Re: [R] column-wise z-scores by group
DISCCRS wrote: Hi, I have a dataset of historical monthly temperature data that is grouped by weather station. I want to create z-scores of the monthly data using a base period of a subset of years. I subset the dataset first to include only data from the years (V2) that make up the base period so I could calculate the appropriate means and standard deviations V1 V2V3 V12 V15 V16 V19 8411084 1978 40.16 63.13 44.06 63.41 63.47 8511084 1979 43.71 60.88 48.09 64.64 62.34 8611084 1980 50.61 61.64 47.93 62.10 63.45 8711084 1981 42.11 63.59 47.29 63.42 63.37 1583 18469 1978 30.78 56.93 34.62 56.40 57.39 1584 18469 1979 33.48 57.68 37.76 58.70 57.30 1585 18469 1980 40.83 54.48 39.27 56.14 57.42 1586 18469 1981 33.33 56.28 37.57 56.20 56.47 2688 25467 1978 52.61 75.51 55.02 68.20 70.70 2689 25467 1979 47.95 74.54 50.70 67.58 70.24 2690 25467 1980 55.12 72.51 56.59 66.49 71.21 2691 25467 1981 56.70 70.33 57.65 69.35 72.16 Then I split the data by group ID (V1) and got the means and std deviations: subsets - split(test,V1) sub.means - data.frame(t(sapply(subsets, mean))) sub.sds - data.frame(t(sapply(subsets, sd, na.rm=T))) Here are the means, for example: V1 V2 V3 V12 V15 V16 V19 11084 11084 1979.5 44.1475 62.3100 46.8425 63.3925 63.1575 18469 18469 1979.5 34.6050 56.3425 37.3050 56.8600 57.1450 25467 25467 1979.5 53.0950 73.2225 54.9900 67.9050 71.0775 How can I approach the next step -- applying the means and std deviations from the two new arrays that I created to the original dataset (by station and by month)? Or should I be using a different approach entirely? There are NAs throughout the dataset. Thanks very much in advance. -Jennife Playing the ball from where it landed, how about nm - as.character(test$V1) (test - sub.means[nm,])/sub.sds[nm,] However, there could be a neater solution by looping ave(V2, V1, FUN=scale) Or, you could apply scale() on each of your split() data and then unsplit(). Just beware that scale() turns things into matrices so you need an as.data.frame step inbetween. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] Reading Data
Hi I don't want to ignore the date row. See basically my first row is the date column and the 1st column is the stocks name. Now using the if loop I have to find prices of stocks corresponding to a date. I hope the problem is clear to you now For example Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08 a1.003.007.003.00 b2.004.004.007.00 c3.008.00655.00 3.00 d4.0023.00 4.005.00 e5.0078.00 6.005.00 Rahul Agarwal Analyst Equities Quantitative Research UBS_ISC, Hyderabad On Net: 19 533 6363 -Original Message- From: Gustaf Rydevik [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 07, 2008 2:19 PM To: Agarwal, Rahul-A Cc: [EMAIL PROTECTED]; r-help@r-project.org Subject: Re: [R] Reading Data On Tue, Oct 7, 2008 at 10:36 AM, [EMAIL PROTECTED] wrote: Hi, I have a data in which the first row is in date format and the first column is in text format and rest all the entries are numeric. Whenever I am trying to read the data using read.table, the whole of my data is converted in to the text format. Please suggest what shall I do because using the numeric data which are prices I need to calculate the return but if these prices are not numeric then calculating return will be a problem regards Rahul Agarwal Analyst Equities Quantitative Research UBS_ISC, Hyderabad On Net: 19 533 6363 Hi, A single column in a data frame can't contain mixed formats. In the absence of example data, would guess one of the following could work : 1) read.table(data.txt,skip=1, header=T) ## If you have headers 2) read.table(data.txt, header=T) ## If the date row is supposed to be variable names. 3) read.table(data.txt,skip=1) ## If there are no headers, and you want to ignore the date regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik [[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] Reading Data
Hi
What if I need to take dates and stock names from these table...i mean I need
to read in this table and then use if function and extratc the data from FOO
Identifier weight Start_Date End_Date
a 6.7631Jan06 31Jan07
g 2.8628Feb06 28Feb07
e 22.94 31Mar06 30Mar07
c 30.05 28Apr06 30Apr07
t 20.55 31May06 31May07
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philipp Pagel
Sent: Tuesday, October 07, 2008 2:52 PM
To: r-help@r-project.org
Subject: Re: [R] Reading Data
For example
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1.003.007.003.00
b 2.004.004.007.00
c 3.008.00655.00 3.00
d 4.0023.00 4.005.00
e 5.0078.00 6.005.00
OK - this may be what you want:
foo - read.table('q.tbl', header=T, check.names=F, row.names=1)
str(foo)
'data.frame': 5 obs. of 4 variables:
$ 30-Jan-08: num 1 2 3 4 5
$ 28-Feb-08: num 3 4 8 23 78
$ 31-Mar-08: num 7 4 655 4 6
$ 30-Apr-08: num 3 7 3 5 5
foo
30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1 3 7 3
b 2 4 4 7
c 3 8 655 3
d 423 4 5
e 578 6 5
foo['31-Mar-08']
31-Mar-08
a 7
b 4
c 655
d 4
e 6
foo['d', '31-Mar-08']
[1] 4
Maybe row.names is not what you want - but the example sould get you going.
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München
Wissenschaftszentrum Weihenstephan 85350 Freising, Germany
http://mips.gsf.de/staff/pagel
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[R] Fitting weibull, exponential and lognormal distributions to left-truncated data.
Dear All, I have two questions regarding distribution fitting. I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: fitdistr-(x,weibull) However, this does not take into consideration the truncation at x=1. I read another posting in this forum that suggested using the argument lower to truncate the distribution fitting. However, this does not seem to be working. For example, when I attempt to fit a weibull distribution truncated at x=1 using lower, it seems to set the best-fit shape parameter at 1: fitdistr(x,weibull,lower=1) shapescale 1. 9.87964337 (0.02358731) (0.40649570) ##I have tried this on other datasets also truncated at x=1 and get the same result (i.e. shape=1). Does anyone know how to successfully fit the exponential, weibull and lognormal distributions to truncated data? Secondly, as my datasets are large (1000 data points) assessing the fit of the distribution with kolmogorov smirnov goodness of fit tests is routinely showing statistical significance for all distributions. Therefore, I would like to plot the observed data with the theoretical best fit distributions (weibull, exponential and lognormal) to visually assess which fits the observed data best. So far I have been doing this as follows: fitdistr(x,weibull) shapescale a b D1-density(x) ##density distribution of observed data D2-density(rweibull(1500,shape=a,scale=b)) ##density of a random variable following the theoretical best fit weibull distribution with shape parameter =a, scale parameter = b. plot(range(D1$x),range(D1$y,D2$y),type=n,xlab=x,ylab=Density) lines(D1,col=red) lines(D2,col=blue) This successfully plots the two density curves on the same graph, but it plots data below the x=1 threshold - even for the observed data! I have tried limiting the scale of x-axis using xlim=c(1,150) but the graph still plots the origin of the graph as (0,0). I can only get different origins if I limit x more extremely e.g. xlim=c(50,150). Does anyone know how I can successfully change the origin of the graph to (1,0)? Sorry for the long e-mail! Any help would be greatly appreciated. Regards, Lauren This message has been checked for viruses but the contents of an attachment may still contain software viruses, which could damage your computer system: you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ 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] Automatic code diagramming for R?
Ben Bryant wrote: Greetings - Is anyone aware of an automatic code diagrammer/flow chart creator that works for the R language (either a contributed package, or external software)? I need to explain some code structure of a package I'm working on to non-R users, and would find it extremely helpful to have such a program similar to, for example, Visustin ( http://www.aivosto.com/visustin.html ). I can do it by hand (possibly with the help of the 'diagram' package) but it seems like automated capabilities for recognizing nested structures and argument-passing would be of good general use to package developers. When you document your package using Roxygen, you can generate callgraphs (using the @callGraph and/or @callGraphPrimitives tags). The roxygen package homepage can be found at http://roxygen.org/ and the vignette has an example of a call graph (and how to generate it). HTH, Tobias Much appreciated, -Ben Bryant [[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.
[R] Mirror Image on Biplot Graphic
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package). I need to
rotate on y axis the lines and symbols of constrained and sites representation.
If I do that on an image editor, I rotate everything, including titles, axes
labels and positions. I just need to rotate the inner par and keep the
variables names (constrained) and symbols in the new positions but with the
right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading Data
For example
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1.003.007.003.00
b 2.004.004.007.00
c 3.008.00655.00 3.00
d 4.0023.00 4.005.00
e 5.0078.00 6.005.00
OK - this may be what you want:
foo - read.table('q.tbl', header=T, check.names=F, row.names=1)
str(foo)
'data.frame': 5 obs. of 4 variables:
$ 30-Jan-08: num 1 2 3 4 5
$ 28-Feb-08: num 3 4 8 23 78
$ 31-Mar-08: num 7 4 655 4 6
$ 30-Apr-08: num 3 7 3 5 5
foo
30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1 3 7 3
b 2 4 4 7
c 3 8 655 3
d 423 4 5
e 578 6 5
foo['31-Mar-08']
31-Mar-08
a 7
b 4
c 655
d 4
e 6
foo['d', '31-Mar-08']
[1] 4
Maybe row.names is not what you want - but the example sould get you going.
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://mips.gsf.de/staff/pagel
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting weibull, exponential and lognormal distributions to left-truncated data.
I have several datasets, all left-truncated at x=1, that I am
attempting
to fit distributions to (lognormal, weibull and exponential). I had
been using fitdistr in the MASS package as follows:
A possible solution is to use the survreg() in the survival package
without specifying the covariates, i.e.
library(survival)
survreg(Surv(..)~1, dist=weibull)
where Surv(..) accepts information about times, censoring/truncation
variables and dist allows to specify alternative distributions.
See ?Surv e ?survreg
The survival package is mostly targeted at right-censored data. The NADA
package provides wrappers for many of the survival routines so they work
with left-censored data.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] Reading Data
Perhaps it would be easier to try something like this:
df1 - read.table(filename, ... ) # All columns read as characters
df1 - t(df1) # Transpose df1
write.table - (df1, newfile, ...) # Write the transposed data
# to a new file
df2 - read.table(newfile, header=TRUE, ...) # Read in transposed
data
I doubt it's the most computation-time-efficient, but conceptually it's
pretty quick.
Benjamin
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 8:28 AM
To: [EMAIL PROTECTED]; r-help@r-project.org
Subject: [R] FW: Reading Data
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
hi let me explain you the problem
we have a database which is in this format
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a1.003.007.003.00
b2.004.004.007.00
c3.008.00655.00 3.00
d4.0023.00 4.005.00
e5.0078.00 6.005.00
and we have a query which is in this format
Identifier weight Start_Date End_Date
a6.7631-Jan-06 31-Jan-07
e2.8628-Feb-06 28-Feb-07
f22.94 31-Mar-06 30-Mar-07
y30.05 28-Apr-06 30-Apr-07
h20.55 31-May-06 31-May-07
d6.76
f2.86
r22.94
okay now my task is to calculate returns for all the indentifiers for
a respective start and end date from table 1.
now length of start date and end date column is same and that of weight
and identifier is same.
i hope everything is clear now.
let me also send you the code that i have written but in my code i have
problem with the date format and also with the stocks name
data=read.table(H:/Rahul/london/david/rexcel/price.txt)
query=read.table(H:/Rahul/london/david/rexcel/prac.txt,header=TRUE)
data=as.matrix(data)
instrument=data[,1]
date=data[1,]
query=as.matrix(query)
q_ins=query[,1]
wt=query[,2]
q_sd=query[,3]
q_ed=query[,4]
returns=function(I,SD,ED){
p=rep(0,2)
for(i in 2:length(instrument))
{
if(instrument[i]==I)
{
for(j in 2:length(date))
{
if(date[j]==SD)
p[1]=data[i,j]
}
for(j in 2:length(date))
{
if(date[j]==ED)
p[2]=data[i,j]
}
}
returns=(p[2]/p[1])-1
}
#print(p)
#print(I)
return(returns)
}
##The original Funda##
matrix_ret=matrix(0,length(q_ins),length(q_sd))
for(i in 1:length(q_sd))
{
for(j in 1:length(q_ins))
{
matrix_ret[j,i]=returns(q_ins[j],q_sd[i],q_ed[i])
}
}
#Removing NA from the matrix
matrix_ret1=sapply(X=matrix_ret, FUN=function(x)
ifelse(is.na(x),0.00,x))
matrix_ret=matrix(matrix_ret1,length(q_ins),length(q_sd))
wt_ret=matrix(0,length(q_sd),1)
for(i in 1:length(q_sd))
{
for(j in 1:length(q_ins))
{
wt_ret[i]=wt_ret[i]+(wt[j]*matrix_ret[j,i])
}
}
result=cbind(q_ed,wt_ret)
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
-Original Message-
From: Gustaf Rydevik [mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] ]
Sent: Tuesday, October 07, 2008 2:19 PM
To: Agarwal, Rahul-A
Cc: [EMAIL PROTECTED]; r-help@r-project.org
Subject: Re: [R] Reading Data
On Tue, Oct 7, 2008 at 10:36 AM, [EMAIL PROTECTED] wrote:
Hi,
I have a data in which the first row is in date format and the first
column is in text format and rest all the entries are numeric.
Whenever I am trying to read the data using read.table, the whole of
my data is converted in to the text format.
Please suggest what shall I do because using the numeric data which
are prices I need to calculate the return but if these prices are not
numeric then calculating return will be a problem
regards
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
Hi,
A single column in a data frame can't contain mixed formats.
In the absence of example data, would guess one of the following could
work :
1)
read.table(data.txt,skip=1, header=T) ## If you have headers
2)
read.table(data.txt, header=T) ## If the date row is supposed to be
variable names.
3)
read.table(data.txt,skip=1) ## If there are no headers, and you want
to ignore the date
regards,
Gustaf
--
Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik
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PLEASE do
Re: [R] lmer: random factor nested in a fixed factor
hi folks, I believe that the construction y ~ A + (A|B) (where A is a categorical variable for a fixed effect and B is a categorical variable for a random effect) will generate random effects for all of the fixed effect coefficients. That is, in addition fitted a systematic fixed effect coefficient for intercept and relevant added effects of levels of A, it will fit corresponding random coefficients for each level of B, including random intercepts, and relevant added random effects of A, given B (i.e. for each separate B). Thus if you think that each family has, in addition to a different overall mean, also has a different response to levels of A (and you have the replication to estimate them), then you should include (A|B) and see if it is better than simply (1|B). Hank On Oct 6, 2008, at 11:02 AM, Christian Ritz wrote: Dear Agnes, I think your model specification should look like this: YourModel1 - lmerlmer(y ~ poptype*matingtype + (1|poptype:pop) + (1| poptype:fam), data = ...) The 1 in front of | refers to models that are random intercepts models as opposed to general random coefficients models in which case 1 would need to be replaced by a variable that is quantitative. So, the (poptype|/pop/fam) construction is definitely not relevant to your problem, unless poptype is a quantitative variable... The combined factor poptype:pop corresponds to the factor pop, but taking the nesting structure into account. Similar for the construction poptype:fam. From the summary output of the lmer() fit you should be able to check whether or not the correct number of groups are used for these random factors. Kind regards Christian __ 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. Hank Stevens, Associate Professor 338 Pearson Hall Botany Department Miami University Oxford, OH 45056 Office: (513) 529-4206 Lab: (513) 529-4262 FAX: (513) 529-4243 http://www.cas.muohio.edu/~stevenmh/ http://www.cas.muohio.edu/ecology http://www.muohio.edu/botany/ If the stars should appear one night in a thousand years, how would men believe and adore. -Ralph Waldo Emerson, writer and philosopher (1803-1882) __ 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] Mirror Image on Biplot Graphic
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and sites
representation.
Easiest is to multiply the axis you want to invert by -1. Something like the
following, where my.cca is the orginal object and yax = obj[, 2] (xax being
obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package). I
need to rotate on y axis the lines and symbols of constrained and sites
representation. If I do that on an image editor, I rotate everything,
including titles, axes labels and positions. I just need to rotate the
inner par and keep the variables names (constrained) and symbols in the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
[[alternative HTML version deleted]]
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View this message in context:
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Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different ordination
packages, in most of which this is possible). What you need to do with vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something like
the following, where my.cca is the orginal object and yax = obj[, 2] (xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package). I
need to rotate on y axis the lines and symbols of constrained and sites
representation. If I do that on an image editor, I rotate everything,
including titles, axes labels and positions. I just need to rotate the
inner par and keep the variables names (constrained) and symbols in the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
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Re: [R] R on the Freakonomics Blog
Quotes I probably have an unhealthy attraction to the powers of Excel. I taught my daughter how to use it when she was 7. When I teach corporate finance, I try to make sure that my law students come away from the course knowing how to crunch in Excel. And R is probably not kept up to speed on the cutting-edge empirical methods as quickly as the traditional packages. Yep, I'd say an unhealthy attraction to Excel and an stunning lack of knowledge about R. I note that many contributers seem to be pointing out that R is just a bit better than the author seems to think it is. --- On Mon, 10/6/08, Scillieri, John [EMAIL PROTECTED] wrote: From: Scillieri, John [EMAIL PROTECTED] Subject: [R] R on the Freakonomics Blog To: r-help@r-project.org Received: Monday, October 6, 2008, 4:26 PM http://freakonomics.blogs.nytimes.com/2008/10/06/free-super-crunching-so ftware/ This e-mail and any attachments are confidential, may contain legal, professional or other privileged information, and are intended solely for the addressee. If you are not the intended recipient, do not use the information in this e-mail in any way, delete this e-mail and notify the sender. CEG-IP2 __ 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. __ [[elided Yahoo spam]] __ 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] R and computer heat
Hi, I noticed the temperature of my laptop rises sharply during execution of a long R script that generates several hundred plots, all of them saved to files. No screen output. Temps reached above 90 Celsius degrees in the box and above 80 C deg in the processor. The machine turns on cooler at maximum speed and exhaled air is really hot. Tried similar operations (batch graphic and music format conversion) and temp rises were usual. System: laptop, Turion 64, Kubuntu Linux, Xorg X server 1.4.0.90, KDE 3.5.9, R 2.7.2 compiled with MBCS, PCRE, etc. Could it be fake due to an interaction of some R piece with system monitors? Alexandre -- Alexandre Santos Aguiar, MD, SCT -- website: http://spsconsultoria.com/ Phone: +55-11-3717-4866 (SP) Phone: +55-21-3717-4866 (RJ) Voicemail: +55-11-2157-6891 (SP) Fax: +55-11-5549-8760 (SP) -- signature.asc Description: This is a digitally signed message part. __ 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 validate model?
Hi! I am working on scorecard model and I have arrived at the regression equation. I have used logistic regression using R. My question is how do I validate this model? I do have hold out sample of 5000 customers. Please guide me. Problem is I had never used Logistic regression earlier neither I am used to credit scoring models. Thanks in advance Maithili __ 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] Reading Data
What if I need to take dates and stock names from these table...i mean I need to read in this table and then use if function and extratc the data from FOO Not sure what exactly the problem is. Can you give an example? Of course the first thing would be to make sure that dates are in the same format... And so far I don't see a good reason to use 'if' - R has extremely flexible indexing capabilities. While using the command foo['31-Mar-08'] in the below given example: On the left hand side I am getting the count and not alphabetsplease suggest Maybe your stock names in the file were numbers? Maybe you left out the row.names parameter? Why m I getting this message data=read.table(H:/Rahul/london/david/rexcel/price1.txt,header=T,check.names=F,row.names=1) Error in read.table(H:/Rahul/london/david/rexcel/price1.txt, header = T, : duplicate 'row.names' are not allowed The message means what it says: row names need to be unique. Apparently yours are not. I.e. You wave multiple rows for the same stock identifier. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ 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] Stacked area chart and legends
mbr wrote: I have found lots of good advice on this forum about stacked area charts but I’ve run into problems with the 2 recommended options: stackploy in plotrix or qplot in ggplot2. I have a many page report that will be in a 2x2 page format “par(mfrow=c(2,2))”” and need one of the page components to be a stacked chart. I’d prefer to use stackpoly, if possible, but I’m stuck on how to do the legend in hopefully a relatively simple way. I just need a simple legend to be outside the plot area (b/c the plot area is full as I’m doing it as 100% stacked area chart format) on the right side with the little box with the color and the series name next to it – same/similar to Excel format. The qplot approach would work as well and seems like the legend is built into the default, but the graph always ends up on its own page and can’t figure out how to force it into one of boxes of my 2x2 page layout. I like the looks of the stackpoly more as consistent with other graphs I have on report but qplot stacked chart would be fine if I could get it into the layout. Hi Matt, The easy way is to make your device wider than it is high: # use the screen device for the example x11(width=10,height=7) #leave lots of room on the right side of the plot: par(mar=c(5,4,4,10)) #then plot, stop clipping to the plot region, add the legend # and restore clipping stackpoly(matrix(abs(rnorm(30)),nrow=10),stack=TRUE, main=Look ma. just like Excel!) par(xpd=TRUE) legend(11,3,1:5,fill=1:5) par(xpd=FALSE) 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] Reading Data
On Tue, Oct 7, 2008 at 10:36 AM, [EMAIL PROTECTED] wrote: Hi, I have a data in which the first row is in date format and the first column is in text format and rest all the entries are numeric. Whenever I am trying to read the data using read.table, the whole of my data is converted in to the text format. Please suggest what shall I do because using the numeric data which are prices I need to calculate the return but if these prices are not numeric then calculating return will be a problem regards Rahul Agarwal Analyst Equities Quantitative Research UBS_ISC, Hyderabad On Net: 19 533 6363 Hi, A single column in a data frame can't contain mixed formats. In the absence of example data, would guess one of the following could work : 1) read.table(data.txt,skip=1, header=T) ## If you have headers 2) read.table(data.txt, header=T) ## If the date row is supposed to be variable names. 3) read.table(data.txt,skip=1) ## If there are no headers, and you want to ignore the date regards, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik __ 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 and computer heat
Why don't you hack a little script that looks at system load, temperature and date/time and writes it somewhere. Then you can load it into R and plot it nicely :-))-O Even compare it witht he desktop :-)-O el on 10/7/08 12:21 PM Alexandre Aguiar said the following: Hi, I noticed the temperature of my laptop rises sharply during execution of a long R script that generates several hundred plots, all of them saved to files. No screen output. Temps reached above 90 Celsius degrees in the box and above 80 C deg in the processor. The machine turns on cooler at maximum speed and exhaled air is really hot. Tried similar operations (batch graphic and music format conversion) and temp rises were usual. System: laptop, Turion 64, Kubuntu Linux, Xorg X server 1.4.0.90, KDE 3.5.9, R 2.7.2 compiled with MBCS, PCRE, etc. Could it be fake due to an interaction of some R piece with system monitors? Alexandre __ 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. Eberhard W. Lisse \/ Obstetrician Gynaecologist (Saar) [EMAIL PROTECTED] el108-ARIN / * | Telephone: +264 81 124 6733 (cell) PO Box 8421 \ / Please do NOT email to this address Bachbrecht, Namibia ;/if it is DNS related in ANY way __ 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 store the results of multiple iterations of 'aov' in a data.frame?
A follow-up to Need to calculate within- and between-run CV I have a dataset of 210 Samples, of which each was run several times in several consecutive days. A dataset for one sample is below: qu.s - structure(list(Sample = c(44L, 44L, 44L, 44L, 44L, 44L, 44L, 44L, 44L, 44L), Run = c(1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L ), Rep = c(1, 2, 1, 2, 3, 4, 1, 2, 3, 4), value = c(120L, 107L, 117L, 124L, 118L, 127L, 110L, 113L, 109L, 113L)), .Names = c(Sample, Run, Rep, value), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L), class = data.frame) and the code I used is below: a=aov(value ~ Run, data=qu.s) I assume that 'Residual' is the variance between replicates in my data. Since I have to run the above code 210 times, my question is: how to store the results of 'aov' in a separate data.frame? For each iteration of 'aov' I need to store: the _sums of squares_ for Run and Residual, the _N_ of replicates and the _mean_. Later I want to use it to calculate the coefficients of variation (CV) for each sample. I looked at the structure of aov object, but the sums of squares are not listed there, though 'summary(a)' prints them. Please help. -- Michal Figurski Michal Figurski wrote: Dear R-helpers, I have a dataset named qu, organized as follows: SampleRunReplicateValue 11125 11240 11333 11429 12137 12244 12345 13125 13240 14133 14229 14325 2 ... Basically, a sample was run on an assay multiple times within a single day. Each of these results is Replicate. Then run was repeated several times in consecutive days - variable Run. There are 210 such samples. I need to actually calculate the CV for each sample: - within run (between replicates) - that's easy to do in Excel - between run - that's the problem. I was thinking of using either 'aov' or 'lme' to solve this. However, I don't know how to interpret the output. For example, a summary output from aov(Value~Run+Replicate, subset(qu,Sample==79))' for one sample was: Df Sum Sq Mean Sq F value Pr(F) Run 1 4.000 4.000 0.3214 0.6104 Replicate1 73.500 73.500 5.9062 0.0933 . Residuals3 37.333 12.444 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Do you guys think this is correct approach? How do I extract these numbers (sum of squares) to store in a separate dataframe for further calculations? And how should I interpret the Residual in this setting? I will appreciate your comments. __ 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] Mirror Image on Biplot Graphic
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[, :
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different ordination
packages, in most of which this is possible). What you need to do with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something like
the following, where my.cca is the orginal object and yax = obj[, 2]
(xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package). I
need to rotate on y axis the lines and symbols of constrained and sites
representation. If I do that on an image editor, I rotate everything,
including titles, axes labels and positions. I just need to rotate the
inner par and keep the variables names (constrained) and symbols in the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total
Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
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[R] Selecting numeric data from a data frame
Hi I have a data frame containing numeric and factor data. I want to extract the numeric data only. How can this be done? Thanks Meir Meir Preiszler - Research Engineer I t a m a r M e d i c a l Ltd. Caesarea, Israel: Tel: +(972) 4 617 7000 ext 232 Fax: +(972) 4 627 5598 Cell: +(972) 54 699 9630 Email: [EMAIL PROTECTED] Web: www.Itamar-medical.com * 88---8--- This E-mail is confidential information of Itamar medical Ltd. It may also be legally privileged. If you are not the addressee you may not copy, forward, disclose or use any part of it. If you have received this message in error, please delete it and all copies from your system and notify the sender immediately by return E-mail. Internet communications cannot be guaranteed to be timely, secure, error or virus-free. The sender does not accept liability for any errors or omissions. Before printing this email , kindly think about the environment. Itamar Medical Ltd. MIS Yan Malgin. 88---8--- [[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 validate model?
Maithili Shiva wrote: Hi! I am working on scorecard model and I have arrived at the regression equation. I have used logistic regression using R. My question is how do I validate this model? I do have hold out sample of 5000 customers. Please guide me. Problem is I had never used Logistic regression earlier neither I am used to credit scoring models. Thanks in advance Maithili The holdout sample of 5000 may be too small. You didn't specify the size of the training sample. I assume it is 50,000 or larger. 50-fold repeats of 10-fold cross-validation is better. With any resampling technique you need to repeat all exploratory and modeling steps inside the loop. Frank __ 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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Selecting numeric data from a data frame
Meir Preiszler wrote: Hi I have a data frame containing numeric and factor data. I want to extract the numeric data only. How can this be done? Thanks Meir d[sapply(d,is.numeric)] -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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 and computer heat
Alexandre Aguiar wrote: Hi, I noticed the temperature of my laptop rises sharply during execution of a long R script that generates several hundred plots, all of them saved to files. No screen output. Temps reached above 90 Celsius degrees in the box and above 80 C deg in the processor. The machine turns on cooler at maximum speed and exhaled air is really hot. Tried similar operations (batch graphic and music format conversion) and temp rises were usual. System: laptop, Turion 64, Kubuntu Linux, Xorg X server 1.4.0.90, KDE 3.5.9, R 2.7.2 compiled with MBCS, PCRE, etc. Could it be fake due to an interaction of some R piece with system monitors? Alexandre My guess is that it the graphics card which is hard at work. Tom __ 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] fitting a curve to data points
carlos.grohmann at gmail.com writes: Hello all. This is likely to be a silly question, but I have a set of data points and I want to fit a curve to it, like this: http://www.igc.usp.br/pessoais/guano/temp/curve.png While you could use a loess-curve, interpreting the rising branch a the end is probably close to cheating. I would simply guess that this is an exp(-t/t0) curve with a few outliers. You should really have good reasons with so few datapoints in the upward branch. Dieter __ 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] S3 vs S4 class
A simple example:
temp - as.Date('2001-10-5')
class(temp)
Date
How do I tell if Date is implemented as an S3 or an S4 class type?
I tried ?class, but didn't see anything obvious. Functions like inherit() work
equally well with both types (which is a good thing).
I expect the answer will be obvious - once someone points it out to me.
Terry T
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Re: [R] plyr package: passing further arguments fail
Thanks Hadley, for some reason I didn't see your email until now. It
works fine with the development version,
library(plyr)
df - data.frame(a=1:10 , b=1:10)
foo1 - function(a, b, cc=0, d=0){
a + b + cc + d
}
mdply(data. = df, foo1, cc=1, d=2)
I think using . prefixes is a safer option (and possibly more
consistent with R customs), although that means the current code using
plyr will have to be modified.
Congratulations on the new ggplot2 release too!
Thanks,
baptiste
On 5 Oct 2008, at 14:39, hadley wickham wrote:
On Sun, Oct 5, 2008 at 8:02 AM, Auguie, Baptiste
[EMAIL PROTECTED] wrote:
Dear list and Hadley,
The new plyr package seems to provide a clean and consistent way to
apply a function on several arguments. However, I don't understand
why the following example does not work like the standard mapply,
library(plyr)
df - data.frame(a=1:10 , b=1:10)
foo1 - function(a, b, cc=0, d=0){
a + b + cc + d
}
mdply(df, foo1, cc=1) # fine
mdply(df, foo1, d=1) # fails
mdply(df, foo1, cc=1, d=2) # fails
Unfortunately this bug is R's partial name matching: d = 2 - data. =
2. You should be able to fix this by manually specifying mdply(data.
= df, foo1, cc=1, d=2) but there are some bugs in the current version
that prevent this from happening. I've fixed this in the development
version, available from http://github.com/hadley/plyr (click the
download link)
However, the whole point of plyr is that you should have to think
about this kind of thing, so I'll revisit my naming scheme - probably
to use . prefixes instead of suffixes.
Hadley
--
http://had.co.nz/
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] S3 vs S4 class
On 07/10/2008 10:41 AM, Terry Therneau wrote:
A simple example:
temp - as.Date('2001-10-5')
class(temp)
Date
How do I tell if Date is implemented as an S3 or an S4 class type?
I tried ?class, but didn't see anything obvious. Functions like inherit() work
equally well with both types (which is a good thing).
I expect the answer will be obvious - once someone points it out to me.
You could use the isS4 function:
temp - as.Date('2001-10-5')
isS4(temp)
[1] FALSE
Duncan Murdoch
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Re: [R] Reading Data
Bang on!!
Thanks for the help
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philipp Pagel
Sent: Tuesday, October 07, 2008 2:52 PM
To: r-help@r-project.org
Subject: Re: [R] Reading Data
For example
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1.003.007.003.00
b 2.004.004.007.00
c 3.008.00655.00 3.00
d 4.0023.00 4.005.00
e 5.0078.00 6.005.00
OK - this may be what you want:
foo - read.table('q.tbl', header=T, check.names=F, row.names=1)
str(foo)
'data.frame': 5 obs. of 4 variables:
$ 30-Jan-08: num 1 2 3 4 5
$ 28-Feb-08: num 3 4 8 23 78
$ 31-Mar-08: num 7 4 655 4 6
$ 30-Apr-08: num 3 7 3 5 5
foo
30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1 3 7 3
b 2 4 4 7
c 3 8 655 3
d 423 4 5
e 578 6 5
foo['31-Mar-08']
31-Mar-08
a 7
b 4
c 655
d 4
e 6
foo['d', '31-Mar-08']
[1] 4
Maybe row.names is not what you want - but the example sould get you going.
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München
Wissenschaftszentrum Weihenstephan 85350 Freising, Germany
http://mips.gsf.de/staff/pagel
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
[R] using assign with lists
I am performing many permutations on a data-set with each permutation
producing a variable number of results. I thought that the best way to keep
track of all this in one object would be with a list ('res.lst'). To address
these variable results for each permutation I attempted to construct this
list using 'assign'. There is even more nesting than indicated below, but
this is a simple example that, if addressed, will fit answer my question.
The below code chunk clearly does not produce the desired results because,
instead of assigning a new vector to the list, it creates a new variable
'res.list$contrast.i.j' . In the last two lines I show what I really want to
happen. Can I use assign in this context by using it differently?
Thanks, Mark
res.lst - list()
for (i in 1:2){
for (j in 1:2){
assign(paste(res.lst$contrast, i, j, sep = .), paste(i,j,sep=.))
}
}
res.lst
ls(pattern = res.lst..?)
res.lst$contrast.5.5 - 5.5
res.lst
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Home
Skype: mkimpel
**
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Re: [R] lme and lmer df's and F-statistics again
Julia S. julia.schroeder at gmail.com writes: Now, I did that in my article and I got a response from a reviewer that I additionally should give the degrees of freedom, and the F-statistics. From what I read here, that would be incorrect to do, and I sort of intuitively also understand why (at least I think I do). ... Well, writing on my rebuttal, I find myself being unable to explain in a few, easy to understand (and, at the same time, correct) sentences stating that it is not a good idea to report (most likely wrong) dfs and F statistics. Can somebody here help me out with a correct explanation for a laymen? Feeling with you, and hoping some day this will be resolved. I am sure you have read Douglas Bates' http://finzi.psych.upenn.edu/R/Rhelp02a/archive/76742.html but I thought this was temporary. The only workaround I have is not to use lmer for gaussian models. Dieter __ 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] read.spss: variable.labels
Hi,
how can I attach variable labels originally read by read.spss() to the
resulting variables?
pre
X - read.spss('data.sav', use.value.labels = TRUE, to.data.frame =
TRUE, trim.factor.names = TRUE, trim_values = TRUE, reencode = UTF-8)
names(X) - tolower(names(X))
attach(X)
/pre
Thank you
Sören
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Re: [R] R and computer heat
Alexandre Aguiar asaguiar at spsconsultoria.com writes:
I noticed the temperature of my laptop rises sharply during execution of a
long R script that generates several hundred plots, all of them saved to
files. No screen output. Temps reached above 90 Celsius degrees in the box
and above 80 C deg in the processor. The machine turns on cooler at maximum
speed and exhaled air is really hot. Tried similar operations (batch
graphic and music format conversion) and temp rises were usual.
Compare it with a while(TRUE){} loop in any not to agressively optimizing
language of your choice. Heat simply shows that R is working hard.
Dieter
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Re: [R] How to validate model?
Hi Frank, Thanks for your feedback! But I think we are talking about two different things. 1) Validation: The generalization performance of the classifier. See, for example, Studies on the Validation of Internal Rating Systems by BIS. 2) Calibration: Correct calibration of a PD rating system means that the calibrated PD estimates are accurate and conform to the observed default rates. See, for instance, An Overview and Framework for PD Backtesting and Benchmarking, by Castermans et al. Frank, you are referring the #1 and I am referring to #2. Nonetheless, I would never create a rating system if my model doesn't discriminate better than a coin toss. Regards, Pedro -Original Message- From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 07, 2008 11:02 AM To: Rodriguez, Pedro Cc: [EMAIL PROTECTED]; r-help@r-project.org Subject: Re: [R] How to validate model? [EMAIL PROTECTED] wrote: Usually one validates scorecards with the ROC curve, Pietra Index, KS test, etc. You may be interested in the WP 14 from BIS (www.bis.org). Regards, Pedro No, the validation should be done using an absolute reliability (calibration) curve. You need to verify that at all levels of predicted risk there is agreement with the true probability of failure. An ROC curve does not do that, and I doubt the others do. A resampling-corrected loess calibration curve is a good approach as implemented in the Design package's calibrate function. Frank -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Maithili Shiva Sent: Tuesday, October 07, 2008 8:22 AM To: r-help@r-project.org Subject: [R] How to validate model? Hi! I am working on scorecard model and I have arrived at the regression equation. I have used logistic regression using R. My question is how do I validate this model? I do have hold out sample of 5000 customers. Please guide me. Problem is I had never used Logistic regression earlier neither I am used to credit scoring models. Thanks in advance Maithili __ 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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically you
should use the extractor function scores() to extract what you want from the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot, reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different ordination
packages, in most of which this is possible). What you need to do with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something
like
the following, where my.cca is the orginal object and yax = obj[, 2]
(xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package).
I
need to rotate on y axis the lines and symbols of constrained and
sites
representation. If I do that on an image editor, I rotate everything,
including titles, axes labels and positions. I just need to rotate the
inner par and keep the variables names (constrained) and symbols in
the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total
Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos do Mar/UFPR
Laboratório de Micropaleontologia
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http://www.nabble.com/Mirror-Image-on-Biplot-Graphic-tp19857549p19858268.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide
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Re: [R] Reading Data
Why m I getting this message
data=read.table(H:/Rahul/london/david/rexcel/price1.txt,header=T,check.names=F,row.names=1)
Error in read.table(H:/Rahul/london/david/rexcel/price1.txt, header = T, :
duplicate 'row.names' are not allowed
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Philipp Pagel
Sent: Tuesday, October 07, 2008 2:52 PM
To: r-help@r-project.org
Subject: Re: [R] Reading Data
For example
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1.003.007.003.00
b 2.004.004.007.00
c 3.008.00655.00 3.00
d 4.0023.00 4.005.00
e 5.0078.00 6.005.00
OK - this may be what you want:
foo - read.table('q.tbl', header=T, check.names=F, row.names=1)
str(foo)
'data.frame': 5 obs. of 4 variables:
$ 30-Jan-08: num 1 2 3 4 5
$ 28-Feb-08: num 3 4 8 23 78
$ 31-Mar-08: num 7 4 655 4 6
$ 30-Apr-08: num 3 7 3 5 5
foo
30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a 1 3 7 3
b 2 4 4 7
c 3 8 655 3
d 423 4 5
e 578 6 5
foo['31-Mar-08']
31-Mar-08
a 7
b 4
c 655
d 4
e 6
foo['d', '31-Mar-08']
[1] 4
Maybe row.names is not what you want - but the example sould get you going.
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München
Wissenschaftszentrum Weihenstephan 85350 Freising, Germany
http://mips.gsf.de/staff/pagel
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting weibull, exponential and lognormal distributions to left-truncated data.
On Tue, 7 Oct 2008, [EMAIL PROTECTED] wrote: I have several datasets, all left-truncated at x=1, that I am attempting to fit distributions to (lognormal, weibull and exponential). I had been using fitdistr in the MASS package as follows: A possible solution is to use the survreg() in the survival package without specifying the covariates, i.e. library(survival) survreg(Surv(..)~1, dist=weibull) where Surv(..) accepts information about times, censoring/truncation variables and dist allows to specify alternative distributions. See ?Surv e ?survreg The survival package is mostly targeted at right-censored data. The NADA package provides wrappers for many of the survival routines so they work with left-censored data. Left-censoring and left-truncation are not the same thing. With left-censoring you see that you had observations 1, and with left-truncation you do not (at least how the terms are usually applied: occasionally the meanings are reversed). For left-truncation it is relatively easy, e.g. ltwei - function(x, shape, scale = 1, log = FALSE) dweibull(x, shape, scale, log)/pweibull(1, shape, scale, lower=FALSE) and use this in fitdistr. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] Mac crash- Probably memory problem
Dear all, I am running a code using bootstraps for estimating standard errors but the mac crashes. When I use small number of bootstraps (100) it works fine but if I increase that number it crashes. Thanks in advance dimitris The code, the error and my mac characteristics are the following: ##code# qr.1- rq(y~factor(year)+factor(state)+x1+I(x^2)+I(x^3), tau=0.05, data=data1) s.qr1.05 - summary(qr.1, se=boot, R=1000)$coefficient ##error# *** caught segfault *** address 0x1de9e000, cause 'memory not mapped' Traceback: 1: .Fortran(xys, as.integer(m), as.integer(n), as.integer(p), as.integer(R), as.integer(m + 5), as.integer(p + 2), as.double(x), as.double(y), as.double(tau), as.double(tol), flag = integer(R), coef = double(p * R), resid = double(m), integer(m), double((m + 5) * (p + 2)), double(m), xx = double(m * p), yy = double(m), as.integer(s), PACKAGE = quantreg) 2: boot.rq.xy(x, y, s, tau) 3: boot.rq(x, y, tau, ...) 4: summary.rq(qr.05.nox, se = boot, R = 1000) 5: summary(qr.05.nox, se = boot, R = 1000) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: ###characteristics# Mac OS X version 10.5.4 Model Name: MacBook Model Identifier: MacBook4,1 Processor Name: Intel Core 2 Duo Processor Speed: 2.4 GHz Number Of Processors: 1 Total Number Of Cores:2 L2 Cache: 3 MB Memory: 4 GB Bus Speed:800 MHz Boot ROM Version: MB41.00C1.B00 -- View this message in context: http://www.nabble.com/Mac-crash--Probably-memory-problem-tp19860892p19860892.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] panel.groups: use group.number to define colors
Dear list,
I've been trying this for a few hours and I just don't understand how
lattice works with groups and subscripts.
Consider the following example,
xx - seq(1, 10, length=100)
x - rep(xx, 4)
y - c(cos(xx), sin(xx), xx, xx^2/10)
fact - factor(rep(c(cos, sin, id, square), each=100))
fact2 - factor(rep(c(periodic, not periodic), each=100))
my.df - data.frame(x=x, y=y, fact = fact, fact2 = fact2)
head(my.df)
myColors - c(2, 4)
xyplot(y ~ x | fact, data = my.df, groups = fact2,
panel = panel.superpose,
panel.groups = function(..., group.number) {
panel.xyplot(...)
#panel.xyplot( ..., col=myColors[group.number]) # error
})
My aim is to assign a custom color to each group, but for some reason
the col parameter is already given to panel.xyplot and I can't find
where it gets the values from.
Many thanks,
baptiste
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] horizontal boxplot + xlim
Antje niederlein-rstat at yahoo.de writes: I get a strange behaviour of a boxplot with the following code. There seems to be a problem with the xlim-parameter. Did I do anything wrong? What else can I do to force the boxplot to have a defined x-range? x - rnorm(100) boxplot(x, notch=TRUE, xlab=parameter, xlim - c(-4,4), horizontal = TRUE) This should be just fine if you say boxplot(x, notch=TRUE, xlab=parameter, xlim=c(-4,4), horizontal = TRUE) instead. - is only for assignment. You must use = for setting arguments to a function. Ben Bolker __ 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] using assign with lists
Here is something to try:
result - list()
for (i in 1:2){
+ for (j in 1:2){
+ result[[paste(i, j, sep=|)]] - rep(i,j)
+ }
+ }
result
$`1|1`
[1] 1
$`1|2`
[1] 1 1
$`2|1`
[1] 2
$`2|2`
[1] 2 2
On Tue, Oct 7, 2008 at 11:01 AM, Mark Kimpel [EMAIL PROTECTED] wrote:
I am performing many permutations on a data-set with each permutation
producing a variable number of results. I thought that the best way to keep
track of all this in one object would be with a list ('res.lst'). To address
these variable results for each permutation I attempted to construct this
list using 'assign'. There is even more nesting than indicated below, but
this is a simple example that, if addressed, will fit answer my question.
The below code chunk clearly does not produce the desired results because,
instead of assigning a new vector to the list, it creates a new variable
'res.list$contrast.i.j' . In the last two lines I show what I really want to
happen. Can I use assign in this context by using it differently?
Thanks, Mark
res.lst - list()
for (i in 1:2){
for (j in 1:2){
assign(paste(res.lst$contrast, i, j, sep = .), paste(i,j,sep=.))
}
}
res.lst
ls(pattern = res.lst..?)
res.lst$contrast.5.5 - 5.5
res.lst
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Home
Skype: mkimpel
**
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting text from html code using the RCurl package.
Hi Tony --
Tony Breyal [EMAIL PROTECTED] writes:
Dear R-help,
I want to download the text from a web page, however what i end up
with is the html code. Is there some option that i am missing in the
RCurl package? Or is there another way to achieve this? This is the
code i am using:
library(RCurl)
html.file - getURI(my.url, ssl.verifyhost = FALSE, ssl.verifypeer = FALSE,
followlocation = TRUE)
print(html.file)
I thought perhaps the htmlTreeParse() function from the XML package
might help, but I just don't know what to do next with it:
library(XML)
htmlTreeParse(html.file)
Many thanks for any help you can provide,
Sounds like you're on the right track. One way is to parse the html
file into its 'internal' representation, and then use xpathApply to
extract relevant information (e.g., the third 'p' (paragraph) element
from the XML mark-up
html = htmlTreeParse(getURL(my.url), useInternal=TRUE)
Opening and ending tag mismatch: td and font
Unexpected end tag : p
Unexpected end tag : form
xpathApply(html, //p[3], xmlValue)
[[1]]
[1] You can subscribe to the list, or change your existing\r\n\t
subscription, in the sections below.\r\n\t
the 'xpath' is the path from the root of the document through various
nested tags to tags of the specified type. //p, says 'start at the
root ('/') and look in all sub-nodes (that this '//') for an 'p'
tag. ?xpathApply. is a good starting place, as is
http://www.w3.org/TR/xpath, especially
http://www.w3.org/TR/xpath#path-abbrev
Martin
Tony Breyal
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom.
1252;LC_MONETARY=English_United Kingdom.
1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods
base
other attached packages:
[1] XML_1.94-0 RCurl_0.9-4
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--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
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PO Box 19024 Seattle, WA 98109
Location: Arnold Building M2 B169
Phone: (206) 667-2793
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Re: [R] Load a program at the front end
Thanks a lot for the suggestion! Unfortunately R --no-save prog.R does not work well with my situation because prog.R contain lines such as readline() and () that require user response in the middle of the execution. I also tried other options such as R -f prog.R and R --interactive prog.R, and they all failed. Any other suggestions? Thanks, Gang On Mon, Oct 6, 2008 at 10:12 PM, Bernardo Rangel Tura [EMAIL PROTECTED] wrote: Em Qui, 2008-10-02 às 14:36 -0400, Gang Chen escreveu: I want to run a R program, prog.R, interactively. My question is, is there a way I can start prog.R on the shell terminal when invoking R, instead of using source() inside R? TIA, Gang Hi Gang I my system just only type: R --no-save prog.R platform x86_64-unknown-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status Patched major 2 minor 7.2 year 2008 month 09 day11 svn rev46532 language R version.string R version 2.7.2 Patched (2008-09-11 r46532) -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ 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] sort a list?
I am trying to sort a list and the data is obiously not in the right format. I am trying: x - list() x[[A]] - 1 x[[B]] - 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin __ 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] R squared value for a line on a plot
Does anyone know how to retrieve the R squared value for a line on a graph? Thanks Georgina __ 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] vectorization of a loop for mahalanobis distance calculation
Dear all,
We have a data frame x with n people as rows and k variables as columns.
Now, for each person (i.e., each row) we want to calculate a distance
between him/her and EACH other person in x. In other words, we want to
create a n x n matrix with distances (with zeros in the diagonal).
However, we do not want to calculate Euclidian distances. We want to calculate
Mahalanobis distances, which take into account the covariance among
variables.
Below is the piece of code we wrote (covmat in the function below is the
variance-covariance matrix among variables in Data that has to be fed into
mahalonobis function we are using).
mahadist = function(x, covmat) {
dismat = matrix(0,ncol=nrow(x),nrow=nrow(x))
for (i in 1:nrow(x)) {
dismat[i,] = mahalanobis(as.matrix(x), as.matrix(x[i,]), covmat)^.5
}
return(dismat)
}
This piece of code works, but it is very slow. We were wondering if it's at
all possible to somehow vectorize this function. Any help would be greatly
appreciated.
Thanks,
Frank
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[R] Algorithm = port convergence codes
Hello all, I am fitting a Gamma distribution to some data I have using nls(). The function obviously runs into issues when using a 0 as a parameter value. I understand the line alg = port can be used to set the lower bounds to prevent this from happening. When I run the code I get the following: Algorithm port, convergence message: relative convergence (4). I have attempted changing the starting parameter values and the paramter estimates appear robust. The R help file mentions the code is 0 for converence. Should I be concerned here that nls() is returning a local optimum here, or is this quite normal? Thanks in advance! Rob PS Please try to ignore the fact a distribution is being fitted to too few data points! ##Code Acancellata.caughtintraps = c(6,2,1,2,0,0,0,0,0,0,0) Acancellata.released = 1250 distances = c(2,3,4,5,7.5,10,12.5,15,17.5,20,25) totalseedscaught = sum(Acancellata.caughtintraps) Pcatch = totalseedscaught/Acancellata.released ## Calculate cumulative seed numbers reaching each distance cumsumcaught - rev(cumsum(rev(Acancellata.caughtintraps))) i = numeric(11) i - distances ## Gamma distribution gamma.nls - nls(cumsumcaught ~ Pcatch*Acancellata.released* pgamma(distances,shape=a,scale=c,lower.tail=FALSE), ## 'alg=port' is required for lower bounds ## alg=port, start = list(a=5,c=1), lower=list(a=0.1,b=0.1)) print(gamma.nls) print(summary(gamma.nls)) ## Output [1] Gamma Nonlinear regression model model: cumsumcaught ~ Pcatch * Acancellata.released * pgamma(distances, shape = a, scale = c, lower.tail = FALSE) data: parent.frame() a c 7.5731 0.4399 residual sum-of-squares: 4.176 Algorithm port, convergence message: relative convergence (4) Formula: cumsumcaught ~ Pcatch * Acancellata.released * pgamma(distances, shape = a, scale = c, lower.tail = FALSE) Parameters: Estimate Std. Error t value Pr(|t|) a 7.5731 2.3786 3.184 0.0111 * c 0.4399 0.1415 3.109 0.0125 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.6812 on 9 degrees of freedom Algorithm port, convergence message: relative convergence (4) -- View this message in context: http://www.nabble.com/Algorithm-%3D-%22port%22-convergence-codes-tp19860224p19860224.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] R and Multi threading
I will preface this message by saying that I am not an R developer and no very little about R...but here is my situation: One of my users has developed a model for analysing commodity prices. At the moment when he runs this model on his daily data set it takes roughly 5 hours to complete. He is using a quad core PC with 2gb of RAM. The R process only uses 1 core..i.e. the overall CPU usage tops out at around 25%. This has been a managable situation for a while, but he would now like to run this model on 5 years of historical data. He has a colleague who ran the model on a 16 core Redhat Linux box, but it took even longer to run. He has asked me for assistance in speeding up this process. I have a couple of questions: 1) Is is possible to run the Windows version of R across all four processors? 2) I was under the impression that R for Linux supported multi-threading by default. Am I correct in this assumption? If not, is it possible for Linux R to multi thread, and how do I go about configuring this? Apologies for the lack of detailed info in this post. I work in trade floor support and engineering and we dont really have much demand for this kind of heavy duty computational work so I am learning as I investigate this issue. Regards pejpm -- View this message in context: http://www.nabble.com/R-and-Multi-threading-tp19861868p19861868.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] Load a program at the front end
Something like env R_PROFILE=prog.R R may work for you. You may need to call .First.sys at the beginning of prog.R to get default packages loaded. luke On Tue, 7 Oct 2008, Gang Chen wrote: Thanks a lot for the suggestion! Unfortunately R --no-save prog.R does not work well with my situation because prog.R contain lines such as readline() and () that require user response in the middle of the execution. I also tried other options such as R -f prog.R and R --interactive prog.R, and they all failed. Any other suggestions? Thanks, Gang On Mon, Oct 6, 2008 at 10:12 PM, Bernardo Rangel Tura [EMAIL PROTECTED] wrote: Em Qui, 2008-10-02 às 14:36 -0400, Gang Chen escreveu: I want to run a R program, prog.R, interactively. My question is, is there a way I can start prog.R on the shell terminal when invoking R, instead of using source() inside R? TIA, Gang Hi Gang I my system just only type: R --no-save prog.R platform x86_64-unknown-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status Patched major 2 minor 7.2 year 2008 month 09 day11 svn rev46532 language R version.string R version 2.7.2 Patched (2008-09-11 r46532) -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ 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. -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: [EMAIL PROTECTED] Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu__ 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] vectorization of a loop for mahalanobis distance calculation
distance() from the ecodist package will calculate Mahalanobis distances. Sarah -- 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] Extracting text from html code using the RCurl package.
I gather you are using Windows and in that case you could use RDCOMClient or rcom to get it via Internet Explorer, e.g. library(RDCOMClient) ie - COMCreate(InternetExplorer.Application) URL - https://stat.ethz.ch/mailman/listinfo/r-help; ie$Navigate(URL) while(ie[[Busy]]) Sys.sleep(1) txt - ie[[document]][[body]][[innerText]] ie$Quit() You may need to run this in elevated mode if you are Vista. On Mon, Oct 6, 2008 at 11:45 AM, Tony Breyal [EMAIL PROTECTED] wrote: Dear R-help, I want to download the text from a web page, however what i end up with is the html code. Is there some option that i am missing in the RCurl package? Or is there another way to achieve this? This is the code i am using: library(RCurl) my.url - 'https://stat.ethz.ch/mailman/listinfo/r-help' html.file - getURI(my.url, ssl.verifyhost = FALSE, ssl.verifypeer = FALSE, followlocation = TRUE) print(html.file) I thought perhaps the htmlTreeParse() function from the XML package might help, but I just don't know what to do next with it: library(XML) htmlTreeParse(html.file) Many thanks for any help you can provide, Tony Breyal sessionInfo() R version 2.7.2 (2008-08-25) i386-pc-mingw32 locale: LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United Kingdom. 1252;LC_MONETARY=English_United Kingdom. 1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] XML_1.94-0 RCurl_0.9-4 __ 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] sort a list?
Try this: x[order(unlist(x), decreasing = TRUE)] On Tue, Oct 7, 2008 at 1:28 PM, [EMAIL PROTECTED] wrote: I am trying to sort a list and the data is obiously not in the right format. I am trying: x - list() x[[A]] - 1 x[[B]] - 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] sort a list?
Exactly what were your expections from sorting the list? What did you expect the answer to look like? You can 'unlist' the list and then sort the elements: x[[A]] - 1:10 x[[B]] - 4:12 sort(unlist(x)) A1 A2 A3 A4 B1 A5 B2 A6 B3 A7 B4 A8 B5 A9 B6 A10 B7 B8 B9 1 2 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 12 Is this what you want? On Tue, Oct 7, 2008 at 12:28 PM, [EMAIL PROTECTED] wrote: I am trying to sort a list and the data is obiously not in the right format. I am trying: x - list() x[[A]] - 1 x[[B]] - 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin __ 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 Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Creating GUIs for R
Hi all, I have looked around for help on creating GUIs for R, but haven't found anything. I would be interested in any advice or webpages that have information on the best language, tutorials etc. for creating simple GUIs. Mainly I want to do this as a heuristic exercise. Thanks for any help. Wade Wall [[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] sort a list?
Since objects of class list in R can be made up of heterogeneous objects, sorting them does not make much sense. For example, does A, come before or after 1000, does a linear model summary come before or after pi? If your data are all numeric, store them as a numeric vector, where sort works. Vectors can be named in R, as in your example. Also, 'order' does something different than 'sort'. try the following: x - c(a = 1, b = 4, c = 2) sort(x) order(x) Hope that helps, Erik [EMAIL PROTECTED] wrote: I am trying to sort a list and the data is obiously not in the right format. I am trying: x - list() x[[A]] - 1 x[[B]] - 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin __ 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] R squared value for a line on a plot
Georgina Sarah Humphreys g.humphreys.1 at research.gla.ac.uk writes: Does anyone know how to retrieve the R squared value for a line on a graph? Thanks Georgina We need a LOT more information to answer your question ... please read the posting guide and tell us in a bit more detail what you're trying to do. It sounds like you have a graphical figure of a regression line in a paper and want to know the R^2, which is basically impossible unless you have the original data, or unless the plot shows the data points and you can retrieve by digitizing (look for the g3data software tool). Ben Bolker __ 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] vectorization of a loop for mahalanobis distance calculation
Hi Frank,
If the way distance() calculates the Mahalanobis distance meets your
needs other than the covariance specification, you can tweak that
_very_ easily. If you use fix(distance) at the command line, you can
edit the source.
change the first line to:
function (x, method = euclidean, icov)
and under method 4, change the icov calculation to:
if(missing(icov)) {
icov - solve(cov(x))
}
Alternatively, here's a simplified distanceM function with everything but the
relevant bits deleted. You'll still need to have ecodist loaded.
distanceM - function (x, method = mahalanobis, icov)
{
paireddiff - function(x) {
N - nrow(x)
P - ncol(x)
A - numeric(N * N * P)
A - .C(pdiff, as.double(as.vector(t(x))), as.integer(N),
as.integer(P), A = as.double(A), PACKAGE = ecodist)$A
A - array(A, dim = c(N, N, P))
A
}
x - as.matrix(x)
N - nrow(x)
P - ncol(x)
if(missing(icov)) {
icov - solve(cov(x))
}
A - paireddiff(x)
A1 - apply(A, 1, function(z) (z %*% icov %*% t(z)))
D - A1[seq(1, N * N, by = (N + 1)), ]
D - D[col(D) row(D)]
attr(D, Size) - N
attr(D, Labels) - rownames(x)
attr(D, Diag) - FALSE
attr(D, Upper) - FALSE
attr(D, method) - METHODS[method]
class(D) - dist
D
}
Sarah
On Tue, Oct 7, 2008 at 1:05 PM, Frank Hedler [EMAIL PROTECTED] wrote:
Dear all,
we just realized something. Sarah's distance function - indeed - calculates
mahalanobis distance very well. However, it uses the
observed variance-covariance matrix by default.
What we actually need (sorry for not stating it clearly in to be able to
specify which variance-covariance matrix goes into that calculation.
On Tue, Oct 7, 2008 at 12:44 PM, Sarah Goslee [EMAIL PROTECTED]
wrote:
distance() from the ecodist package will calculate Mahalanobis distances.
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating GUIs for R
Wade, What type of GUI do you want? Do you want a full GUI that the user runs to do everything (that uses R as the computational engine)? Look at R commander, JGR, and the R plugin for Excel as possible examples. Are you more interested in a simple GUI for one specific plot or function to run from within R? Some examples include the tkexamp function in the TeachingDemos package (or many of the functions in that package for examples), the fgui package, the playwith package, the tkrplot package, and the details of the tcltk and RGtk2 packages. There are other packages that help with web interfaces (not GUIs, but still show using R as a back end). I don't know of any specific documents on how to write either type of gui, but the above can give examples to start from. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Wade Wall Sent: Tuesday, October 07, 2008 10:56 AM To: [EMAIL PROTECTED] Subject: [R] Creating GUIs for R Hi all, I have looked around for help on creating GUIs for R, but haven't found anything. I would be interested in any advice or webpages that have information on the best language, tutorials etc. for creating simple GUIs. Mainly I want to do this as a heuristic exercise. Thanks for any help. Wade Wall [[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] R squared value for a line on a plot
Georgina Sarah Humphreys g.humphreys.1 at research.gla.ac.uk writes: Does anyone know how to retrieve the R squared value for a line on a graph? The balloon rule http://www.jstor.org/pss/2683562 Dieter __ 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] Creating GUIs for R
Seek and ye shall find ... Check the RGUI's link on the other web page on CRAN. If you are on Windows, there is some simple built-in GUI functionality. ?winMenuAdd, ?select.list and the links therein will get you started there. Cheers, Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wade Wall Sent: Tuesday, October 07, 2008 9:56 AM To: [EMAIL PROTECTED] Subject: [R] Creating GUIs for R Hi all, I have looked around for help on creating GUIs for R, but haven't found anything. I would be interested in any advice or webpages that have information on the best language, tutorials etc. for creating simple GUIs. Mainly I want to do this as a heuristic exercise. Thanks for any help. Wade Wall [[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] panel.groups: use group.number to define colors
Not exactly sure what you want to do, but ...
In your example, you do not need groups, since the color doesn't change
within the levels of the conditioning variable (fact). Hence you can use the
panel.number() function to choose the plotting color of each panel, like
this:
## .. continuing with your example
myColors - rep(c(2,4),2)
xyplot(y ~ x | fact, data = my.df,
panel = function(...){
panel.xyplot(...,col=myColors[panel.number()])
}
)
If you actually **need** groups (to color different subsets of the data
within a panel differently). it does get a bit more complicated.
Incidentally, note that col is already a formal argument of panel.superpose
and was therefore picked up in the ... argument of the panel.groups
function -- that's why you got the error you did when you repeated col
explicitly in the panel.xyplot call. By default, it's values are those of
trellis.par.get(superpose.symbol) I believe. Also, group.number appears to
be undefined in your code.
HTH
Cheers,
Bert Gunter
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating GUIs for R
Sorry that my post wasn't very clear. What I am wanting to do is learn to build some simple GUIs for a limited number of functions. Basically, I am envisioning a screen with check boxes, a drop down menu etc. that users could select to run analyses on imported data. I have worked with VB before, but it has been several years and I am not sure how it interfaces with R. On Tue, Oct 7, 2008 at 1:20 PM, Bert Gunter [EMAIL PROTECTED] wrote: Seek and ye shall find ... Check the RGUI's link on the other web page on CRAN. If you are on Windows, there is some simple built-in GUI functionality. ?winMenuAdd, ?select.list and the links therein will get you started there. Cheers, Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wade Wall Sent: Tuesday, October 07, 2008 9:56 AM To: [EMAIL PROTECTED] Subject: [R] Creating GUIs for R Hi all, I have looked around for help on creating GUIs for R, but haven't found anything. I would be interested in any advice or webpages that have information on the best language, tutorials etc. for creating simple GUIs. Mainly I want to do this as a heuristic exercise. Thanks for any help. Wade Wall [[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] vectorization of a loop for mahalanobis distance calculation
Dear all,we just realized something. Sarah's distance function - indeed -
calculates mahalanobis distance very well. However, it uses the
observed variance-covariance matrix by default.
What we actually need (sorry for not stating it clearly in to be able to
specify which variance-covariance matrix goes into that calculation.
On Tue, Oct 7, 2008 at 12:44 PM, Sarah Goslee [EMAIL PROTECTED]wrote:
distance() from the ecodist package will calculate Mahalanobis distances.
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
ORIGINAL request:
Dear all,
We have a data frame x with n people as rows and k variables as columns.
Now, for each person (i.e., each row) we want to calculate a distance
between him/her and EACH other person in x. In other words, we want to
create a n x n matrix with distances (with zeros in the diagonal).
However, we do not want to calculate Euclidian distances. We want to calculate
Mahalanobis distances, which take into account the covariance among
variables.
Below is the piece of code we wrote (covmat in the function below is the
variance-covariance matrix among variables in Data that has to be fed into
mahalonobis function we are using).
mahadist = function(x, covmat) {
dismat = matrix(0,ncol=nrow(x),nrow=nrow(x))
for (i in 1:nrow(x)) {
dismat[i,] = mahalanobis(as.matrix(x), as.matrix(x[i,]), covmat)^.5
}
return(dismat)
}
This piece of code works, but it is very slow. We were wondering if it's at
all possible to somehow vectorize this function.
Any help would be greatly appreciated.
Thanks,
Frank
[[alternative HTML version deleted]]
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Re: [R] Mirror Image on Biplot Graphic
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically you
should use the extractor function scores() to extract what you want from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot, reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your
plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different
ordination
packages, in most of which this is possible). What you need to do with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something
like
the following, where my.cca is the orginal object and yax = obj[, 2]
(xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan package).
I
need to rotate on y axis the lines and symbols of constrained and
sites
representation. If I do that on an image editor, I rotate everything,
including titles, axes labels and positions. I just need to rotate
the
inner par and keep the variables names (constrained) and symbols in
the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
plot.cca(CAPpotiFT,type='none',display=c('bp','sites'),main='Total
Fauna
Sites x Environment')
text.cca(CAPpotiFT,dis='cn',col='black',cex=0.6,lwd=0.5,lty='dotted')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='1',],pch=21)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='2',],pch=20)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='3',],pch=22,bg='gray')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='4',],pch=24)
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='5',],pch=24,bg='black')
points(ScoresCAPFT$sites[FatoresRep$BiofaciesBC=='6',],pch=25,bg='gray')
Thanks in advice.
___
MSc. Rodrigo Aluizio
Centro de Estudos
[R] Statistically significant in linear and non-linear model
Hi, I have a question to ask. if in a linear regression model, the independent variables are not statistically significant, is it necessary to test these variables in a non-linear model? Since most of non-linear form of a variable can be represented to a linear combination using Taylor's theorem, so I wonder whether the non-linear form is also not statistically significant in such a situation. Best Regards Hsiao-nan Cheung 2008/10/08 [[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] sort a list?
THank you. I just didn't know the rules. In other languages it is possible to pass in a 'compare' function so the sort is defined by the function. I guess I was a) wondering why it failed, and b) if there was a similar work around to sort generic lists. Also I was specifically addressing the list that I gave in the example not a generic list. I think I have a solution. Thank you. Kevin Erik Iverson [EMAIL PROTECTED] wrote: Since objects of class list in R can be made up of heterogeneous objects, sorting them does not make much sense. For example, does A, come before or after 1000, does a linear model summary come before or after pi? If your data are all numeric, store them as a numeric vector, where sort works. Vectors can be named in R, as in your example. Also, 'order' does something different than 'sort'. try the following: x - c(a = 1, b = 4, c = 2) sort(x) order(x) Hope that helps, Erik [EMAIL PROTECTED] wrote: I am trying to sort a list and the data is obiously not in the right format. I am trying: x - list() x[[A]] - 1 x[[B]] - 2 order(x) But am getting: Error in order(x) : unimplemented type 'list' in 'orderVector1' How should I change the list so that it can be sorted? What kinds of objects (classes of objects) can be sorted? Thank you. Kevin __ 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] Programing and writing function help
Hello R users My goal is to use R to write functions for and automate a series of analyses i would like to do on a large data set. The calculations are not very difficult in themselves, however they will be very time consuming (Plus I think R will be extremely useful and this is another excuse to learn how to program). I have a vector of 20 values x - c(20,18, 45, 16, 47, 47, 15, 26, 14,14,12,16,35,27,18,94,16,26,26,30) 1. I want to select random pairs from this data set but do it without replacement exhaustively I know i can select random pairs without replacement using sample(N,n,replace=F) However i am wondering if there is any way to get 10 random pairs from this data set without repeating any of the data points that is to say if i got a (20, 94) for one pair, i would like to get 9 other pairs from the data without again getting 20 or 94? 2. The second thing i would like to do is be able to select all possible pairs of numbers and calculate each pairs variance. I think i will need a for loop here but I am unsure of how to do the programing. I am reading two books on S and s-plus now and hope they are some help. I thought i would also post on this list and get some expert advice for the more experienced R users Thank-you very much Stephen Cole Marine Ecology Lab Saint Francis Xavier University [[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] Load a program at the front end
Thanks for the suggestion, Luke! Your approach would load prog.R whenever invoking R, but that is not exactly what I want. Basically I can run prog.R through source(~/someDir/prog.R) inside R. Also since prog.R involves many lines of readline() and tclvalue() in tcltk package, the program requires the user responses in the middle of the execution. If possible, I would like to create an executable shell script RunProg with one line or two inside like R ... ~/someDir/prog.R If this is feasible, whenever I want to run prog.R, I can simply type the following at the terminal prompt: RunProg and I would still be able to interact with the user. Is this something doable? Thanks, Gang On Tue, Oct 7, 2008 at 12:43 PM, Luke Tierney [EMAIL PROTECTED] wrote: Something like env R_PROFILE=prog.R R may work for you. You may need to call .First.sys at the beginning of prog.R to get default packages loaded. luke On Tue, 7 Oct 2008, Gang Chen wrote: Thanks a lot for the suggestion! Unfortunately R --no-save prog.R does not work well with my situation because prog.R contain lines such as readline() and () that require user response in the middle of the execution. I also tried other options such as R -f prog.R and R --interactive prog.R, and they all failed. Any other suggestions? Thanks, Gang On Mon, Oct 6, 2008 at 10:12 PM, Bernardo Rangel Tura [EMAIL PROTECTED] wrote: Em Qui, 2008-10-02 às 14:36 -0400, Gang Chen escreveu: I want to run a R program, prog.R, interactively. My question is, is there a way I can start prog.R on the shell terminal when invoking R, instead of using source() inside R? TIA, Gang Hi Gang I my system just only type: R --no-save prog.R platform x86_64-unknown-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status Patched major 2 minor 7.2 year 2008 month 09 day11 svn rev46532 language R version.string R version 2.7.2 Patched (2008-09-11 r46532) -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ 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. -- Luke Tierney Chair, Statistics and Actuarial Science Ralph E. Wareham Professor of Mathematical Sciences University of Iowa Phone: 319-335-3386 Department of Statistics andFax: 319-335-3017 Actuarial Science 241 Schaeffer Hall email: [EMAIL PROTECTED] Iowa City, IA 52242 WWW: http://www.stat.uiowa.edu __ 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] Statistically significant in linear and non-linear model
Well here is one example where x is not significant in a linear regression: x - seq( -1,1, 0.1 ) y - x^2 + rnorm(21,0,.1) summary(lm(y~x)) Would you really want to dismiss any relationship (non-linear) between x and y based on the above p-value? -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Hsiao-nan Cheung Sent: Tuesday, October 07, 2008 11:47 AM To: R-help Subject: [R] Statistically significant in linear and non-linear model Hi, I have a question to ask. if in a linear regression model, the independent variables are not statistically significant, is it necessary to test these variables in a non-linear model? Since most of non-linear form of a variable can be represented to a linear combination using Taylor's theorem, so I wonder whether the non-linear form is also not statistically significant in such a situation. Best Regards Hsiao-nan Cheung 2008/10/08 [[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] Creating GUIs for R
I think prior posters addressed that or maybe what you really want is a command that accepts your function and automatically constructs a GUI front end for it. In that case see ggenericwidget in the gWidgets package and the fgui package. You don't need VB at all for this. On Tue, Oct 7, 2008 at 1:32 PM, Wade Wall [EMAIL PROTECTED] wrote: Sorry that my post wasn't very clear. What I am wanting to do is learn to build some simple GUIs for a limited number of functions. Basically, I am envisioning a screen with check boxes, a drop down menu etc. that users could select to run analyses on imported data. I have worked with VB before, but it has been several years and I am not sure how it interfaces with R. On Tue, Oct 7, 2008 at 1:20 PM, Bert Gunter [EMAIL PROTECTED] wrote: Seek and ye shall find ... Check the RGUI's link on the other web page on CRAN. If you are on Windows, there is some simple built-in GUI functionality. ?winMenuAdd, ?select.list and the links therein will get you started there. Cheers, Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wade Wall Sent: Tuesday, October 07, 2008 9:56 AM To: [EMAIL PROTECTED] Subject: [R] Creating GUIs for R Hi all, I have looked around for help on creating GUIs for R, but haven't found anything. I would be interested in any advice or webpages that have information on the best language, tutorials etc. for creating simple GUIs. Mainly I want to do this as a heuristic exercise. Thanks for any help. Wade Wall [[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. __ 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] Statistically significant in linear and non-linear model
On 07-Oct-08 17:46:52, Hsiao-nan Cheung wrote: Hi, I have a question to ask. if in a linear regression model, the independent variables are not statistically significant, is it necessary to test these variables in a non-linear model? Since most of non-linear form of a variable can be represented to a linear combination using Taylor's theorem, That depends on the coefficients in the Taylor's series expansion. It is quite possible to have the linear coefficient zero, and the quadratic coefficient non-zero. so I wonder whether the non-linear form is also not statistically significant in such a situation. Best Regards Hsiao-nan Cheung 2008/10/08 Example: X - 0.2*((-10):10) Y - 0.5*(X^2) + 0.2*rnorm(21) X2 - X^2 [A] Linear regression, Y on X: summary(lm(Y ~ X))$coef # Estimate Std. Error t value Pr(|t|) # (Intercept) 0.72840442 0.1554215 4.6866382 0.0001606966 # X 0.06570652 0.1283351 0.5119919 0.6145564688 So the coefficient of X is not significant. [B] Quadratic regression, Y on X and X^2: summary(lm(Y ~ X + X2))$coef #Estimate Std. Error t value Pr(|t|) # (Intercept) 0.003425041 0.07203265 0.04754846 9.625997e-01 # X 0.065706524 0.03957727 1.66020864 1.141924e-01 # X2 0.494304121 0.03666239 13.48259513 7.570563e-11 So the coefficient of X is still not significant (P = 0.14), but the coefficient of X^2 is *highly* significant! So it all depends ... of course the original coefficients (Taylor) could be anything. Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 07-Oct-08 Time: 19:16:04 -- XFMail -- __ 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] panel.groups: use group.number to define colors
Hi, and thanks for your email,
I realise my example was not very good. The actual dataset I'm trying
to plot is rather big and this oversimplified example did not make
much sense.
I actually do need to color different subsets of the data differently
in each panel, that's why I thought of using panel.groups. Here's a
more realistic example:
x - seq(1, 2*pi, length=100)
numberOfCurves - 20
y1 - sapply(seq(0, pi/2, length=numberOfCurves), function(phi) sin(x
+phi))
y2 - sapply(seq(0, pi/2, length=numberOfCurves), function(phi) cos(x
+phi))
y - cbind(y1, y2)
fact - factor(rep(c(cos, sin), each=numberOfCurves*100))
fact2 - factor(rep(seq(0, pi, length=numberOfCurves), each=100,
length=2*numberOfCurves*100))
my.df - data.frame(x=rep(x, length=800), y=as.vector(y), fact =
fact, fact2 = fact2)
head(my.df)
myColors - c(grey, grey, red, rep(grey, ncol(y)-3))
xyplot(y ~ x | fact, data = my.df, groups = fact2, type=l,
par.settings=list(superpose.line=list(col=myColors, lwd=2)),
panel = panel.superpose,
panel.groups = function(..., group.number) {
panel.xyplot(...)
})
Two things I don't like about my approach:
- I'd rather select the colors in the panel function than set a
specific palette in par.settings, as it's not obvious to me what the
order of the plotting will be. This is where I fail to use
group.number correctly
- the purpose of the red line is make this particular curve stand out
from the mess of grey curves. However, they partially cover it and I
don't really know how to change the plotting order (or replot the red
one only on top if it's any easier)
Hope this is a bit clearer,
Best regards,
baptiste
On 7 Oct 2008, at 18:25, Bert Gunter wrote:
Not exactly sure what you want to do, but ...
In your example, you do not need groups, since the color doesn't
change
within the levels of the conditioning variable (fact). Hence you can
use the
panel.number() function to choose the plotting color of each panel,
like
this:
## .. continuing with your example
myColors - rep(c(2,4),2)
xyplot(y ~ x | fact, data = my.df,
panel = function(...){
panel.xyplot(...,col=myColors[panel.number()])
}
)
If you actually **need** groups (to color different subsets of the
data
within a panel differently). it does get a bit more complicated.
Incidentally, note that col is already a formal argument of
panel.superpose
and was therefore picked up in the ... argument of the panel.groups
function -- that's why you got the error you did when you repeated col
explicitly in the panel.xyplot call. By default, it's values are
those of
trellis.par.get(superpose.symbol) I believe. Also, group.number
appears to
be undefined in your code.
HTH
Cheers,
Bert Gunter
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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and provide commented, minimal, self-contained, reproducible code.
[R] weighted quantiles
I have a set of values and their corresponding weights. I can use the function weighted.mean to calculate the weighted mean, I would like to be able to similarly calculate the weighted median and quantiles? Is there a function in R that can do this? thanks, Spencer [[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] Programing and writing function help
Here is one way that might work for you: x - c(20,18, 45, 16, 47, 47, 15, 26, 14,14,12,16,35,27,18,94,16,26,26,30) # create a matrix of pairs matrix(sample(x), ncol=2) [,1] [,2] [1,] 14 16 [2,] 27 20 [3,] 26 16 [4,] 47 15 [5,] 35 47 [6,] 16 18 [7,] 14 12 [8,] 18 26 [9,] 45 26 [10,] 94 30 # all possible pairs # create a matrix of the dimensions x*x and the the indices of the lower triangle ind - which(lower.tri(matrix(1,length(x), length(x))), arr.ind=TRUE) # create the pairs x.pairs - cbind(x[ind[,1]], x[ind[,2]]) str(x.pairs) num [1:190, 1:2] 18 45 16 47 47 15 26 14 14 12 ... head(x.pairs) [,1] [,2] [1,] 18 20 [2,] 45 20 [3,] 16 20 [4,] 47 20 [5,] 47 20 [6,] 15 20 On Tue, Oct 7, 2008 at 1:50 PM, Stephen Cole [EMAIL PROTECTED] wrote: Hello R users My goal is to use R to write functions for and automate a series of analyses i would like to do on a large data set. The calculations are not very difficult in themselves, however they will be very time consuming (Plus I think R will be extremely useful and this is another excuse to learn how to program). I have a vector of 20 values x - c(20,18, 45, 16, 47, 47, 15, 26, 14,14,12,16,35,27,18,94,16,26,26,30) 1. I want to select random pairs from this data set but do it without replacement exhaustively I know i can select random pairs without replacement using sample(N,n,replace=F) However i am wondering if there is any way to get 10 random pairs from this data set without repeating any of the data points that is to say if i got a (20, 94) for one pair, i would like to get 9 other pairs from the data without again getting 20 or 94? 2. The second thing i would like to do is be able to select all possible pairs of numbers and calculate each pairs variance. I think i will need a for loop here but I am unsure of how to do the programing. I am reading two books on S and s-plus now and hope they are some help. I thought i would also post on this list and get some expert advice for the more experienced R users Thank-you very much Stephen Cole Marine Ecology Lab Saint Francis Xavier University [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Mirror Image on Biplot Graphic
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you are
reversing the axes you want. In your original post you mention axis 2, i.e.
y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically you
should use the extractor function scores() to extract what you want from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing
axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot,
reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your
plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] - scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different
ordination
packages, in most of which this is possible). What you need to do with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something
like
the following, where my.cca is the orginal object and yax = obj[, 2]
(xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan
package).
I
need to rotate on y axis the lines and symbols of constrained and
sites
representation. If I do that on an image editor, I rotate
everything,
including titles, axes labels and positions. I just need to rotate
the
inner par and keep the variables names (constrained) and symbols in
the
new positions but with the right direction.
So, is there on R a way to do that while creating the image?
Here is the code that generate the graphic.
Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you are
reversing the axes you want. In your original post you mention axis 2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically you
should use the extractor function scores() to extract what you want from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing
axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot,
reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want
reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your
plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different
ordination
packages, in most of which this is possible). What you need to do
with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2] * -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained
and
sites representation.
Easiest is to multiply the axis you want to invert by -1. Something
like
the following, where my.cca is the orginal object and yax = obj[, 2]
(xax
being obj[, 1]). Obviously, copying isn't necessary.
mynew.cca - my.cca
mynew.cca$scores$species[, 2] - my.cca$scores$species[, 2] * -1
mynew.cca$scores$sites[, 2] - my.cca$scores$sites[, 2] * -1
mynew.cca$scores$centroids[, 2] - my.cca$scores$centroids[, 2] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
He everybody,
Well I have a biplot CCA-like origined from plot.cca (vegan
package).
I
need to rotate on y axis the lines and symbols of constrained and
sites
representation. If I do that on an image editor, I rotate
everything,
Re: [R] weighted quantiles
sj wrote: I have a set of values and their corresponding weights. I can use the function weighted.mean to calculate the weighted mean, I would like to be able to similarly calculate the weighted median and quantiles? Is there a function in R that can do this? thanks, Spencer library(Hmisc) ?wtd.quantile __ 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] read.spss: variable.labels
[EMAIL PROTECTED] wrote:
Hi,
how can I attach variable labels originally read by read.spss() to the
resulting variables?
pre
X - read.spss('data.sav', use.value.labels = TRUE, to.data.frame =
TRUE, trim.factor.names = TRUE, trim_values = TRUE, reencode = UTF-8)
names(X) - tolower(names(X))
attach(X)
/pre
Thank you
Sören
library(Hmisc)
?spss.get
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
__
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] weighted quantiles
Try the linear quantile regression function rq() in the quantreg package. For 1 sample estimates, your model would have just an intercept term. There is a weight argument. quantiles.out - rq(y ~ 1, data=mydata, tau=1:9/10, weight=myweights) would yield the 0.10, 0.20, ..., 0.80, 0.90 weighted quantile estimates. Brian Brian S. Cade, PhD U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: [EMAIL PROTECTED] tel: 970 226-9326 sj [EMAIL PROTECTED] Sent by: [EMAIL PROTECTED] 10/07/2008 12:38 PM To r-help [EMAIL PROTECTED] cc Subject [R] weighted quantiles I have a set of values and their corresponding weights. I can use the function weighted.mean to calculate the weighted mean, I would like to be able to similarly calculate the weighted median and quantiles? Is there a function in R that can do this? thanks, Spencer [[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] Mirror Image on Biplot Graphic
Sorry for the mistake, I'm trying to rotate the CAP 1 axis not 2!
Ah well, I tried the this other way, but it gave me an error.
Something in the syntax, I tried to change something but didn't fix the
error.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1
Erro: unexpected ',' in for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of limits)
Well, it needs lots of patience...
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 3:50 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2]
* -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you are
reversing the axes you want. In your original post you mention axis 2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically you
should use the extractor function scores() to extract what you want
from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing
axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot,
reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want
reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your
plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the suggestion:
CAPpotiFTI-CAPpotiFT
scores(CAPpotiFTI)$species[,2]-scores(CAPpotiFTI)$species[,2]*-1
scores(CAPpotiFTI)$sites[,2]-scores(CAPpotiFTI)$sites[,2]*-1
Any ideas? I'm trying similar things but without success.
--
Sent: Tuesday, October 07, 2008 10:34 AM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry, that will not return a result (I use several different
ordination
packages, in most of which this is possible). What you need to do
with
vegan
is the following:
scores(mynew.cca)$species[, 2] - scores(mynew.cca)$species[, 2]
* -1
You will be able to do the rest.
Regards, Mark.
Mark Difford wrote:
Hi Rodrigo,
I need to rotate on y axis the lines and symbols of constrained
and
sites representation.
Easiest is to multiply the axis you want to invert by -1.
[R] RBloomberg - Converting international stock prices into $US
To all: I'm using RBloomberg to pull historical equity prices across a range of international markets. Bloomberg defaults to returning stock prices to R in local currency, for example, blpGetData(conn, ALUA AR Equity, PX_LAST,start=09/30/08, end=09/30/08) returns a stock price in Argentine Peso's. If ones use the BLPH function directly in Excel to do this pull, you can change the currency base using the CCY parameter set to USD. How would I do something comparable in R with the blpGetHistoricalData function of RBloomberg?? Obviously, I could download exchange rates and do the calculations myself but was hoping to avoid this. Thanks very much in advance, Dan LaPushin [EMAIL PROTECTED] [[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 validate model?
[EMAIL PROTECTED] wrote: Hi Frank, Thanks for your feedback! But I think we are talking about two different things. 1) Validation: The generalization performance of the classifier. See, for example, Studies on the Validation of Internal Rating Systems by BIS. I didn't think the desire was for a classifier but instead was for a risk predictor. If prediction is the goal, classification methods or accuracy indexes based on classifications do not work very well. 2) Calibration: Correct calibration of a PD rating system means that the calibrated PD estimates are accurate and conform to the observed default rates. See, for instance, An Overview and Framework for PD Backtesting and Benchmarking, by Castermans et al. I'm unclear on what you mean here. Correct calibration of a predictive system means that the UNcalibrated estimates are accurate (i.e., they don't need any calibration). (What is PD?) Frank, you are referring the #1 and I am referring to #2. Nonetheless, I would never create a rating system if my model doesn't discriminate better than a coin toss. For sure Frank Regards, Pedro -Original Message- From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 07, 2008 11:02 AM To: Rodriguez, Pedro Cc: [EMAIL PROTECTED]; r-help@r-project.org Subject: Re: [R] How to validate model? [EMAIL PROTECTED] wrote: Usually one validates scorecards with the ROC curve, Pietra Index, KS test, etc. You may be interested in the WP 14 from BIS (www.bis.org). Regards, Pedro No, the validation should be done using an absolute reliability (calibration) curve. You need to verify that at all levels of predicted risk there is agreement with the true probability of failure. An ROC curve does not do that, and I doubt the others do. A resampling-corrected loess calibration curve is a good approach as implemented in the Design package's calibrate function. Frank -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Maithili Shiva Sent: Tuesday, October 07, 2008 8:22 AM To: r-help@r-project.org Subject: [R] How to validate model? Hi! I am working on scorecard model and I have arrived at the regression equation. I have used logistic regression using R. My question is how do I validate this model? I do have hold out sample of 5000 customers. Please guide me. Problem is I had never used Logistic regression earlier neither I am used to credit scoring models. Thanks in advance Maithili __ 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. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Mirror Image on Biplot Graphic
Hi Rodrigo,
Yes it does, but this will work. You have erroneously used a , in your
call, it should be a ., if anything. You have also inserted cca... And
the axes being switched must be conformant:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1 ## you are
writing axis 1 over axis 2; and there is no ...,cca...
Erro: unexpected ',' in for(i in c(2:8,15)) ## unexpected
comma
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
## Using your example to switch axis 1. Make sure that CAPpotiFT exists and
is your
## original object. Copy and paste the lines below into your R console
window and run
CAPpotiFTI - CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] * -1
Cheers, Mark.
Rodrigo Aluizio wrote:
Sorry for the mistake, I'm trying to rotate the CAP 1 axis not 2!
Ah well, I tried the this other way, but it gave me an error.
Something in the syntax, I tried to change something but didn't fix the
error.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1
Erro: unexpected ',' in for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of limits)
Well, it needs lots of patience...
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 3:50 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You
need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2]
* -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you are
reversing the axes you want. In your original post you mention axis 2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically
you
should use the extractor function scores() to extract what you want
from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing
axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot,
reversing
orientations along the way. So do (my.cca is your original object):
Temp.data - scores(my.cca, display=c(sites, species, cn))
## Reverse yaxes
Temp.data$sites[, 2] - Temp.data$sites[, 2] * -1
Temp.data$species[, 2] - Temp.data$species[, 2] * -1
Temp.data$centroid[, 2] - Temp.data$centroid[, 2] * -1
Do this for all the things you are plotting whose axes you want
reversed.
Check this against your original plot to ensure that the mirror isn't
broken.
You can then use the Temp.data with the scores() function to make your
plot.
## Mock e.g.
points(scores(Temp.data, display=sites), pch=21)
Regards, Mark.
Rodrigo Aluizio wrote:
Thanks a lot for your help (again) Mark.
I adapted your suggestion to my analysis.
But I'm getting and error when trying to apply the new values.
The error is:
Erro em scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
não foi posssível encontrar a função scores-
Translating:
Error at scores(CAPpotiFTI)$species[, 2] -
scores(CAPpotiFTI)$species[,
:
it's not possible to find the function scores-
Bellow how I applied the
Re: [R] Ecological Niche Modelling on R
that is a pretty big question. And I am not qualified to answer it, but my suggestion- from figuring out how to use other things- is to find out how other people have modeled ecological niches- not necessarily with R, and then try and find out functions that preform the pieces or the whole analysis. I believe there is a book out there about modeling ecological niches with R. I did a google search for that and this is what I found: http://www.zoology.ufl.edu/bolker/emdbook/. I don't know if this helps, but maybe it will. Good luck. On Tue, Oct 7, 2008 at 12:46 PM, milton ruser [EMAIL PROTECTED] wrote: Dear all, I have strong interest on Ecological Niche Model, which in general use a set of environmental variables (continuous, categorical etc) and Presence (or Presense/Absence) records for species. I think that grasp and adehabitat packages could help me on these tasks. My input layers are on ASC format, and the record of species is a data-frame with X, Y, name of species. Note that in general we don´t have real absence data. Could someone give some advice how can I use grasp and/or adehabitat to generate my predictive models? Best regards, Miltinho AStronauta brazil [[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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] panel.groups: use group.number to define colors
On Tue, Oct 7, 2008 at 8:54 AM, baptiste auguie [EMAIL PROTECTED] wrote:
Dear list,
I've been trying this for a few hours and I just don't understand how
lattice works with groups and subscripts.
Consider the following example,
xx - seq(1, 10, length=100)
x - rep(xx, 4)
y - c(cos(xx), sin(xx), xx, xx^2/10)
fact - factor(rep(c(cos, sin, id, square), each=100))
fact2 - factor(rep(c(periodic, not periodic), each=100))
my.df - data.frame(x=x, y=y, fact = fact, fact2 = fact2)
head(my.df)
myColors - c(2, 4)
xyplot(y ~ x | fact, data = my.df, groups = fact2,
panel = panel.superpose,
panel.groups = function(..., group.number) {
panel.xyplot(...)
#panel.xyplot( ..., col=myColors[group.number]) #
error
})
My aim is to assign a custom color to each group, but for some reason the
col parameter is already given to panel.xyplot and I can't find where it
gets the values from.
It get's the values from the 'panel.superpose' function, of which
'panel.groups' is an argument. Both the documentation and source code
for 'panel.superpose' should make this clear.
From what I understand, what you want should be as simple as (with a
small correction to your example):
xx - seq(1, 10, length=100)
x - rep(xx, 4)
y - c(cos(xx), sin(xx), xx, xx^2/10)
fact - factor(rep(c(cos, sin, id, square), each=100))
fact2 - factor(rep(c(periodic, not periodic), each=200))
my.df - data.frame(x=x, y=y, fact = fact, fact2 = fact2)
myColors - c(2, 4)
xyplot(y ~ x | fact, data = my.df, groups = fact2, col = myColors, type = l)
The use of 'par.settings' is not compulsory, but would help if you
needed to add a legend; e.g.,
xyplot(y ~ x | fact, data = my.df, groups = fact2, type = l,
par.settings = simpleTheme(col = myColors),
auto.key = list(lines = TRUE, points = FALSE))
If you insist on writing your own panel function, what you need is
xyplot(y ~ x | fact, data = my.df, groups = fact2,
panel = panel.superpose,
panel.groups = function(..., col.line, type, group.number) {
panel.xyplot(..., type = l,
col.line = myColors[group.number])
})
As Bert pointed out, you are responsible for ensuring that argument
names are not repeated by capturing and overriding them.
All of this applies to your second, more complicated, example as well.
As for the order in which the groups are plotted, it is the order of
levels(fact2), which seemed to me the most obvious (or at least the
least surprising) choice. You are free to specify the order when you
create the factor; see ?factor to learn how.
-Deepayan
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mac crash- Probably memory problem
not reproducible. On Tue, Oct 7, 2008 at 11:35 AM, dimitris kapetanakis [EMAIL PROTECTED] wrote: Dear all, I am running a code using bootstraps for estimating standard errors but the mac crashes. When I use small number of bootstraps (100) it works fine but if I increase that number it crashes. Thanks in advance dimitris The code, the error and my mac characteristics are the following: ##code# qr.1- rq(y~factor(year)+factor(state)+x1+I(x^2)+I(x^3), tau=0.05, data=data1) s.qr1.05 - summary(qr.1, se=boot, R=1000)$coefficient ##error# *** caught segfault *** address 0x1de9e000, cause 'memory not mapped' Traceback: 1: .Fortran(xys, as.integer(m), as.integer(n), as.integer(p), as.integer(R), as.integer(m + 5), as.integer(p + 2), as.double(x), as.double(y), as.double(tau), as.double(tol), flag = integer(R), coef = double(p * R), resid = double(m), integer(m), double((m + 5) * (p + 2)), double(m), xx = double(m * p), yy = double(m), as.integer(s), PACKAGE = quantreg) 2: boot.rq.xy(x, y, s, tau) 3: boot.rq(x, y, tau, ...) 4: summary.rq(qr.05.nox, se = boot, R = 1000) 5: summary(qr.05.nox, se = boot, R = 1000) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: ###characteristics# Mac OS X version 10.5.4 Model Name: MacBook Model Identifier: MacBook4,1 Processor Name: Intel Core 2 Duo Processor Speed: 2.4 GHz Number Of Processors: 1 Total Number Of Cores:2 L2 Cache: 3 MB Memory: 4 GB Bus Speed:800 MHz Boot ROM Version: MB41.00C1.B00 -- View this message in context: http://www.nabble.com/Mac-crash--Probably-memory-problem-tp19860892p19860892.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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] panel.groups: use group.number to define colors
Many thanks, I think I got the spirit of 'capturing and overriding'
the arguments which was the bit i was missing. It's much clearer now
with a working example.
Thanks again,
baptiste
On 7 Oct 2008, at 21:19, Deepayan Sarkar wrote:
On Tue, Oct 7, 2008 at 8:54 AM, baptiste auguie [EMAIL PROTECTED]
wrote:
Dear list,
I've been trying this for a few hours and I just don't understand how
lattice works with groups and subscripts.
Consider the following example,
xx - seq(1, 10, length=100)
x - rep(xx, 4)
y - c(cos(xx), sin(xx), xx, xx^2/10)
fact - factor(rep(c(cos, sin, id, square), each=100))
fact2 - factor(rep(c(periodic, not periodic), each=100))
my.df - data.frame(x=x, y=y, fact = fact, fact2 = fact2)
head(my.df)
myColors - c(2, 4)
xyplot(y ~ x | fact, data = my.df, groups = fact2,
panel = panel.superpose,
panel.groups = function(..., group.number) {
panel.xyplot(...)
#panel.xyplot( ...,
col=myColors[group.number]) #
error
})
My aim is to assign a custom color to each group, but for some
reason the
col parameter is already given to panel.xyplot and I can't find
where it
gets the values from.
It get's the values from the 'panel.superpose' function, of which
'panel.groups' is an argument. Both the documentation and source code
for 'panel.superpose' should make this clear.
From what I understand, what you want should be as simple as (with a
small correction to your example):
xx - seq(1, 10, length=100)
x - rep(xx, 4)
y - c(cos(xx), sin(xx), xx, xx^2/10)
fact - factor(rep(c(cos, sin, id, square), each=100))
fact2 - factor(rep(c(periodic, not periodic), each=200))
my.df - data.frame(x=x, y=y, fact = fact, fact2 = fact2)
myColors - c(2, 4)
xyplot(y ~ x | fact, data = my.df, groups = fact2, col = myColors,
type = l)
The use of 'par.settings' is not compulsory, but would help if you
needed to add a legend; e.g.,
xyplot(y ~ x | fact, data = my.df, groups = fact2, type = l,
par.settings = simpleTheme(col = myColors),
auto.key = list(lines = TRUE, points = FALSE))
If you insist on writing your own panel function, what you need is
xyplot(y ~ x | fact, data = my.df, groups = fact2,
panel = panel.superpose,
panel.groups = function(..., col.line, type, group.number) {
panel.xyplot(..., type = l,
col.line = myColors[group.number])
})
As Bert pointed out, you are responsible for ensuring that argument
names are not repeated by capturing and overriding them.
All of this applies to your second, more complicated, example as well.
As for the order in which the groups are plotted, it is the order of
levels(fact2), which seemed to me the most obvious (or at least the
least surprising) choice. You are free to specify the order when you
create the factor; see ?factor to learn how.
-Deepayan
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lme and lmer df's and F-statistics again
You may find this site useful: http://lme4.r-forge.r-project.org/bib/lme4bib.html On Tue, Oct 7, 2008 at 9:08 AM, Dieter Menne [EMAIL PROTECTED] wrote: Julia S. julia.schroeder at gmail.com writes: Now, I did that in my article and I got a response from a reviewer that I additionally should give the degrees of freedom, and the F-statistics. From what I read here, that would be incorrect to do, and I sort of intuitively also understand why (at least I think I do). ... Well, writing on my rebuttal, I find myself being unable to explain in a few, easy to understand (and, at the same time, correct) sentences stating that it is not a good idea to report (most likely wrong) dfs and F statistics. Can somebody here help me out with a correct explanation for a laymen? Feeling with you, and hoping some day this will be resolved. I am sure you have read Douglas Bates' http://finzi.psych.upenn.edu/R/Rhelp02a/archive/76742.html but I thought this was temporary. The only workaround I have is not to use lmer for gaussian models. Dieter __ 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] vectorized sub, gsub, grep, etc.
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement
for every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
-John
Confidentiality Notice: This e-mail message, including a...{{dropped:8}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mirror Image on Biplot Graphic
Well this time I have to assume, I'm not understanding what is wrong now.
And I have to say: 'Thank You for your patience', cause I'm going crazy
here! :-)
Mark, I tried this:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] * -1
and got this:
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of limits)
Here is the script until this point
library(vegan)
library(xlsReadWrite)
#FT--#
PotiAbioFT-read.xls('FatorialReplica.xls',sheet=4,rowNames=T)
PotiBioFT-read.xls('FatorialReplica.xls',sheet=6,rowNames=T)
attach(PotiAbioFT)
LogPotiBioFT-log(PotiBioFT+1)
CAPpotiFT-capscale(t(LogPotiBioFT)~Envoronmental+Variables,dist=bray,add=T)
PermCAPFT-anova.cca(CAPpotiFT,alpha=0.05,model='full',first=F,permutations=999)
PermCAPFT
summary(CAPpotiFT)
# Rotating the axis
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] * -1
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 5:17 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Yes it does, but this will work. You have erroneously used a , in your
call, it should be a ., if anything. You have also inserted cca... And
the axes being switched must be conformant:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1 ## you are
writing axis 1 over axis 2; and there is no ...,cca...
Erro: unexpected ',' in for(i in c(2:8,15)) ## unexpected
comma
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
## Using your example to switch axis 1. Make sure that CAPpotiFT exists
and
is your
## original object. Copy and paste the lines below into your R console
window and run
CAPpotiFTI - CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1]
* -1
Cheers, Mark.
Rodrigo Aluizio wrote:
[[elided Yahoo spam]]
Ah well, I tried the this other way, but it gave me an error.
Something in the syntax, I tried to change something but didn't fix the
error.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1
Erro: unexpected ',' in for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of
limits)
Well, it needs lots of patience...
--
Sent: Tuesday, October 07, 2008 3:50 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You
need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2]
* -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you
are
reversing the axes you want. In your original post you mention axis 2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+ Species x Environment')
Erro em match.arg(display) : 'arg' must be of length 1
or
text.cca(scores(CAPpotiFTI$biplot),col=323232,cex=0.6,lwd=2,lty='dotted')
Doesn't return an error, but it's not a mirror image as expected, it's
nonsense.
--
Sent: Tuesday, October 07, 2008 12:22 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Sorry again: the structure vegan uses is quite complex. Technically
you
should use the extractor function scores() to extract what you want
from
the
object, e.g. scores(my.cca, display=species) or scores(my.cca,
display=cn) [centroids]. I did work out a quicker way of reversing
axes,
but that seems to have gone walking...and I am not using vegan at the
moment.
Quickest hack is to make a copy of the objects you want to plot,
reversing
orientations along the way. So do (my.cca is your original object):
Temp.data -
Re: [R] vectorized sub, gsub, grep, etc.
Hi John,
Wouldn't you get the same with just mapply(sub, patt, repl, X) ?
Nael
On Tue, Oct 7, 2008 at 9:58 PM, Thaden, John J [EMAIL PROTECTED] wrote:
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement
for every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
-John
Confidentiality Notice: This e-mail message, including a...{{dropped:8}}
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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Re: [R] vectorized sub, gsub, grep, etc.
John,
Try the following:
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
b cda
aB CD ef
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Thaden, John J
Sent: Tuesday, October 07, 2008 3:59 PM
To: r-help@r-project.org
Cc: [EMAIL PROTECTED]
Subject: [R] vectorized sub, gsub, grep, etc.
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement for
every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
-John
Confidentiality Notice: This e-mail message, including
a...{{dropped:8}}
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple hist(ograms) - One plot
Deepayan, Thanks for the advice. -Micahel On Fri, Oct 3, 2008 at 1:24 AM, Deepayan Sarkar [EMAIL PROTECTED]wrote: On 10/2/08, Michael Just [EMAIL PROTECTED] wrote: Dieter and Thierry: Per you suggestions I have tried: ggplot2 from Thierry: p - ggplot(dat, aes(x=bbContag, y=..density..)) + geom_histogram() p + facet_grid(. ~ sc_recov %in% c(21,31,41)) But get the followinng error: Error in check_formula(formula, varnames) : Formula contains variables not in list of known variables If I don't make selections it works, but makes that very busy plot. histogram from Dieter: histogram(~bbContag | sc_recov %in% c(21,31,41), data=dat) This produces a plot with only two histograms side by side, but I have chosen three groups, where is the third? I tried only choosing two groups and it plotted two histograms different than when I tried choosing three. Actually, Dieter's suggestion was to use the subset argument, which would go something like histogram(~bbContag | sc_recov, data=dat, subset = (sc_recov %in% c(21,31,41))) Given that your 'sc_recov' seems to be a numeric variable, you will probably get better annotation with histogram(~bbContag | factor(sc_recov), data=dat, subset = (sc_recov %in% c(21,31,41))) -Deepayan Please advise, Thank you for your continued help, Michael Just On Thu, Oct 2, 2008 at 5:21 AM, ONKELINX, Thierry [EMAIL PROTECTED]wrote: Michael, Use %in% to select multiple cases: dat[dat$sc_recov %in% c(21, 31, 41), ] ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto: [EMAIL PROTECTED] Namens Michael Just Verzonden: donderdag 2 oktober 2008 10:59 Aan: r-help@r-project.org CC: [EMAIL PROTECTED] Onderwerp: Re: [R] Multiple hist(ograms) - One plot Dieter Menne: Thanks for the suggestion and link, it looks good. I think my trouble now is lack of basic R knowledge. Cheers, Michael For example, I tried: histogram(~bbContag | sc_recov, data=dat) This would work fine if I could select cases from sc_recov. How can I select more than one case? I know I can do: sc_recov.21 - dat[dat$sc_recov=21,] but how could I select all cases where sc_recov = 21, 31 or 41? On Thu, Oct 2, 2008 at 3:33 AM, Michael Just [EMAIL PROTECTED] wrote: Hello, If I use: p - ggplot(dat, aes(x=bbContag, y=..density..)) + geom_histogram() and then: p + facet_grid(. ~ sc_recov) Its a little crazy because I have 48 different values in 'sc_recov'. Instead I want to select cases from 'sc_recov' and only use three at a time: I tried: p + facet_grid(sc_recov==21 ~.) Error in check_formula(formula, varnames) : Formula contains variables not in list of known variables How can I select cases? Any ideas or suggestions? Thanks, M Just On Thu, Oct 2, 2008 at 2:42 AM, ONKELINX, Thierry [EMAIL PROTECTED] wrote: Dear Michael, Try ggplot2. Use something like install.packages(ggplot2) library(ggplot2) recov - 0:2 n - 1000 all - data.frame(bbED = rnorm(3 * n, mean = recov), recov = factor(rep(recov, n))) ggplot(data = all, aes(x = bbED)) + geom_histogram() + facet_grid(. ~ recov) ggplot(data = all, aes(x = bbED)) + geom_histogram() + facet_grid(recov ~ .) ggplot(data = all, aes(x = bbED, colour = recov)) + geom_density() You'll find more information on ggplot2 at http://had.co.nz/ggplot2/ HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance
Re: [R] Multiple hist(ograms) - One plot
Thierry, Thanks. This worked. Cheers, Michael On Fri, Oct 3, 2008 at 2:46 AM, ONKELINX, Thierry [EMAIL PROTECTED]wrote: Michael, You get this error because you make the subset at the wrong place. Try p - ggplot(dat[dat$sc_recov %in% c(21,31,41), ], aes(x=bbContag, y=..density..)) + geom_histogram() p + facet_grid(. ~ sc_recov) or subdat - dat[dat$sc_recov %in% c(21,31,41), ] p - ggplot(subdat, aes(x=bbContag, y=..density..)) + geom_histogram() p + facet_grid(. ~ sc_recov) HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -- *Van:* Michael Just [mailto:[EMAIL PROTECTED] *Verzonden:* vrijdag 3 oktober 2008 6:52 *Aan:* ONKELINX, Thierry *CC:* r-help@r-project.org; [EMAIL PROTECTED] *Onderwerp:* Re: [R] Multiple hist(ograms) - One plot Dieter and Thierry: Per you suggestions I have tried: ggplot2 from Thierry: p - ggplot(dat, aes(x=bbContag, y=..density..)) + geom_histogram() p + facet_grid(. ~ sc_recov %in% c(21,31,41)) But get the followinng error: Error in check_formula(formula, varnames) : Formula contains variables not in list of known variables If I don't make selections it works, but makes that very busy plot. histogram from Dieter: histogram(~bbContag | sc_recov %in% c(21,31,41), data=dat) This produces a plot with only two histograms side by side, but I have chosen three groups, where is the third? I tried only choosing two groups and it plotted two histograms different than when I tried choosing three. Please advise, Thank you for your continued help, Michael Just On Thu, Oct 2, 2008 at 5:21 AM, ONKELINX, Thierry [EMAIL PROTECTED] wrote: Michael, Use %in% to select multiple cases: dat[dat$sc_recov %in% c(21, 31, 41), ] ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Michael Just Verzonden: donderdag 2 oktober 2008 10:59 Aan: r-help@r-project.org CC: [EMAIL PROTECTED] Onderwerp: Re: [R] Multiple hist(ograms) - One plot Dieter Menne: Thanks for the suggestion and link, it looks good. I think my trouble now is lack of basic R knowledge. Cheers, Michael For example, I tried: histogram(~bbContag | sc_recov, data=dat) This would work fine if I could select cases from sc_recov. How can I select more than one case? I know I can do: sc_recov.21 - dat[dat$sc_recov=21,] but how could I select all cases where sc_recov = 21, 31 or 41? On Thu, Oct 2, 2008 at 3:33 AM, Michael Just [EMAIL PROTECTED] wrote: Hello, If I use: p - ggplot(dat, aes(x=bbContag, y=..density..)) + geom_histogram() and then: p + facet_grid(. ~ sc_recov) Its a little crazy because I have 48 different values in 'sc_recov'. Instead I want to select cases from 'sc_recov' and only use three at a time: I tried: p + facet_grid(sc_recov==21 ~.) Error in check_formula(formula, varnames) : Formula contains variables not in list of known variables How can I select cases? Any ideas or suggestions? Thanks, M Just On Thu, Oct 2, 2008 at 2:42 AM, ONKELINX, Thierry [EMAIL PROTECTED] wrote: Dear Michael, Try ggplot2. Use something like install.packages(ggplot2) library(ggplot2) recov - 0:2 n - 1000 all - data.frame(bbED = rnorm(3 * n, mean = recov), recov = factor(rep(recov, n))) ggplot(data = all, aes(x = bbED)) + geom_histogram() +
[R] Factor tutorial?
This is probably a very basic question. I want to understand factors but I am not sure where to turn. Looking up factor in the Chambers book doesn't even show up in the index. Maybe I am just slow but ?factor doesn't help either. Would someone please point me to a very basic tutorial where I can see what the usefullness of factors is (so far they have just gotten in the way). Thank you. Kevin __ 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] Ecological Niche Modelling on R
stephen sefick ssefick at gmail.com writes: that is a pretty big question. And I am not qualified to answer it, but my suggestion- from figuring out how to use other things- is to find out how other people have modeled ecological niches- not necessarily with R, and then try and find out functions that preform the pieces or the whole analysis. I believe there is a book out there about modeling ecological niches with R. I did a google search for that and this is what I found: http://www.zoology.ufl.edu/bolker/emdbook/. [snip] On Tue, Oct 7, 2008 at 12:46 PM, milton ruser milton.ruser at gmail.com wrote: Dear all, I have strong interest on Ecological Niche Model, which in general use a set of environmental variables (continuous, categorical etc) and Presence (or Presense/Absence) records for species. I think that grasp and adehabitat packages could help me on these tasks. [snip] Note that in general we don´t have real absence data. It's very kind of Stephen to plug my book, but it's not what you're looking for. You need to read more about this general topic, and about the particular packages: try http://www.unine.ch/CSCF/grasp/grasp-r/index.html http://www.unine.ch/CSCF/grasp/ Based on downloading grasp , it doesn't look as though it will handle presence-only data, though -- you may need to look further. It doesn't look like adehabitat is what you want. From Calenge, Clement. 2006. The package adehabitat for the R software: A tool for the analysis of space and habitat use by animals. Ecological Modelling 197, no. 3-4 (August 25): 516-519. doi:10.1016/j.ecolmodel.2006.03.017. ' ... the “adehabitat” package for the R software, which offers basic GIS (Geographic Information System) functions, methods to analyze radio-tracking data and habitat selection by wildlife, and interfaces with other R packages.' General advice about I want to do X in R -- (expanding on Stephen's advice above): 1. read about X in general (perhaps you have already done this); 2. search for R packages and functions that do what you want (you've already done this, although you misidentified adehabitat 3. install those packages and see what they do. Look at the documentation included with the packages, including any citations referenced. Try the examples. 4. If you don't know enough R to understand the examples or how to get your data into R, back up and read the introductory R documentation. If you get stuck again, you might want to try this query on the r-sig-eco mailing list. cheers Ben Bolker __ 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] Mirror Image on Biplot Graphic
Hi Rodrigo,
This is because [I now see] you are working on a capscale object, not a cca
object, which is what I thought you were using. It still works, though, even
with c(2:8,15), though DF #15 doesn't exist in capscale objects. To show you
that this is so, run the example below, which uses a data set from vegan. I
have dropped #15 from the selection set.
## Ex. of switching axes of capscale object
windows(); par(mfrow=c(2,2))
data(varespec)
data(varechem)
vare.cap - capscale(varespec ~ N + P + K + Condition(Al), varechem,
dist=bray)
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 1]-vare.cap$CCA[[i]][, 1] * -1
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 2]-vare.cap$CCA[[i]][, 2] * -1
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 1:2]-vare.cap$CCA[[i]][, 1:2] * -1
plot(vare.cap)
Cheers, Mark.
Rodrigo Aluizio wrote:
Well this time I have to assume, I'm not understanding what is wrong now.
And I have to say: 'Thank You for your patience', cause I'm going crazy
here! :-)
Mark, I tried this:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] *
-1
and got this:
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of limits)
Here is the script until this point
library(vegan)
library(xlsReadWrite)
#FT--#
PotiAbioFT-read.xls('FatorialReplica.xls',sheet=4,rowNames=T)
PotiBioFT-read.xls('FatorialReplica.xls',sheet=6,rowNames=T)
attach(PotiAbioFT)
LogPotiBioFT-log(PotiBioFT+1)
CAPpotiFT-capscale(t(LogPotiBioFT)~Envoronmental+Variables,dist=bray,add=T)
PermCAPFT-anova.cca(CAPpotiFT,alpha=0.05,model='full',first=F,permutations=999)
PermCAPFT
summary(CAPpotiFT)
# Rotating the axis
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] *
-1
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 5:17 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Yes it does, but this will work. You have erroneously used a , in your
call, it should be a ., if anything. You have also inserted cca...
And
the axes being switched must be conformant:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1 ## you are
writing axis 1 over axis 2; and there is no ...,cca...
Erro: unexpected ',' in for(i in c(2:8,15)) ## unexpected
comma
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
## Using your example to switch axis 1. Make sure that CAPpotiFT exists
and
is your
## original object. Copy and paste the lines below into your R console
window and run
CAPpotiFTI - CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1]
* -1
Cheers, Mark.
Rodrigo Aluizio wrote:
[[elided Yahoo spam]]
Ah well, I tried the this other way, but it gave me an error.
Something in the syntax, I tried to change something but didn't fix the
error.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1
Erro: unexpected ',' in for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of
limits)
Well, it needs lots of patience...
--
Sent: Tuesday, October 07, 2008 3:50 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You
need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2]
* -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be
very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you
are
reversing the axes you want. In your original post you mention axis 2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2] * -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1] * -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
CAPpotiFTI-scores(CAPpotiFT, display=c('bp','species','cn','sites')
CAPpotiFTI$species[,1]-CAPpotiFTI$species[,1]*-1
CAPpotiFTI$cn[,1]-CAPpotiFTI$cn[,1]*-1
CAPpotiFTI$sites[,1]-CAPpotiFTI$sites[,1]*-1
CAPpotiFTI$bp[,1]-CAPpotiFTI$bp[,1]*-1
plot.cca(CAPpotiFTI,type='none',display=c('bp','species'),main='Total
Fauna
+
Re: [R] lme and lmer df's and F-statistics again
Well, writing on my rebuttal, I find myself being unable to explain in a few, easy to understand (and, at the same time, correct) sentences stating that it is not a good idea to report (most likely wrong) dfs and F statistics. Without pretending to be able to discuss the details, may I nevertheless ask WHY one should assume that an easy to understand (and, at the same time, correct) answer to the question exists? For example, I would not begin to presume that an easy to understand (and, at the same time, correct answer exists for why an electron can simultaneously have the properties of a particle (photoelectric effect) and a wave (2 slit interference patterns)-- or to the question of why is the 2nd law of thermodynamics equivalent to information loss? -- or to how the Krebs cycle works or the nature of Benzene rings. It has never ceased to amaze me that many casual (in the sense of not having training at, say, the graduate statistics level)users of statistics automatically assume that all statistical principles are fundamentally simple and at least easily comprehensible at a conceptual level by someone with only minimal (or no!) background in the discipline. The extreme manifestation of this is the popular view of a statistician as someone who expertly compiles and tracks baseball records! While I would readily admit that there is much that we can and should do to make our discipline more accessible and useful, I am still offended by those whose attitude is, as appears to be the case here, that even the most technical aspects of the discipline can be made manifest to anyone with half a brain and a stat 101 course under their belt. I think we owe Doug Bates a little more respect than that! Cheers, Bert Gunter __ 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] Mirror Image on Biplot Graphic
Exactly. It's solved. Just remove #15.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8))CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
It seems that I've to pay more attention and study more about R language.
Thanks a lot once more.
Rodrigo.
--
From: Mark Difford [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 6:34 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
This is because [I now see] you are working on a capscale object, not a
cca
object, which is what I thought you were using. It still works, though,
even
with c(2:8,15), though DF #15 doesn't exist in capscale objects. To show
you
that this is so, run the example below, which uses a data set from vegan.
I
have dropped #15 from the selection set.
## Ex. of switching axes of capscale object
windows(); par(mfrow=c(2,2))
data(varespec)
data(varechem)
vare.cap - capscale(varespec ~ N + P + K + Condition(Al), varechem,
dist=bray)
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 1]-vare.cap$CCA[[i]][, 1] * -1
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 2]-vare.cap$CCA[[i]][, 2] * -1
plot(vare.cap)
for(i in c(2:8)) vare.cap$CCA[[i]][, 1:2]-vare.cap$CCA[[i]][, 1:2] * -1
plot(vare.cap)
Cheers, Mark.
Rodrigo Aluizio wrote:
Well this time I have to assume, I'm not understanding what is wrong now.
And I have to say: 'Thank You for your patience', cause I'm going crazy
here! :-)
Mark, I tried this:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] *
-1
and got this:
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of
limits)
Here is the script until this point
library(vegan)
library(xlsReadWrite)
#FT--#
PotiAbioFT-read.xls('FatorialReplica.xls',sheet=4,rowNames=T)
PotiBioFT-read.xls('FatorialReplica.xls',sheet=6,rowNames=T)
attach(PotiAbioFT)
LogPotiBioFT-log(PotiBioFT+1)
CAPpotiFT-capscale(t(LogPotiBioFT)~Envoronmental+Variables,dist=bray,add=T)
PermCAPFT-anova.cca(CAPpotiFT,alpha=0.05,model='full',first=F,permutations=999)
PermCAPFT
summary(CAPpotiFT)
# Rotating the axis
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1] *
-1
--
Sent: Tuesday, October 07, 2008 5:17 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Yes it does, but this will work. You have erroneously used a , in your
call, it should be a ., if anything. You have also inserted cca...
And
the axes being switched must be conformant:
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1 ## you are
writing axis 1 over axis 2; and there is no ...,cca...
Erro: unexpected ',' in for(i in c(2:8,15)) ## unexpected
comma
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
## Using your example to switch axis 1. Make sure that CAPpotiFT exists
and
is your
## original object. Copy and paste the lines below into your R console
window and run
CAPpotiFTI - CAPpotiFT
for(i in c(2:8,15)) CAPpotiFTI$CCA[[i]][, 1]-CAPpotiFTI$CCA[[i]][, 1]
* -1
Cheers, Mark.
Rodrigo Aluizio wrote:
[[elided Yahoo spam]]
Ah well, I tried the this other way, but it gave me an error.
Something in the syntax, I tried to change something but didn't fix the
error.
CAPpotiFTI-CAPpotiFT
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,cca$CCA[[i]][,1]*-1
Erro: unexpected ',' in for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,2]-CAPpotiFTI,
for(i in c(2:8,15))
CAPpotiFTI$CCA[[i]][,1]-CAPpotiFTI$CCA[[i]][,1]*-1
Erro em CAPpotiFTI$CCA[[i]] : índice fora de limites (index out of
limits)
Well, it needs lots of patience...
--
Sent: Tuesday, October 07, 2008 3:50 PM
To: r-help@r-project.org
Subject: Re: [R] Mirror Image on Biplot Graphic
Hi Rodrigo,
Again an error, as that doesn't touch one of the data structures. You
need
to extend the range to include #15, as below:
## This does axis 2
mynew.cca - my.cca
for (i in c(2:8,15)) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][,
2]
* -1
Cheers, Mark.
Mark Difford wrote:
Hi Rodrigo,
I looked through my scripts and found my old hack. __WARNING__: Be
very
careful to check that
the mirror isn't broken and that all is in order! Make sure that you
are
reversing the axes you want. In your original post you mention axis
2,
i.e. y-axis. Below you have used axis 1.
## This does axis 2
mynew.cca - my.cca
for (i in 2:8) mynew.cca$CCA[[i]][, 2] - mynew,cca$CCA[[i]][, 2]
* -1
plot(mynew.cca)
## This does axis 1
for (i in 2:8) mynew.cca$CCA[[i]][, 1] - mynew,cca$CCA[[i]][, 1]
* -1
Regards, Mark.
Rodrigo Aluizio wrote:
Ok Mark, it worked for the species.
I still get an error if I try with biplot (cn,sp), but it's not a
problem,
repositioning the species is enough.
Thanks once again.
Just for future consults the error that still remains:
