columns)

[John Kane]

Kristi,
Your data has come through in a totally unreadable layout.

Have a look at ?dput as a handy way to present sample data.  Just run 
dput(mydata), copy the output and paste into the email.  

John Kane
Kingston ON Canada


 -Original Message-
 From: kristi.glo...@hotmail.com
 Sent: Thu, 10 May 2012 12:50:43 -0300
 To: r-help@r-project.org
 Subject: [R] averaging two tables (rows with columns)
 
 
 Hi R user,I am struggling to figure out on how I can calculate the
 average from the two tables in R. Any one can help me? really your help
 would be grateful- I am spending so much time to figure it out. It should
 not be so hard, I think.
 I have very big data but I have created a hypothetical data for
 simplification.
 for example
 I have : table 1
 
 
 
 
 table 1: species occurance data
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 
 
 Plot1
 
 
 1
 
 
 0
 
 
 1
 
 
 0
 
 
 
 
 Plot2
 
 
 0
 
 
 1
 
 
 1
 
 
 0
 
 
 
 
 Plot3
 
 
 0
 
 
 0
 
 
 0
 
 
 1
 
 
 
 
 Plot4
 
 
 1
 
 
 0
 
 
 1
 
 
 0
 
 
 
 
 Table 2
 
 
 
 table 2. species tolerance data
 
 
 
 
 
 
 
 
 
 
 
 
 EnviA
 
 
 EnviB
 
 
 EnviC
 
 
 
 
 speciesX
 
 
 0.21
 
 
 0.4
 
 
 0.17
 
 
 
 
 speciesY
 
 
 0.1
 
 
 0.15
 
 
 0.18
 
 
 
 
 speciesXX
 
 
 0.14
 
 
 0.16
 
 
 0.19
 
 
 
 You may noticed that table 2 does not have species Z which was in table
 1.
 
 
 Now I want to get the average value of species tolerance in each plot
 based on each environmental value (EnviA or EnviB etc)The example of the
 out come (final table I was looking for it) Results table 1a: average
 species tolerance in each plot based on EnviA
 
 
 
 
 Result Table 3. Average species tolerance in each plot based on EnviA
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 Average
 
 
 
 
 Plot1
 
 
 0.21
 
 
 NA
 
 
 Nodata
 
 
 0.14
 
 
 0.175
 
 
 
 
 Plot2
 
 
 NA
 
 
 0.1
 
 
 Nodata
 
 
 NA
 
 
 0.1
 
 
 
 
 Plot3
 
 
 NA
 
 
 NA
 
 
 Nodata
 
 
 0.14
 
 
 0.14
 
 
 
 
 Plot4
 
 
 0.21
 
 
 NA
 
 
 Nodata
 
 
 NA
 
 
 0.21
 
 
 
 
 Result table 1b: average species tolerance in plot based on EnviB
 
 
 
 Table 4. Average species tolerance in each plot based on EnviB
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 Average
 
 
 
 
 Plot1
 
 
 0.4
 
 
 NA
 
 
 Nodata
 
 
 0.16
 
 
 0.28
 
 
 
 
 Plot2
 
 
 NA
 
 
 0.15
 
 
 Nodata
 
 
 NA
 
 
 0.15
 
 
 
 
 Plot3
 
 
 NA
 
 
 NA
 
 
 Nodata
 
 
 0.16
 
 
 0.16
 
 
 
 
 Plot4
 
 
 0.4
 
 
 NA
 
 
 Nodata
 
 
 NA
 
 
 0.4
 
 
 
 
 Would any one help me how I can calculate these?Thanks
 Kristi Golver==
 
   [[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.


FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop!

__
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] PLS Q2 value?

[Charles Determan Jr]

Greetings R users,

My interest in the Q2cum score comes my endeavor to replicate SIMCAP
PLS-DA analysis in R.  I use the exact same dataset.  After doing the
analysis in R, I can get the exact same R2Ycum.  However, the Q2cum is
significantly off.  Adding the Q2cum of the 1st and 2nd component
comes close but that seems unconvincing and I don't understand why the
3rd component Q2cum would be so far off.  Below is the code and
output:

pls.reg.new=plsreg2(newserumvars.sc, newtimematrix, nc=3)
pls.reg.new$expvar  #this provides the exact R2Y I get in SIMCAP
pls.reg.new$Q2cum

The Q2cum values should be close to
component 1 = .2149
component 2 = .4578
component 3 = .6627 #value of concern

All of the R output values are close for the respective class except
for the final Q2cum value:

   Q2cum.B Q2cum.FR8 Q2cum.S45  Q2cum
t1    -0.0546        0.3992        0.2973   0.2140
t2 0.2863        0.3824        0.7039   0.4573
t3     0.5839    0.7430        0.6980   0.3735  #not close
Done.


I have read that the plspm package used LOO cross validation.  SIMCAP
uses the same method, however it leaves '7' out instead.  Perhaps this
might be an issue that results in this difference?

Any insight or assistance would be most appreciated.  I certainly
appreciate any time you take to answer my question.

Regards,

Charles Determan

__
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] averaging two tables (rows with columns)

[Petr PIKAL]

Hi

as already mentioned your data can not be deciphered. Use

dput(table1) for sending usable data.

From what you describe probably

?aggregate can be used.
But without suitable data you hardly get any advice.

Regards
Petr

 
 
 Hi R user,I am struggling to figure out on how I can calculate the 
average
 from the two tables in R. Any one can help me? really your help  would 
be 
 grateful- I am spending so much time to figure it out. It should not be 
so
 hard, I think. 
 I have very big data but I have created a hypothetical data for 
simplification. 
 for example 
 I have : table 1
 
 
 
 
 table 1: species occurance data
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 
 
 Plot1
 
 
 1
 
 
 0
 
 
 1
 
 
 0
 
 
 
 
 Plot2
 
 
 0
 
 
 1
 
 
 1
 
 
 0
 
 
 
 
 Plot3
 
 
 0
 
 
 0
 
 
 0
 
 
 1
 
 
 
 
 Plot4
 
 
 1
 
 
 0
 
 
 1
 
 
 0
 
 
 
 
 Table 2
 
 
 
 table 2. species tolerance data 
 
 
 
 
 
 
 
 
 
 
 
 
 EnviA
 
 
 EnviB
 
 
 EnviC
 
 
 
 
 speciesX
 
 
 0.21
 
 
 0.4
 
 
 0.17
 
 
 
 
 speciesY
 
 
 0.1
 
 
 0.15
 
 
 0.18
 
 
 
 
 speciesXX
 
 
 0.14
 
 
 0.16
 
 
 0.19
 
 
 
 You may noticed that table 2 does not have species Z which was in table 
1.
 
 
 Now I want to get the average value of species tolerance in each plot 
 based on each environmental value (EnviA or EnviB etc)The example of the 

 out come (final table I was looking for it) Results table 1a: average 
 species tolerance in each plot based on EnviA
 
 
 
 
 Result Table 3. Average species tolerance in each plot based on EnviA
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 Average
 
 
 
 
 Plot1
 
 
 0.21
 
 
 NA
 
 
 Nodata
 
 
 0.14
 
 
 0.175
 
 
 
 
 Plot2
 
 
 NA
 
 
 0.1
 
 
 Nodata
 
 
 NA
 
 
 0.1
 
 
 
 
 Plot3
 
 
 NA
 
 
 NA
 
 
 Nodata
 
 
 0.14
 
 
 0.14
 
 
 
 
 Plot4
 
 
 0.21
 
 
 NA
 
 
 Nodata
 
 
 NA
 
 
 0.21
 
 
 
 
 Result table 1b: average species tolerance in plot based on EnviB
 
 
 
 Table 4. Average species tolerance in each plot based on EnviB
 
 
 
 
 
 
 
 
 speciesX
 
 
 speciesY
 
 
 speciesZ
 
 
 speciesXX
 
 
 Average
 
 
 
 
 Plot1
 
 
 0.4
 
 
 NA
 
 
 Nodata
 
 
 0.16
 
 
 0.28
 
 
 
 
 Plot2
 
 
 NA
 
 
 0.15
 
 
 Nodata
 
 
 NA
 
 
 0.15
 
 
 
 
 Plot3
 
 
 NA
 
 
 NA
 
 
 Nodata
 
 
 0.16
 
 
 0.16
 
 
 
 
 Plot4
 
 
 0.4
 
 
 NA
 
 
 Nodata
 
 
 NA
 
 
 0.4
 
 
 
 
 Would any one help me how I can calculate these?Thanks
 Kristi Golver==
 
[[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
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] Help with stemDocument

[Alekseiy Beloshitskiy]

Hi Triss, 

If you need to stem just one text in the Corupus use a[[n]]-stemDocument

Best,
-Alex

From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] on behalf of 
Triss.Ashton [triss.ash...@unt.edu]
Sent: 02 May 2012 21:09
To: r-help@r-project.org
Subject: Re: [R] Help with stemDocument

I am having a problem with stemDocuments also.  I can make it work by moving
the data into a Corpus by using:

  a - Corpus(VectorSource(df$text)) # create corpus object
  a - tm_map(a, stemDocument, language = english)

but it is horrably slow.  I want to stem outside the Corpus object like:

df$text - stemDocument(df$text, language = english)

but it returns the original text.

In fact, using the example in the tm package documentation does not work
either:

 data(crude)
 crude[[1]]
Diamond Shamrock Corp said that
effective today it had cut its contract prices for crude oil by
1.50 dlrs a barrel.
The reduction brings its posted price for West Texas
Intermediate to 16.00 dlrs a barrel, the copany said.
The price reduction today was made in the light of falling
oil product prices and a weak crude oil market, a company
spokeswoman said.
Diamond is the latest in a line of U.S. oil companies that
have cut its contract, or posted, prices over the last two days
citing weak oil markets.
 Reuter
 stemDocument(crude[[1]], language = english) # specify language
Diamond Shamrock Corp said that
effective today it had cut its contract prices for crude oil by
1.50 dlrs a barrel.
The reduction brings its posted price for West Texas
Intermediate to 16.00 dlrs a barrel, the copany said.
The price reduction today was made in the light of falling
oil product prices and a weak crude oil market, a company
spokeswoman said.
Diamond is the latest in a line of U.S. oil companies that
have cut its contract, or posted, prices over the last two days
citing weak oil markets.
 Reuter
 stemDocument(crude[[1]]) # language not specified
Diamond Shamrock Corp said that
effective today it had cut its contract prices for crude oil by
1.50 dlrs a barrel.
The reduction brings its posted price for West Texas
Intermediate to 16.00 dlrs a barrel, the copany said.
The price reduction today was made in the light of falling
oil product prices and a weak crude oil market, a company
spokeswoman said.
Diamond is the latest in a line of U.S. oil companies that
have cut its contract, or posted, prices over the last two days
citing weak oil markets.
 Reuter



--
View this message in context: 
http://r.789695.n4.nabble.com/Help-with-stemDocument-tp4554523p4604022.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
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] Matrix heatmap

[Petr PIKAL]

Hi

what is wrong with

 heatmap(as.matrix(test), col=my.colors(25))

with test from your dput

Regards
Petr

 The heat map generated the correct result:
 
 library(gplots)
 arq -read.table(l)
 matrix_l -data.matrix(arq)
 my.colors -
 colorRampPalette(c
 
(gray0,gray10,gray20,gray30,gray40,gray50,gray60,gray80,gray90,gray100))
 heatmap.2(matrix_l,dendrogram=none, Rowv=NA, Colv=NA, 
col=my.colors(256)) 
 
 --
 
 Now I have the following file with 5 data, similar to the above:
 
RF2   RF00013   RF00100   RF00381   RF00434   RF00453   RF00165 
 RF00496   RF00497
 RF00014   RF00048   RF00234   RF00163   RF8   RF00094   RF00032 
 RF00028   RF00216
 RF00487   RF00209   RF00465   RF00485   RF00363   RF00366
 RF2   63   7   5   7   17   12   14   5   23   3   56   14   72   84 
 
 15   64   20   0   1   8   6   65   3   4
 RF00013   45   7   4   6   17   12   14   5   23   3   56   12   60   84 
 
 15   64   20   0   0   2   2   65   3   4
 RF00100   22   1   5   3   2   9   0   0   0   0   5   0   16   8   1 0 
 0   0   0   0   0   26   2   3
 RF00381   63   7   5   13   17   11   3   5   18   3   56   14   33   12 
 
 2   15   4   18   12   25   11   69   3   4
 RF00434   2   0   0   3   17   11   14   5   23   3   55   12   59   84  

 15   64   20   0   0   0   0   40   1   3
 RF00453   3   1   0   2   16   12   13   3   7   0   45   12   42   78 
 15   53   20   0   0   0   0   33   2   0
 RF00165   0   0   0   2   10   1   14   1   7   0   44   12   38   68 13
 48   20   0   0   0   0   18   0   0
 RF00496   0   0   0   0   0   0   1   5   6   0   0   0   4   2   0   0  

 0   0   0   0   0   0   0   0
 RF00497   0   0   0   3   10   0   12   5   23   3   40   8   37   77 15
 64   20   0   0   0   0   20   0   0
 RF00014   0   0   0   0   0   0   0   0   8   3   6   0   0   0   0   0  

 0   0   0   0   0   0   0   0
 RF00048   3   1   0   3   17   10   14   5   23   3   56   12   59   83  

 15   64   20   0   0   0   0   52   3   3
 RF00234   62   7   5   6   17   12   14   5   23   3   56   14   70   84 
 
 15   64   20   0   0   0   1   65   3   4
 RF00163   63   7   5   7   17   12   14   5   23   3   56   14   75   84 
 
 15   64   21   6   1   10   9   65   3   4
 RF8   3   1   0   3   17   12   14   5   23   3   56   12   58   84  

 15   64   20   0   0   0   0   52   3   2
 RF00094   0   0   0   0   0   1   11   0   1   0   0   0   34   73   15  

 49   20   0   0   0   0   12   0   0
 RF00032   0   0   0   3   10   1   14   5   23   3   56   12   43   80 
 15   64   20   0   0   0   0   21   0   0
 RF00028   63   7   5   13   17   12   14   5   23   3   56   14   75 84 
 15   64   30   23   14   25   20   85   3   4
 RF00216   63   7   5   13   17   12   14   5   23   3   56   14   75 84 
 15   64   28   23   14   25   20   85   3   4
 RF00487   63   7   5   13   17   12   14   5   23   3   56   14   75 84 
 15   64   28   20   14   25   16   83   3   4
 RF00209   50   7   5   3   2   2   0   0   0   0   1   2   26   4   0 0 
 1   0   8   25   5   28   3   3
 RF00465   59   7   5   10   7   11   0   0   10   3   11   2   32   9 1 
 3   6   15   5   14   20   63   3   4
 RF00485   63   7   5   13   17   12   14   5   23   3   56   14   75 84 
 15   64   26   17   14   25   19   85   3   4
 RF00363   5   3   0   3   10   1   1   5   20   3   50   12   44   24 5 
 5   0   0   0   0   0   42   3   3
 RF00366   8   2   1   4   14   9   13   5   23   3   52   12   51   68 
 12   8   0   0   0   0   0   48   3   4
 
 Now I have the following file with 5 data, similar to the above:
 Is represented by an array of 25x25 and 10x10 not like the previous
 
 when I give the command dput (arch) it returns me the following output:
 
 structure(list(RF2 = c(63L, 45L, 22L, 63L, 2L, 3L, 0L, 0L, 
 0L, 0L, 3L, 62L, 63L, 3L, 0L, 0L, 63L, 63L, 63L, 50L, 59L, 63L, 
 5L, 8L), RF00013 = c(7L, 7L, 1L, 7L, 0L, 1L, 0L, 0L, 0L, 0L, 
 1L, 7L, 7L, 1L, 0L, 0L, 7L, 7L, 7L, 7L, 7L, 7L, 3L, 2L), RF00100 = c(5L, 

 4L, 5L, 5L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 5L, 5L, 0L, 0L, 0L, 5L, 
 5L, 5L, 5L, 5L, 5L, 0L, 1L), RF00381 = c(7L, 6L, 3L, 13L, 3L, 
 2L, 2L, 0L, 3L, 0L, 3L, 6L, 7L, 3L, 0L, 3L, 13L, 13L, 13L, 3L, 
 10L, 13L, 3L, 4L), RF00434 = c(17L, 17L, 2L, 17L, 17L, 16L, 10L, 
 0L, 10L, 0L, 17L, 17L, 17L, 17L, 0L, 10L, 17L, 17L, 17L, 2L, 
 7L, 17L, 10L, 14L), RF00453 = c(12L, 12L, 9L, 11L, 11L, 12L, 
 1L, 0L, 0L, 0L, 10L, 12L, 12L, 12L, 1L, 1L, 12L, 12L, 12L, 2L, 
 11L, 12L, 1L, 9L), RF00165 = c(14L, 14L, 0L, 3L, 14L, 13L, 14L, 
 1L, 12L, 0L, 14L, 14L, 14L, 14L, 11L, 14L, 14L, 14L, 14L, 0L, 
 0L, 14L, 1L, 13L), RF00496 = c(5L, 5L, 0L, 5L, 5L, 3L, 1L, 5L, 
 5L, 0L, 5L, 5L, 5L, 5L, 0L, 5L, 5L, 5L, 5L, 0L, 0L, 5L, 5L, 5L
 ), RF00497 = c(23L, 23L, 0L, 18L, 23L, 7L, 7L, 6L, 23L, 8L, 23L, 
 23L, 23L, 23L, 1L, 23L, 23L, 23L, 23L, 0L, 10L, 23L, 20L, 23L
 ), RF00014 = c(3L, 3L, 0L, 3L, 3L, 0L, 0L, 0L, 3L, 3L, 3L, 3L, 
 3L, 3L, 0L, 3L, 3L, 3L, 3L, 0L, 3L, 3L, 3L, 3L), RF00048 = c(56L, 
 56L, 5L, 56L, 

[R] Outcome~predictor model evaluation, repeated measurements

[rad mac]

Dear all,


I have simple question regarding how to fit a model (i.e. linear) to the
data.

Say I have 10 subjects with different phenotypes (dependent var Y,
identical for a particular subject) and one predictor variable measured 3
times for each subject (X). By other words:

Y Subj X
1 1 1.2
1 1 1.3
1 1 0.7
3 2 2.1
3 2 2.5
3 2 4
5 3 3
5 3 4
5 3 4
...
20 10 12
20 10 13
20 10 12.5

Subj is a grouping variable.




I would like know the correlation of Y with X (Y~X) and the effect of
within subject variance on this correlation. And thus, overall significance
and correlation.

Will it be valid to fit lm to all combinations of x and y and take an
average values of p and R-squared?

Usually, I estmate the correlation using simple lm between outcome and
averaged predictor (1-to-1, i.e. 20 outcomes versus 20 predictors).
However, I would like to take in account variations associated with
replicated measurements (i.e. the same 20 outcomes versus 20 predictors
replicated say 3 times), and, therefore, evaluate slope and intercept
variabilities. Do mixed model regression analysis suitable for such an
analysis for example using lme function from nlme package? If not, what
kind of analysis is most appropriate? Thank you.

[[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.

Re: [R] averaging two tables (rows with columns)

[Kristi Glover]

Hi John , Petr and R users
I am sorry that my data was not readable formate in the last email. Again I am 
trying to send it. hope this time, that  table can be readable.As I mentioned 
earlier that I was struggling to figure out on how I can calculate the average 
from the two tables in R. Any one can help me? really your helpwould be 
grateful- I am spending so much time to figure it out. It should not be so 
hard, I think.I have very big data but I have created a hypothetical data for 
simplification.for exampleI have : table 1Table 1: species occurrence data 
table1  X speciesX speciesY speciesZ speciesXX1 Plot110 
   1 02 Plot2011 03 plot31  
  00 14 plot4001 0Table 2: 
table 2. species tolerance datatable2  X EnviA EnviB EnviC1  
SpeciesX  0.21  0.40  0.172  SpeciesY  0.10  0.15  0.183 SpeciesXY  0.14  0.16  
0.19You may noticed that table 2 does not have species Z wh!
 ich was in tableTable 3: Now I want to get the average value of species 
tolerance in each plot based on each environmental value (EnviA or EnviB 
etc).The example of the out come (final table I was looking for it).Results 
table 1a: averages species tolerance in each plot based on EnviAsuch as:  
table3.a  X speciesX speciesY speciesZ speciesXX average1 plot1 0.21
   NA   Nodata  0.14   0.1752 plot2   NA  0.1   NodataNA   
0.1003 plot3   NA   NA   Nodata  0.14   0.1404 plot4 0.21   
NA   NodataNA   0.210Table 4 table3.b: Result table 1b: average 
species tolerance in plot based on EnviB  X speciesX speciesY speciesZ 
speciesXX average1 plot1  0.4   NA   Nodata  0.160.282 plot2
   NA 0.15   NodataNA0.153 plot3   NA   NA   Nodata 
 0.160.164 plot4  0.4   NA   NodataNA0.40I hope this 
time the data would be readable formate. Would any one help me how I!
  can calculate these?ThanksKristi Golver==
 To: kristi.glo...@hotmail.com
 CC: r-help@r-project.org
 Subject: Re: [R] averaging two tables (rows with columns)
 From: petr.pi...@precheza.cz
 Date: Thu, 10 May 2012 18:10:33 +0200
 
 Hi
 
 as already mentioned your data can not be deciphered. Use
 
 dput(table1) for sending usable data.
 
 From what you describe probably
 
 ?aggregate can be used.
 But without suitable data you hardly get any advice.
 
 Regards
 Petr

  
[[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] FW: averaging two tables (rows with columns)

[Kristi Glover]

Hi R-usuar,
I still find the table is not in readable formate. I now forwarded the 
email.hope it works. I sent it to my account . It was perfect- but when I sent 
it to R help- then it comes with unreadable formate. This is my last try. If 
not, then I will try to do in Excel manually. cheers,

From: kristi.glo...@hotmail.com
To: kristi.glo...@hotmail.com
Subject: RE: [R] averaging two tables (rows with columns)
Date: Thu, 10 May 2012 14:06:41 -0300





Hi John , Petr and R userI am sorry that my data was not readable formate in 
the last email. Agin I am trying to send it. hope this time, that  table can be 
readable.
As I mentioned earlier that I was struggling to figure out on how I can 
calculate the average from the two tables in R. Any one can help me? really 
your helpwould be grateful- I am spending so much time to figure it out. It 
should not be so hard, I think.I have very big data but I have created a 
hypothetical data for simplification.for example
I have : table 1
Table 1: species occurrence data table1  X speciesX speciesY speciesZ 
speciesXX1 Plot1101 02 Plot201  
  1 03 plot3100 14 plot40   
 01 0

Table 2: table 2. species tolerance data
table2  X EnviA EnviB EnviC1  SpeciesX  0.21  0.40  0.172  
SpeciesY  0.10  0.15  0.183 SpeciesXY  0.14  0.16  0.19

You may noticed that table 2 does not have species Z which was in table

Table 3: Now I want to get the average value of species tolerance in each plot 
based on each environmental value (EnviA or EnviB etc).The example of the out 
come (final table I was looking for it).
Results table 1a: averages species tolerance in each plot based on EnviA
such as: 
 table3.a  X speciesX speciesY speciesZ speciesXX average1 plot1 0.21  
  NA   Nodata  0.14   0.1752 plot2   NA  0.1   Nodata
 NA   0.1003 plot3   NA   NA   Nodata  0.14   0.1404 plot4 
 0.21   NA   NodataNA   0.210Table 4

 table3.b: Result table 1b: average species tolerance in plot based on EnviB   
X speciesX speciesY speciesZ speciesXX average1 plot1  0.4   NA   
 Nodata  0.160.282 plot2   NA 0.15   NodataNA0.153 
 plot3   NA   NA   Nodata  0.160.164 plot4  0.4   NA   
 NodataNA0.40
I hope this time the data would be readable formate. Would any one help me how 
I can calculate these?Thanks
Kristi Golver==



 Date: Thu, 10 May 2012 08:01:11 -0800
 From: jrkrid...@inbox.com
 Subject: RE: [R] averaging two tables (rows with columns)
 To: kristi.glo...@hotmail.com; r-help@r-project.org
 
 Kristi,
 Your data has come through in a totally unreadable layout.
 
 Have a look at ?dput as a handy way to present sample data.  Just run 
 dput(mydata), copy the output and paste into the email.  
 
 John Kane
 Kingston ON Canada
 


  
[[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.

Re: [R] FW: averaging two tables (rows with columns)

[Sarah Goslee]

Kristi,

Several people have already suggested you use dput() to provide your data.

In R,
dput(table1)
dput(table2)

Then copy and paste the output of those commands into your email.

There's no way to reliably copy and paste your raw data: using dput()
is the best way to provide it.

Sarah

On Thu, May 10, 2012 at 1:18 PM, Kristi Glover
kristi.glo...@hotmail.com wrote:

 Hi R-usuar,
 I still find the table is not in readable formate. I now forwarded the 
 email.hope it works. I sent it to my account . It was perfect- but when I 
 sent it to R help- then it comes with unreadable formate. This is my last 
 try. If not, then I will try to do in Excel manually. cheers,

 From: kristi.glo...@hotmail.com
 To: kristi.glo...@hotmail.com
 Subject: RE: [R] averaging two tables (rows with columns)
 Date: Thu, 10 May 2012 14:06:41 -0300





 Hi John , Petr and R userI am sorry that my data was not readable formate in 
 the last email. Agin I am trying to send it. hope this time, that  table can 
 be readable.
 As I mentioned earlier that I was struggling to figure out on how I can 
 calculate the average from the two tables in R. Any one can help me? really 
 your helpwould be grateful- I am spending so much time to figure it out. It 
 should not be so hard, I think.I have very big data but I have created a 
 hypothetical data for simplification.for example
 I have : table 1
 Table 1: species occurrence data table1      X speciesX speciesY speciesZ 
 speciesXX1 Plot1        1        0        1         02 Plot2        0        
 1        1         03 plot3        1        0        0         14 plot4       
  0        0        1         0

 Table 2: table 2. species tolerance data
table2                  X EnviA EnviB EnviC1  SpeciesX  0.21  0.40  0.172  
SpeciesY  0.10  0.15  0.183 SpeciesXY  0.14  0.16  0.19

 You may noticed that table 2 does not have species Z which was in table

 Table 3: Now I want to get the average value of species tolerance in each 
 plot based on each environmental value (EnviA or EnviB etc).The example of 
 the out come (final table I was looking for it).
 Results table 1a: averages species tolerance in each plot based on EnviA
 such as:
 table3.a      X speciesX speciesY speciesZ speciesXX average1 plot1     0.21 
       NA   Nodata      0.14   0.1752 plot2       NA      0.1   Nodata        
 NA   0.1003 plot3       NA       NA   Nodata      0.14   0.1404 plot4     
 0.21       NA   Nodata        NA   0.210Table 4

 table3.b: Result table 1b: average species tolerance in plot based on EnviB  
     X speciesX speciesY speciesZ speciesXX average1 plot1      0.4       NA  
  Nodata      0.16    0.282 plot2       NA     0.15   Nodata        NA    
 0.153 plot3       NA       NA   Nodata      0.16    0.164 plot4      0.4     
   NA   Nodata        NA    0.40
 I hope this time the data would be readable formate. Would any one help me 
 how I can calculate these?Thanks
 Kristi Golver==



 Date: Thu, 10 May 2012 08:01:11 -0800
 From: jrkrid...@inbox.com
 Subject: RE: [R] averaging two tables (rows with columns)
 To: kristi.glo...@hotmail.com; r-help@r-project.org

 Kristi,
 Your data has come through in a totally unreadable layout.

 Have a look at ?dput as a handy way to present sample data.  Just run 
 dput(mydata), copy the output and paste into the email.

 John Kane
 Kingston ON Canada


__
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] FW: averaging two tables (rows with columns)

[Kristi Glover]

oppps,  Now I used 'dput' function. Again I am sending. I am so sorry for 
inconvenience. 
HI R userI am sorry that my data was not readable formate in the last email. 
Agin I am trying to send it. hope this time, that  table can be readable.As I 
mentioned earlier that I was struggling to figure out on how I can calculate 
the average from the two tables in R. Any one can help me? really your 
helpwould be grateful- I am spending so much time to figure it out. It should 
not be so hard, I think.I have very big data but I have created a hypothetical 
data for simplification.for exampleI have : table 1Table 1: species occurrence 
data dput(table1)structure(list(X = structure(1:4, .Label = c(Plot1, 
Plot2, plot3, plot4), class = factor), speciesX = c(1L, 0L, 1L, 0L), 
speciesY = c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(0L, 
0L, 1L, 0L)), .Names = c(X, speciesX, speciesY, speciesZ, speciesXX), 
class = data.frame, row.names = c(NA, -4L))
Table 2: table 2. species tolerance data dput(table2)structure(list(X = 
structure(c(1L, 3L, 2L), .Label = c(SpeciesX, SpeciesXY, SpeciesY), class 
= factor), EnviA = c(0.21, 0.1, 0.14), EnviB = c(0.4, 0.15, 0.16), EnviC = 
c(0.17, 0.18, 0.19)), .Names = c(X, EnviA, EnviB, EnviC), class = 
data.frame, row.names = c(NA, -3L)) You may noticed that table 2 does not 
have species Z which was in tableTable 3: Now I want to get the average value 
of species tolerance in each plot based on each environmental value (EnviA or 
EnviB etc).The example of the out come (final table I was looking for 
it).Results table 3a: averages species tolerance in each plot based on 
EnviAsuch as:   dput(table3a)structure(list(X = structure(1:4, .Label = 
c(plot1, plot2, plot3, plot4), class = factor), speciesX = c(0.21, 
NA, NA, 0.21), speciesY = c(NA, 0.1, NA, NA), speciesZ = structure(c(1L, 1L, 
1L, 1L), .Label = Nodata, class = factor), speciesXX = c(0.14, NA, 0.14, 
NA), av!
 erage = c(0.175, 0.1, 0.14, 0.21)), .Names = c(X, speciesX, speciesY, 
speciesZ, speciesXX, average), class = data.frame, row.names = c(NA, 
-4L))
Table 3b
Result table 3b: average species tolerance in plot based on EnviB  
dput(table3b) structure(list(X = structure(1:4, .Label = c(plot1, 
plot2, plot3, plot4), class = factor), speciesX = c(0.4, NA, NA, 0.4), 
speciesY = c(NA, 0.15, NA, NA), speciesZ = structure(c(1L, 1L, 1L, 1L), .Label 
= Nodata, class = factor), speciesXX = c(0.16, NA, 0.16, NA), average = 
c(0.28, 0.15, 0.16, 0.4)), .Names = c(X, speciesX, speciesY, speciesZ, 
speciesXX, average), class = data.frame, row.names = c(NA, -4L))I hope 
this time the data would be readable formate. Would any one help me how I can 
calculate these?ThanksKristi Golver==


again 

  
[[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.

Re: [R] converting raster image

[R. Michael Weylandt]

write.csv() perhaps? I'm not sure what the csv of a raster image is though...

What does p look like? str(p) will be important -- if the raster codes
(RGB values) are in lists or 3D arrays or something more complex
(i.e., if no grayscale) you might need to put them in multiple files.

Hope this helps,
Michael

On Thu, May 10, 2012 at 11:12 AM, Mintewab Bezabih
mintewab.beza...@economics.gu.se wrote:

 Dear R users,

 I was wondering how I can convert a raster image (that made R through 
 interpolation)  into an ascii or csv format?

 this is the last line of my command

 p - interpolate(r, tpsfit)

 So p is my raster file which I want to convert into ascii or csv

 Many thanks
 Regards
 Mintewab
 __
 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
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] Split the work for many cores

[R. Michael Weylandt]

library(parallel) # Requires R =2.14
mclapply # a parallelized lapply()

You can process in parallel, but I'm not sure if it's safe to do
graphics in parallel: someone else will need to chime in about that.

Best,
Michael

On Thu, May 10, 2012 at 9:12 AM, Alaios ala...@yahoo.com wrote:
 Dear all,
 I am using my code the vgram.matrix of packets fields. I have around 500 
 matrices that I need to pass inside that function and then plot those results.

 Even though my system has 16 cores is quite clear that I am only using one of 
 those.
 Would it be able to skip these 500 tasks to the 16 cores, with each 
 processor having around 4 matrices to process?

 What would you suggest me doing?

 Regards
 Alex

        [[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
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] FW: averaging two tables (rows with columns)

[John Kane]

Kristi
The tables , if read in edit mode are vagely  readable but please just do as 
Petr and I suggest.
Use the dput command.

If you have the table in an R data.frame all you need to do is use the command 
dput(mytable)

I read your two tables into R and did that . Note I called them dat1  dat2.  
table is a command in R and it is best not to use it as a variable name.


Results 

1 dput(dat1)

structure(list(X = c(Plot1, Plot2, plot3, plot4), speciesX = c(1L, 
0L, 1L, 0L), speciesY = c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 
0L, 1L), speciesXX = c(0L, 0L, 1L, 0L)), .Names = c(X, speciesX, 
speciesY, speciesZ, speciesXX), class = data.frame, row.names = c(NA, 
-4L))

1 dput(dat2)

structure(list(X = c(SpeciesX, SpeciesY, SpeciesXY), EnviA = c(0.21, 
0.1, 0.14), EnviB = c(0.4, 0.15, 0.16), EnviC = c(0.17, 0.18, 
0.19)), .Names = c(X, EnviA, EnviB, EnviC), class = data.frame, 
row.names = c(1, 
2, 3))
1 

Now you just need to stick  variable name and an - in front of the dput 
material and paste it into R
Like THIS
 mytable1  - structure(list(X = c(Plot1, Plot2, plot3, plot4), 
speciesX = c(1L, 
0L, 1L, 0L), speciesY = c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 
0L, 1L), speciesXX = c(0L, 0L, 1L, 0L)), .Names = c(X, speciesX, 
speciesY, speciesZ, speciesXX), class = data.frame, row.names = c(NA, 
-4L))

and paste it into R.

Now we probably can get somewhere.

John Kane
Kingston ON Canada


 -Original Message-
 From: kristi.glo...@hotmail.com
 Sent: Thu, 10 May 2012 14:18:50 -0300
 To: r-help@r-project.org
 Subject: [R] FW: averaging two tables (rows with columns)
 
 
 Hi R-usuar,
 I still find the table is not in readable formate. I now forwarded the
 email.hope it works. I sent it to my account . It was perfect- but when I
 sent it to R help- then it comes with unreadable formate. This is my last
 try. If not, then I will try to do in Excel manually. cheers,
 
 From: kristi.glo...@hotmail.com
 To: kristi.glo...@hotmail.com
 Subject: RE: [R] averaging two tables (rows with columns)
 Date: Thu, 10 May 2012 14:06:41 -0300
 
 
 
 
 
 Hi John , Petr and R userI am sorry that my data was not readable formate
 in the last email. Agin I am trying to send it. hope this time, that
 table can be readable.
 As I mentioned earlier that I was struggling to figure out on how I can
 calculate the average from the two tables in R. Any one can help me?
 really your helpwould be grateful- I am spending so much time to figure
 it out. It should not be so hard, I think.I have very big data but I have
 created a hypothetical data for simplification.for example
 I have : table 1
 Table 1: species occurrence data table1  X speciesX speciesY
 speciesZ speciesXX1 Plot1101 02 Plot2
 011 03 plot3100
 14 plot4001 0
 
 Table 2: table 2. species tolerance data
 table2  X EnviA EnviB EnviC1  SpeciesX  0.21  0.40
 0.172  SpeciesY  0.10  0.15  0.183 SpeciesXY  0.14  0.16  0.19
 
 You may noticed that table 2 does not have species Z which was in table
 
 Table 3: Now I want to get the average value of species tolerance in each
 plot based on each environmental value (EnviA or EnviB etc).The example
 of the out come (final table I was looking for it).
 Results table 1a: averages species tolerance in each plot based on EnviA
 such as:
 table3.a  X speciesX speciesY speciesZ speciesXX average1 plot1
 0.21   NA   Nodata  0.14   0.1752 plot2   NA  0.1
 NodataNA   0.1003 plot3   NA   NA   Nodata  0.14
 0.1404 plot4 0.21   NA   NodataNA   0.210Table 4
 
 table3.b: Result table 1b: average species tolerance in plot based on
 EnviB  X speciesX speciesY speciesZ speciesXX average1 plot1
 0.4   NA   Nodata  0.160.282 plot2   NA 0.15
 NodataNA0.153 plot3   NA   NA   Nodata  0.16
 0.164 plot4  0.4   NA   NodataNA0.40
 I hope this time the data would be readable formate. Would any one help
 me how I can calculate these?Thanks
 Kristi Golver==
 
 
 
 Date: Thu, 10 May 2012 08:01:11 -0800
 From: jrkrid...@inbox.com
 Subject: RE: [R] averaging two tables (rows with columns)
 To: kristi.glo...@hotmail.com; r-help@r-project.org
 
 Kristi,
 Your data has come through in a totally unreadable layout.
 
 Have a look at ?dput as a handy way to present sample data.  Just run
 dput(mydata), copy the output and paste into the email.
 
 John Kane
 Kingston ON Canada
 
 
 
   [[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.


GET FREE SMILEYS FOR YOUR IM  EMAIL - Learn more at 

Re: [R] outlier identify in qqplot

[R. Michael Weylandt]

Do you want a qqnorm() instead of a qqplot() ? [Reproducibility also
involves posting the code you used that lead to the error / warning]

This seems to work for me:

# mydata - 
source(http://r.789695.n4.nabble.com/file/n4623493/mydata.txt;)[[1]]
# Have to drop visible attribute

lmmodel - lm(log(vdep) ~ v1 + sqrt(v2) + v3 +v5 + v6 + v7 + v8 + v9 +
v10, data = mydata)

qqnorm(residuals(lmmodel))

# Or if interactive:

qqnormInt - function(..., IDENTIFY = TRUE){
 qqnorm(...) - X
 if(IDENTIFY) return(identify(X))
 invisisble(X)
}


qqnormInt(residuals(lmmodel))

Michael


On Thu, May 10, 2012 at 9:20 AM, agent dunham crossp...@hotmail.com wrote:
 Find the data attached,

 http://r.789695.n4.nabble.com/file/n4623493/mydata.txt mydata.txt

 The model would be /lmmodel - lm(log(vdep) ~ v1 + sqrt(v2) + v3 +v5 + v6 +
 v7 + v8 + v9 + v10, data = mydata)/

 Thanks again,


 u...@host.com

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/outlier-identify-in-qqplot-tp4076587p4623493.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
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] FW: averaging two tables (rows with columns)

[John Kane]

Well I'd suggest learning to use the return key.  It really helps improve the 
readability of a posting.

Three of the for files came through okay but table3a has a problem.  It looks 
like some kind of extraneous charater(s) got into it but a second or so of 
editing makes it work so we have all 4 tables now.



John Kane
Kingston ON Canada


 -Original Message-
 From: kristi.glo...@hotmail.com
 Sent: Thu, 10 May 2012 14:36:51 -0300
 To: r-help@r-project.org
 Subject: Re: [R] FW: averaging two tables (rows with columns)
 
 
 oppps,  Now I used 'dput' function. Again I am sending. I am so sorry for
 inconvenience.
 HI R userI am sorry that my data was not readable formate in the last
 email. Agin I am trying to send it. hope this time, that  table can be
 readable.As I mentioned earlier that I was struggling to figure out on
 how I can calculate the average from the two tables in R. Any one can
 help me? really your helpwould be grateful- I am spending so much time to
 figure it out. It should not be so hard, I think.I have very big data but
 I have created a hypothetical data for simplification.for exampleI have :
 table 1Table 1: species occurrence data dput(table1)structure(list(X =
 structure(1:4, .Label = c(Plot1, Plot2, plot3, plot4), class =
 factor), speciesX = c(1L, 0L, 1L, 0L), speciesY = c(0L, 1L, 0L, 0L),
 speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(0L, 0L, 1L, 0L)), .Names =
 c(X, speciesX, speciesY, speciesZ, speciesXX), class =
 data.frame, row.names = c(NA, -4L))
 Table 2: table 2. species tolerance data dput(table2)structure(list(X =
 structure(c(1L, 3L, 2L), .Label = c(SpeciesX, SpeciesXY, SpeciesY),
 class = factor), EnviA = c(0.21, 0.1, 0.14), EnviB = c(0.4, 0.15,
 0.16), EnviC = c(0.17, 0.18, 0.19)), .Names = c(X, EnviA, EnviB,
 EnviC), class = data.frame, row.names = c(NA, -3L)) You may noticed
 that table 2 does not have species Z which was in tableTable 3: Now I
 want to get the average value of species tolerance in each plot based on
 each environmental value (EnviA or EnviB etc).The example of the out come
 (final table I was looking for it).Results table 3a: averages species
 tolerance in each plot based on EnviAsuch as: 
 dput(table3a)structure(list(X = structure(1:4, .Label = c(plot1,
 plot2, plot3, plot4), class = factor), speciesX = c(0.21, NA, NA,
 0.21), speciesY = c(NA, 0.1, NA, NA), speciesZ = structure(c(1L, 1L, 1L,
 1L), .Label = Nodata, class = factor), speciesXX = c(0.14, NA, 0.14,
 NA), av!
  erage = c(0.175, 0.1, 0.14, 0.21)), .Names = c(X, speciesX,
 speciesY, speciesZ, speciesXX, average), class = data.frame,
 row.names = c(NA, -4L))
 Table 3b
 Result table 3b: average species tolerance in plot based on EnviB
 dput(table3b) structure(list(X = structure(1:4, .Label = c(plot1,
 plot2, plot3, plot4), class = factor), speciesX = c(0.4, NA, NA,
 0.4), speciesY = c(NA, 0.15, NA, NA), speciesZ = structure(c(1L, 1L, 1L,
 1L), .Label = Nodata, class = factor), speciesXX = c(0.16, NA, 0.16,
 NA), average = c(0.28, 0.15, 0.16, 0.4)), .Names = c(X, speciesX,
 speciesY, speciesZ, speciesXX, average), class = data.frame,
 row.names = c(NA, -4L))I hope this time the data would be readable
 formate. Would any one help me how I can calculate these?ThanksKristi
 Golver==
 
 
 again
 
 
   [[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.


Send your photos by email in seconds...
TRY FREE IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if3
Works in all emails, instant messengers, blogs, forums and social networks.

__
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] storage of matrices of diff dimension together

[carol white]

Hi,
It might be a trivial question but how do you store matrices of different 
dimensions read from a file or in a loop together? The best solution might be a 
list but I don't store the first matrix  correctly:

 m = rbind(c(1,2),c(3,4),c(5,6))
t=rbind(c(1,2),c(5,6))


 l = list(m)
 l = list(l,t)    #I assumed that at the begining I don't have m and t at the 
 same time to do list(m,t), but I list them, one after the other
 l
[[1]]
[[1]][[1]]
 [,1] [,2]
[1,]    1    2
[2,]    3    4
[3,]    5    6


[[2]]
  [,1] [,2]
t    1    2
 5    6


thanks

Carol

[[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.

Re: [R] storage of matrices of diff dimension together

[Sarah Goslee]

Hi Carol,

 m1 - rbind(c(1,2),c(3,4),c(5,6))
 m2 - rbind(c(1,2),c(5,6))
 mylist - list(m1)
 c(mylist, list(m2))
[[1]]
 [,1] [,2]
[1,]12
[2,]34
[3,]56

[[2]]
 [,1] [,2]
[1,]12
[2,]56

See the examples in ?c for more information.

Sarah

On Thu, May 10, 2012 at 2:35 PM, carol white wht_...@yahoo.com wrote:
 Hi,
 It might be a trivial question but how do you store matrices of different 
 dimensions read from a file or in a loop together? The best solution might be 
 a list but I don't store the first matrix  correctly:

  m = rbind(c(1,2),c(3,4),c(5,6))
 t=rbind(c(1,2),c(5,6))


  l = list(m)
 l = list(l,t)    #I assumed that at the begining I don't have m and t at the 
 same time to do list(m,t), but I list them, one after the other
 l
 [[1]]
 [[1]][[1]]
  [,1] [,2]
 [1,]    1    2
 [2,]    3    4
 [3,]    5    6


 [[2]]
   [,1] [,2]
 t    1    2
  5    6


 thanks

 Carol


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

__
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] fda problems - version correct?

[Troels Ring]

Dear friends - I am trying to understand fda and working with the 2009 
book from Springer. I run the scripts directly from a new installed fda 
library and include sessionInfo() below. This is from the script 
fdarm-ch09.R  - I seem to have got not fda_2.2.8  but 2.2.7 - is that 
the problem? Where do I get .8 as I just reinstalled? Best wishes
Troels Ring, Nephrology
Aalborg, Denmark

  betavar  = coefvar[2]*harmonics[1]^2 + coefvar[3]*harmonics[2]^2 +
+coefvar[4]*harmonics[3]^2
Error in `^.fd`(harmonics[1], 2) : FDOBJ does not have a spline basis.
  sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-mingw32/x64 (64-bit)

locale:
[1] LC_COLLATE=Danish_Denmark.1252  LC_CTYPE=Danish_Denmark.1252
[3] LC_MONETARY=Danish_Denmark.1252 LC_NUMERIC=C
[5] LC_TIME=Danish_Denmark.1252

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

other attached packages:
[1] fda_2.2.7 zoo_1.7-7

loaded via a namespace (and not attached):
[1] grid_2.15.0lattice_0.20-6 tools_2.15.0

[[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.

Re: [R] converting raster image

[Michael Sumner]

You should provide reproducible code and at the very least declare the
packages you are using. I guess this from the package raster? See
example in ?raster::interpolate:

  ## Thin plate spline interpolation with x and y only
 library(fields)
 r - raster(system.file(external/test.grd, package=raster))
 ra - aggregate(r, 10)
 xy - data.frame(xyFromCell(ra, 1:ncell(ra)))
 v - getValues(ra)
 tps - Tps(xy, v)
 p - raster(r)
 p - interpolate(p, tps)

One way to convert this to tabular format is to coerce to
SpatialGridDataFrame, then write that out with write.table
(write.csv):

require(sp)
head(as.data.frame(as(p, SpatialGridDataFrame)))
 layer s1 s2
1 1823.450 178420 333980
2 1818.925 178460 333980
3 1814.405 178500 333980
4 1809.876 178540 333980
5 1805.324 178580 333980
6 1800.734 178620 333980

There would be functions in raster to do that more directly, I'm just
not that familiar with it.

So,

x - as.data.frame(as(p, SpatialGridDataFrame))

Now x is just a data.frame with every cell stored explicitly with its
centre coordinate, s1 and s2 are the X and Y coordinates, layer
is the raster value.

Cheers, Mike.

On Fri, May 11, 2012 at 1:12 AM, Mintewab Bezabih
mintewab.beza...@economics.gu.se wrote:

 Dear R users,

 I was wondering how I can convert a raster image (that made R through 
 interpolation)  into an ascii or csv format?

 this is the last line of my command

 p - interpolate(r, tpsfit)

 So p is my raster file which I want to convert into ascii or csv

 Many thanks
 Regards
 Mintewab
 __
 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.



-- 
Michael Sumner
Institute for Marine and Antarctic Studies, University of Tasmania
Hobart, Australia
e-mail: mdsum...@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.

[R] Resampling question

[Vikram Chhatre]

Hello -

I have a population of 100 individuals that I would like to bootstrap
10 times, every time removing 5 *different* individuals.

So far, I have done the following:

pop - read.table('mypop.txt', header=FALSE)

replicate(10, sample(pop, 95, replace=FALSE))

I have not actually gone through each of the 10 files created to make
sure no single individual was removed more than once during the 10
bootstraps.  But will the above syntax achieve this object?

Thanks
V

__
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] Barplots inside loop - several data errors, workaround needed

[Lee]

Looking at the documentation for try() I am not sure how it would be best
applied in this situation. My background is not extensively programming.
Would writing a function first be appropriate?

Also, I'm not sure just a simple error catch would solve my first problem.
I do, in fact, need it to plot the barplot based on the table which is
created above. However, R doesn't like the lack of several columns.

Further guidance would be appreciated.

--
all the best,
Lee Mueller
ISA Certified Arborist MI-4148A
Registered Forester #46043





On Wed, May 9, 2012 at 4:30 AM, Jim Holtman jholt...@gmail.com wrote:

 ?try

 Sent from my iPad

 On May 8, 2012, at 22:03, Lee muell...@gmail.com wrote:

  I have a series of data which is managed through a loop. The loop creates
  pivot tables of my data using the *cast* function in the
  *reshape*library. For the most part, the data is all plotted
  correctly.
  Unfortunately, there are a couple of data sets which create errors and
 halt
  the loop.
 
  One of the tables looks like the following:
 
   dbh Black Walnut
  1   8 38.19722
  2  10 48.89244
  3  12 38.19722
 
  When the loop attempts the barplot() function, the following error is
  returned:
  Error in seq_len(p) : argument must be coercible to non-negative integer
 
  In other cases, there is simply no data in the specified set of data.
  Therefore, the table is full of NA's. Obviously, this does not need to be
  plotted, but I cannot simply remove it from the larger database. I need
 my
  loop to continue regardless of if it runs into these issues.
 
  *Question: What can I do to ensure the above single variable table will
  plot correctly? and what can I do to suppress errors on the datasets
 which
  do not have data so the loop continues?*
 
  full code: http://pastebin.com/LB88hpfM
 
  Thank you in advance.
 
  --
  all the best,
  Lee Mueller
  ISA Certified Arborist MI-4148A
  Registered Forester #46043
 
 [[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]]

__
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] Resampling question

[R. Michael Weylandt]

No, it will not, except possibly by chance: if the draws of sample are
IID (and they are supposed to be) there's no reason to expect them not
to overlap.

If you want that -- and I'm not sure it's totally on the level
bootstrapping-wise -- you need to decide which ones to remove all in
one fell swoop:

Something like this:

ToDrop - matrix(sample(100, 50, replace = FALSE), ncol = 10)

apply(ToDrop, 2, function(x) pop[-x])

Michael


On Thu, May 10, 2012 at 7:56 PM, Vikram Chhatre
crypticline...@gmail.com wrote:
 Hello -

 I have a population of 100 individuals that I would like to bootstrap
 10 times, every time removing 5 *different* individuals.

 So far, I have done the following:

 pop - read.table('mypop.txt', header=FALSE)

 replicate(10, sample(pop, 95, replace=FALSE))

 I have not actually gone through each of the 10 files created to make
 sure no single individual was removed more than once during the 10
 bootstraps.  But will the above syntax achieve this object?

 Thanks
 V

 __
 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
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] Resampling question

[Nordlund, Dan (DSHS/RDA)]

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Vikram Chhatre
 Sent: Thursday, May 10, 2012 4:57 PM
 To: r-help@r-project.org
 Subject: [R] Resampling question
 
 Hello -
 
 I have a population of 100 individuals that I would like to bootstrap
 10 times, every time removing 5 *different* individuals.
 
 So far, I have done the following:
 
 pop - read.table('mypop.txt', header=FALSE)
 
 replicate(10, sample(pop, 95, replace=FALSE))
 
 I have not actually gone through each of the 10 files created to make
 sure no single individual was removed more than once during the 10
 bootstraps.  But will the above syntax achieve this object?
 
 Thanks
 V

What you are doing is probably more appropriately called a jack-knife, rather 
than bootstrap.  Be that as it may, your approach will not prevent someone 
being dropped more than once.  You might randomly select 50 people without 
replacement and then drop them 5 at a time from the original population

to_drop - sample(100,50,replace=FALSE)
rep1 - pop[-to_drop[1:5],]
rep2 - pop[-to_drop[6:10],] 
...

I will let you decide how you want automate this.

Dan


Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204


__
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] Barplots inside loop - several data errors, workaround needed

[Steve Lianoglou]

Hi,

On Thu, May 10, 2012 at 8:35 PM, Lee muell...@gmail.com wrote:
 Looking at the documentation for try() I am not sure how it would be best
 applied in this situation. My background is not extensively programming.
 Would writing a function first be appropriate?

 Also, I'm not sure just a simple error catch would solve my first problem.
 I do, in fact, need it to plot the barplot based on the table which is
 created above. However, R doesn't like the lack of several columns.

 Further guidance would be appreciated.

Consider this block of code that you can run in your R workspace:

~
set.seed(123)
random.error - runif(3, 0, 2)  1.5
for (throws.error in random.error) {
  cat(I'm about to try something\n)
  result - try({
cat(  I'm in the middle of trying something\n)
cat(  There is a chance it might result in an error\n)
if (throws.error) {
  stop(Error!)
}
  }, silent=TRUE)
  if (is(result, 'try-error')) {
cat(An error occurred while trying something, but I'm OK\n\n)
  } else {
cat(No error occurred while trying something\n\n)
  }
}

~

and the output it gives:

~
I'm about to try something
  I'm in the middle of trying something
  There is a chance it might result in an error
No error occurred while trying something

I'm about to try something
  I'm in the middle of trying something
  There is a chance it might result in an error
An error occurred while trying something, but I'm OK

I'm about to try something
  I'm in the middle of trying something
  There is a chance it might result in an error
No error occurred while trying something
~

Does that help?

-steve

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

__
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] moving data from one frame to another

[Daniel_55]

Hello,

I am working with two different data frames, and I'd like to move data from
one to the other. Here is the first frame:

 head(five)
  Week  Game.ID VTm VPts HTm HPts HDifferential VDifferential
11  NFL_20050908_OAK@NE OAK   20  NE   3010   -10
21 NFL_20050911_ARI@NYG ARI   19 NYG   4223   -23
31 NFL_20050911_CHI@WAS CHI7 WAS9 2-2
41 NFL_20050911_CIN@CLE CIN   27 CLE   13   -1414
51  NFL_20050911_DAL@SD DAL   28  SD   24-4 4
61 NFL_20050911_DEN@MIA DEN   10 MIA   3424   -24
  VTm.f HTm.f average
1   OAKNE 19.4375
2   ARI   NYG 19.4375
3   CHI   WAS 19.4375
4   CIN   CLE 19.4375
5   DALSD 19.4375
6   DEN   MIA 19.4375

and here is the other (aggregated means from the first frame).

 head(fiveINFO)
  HTm HPts VPts  average
1 ARI 19.87500 19.0 19.43750
2 ATL 24.75000 19.12500 21.93750
3 BAL 19.37500 13.75000 16.56250
4 BUF 16.5 17.37500 16.93750
5 CAR 25.12500 23.27273 24.19886
6 CHI 18.8 14.0 16.38889

For reference, this data is looking at NFL scores. I want to take the
averages in fiveINFO, frame two, and move them to the corresponding team in
the first frame. five is 266 rows long, while fiveINFO is 32 rows — fiveINFO
contains each HTm only once, while five contains each one 8-10 times,
depending on the number of home games each team plays. 

I'm imagining I'll need to use some kind of for loop for this, but
everything I'm doing is striking out. Help?

--
View this message in context: 
http://r.789695.n4.nabble.com/moving-data-from-one-frame-to-another-tp4625193.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] Result of clustering on plot

[greyofthedawn]

Hello! I 'm new to R and need your help in one question...
I did cluster analysis using kmeans function.  I load data from file:

This file also contains headers for objects I want to cluster.
After making a clustering procedure I display result on plot:

And it shows only points (i.e. objects) without their names.
Is there any way to mark points on this plot with their names from file?
Your help would be very appreciated! Thank you!


--
View this message in context: 
http://r.789695.n4.nabble.com/Result-of-clustering-on-plot-tp4625043.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.

Re: [R] Can't read xlsx file into R. Seem, Seem to have XLConnect loaded.

[Martin Studer]

Hi Mike,

as others have already suggested, there might be a problem with quoting.
Otherwise, did you already have a look at the package vignette (
http://cran.r-project.org/web/packages/XLConnect/vignettes/XLConnect.pdf
http://cran.r-project.org/web/packages/XLConnect/vignettes/XLConnect.pdf )
that gives a number of examples on how to read/write Excel files?

Best regards,
Martin

--
View this message in context: 
http://r.789695.n4.nabble.com/Can-t-read-xlsx-file-into-R-Seem-Seem-to-have-XLConnect-loaded-tp469p4624156.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] Error t value matrix

[Trying To learn again]

Hi all,

I want to make the following:

I want to run  a linear regression on each column of a matrix estima on
the correspondent column on the matrix estima2.

You see I want to regress estima[,1] on estima2[,1] this way to all
columns

At the same time I want to make a regression adding each time a new
observation.

You see, the first regression will regress only one observation with one
observation (I now this has no sense in this only one observation step)

the second turn of observation will make

estima[1:2,n] on estima2[1:2,n]  for all n.

Third stimation will make

estima[1:3,n] on estima2[1:3,n]  for all n.

And so on.

Make this, I want to make an output matrix on each t-value associated
with the regressor.

Conclusion my final matrix called t value should include al the t values
on the regression each of them incorporating a new observation, with the
same rows and colums than estima.

I have tried several thing but I cannot achive.

I writte to see if you can guide me¡¡¡

I swear I´m trying.



randz-matrix(rnorm(5000),50,100)

H-matrix(0,50,100)

H[1,]-randz[1,]
for (i in 2:50){
 if(i  26) {
   H[i, ] - 0.6 * H[i-1, ] + randz[i, ]
 } else {
   H[i, ] - H[i-1, ] + randz[i, ]
 }
}


write.table(H, file = datad.txt)
g-read.table(datad.txt)

hy-nrow(g)-1
estima-H[2:nrow(g), ]
estima2-H[ 1:hy, ]

mycoef - function (x,y)
a-estima
b-estima2
f-summary(lm(a~b))
ff-coef(f)
ff[2,t value]
tvalue - sapply (2:ncol(b) , function (i){
y-a[,i]
x-b[,i]
mycoef(x,y)

}
)
print (summary(tvalue))

[[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.

Re: [R] moving data from one frame to another

[Daniel_55]

When I tried doing that, it only merged the averages from column 1 — ARI —
and did not spread the averages for ARI, ATL, BAL. Any ideas on how I could
keep it from making all of the HTms ARI and all of the averages 19.4375? 

--
View this message in context: 
http://r.789695.n4.nabble.com/moving-data-from-one-frame-to-another-tp4625193p4625234.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] Coloring subsets of points of a strip chart by quantiles

[Nicholas Edgington]

Hello All,

I am trying to color five subsets of points in a series of vertical strip
charts by quintiles using the quantile function, but I have not been able to
do this successfully, nor have I been able to find an example of this
online.  I would like to have the E1, E2, and E3 columns plotted on a
single plot (containing the three strip charts).  I am overlaying the strip
chart over a transparent background box plot.

##OK, here's some data (This is probably an inelegant way to do this, but
I'm obviously new to R):

mdat -
matrix(c(round(runif(14,min=50,max=100)),round(runif(14,min=50,max=100)),round(runif(14,min=50,max=100))),
nrow = 14, ncol=3, byrow=F,dimnames = list(c(seq(1,14,by=1)),c(E1, E2,
E3))) ; mdat - as.data.frame(mdat)

mdat

   E1 E2  E3
1  72 96  95
2  56 81  85
3  57 55  95
4  72 83  89
5  61 65  51
6  80 86  85
7  99 85 100
8  50 98  60
9  79 54  63
10 81 58  96
11 60 74  52
12 56 51  80
13 65 51  75
14 90 88  65

boxplot(mdat, col=transparent, main=mdat, boxwex = 0.25, ylab=Score,
cex=2, cex.axis=2, cex.lab=1.5, cex.main=2, las=1, col.main=blue,
col.lab=blue, col.axis=blue)

stripchart(mdat,vertical=T, col=blue, method=stack, pch=21,cex=2, add=T,
bg= rainbow(quantile(mdat, prob = c(.2, .4, .6, .8, 1), na.rm =
T),alpha=0.5))

##Here's the plot, but I don't think that the coloring is behaving as I
would like. (1) I don't think that the quantile function is getting applied
to each E1/2/3 separately. (2) I don't think that the data is getting
cut and colored by (five) ranges that depend upon the quintiles.

Any help would be appreciated.  Thanks.
http://r.789695.n4.nabble.com/file/n4624535/mdat.png 

--
View this message in context: 
http://r.789695.n4.nabble.com/Coloring-subsets-of-points-of-a-strip-chart-by-quantiles-tp4624535.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] survival analysis simulation question

[Grace Ma]

Hi,
I am trying to simulate a regression on survival data under a few
conditions:
1. Under different error distributions
2. Have the error term be dependent on the covariates

But I'm not sure how to specify either conditions. I am using the Design
package to perform the survival analysis using the survreg, bj, coxph
functions.   Any help is greatly appreciated.

This is what I have so far:
survtime  - 10*rexp(500)  #distribution of survival time
cens - ifelse(survtime  10, 0, 1) #indicator for censored/observed
survtime  - pmin(survtime, 10) #new survival time values with censored
info
age - rnorm(200, 40, 10)  #age variable
race - factor(sample(c('a','b'),500,TRUE))  #categorical variable
test - bj(Surv(survtime, cens) ~ rcs(age,5) + race)

[[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] how to do averaging of two tables (rows with columns)

[Kristi Glover]

Hi R user,
 I finally able to send you the table in 
readable format. I have seen that  some of you do send tables in email 
when asking questions, but why i could not send. Any way 
some of you helped me to send you the example table in a readable format. 

now,
 I want to concentrate on my problem. I am trying to get the information
 (dat 3)  from dat 1 and 2 in R. I have very big data but these data are
 just hypothetical data. my data structures are exactly same as dat 1 
and dat 2. I created dat 3 and dat4 manually to show what information I 
wanted to have.  

I am struggling to figure it out how I can do in R. I think it is not 
difficult.  I hope any one can help me. 

dat1 is the table of species occurrence (o means species absence, 1 means 
species presence).

dat1  - structure(list(X = structure(1:4, .Label = c(Plot1, 
Plot2, plot3, plot4), class = factor), speciesX = c(1L, 0L, 1L, 0L), 
speciesY = 
c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(0L, 0L, 1L, 
0L)), .Names = 
c(X, speciesX, speciesY, speciesZ, speciesXX), class = data.frame, 
row.names = 
c(NA, -4L)) 

dat2 is the species tolerances value in each environmental variable

dat2  - structure(list(X = structure(c(1L,

3L, 2L), .Label = c(SpeciesX, SpeciesXX, SpeciesY), class = factor), 
EnviA =

c(0.21, 0.1, 0.14), EnviB = c(0.4, 0.15, 0.16), EnviC = c(0.17, 0.18, 0.19)), 
.Names =

c(X, EnviA, EnviB, EnviC), class = data.frame, row.names = c(NA, -3L))


 ## note (here in dat 2 there is no species Z you can see that )
Now, I want to get the average value of tolerances in each grid. like dat 3

the dat3 is based on the column EnviA.

dat3  -structure(list(X = structure(1:4, .Label = c(plot1,

 plot2, plot3, plot4), class = factor), speciesX = c(0.21, NA, NA,

 0.21), speciesY = c(NA, 0.1, NA, NA), speciesZ = structure(c(1L, 1L, 1L,

 1L), .Label = Nodata, class = factor), speciesXX = c(0.14, NA, 0.14,

 NA), average = c(0.175, 0.1, 0.14, 0.21)), .Names = c(X, speciesX,

 speciesY, speciesZ, speciesXX, average), class = data.frame,

 row.names = c(NA, -4L))


dat4 is same thing as dat3 but here i used EnviB instead of EnviA.

dat4  - structure(list(X = structure(1:4, .Label = c(plot1, plot2, 
plot3, plot4), class =

factor), speciesX = c(0.4, NA, NA, 0.4), speciesY = c(NA, 0.15, NA, NA), 
speciesZ =

structure(c(1L, 1L, 1L, 1L), .Label = Nodata, class = factor), speciesXX = 
c(0.16, NA,

0.16, NA), average = c(0.28, 0.15, 0.16, 0.4)), .Names = c(X, speciesX, 
speciesY,

speciesZ, speciesXX, average), class = data.frame, row.names = c(NA, 
-4L))  

I hope you understand my problem and you can help me.

Thanks 

Kristi



  
[[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] how to combine two tables with different orientation and get mean from the two tables

[Kristi Glover]

Hi R user,
 I finally able to send you the table in 
readable format. I have seen that  some of you do send tables in email 
when asking questions, but why i could not send. Any way 
some of you helped me to send you the example table in a readable format. 

now,
 I want to concentrate on my problem. I am trying to get the information
 (dat 3)  from dat 1 and 2 in R. I have very big data but these data are
 just hypothetical data. my data structures are exactly same as dat 1 
and dat 2. I created dat 3 and dat4 manually to show what information I 
wanted to have.  

I am struggling to figure it out how I can do in R. I think it is not 
difficult.  I hope any one can help me. 

dat1 is the table of species occurrence (o means species absence, 1 means 
species presence).

dat1  - structure(list(X = structure(1:4, .Label = c(Plot1, 
Plot2, plot3, plot4), class = factor), speciesX = c(1L, 0L, 1L, 0L), 
speciesY = 
c(0L, 1L, 0L, 0L), speciesZ = c(1L, 1L, 0L, 1L), speciesXX = c(0L, 0L, 1L, 
0L)), .Names = 
c(X, speciesX, speciesY, speciesZ, speciesXX), class = data.frame, 
row.names = 
c(NA, -4L)) 

dat2 is the species tolerances value in each environmental variable

dat2  - structure(list(X = structure(c(1L,

3L, 2L), .Label = c(SpeciesX, SpeciesXX, SpeciesY), class = factor), 
EnviA =

c(0.21, 0.1, 0.14), EnviB = c(0.4, 0.15, 0.16), EnviC = c(0.17, 0.18, 0.19)), 
.Names =

c(X, EnviA, EnviB, EnviC), class = data.frame, row.names = c(NA, -3L))


 ## note (here in dat 2 there is no species Z you can see that )
Now, I want to get the average value of tolerances in each grid. like dat 3

the dat3 is based on the column EnviA.

dat3  -structure(list(X = structure(1:4, .Label = c(plot1,

 plot2, plot3, plot4), class = factor), speciesX = c(0.21, NA, NA,

 0.21), speciesY = c(NA, 0.1, NA, NA), speciesZ = structure(c(1L, 1L, 1L,

 1L), .Label = Nodata, class = factor), speciesXX = c(0.14, NA, 0.14,

 NA), average = c(0.175, 0.1, 0.14, 0.21)), .Names = c(X, speciesX,

 speciesY, speciesZ, speciesXX, average), class = data.frame,

 row.names = c(NA, -4L))


dat4 is same thing as dat3 but here i used EnviB instead of EnviA.

dat4  - structure(list(X = structure(1:4, .Label = c(plot1, plot2, 
plot3, plot4), class =

factor), speciesX = c(0.4, NA, NA, 0.4), speciesY = c(NA, 0.15, NA, NA), 
speciesZ =

structure(c(1L, 1L, 1L, 1L), .Label = Nodata, class = factor), speciesXX = 
c(0.16, NA,

0.16, NA), average = c(0.28, 0.15, 0.16, 0.4)), .Names = c(X, speciesX, 
speciesY,

speciesZ, speciesXX, average), class = data.frame, row.names = c(NA, 
-4L))  

I hope you understand my problem and you can help me.

Thanks 

Kristi




  
[[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.

Re: [R] Interweaving of two datasets

[lunarossa]

I tried to do using merge, I think now it works (anyway it is your worth), I
wrote:

df1-read.csv(df1.csv,head=T)
attach(df1)
df2-read.csv(df2.csv,head=T)
attach(df2)
join-merge(df1,df2,by.x=name,by.y=name)

Is it correct? Coz I'm very frightened to lose datas or make mistakes 

--
View this message in context: 
http://r.789695.n4.nabble.com/Interweaving-of-two-datasets-tp4608505p4624852.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.

Re: [R] Matrix heatmap

[fjucks]

how do I plot only the data below 10? everything is white for the 0-10 and
10-90 is black ..
those functions which do this?
was bad for such basic questions, but I started tinkering with R is 6 days

--
View this message in context: 
http://r.789695.n4.nabble.com/Matrix-heatmap-tp4619084p4625021.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.