[R] tuning parameter in lasso
Dear All, I understand in package lasso2 or lars, the default way to select the tuning parameter is the CV. How can I manually set the tuning parameter $\lambda$? I cannot figure out this according to the help file. Thank you. Huang [[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] Teach me how to transpose in R
Hyo Lee wrote: Teach me how to deal with this problem. Thank you very much. -Hyo A good thing to try if you're stuck finding the right function in R is searching with ??. For example, if you type: ??transpose The base routine t(), which performs a matrix transpose, is one of the first things that comes up. Hope that helps! -Charlie - Charlie Sharpsteen Undergraduate Environmental Resources Engineering Humboldt State University -- View this message in context: http://www.nabble.com/Teach-me-how-to-transpose-in-R-tp25630869p25631393.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] QQ plotting of various distributions...
Thanks for the answer. Now, only problem is to to get parameter(s) of a given function. For gamma, I shall try with gammafit() from mhsmm package. Also, I shall look for others appropriate parameter estimates. Will use SuppDists too. Best, PM Sunil Suchindran wrote: #same shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=6,scale=2) plot(sort(some_data),sort(test_data)) # You can also use qqplot(some_data,test_data) abline(0,1) # different shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=4,scale=2) plot(sort(some_data),sort(test_data)) abline(0,1) It is helpful to assess the sampling variability, by creating repeated sets of test_data, and plotting all of these along with your observations to create a confidence envelope. The SuppDists provides Inverse Gauss. On Thu, Sep 17, 2009 at 11:46 AM, Petar Milin pmi...@ff.uns.ac.rs wrote: Hello! I am trying with this question again: I would like to test few distributional assumptions for some behavioral response data. There are few theories about true distribution of those data, like: normal, lognormal, gamma, ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would be via qq-plot, to show to students differences. First two are trivial: qqnorm(dat$X) qqnorm(log(dat$X)) Then, things are getting more hairy. I am not sure how to make plots for the rest. I tried gamma with: qqmath(~ X, data=dat, distribution=function(X) � qgamma(X, shape, scale)) Which should be the same as: plot(qgamma(ppoints(dat$X), shape, scale), sort(dat$X)) Shape and scale parameters I got via mhsmm package that has gammafit() for shape and scale parameters estimation. Am I on right track? Does anyone know how to plot the rest: ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian)? Thanks, PM __ R-help@r-project.org mailto: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 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] need help in mixed poisson regression
Dear all, I have a mixed Poisson regression model . my function is: ## P(N=k)=[(a+2)/a ][(lambda^k)]*[((a+2/a+1)+lambda)^(-k-1)-(a+2+lambda)^(-k-1)]; lambda=a0+a1*x1+a2*x2 This is same as a Poisson-inverse gaussian or Poisson-lognormal regression.I want to estimate the parameters by fitting this model on my data and also I need loglike value. Since some of the explanatory variables (x1,x2,...) may be nominal or ordinal; which package in R can be used? I am looking for any answer. with the best wishes Zamani.H [[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] for cycle with uncontinuous numbers
Hi nice people,
I would like to do a for cycle but i wish it to assume only the numers 50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
Could anybody help me?
Thanks in advance
Marcio
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Re: [R] Teach me how to transpose in R
Hi Hyo,
I am kinda of new in R but I think if you use
x - as.matrix(data2) #with the numer of collumns you wish
y - t(x)
it should work
Hope I´ve helped
Márcio
Hyo Lee wrote:
Hi guys,
I need your help!!
My goal is to make a csv file from ncdf file.
This is the code i've used :
hyo=open.ncdf(C:/CRUTEM3.nc)
hyo
[1] file C:/CRUTEM3.nc has 4 dimensions:
[1] longitude Size: 72
[1] latitude Size: 36
[1] unspecified Size: 1
[1] t Size: 1916
[1]
[1] file C:/CRUTEM3.nc has 1 variables:
[1] float temp[longitude,latitude,unspecified,t] Longname:Temperature T
Missval:2.0004008175e+20
data2=get.var.ncdf(hyo)
write.csv(data2,file=C:/ple.csv)
But the problem is, I expected this data would be 17000 * 72 (row* col) ;
but, it is the other way around. 72*17000
Because the maximum col number in excel is 16383, this cvs file doesn't
show
all data. Obviously, I need to transpose the matrix..
I tried to use transpose function but failed.
bbb=t(data2)
Error in t.default(data2) : argument is not a matrix
ccc=t(hyo)
ccc
[1] file has dimensions:
Error in if (nc$ndims 0) for (i in 1:nc$ndims) { :
argument is of length zero
Teach me how to deal with this problem.
Thank you very much.
-Hyo
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Re: [R] for cycle with uncontinuous numbers
Hi nice people,
:-)
I would like to do a for cycle but i wish it to assume only the numers 50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
If you would use a reproducible code example we
could point out where the error comes from;
otherwise we can only tell this should work;
try
x - c(50,100,200,300,900,2343)
for (i in x){
cat(i^2, \n)
}
# 2500
# 1
# 4
# 9
# 81
# 5489649
HTH,
Tobias
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[R] Select.spatial on spplots
Hi everyone. I posted this on R-sig-geo but got no response. Can select.spatial() be used in an existing spplot? I have tried selecting points (eq) from a plot generated from sp. However, when I invoke select.spatial(eq). It generates only the points without the background containing the faults. I need the background to select which earthquakes coalesce on which fault. Is there an alternative? eq.pts-list(sp.points,eq, col=blue, lwd=0.5, pch = 4) spplot(faults,Dip, xlim = c(11,12), ylim = c(376,389), sp.layout=list(eq.pts), col = heat.colors(3)) select.spatial(eq) Cheers, Julius Tesoro __ 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] multiple lattice, xyplot levelplot on same page
Ky,
The error you report is a problem with the c() method for trellis
plots: it allows for a different panel function in each panel, but
only one prepanel function. Because of this, the
prepanel.default.levelplot function is not getting what it expects to
recompute the panel dimensions. It may be possible to extend the
function to handle this, but for now you have a few options:
(1) Set subscripts = TRUE in the xyplot:
p1 - update(p1, subscripts = TRUE)
update(c(p2, p1, x.same = TRUE), layout = c(1, 2))
(2) Put the xyplot first, but then re-order the panels by indexing:
update(c(p1, p2, x.same = TRUE), layout = c(1, 2))[2:1]
(3) Take out the x.same=TRUE argument, if you can set the x scales to
be the same beforehand. In your example, the xyplot and levelplot had
the same x axis limits anyway, but presumably that is not the case in
your real problem.
update(c(p2, p1), layout = c(1, 2))
(4) Use a completely different mechanism: draw the two plots in the
desired position (after setting the x axes to be identical beforehand,
if desired). See ?plot.trellis
This latter mechanism was designed in to trellis from the
beginning, whereas the c() method is basically a hack. It also has the
advantage of retaining complete control over each plot independently.
By the way, I can't understand what you are trying to do with the
par.settings$layout.width argument. It doesn't seem to have any
effect. Ditto for the 'y = c(1/4, 3/4))' argument.
Also, I seem to have lost control of par settings such as las = 1
This is a base graphics (?par) parameter, which does not apply to
grid or lattice graphics. The equivalent of las in lattice is the
scales$rot argument. See ?xyplot
Hope that helps
-Felix
2009/9/26 Ky Mathews k.math...@usyd.edu.au:
Dear R-users,
I'd like to place an xyplot() at the top of a page and a levelplot() at the
bottom of the same page, and have the x-axes be the same.
I've come close to finding a solution through Rarchive, and can produce an
upside-down version of what I'd like (levelplot() on the top - see code
below).
However, the following error occurs when I try and plot the xyplot() at the
top:
Error in prepanel.default.function(x = 0:10, y = c(0, 1, 4, 9, 16, 25, :
element 1 is empty;
the part of the args list of 'length' being evaluated was:
(subscripts)
Any pointers in the right direction would be much appreciated.
#OS: Windows XP 2002 SP3; R: 2.9.2; lattice 0.17-25; latticeExtra 0.6-1
Thanks and regards,
Ky
###
#Rcode for xyplot and lattice plot on the same page.
library(lattice)
library(latticeExtra)
#xyplot
x1 - 0:10
x2 - x1^2
p1 - xyplot(x2 ~ x1
, par.settings = list(layout.width = list(panel=1, ylab = 2
, axis.left =1.0, left.padding=1
, ylab.axis.padding=1, axis.panel=1)))
#levelplot
y.df - data.frame(y1 = rep(x1, times = 3)
, y2 = rep(c('E1', 'E2', 'E3'), each = length(x1))
, y3 = c(x1, x1+2, x1-1))
p2 - levelplot(y3 ~ y1*y2, data = y.df,
, par.settings = list(layout.width = list(panel=1, ylab = 2
, axis.left =1.0, left.padding=1
, ylab.axis.padding=1, axis.panel=1)))
#Printing the plots on the same page
#This is what I found on an Rarchive post (thank-you)
#it works if the levelplot (p2) is at the top of the page
#i.e.
update(c(p1, p2, x.same = TRUE)
, layout = c(1, 2)
, ylab = list(c(p1, p2)
, y = c(1/4, 3/4))
, par.settings = list(layout.heights = list(panel = c(1, 1
#however, the following error appears if the order is reversed (which is
what I would like)
update(c(p2, p1, x.same = TRUE)
, layout = c(1, 2)
, ylab = list(c(p2, p1)
, y = c(1/4, 3/4))
, par.settings = list(layout.heights = list(panel = c(1, 1
The following error appears:
#Error in prepanel.default.function(x = 0:10, y = c(0, 1, 4, 9, 16, 25, :
# element 1 is empty;
# the part of the args list of 'length' being evaluated was:
# (subscripts)
Also, I seem to have lost control of par settings such as las = 1
#---
Dr Ky L. Mathews
Co-ordinator, CIMMYT ICARDA Communications Project
Research Fellow,
Plant Breeding Institute, The University of Sydney, Australia
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--
Felix Andrews / 安福立
Postdoctoral Fellow
Integrated Catchment Assessment and Management (iCAM) Centre
Fenner School of Environment and Society [Bldg 48a]
The Australian National University
Canberra ACT 0200 Australia
M: +61 410 400 963
T: +
Re: [R] Fw: Re: Multiple Normal Curves
Hi Jim Sorry to bother you but i don't seem to understand what you are saying. I have the simulated data as follows: Hypermarket - matrix(rnorm(100, mean=5, sd=5000)) Supermarket - matrix(rnorm(400, mean=34000, sd=3000)) Minimarket - matrix(rnorm(1000, mean=1,sd=2000)) Cornershop - matrix(rnorm(1500, mean=2500, sd=500)) Spazashop - matrix(rnorm(2000, mean=1000, sd=250)) dat=data.frame(type=c(rep(Hypermarket,100), rep(Supermarket,400), rep(Minimarket,1000),rep(Cornershop,1500), rep(Spazashop,2000)), value=c(Hypermarket, Supermarket, Minimarket, Cornershop,Spazashop)) Now, i want to do multiple histograms and normal curves on the same plot as you rightfully pointed out, but i cannot plot all of them even when i have done the following i can only do one. Hypermarket-rnorm(100, mean=5, sd=3000) par(las=1) hist(Hypermarket, breaks=seq(0, 6, 500), freq=F) x- seq(0, 6, 1000) lines(x, dnorm(x, 5, 3000)) As for the code below, i dicided to leave it as i does not allow me to do more than three curves. I would appreciate all the help. Kabeli --- On Thu, 24/9/09, Jim Lemon j...@bitwrit.com.au wrote: From: Jim Lemon j...@bitwrit.com.au Subject: Re: [R] Fw: Re: Multiple Normal Curves To: KABELI MEFANE kabelimef...@yahoo.co.uk, r-help@r-project.org Date: Thursday, 24 September, 2009, 12:53 PM On 09/24/2009 08:57 PM, KABELI MEFANE wrote: Sorry about the subject --- On Thu, 24/9/09, KABELI MEFANEkabelimef...@yahoo.co.uk wrote: From: KABELI MEFANEkabelimef...@yahoo.co.uk Subject: Re: [R] Multiply Normal Curves To: R-help@r-project.org Date: Thursday, 24 September, 2009, 11:48 AM R -helpers i have been trying to do this problem without must success,i managed to do a graph for x, but it is not what i want to define,(i want to specify number of observations as well). I have also been able to do simple rendom sample. data.frame(ID=c(1,2,3),mu=c(1,34000,5),sigma=c(2000,3000,5000)) curve(dnorm(x,mean=parms$mu[1],sd=parms$sigma[1]),from=2000, to=8, ylab=density, col=red) curve(dnorm(x,mean=parms$mu[2],sd=parms$sigma[2]),from=1000, to=8, ylab=density, col=blue, add=TRUE) curve(dnorm(x,mean=parms$mu[3],sd=parms$sigma[3]),from=1000, to=8, ylab=density, col=forestgreen, add=TRUE) ### R-helpers I have been learning a little bit of R. I am simulating and i want to draw a normal curve for all my variables so that i will see the overlaps and reduce them, after that i want to draw a gragh of all the values that are in the data frame to see if it follows a normal distribution also. Lastly i will try to sample from this data. Please help and make suggestions. Hi Kabeli, I think you want to get multiple histograms and normal curves on the same plot. You can do something like that if you get a table of frequencies for each of your three sets of values using cut or hist, combine these vectors of frequencies into a matrix and pass this to barplot. Then draw your normal curves using curve on top of the grouped bars. Jim [[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] QQ plotting of various distributions...
The supposed example of a Q-Q plot is most certainly not how to make a Q-Q plot. I don't even know where to start First off, the two Q:s in the title of the plot stand for quantile, not random. The answer supplied simply plots two sorted samples of a distribution against each other. While this may resemble the general shape of a QQ plot, that is where the similarities end. Some general advice: be careful who you take advice from on the internet. The Wikipedia entry for Q-Q plot may be a good start if you don't know what a Q-Q plot is, although you should also use it with caution. Lets say you have some samples that may be normally distributed: set.seed(1) x - rnorm(30) # now try with R's built in function qqnorm(x, xlim = c(-3, 3), ylim = c(-3, 3)) # Now try Sunil's Q-Q plot method, but for rnorm # rather than rgamma some_data - x test_data - rnorm(30) points(sort(some_data),sort(test_data), col = blue) # Note that the points are NOT the same! This should have been obvious for the simple reason that the QQ plot should not be influenced by the random number generator that you are using! A QQ plot is uniquely reproducible. The more general (and correct) way to get the QQ plot involves choosing a plotting position and the quantile function (e.g. qnorm or qgamma functions in R) of the pertinent distribution: # Sort the data: x.s - sort(x) n - length(x) # Plotting position (must be careful here in general!) p - ppoints(n) # Compute the quantile x.q - qnorm(p) points(x.q, x.s, col = red) # and they fall exactly on the points generated by qqnorm(). Now, you should be able to generalize this for any distribution. Hope this helps. Eric Thompson 2009/9/27 Petar Milin pmi...@ff.uns.ac.rs: Thanks for the answer. Now, only problem is to to get parameter(s) of a given function. For gamma, I shall try with gammafit() from mhsmm package. Also, I shall look for others appropriate parameter estimates. Will use SuppDists too. Best, PM Sunil Suchindran wrote: #same shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=6,scale=2) plot(sort(some_data),sort(test_data)) # You can also use qqplot(some_data,test_data) abline(0,1) # different shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=4,scale=2) plot(sort(some_data),sort(test_data)) abline(0,1) It is helpful to assess the sampling variability, by creating repeated sets of test_data, and plotting all of these along with your observations to create a confidence envelope. The SuppDists provides Inverse Gauss. On Thu, Sep 17, 2009 at 11:46 AM, Petar Milin pmi...@ff.uns.ac.rs wrote: Hello! I am trying with this question again: I would like to test few distributional assumptions for some behavioral response data. There are few theories about true distribution of those data, like: normal, lognormal, gamma, ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would be via qq-plot, to show to students differences. First two are trivial: qqnorm(dat$X) qqnorm(log(dat$X)) Then, things are getting more hairy. I am not sure how to make plots for the rest. I tried gamma with: qqmath(~ X, data=dat, distribution=function(X) � qgamma(X, shape, scale)) Which should be the same as: plot(qgamma(ppoints(dat$X), shape, scale), sort(dat$X)) Shape and scale parameters I got via mhsmm package that has gammafit() for shape and scale parameters estimation. Am I on right track? Does anyone know how to plot the rest: ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian)? Thanks, PM __ r-h...@r-project.org mailto: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 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-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] Lattice, stripplot (xyplot), plotting data with median line, numeric x-axis
Hi David, stripplot is for numeric vs categorical data (and is a thin wrapper around xyplot). Just change stripplot to xyplot and it will work. -Felix 2009/9/27 Afshartous, David dafshart...@med.miami.edu: All, On p.52 of Deepayan Sarkar's Lattice book there is a nice plot of showing residuals with median lines superimposed or various groups: library(lattice) stripplot(sqrt(abs(residuals(lm(yield~variety+year+site ~ site, data = barley, groups = year, jitter.data = TRUE, type = c(p, a), fun = median) Suppose we wanted to make a similar plot for a numeric x-axis. Is there any way to do this with stripplot or does one have to xyplot and presumably panel functionality to get the median line? This does not work: barley$site.numeric =as.numeric(barley$site) stripplot(sqrt(abs(residuals(lm(yield~variety+year+site ~ site.numeric, data = barley, groups = year, jitter.data = TRUE, type = c(p, a), fun = median) Any tips much appreciated. For my data I had made my x-axis a factor but forgot that this doesn't work since the intervals are not equally spaced. Thanks! David __ 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. -- Felix Andrews / 安福立 Postdoctoral Fellow Integrated Catchment Assessment and Management (iCAM) Centre Fenner School of Environment and Society [Bldg 48a] The Australian National University Canberra ACT 0200 Australia M: +61 410 400 963 T: + 61 2 6125 1670 E: felix.andr...@anu.edu.au CRICOS Provider No. 00120C -- http://www.neurofractal.org/felix/ __ 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.text question
Use packet.number() to identify which data subset you are dealing with
(inside the panel function):
xyplot(y~x|a,
panel=function(...){
panel.loess(...)
panel.text(0,2,label=c('best','better','bad','worst')[packet.number()])
})
There is also panel.number() which is similar but refers to the panel
position rather than data set. The difference shows up if you
re-arrange the panels from the default ordering.
2009/9/27 Osman Al-Radi osman.al.r...@gmail.com:
Hello,
Thanks for your suggestion. It works in my simplified example. However, it
didn't work in my real code. It is probably because I neglected to include
the group argument in the example. I apologize for that.
Below is the real code, if you need the actual data I can include it too.
# this works well
xyplot(PaCO2~time|group, group=animal,layout=c(3,1,1),aspect=1,
panel=function(...){
panel.loess(...)
panel.superpose(...)}
,data=pig,subset=time5 time181,
xlab='Time (minutes)',
ylab='PaCO2 (mmHg)')
# this gives the following error Error in using packet 1 data, X argument
missing with no #default in each of the plot panel
xyplot(PaCO2~time|group, group=animal,layout=c(3,1,1),aspect=1,
panel=function(x,y,subscripts,...){
panel.loess(...)
panel.superpose(...)
panel.text(100,110,label=c(' ','p=0.007','p=0.006')[tail(subscripts,
1)])}
,data=pig, subset=time5 time181,
xlab='Time (minutes)',
ylab='PaCO2 (mmHg)')
Thanks tremendously for your help. I don't know why its soo hard just to
add some text!
Osman
Osman O. Al-Radi, MD, MSc, FRCSC
Staff Cardiovascular Surgeon
Co-medical director, Tissue Bank
The Hospital for Sick Children
University of Toronto, Canada
On Thu, Sep 24, 2009 at 2:18 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
xyplot(y ~ x | a,
panel=function(x, y, subscripts, ...){
panel.loess(x, y)
panel.text(0, 2,
label=c('best','better','bad','worst')[tail(subscripts, 1)/100])
})
On Thu, Sep 24, 2009 at 2:45 PM, Osman Al-Radi osman.al.r...@gmail.com
wrote:
Dear R-help,
I would like to add text to each of four panels in a plot generated by
xyplot in lattice library. A sample code is given below, the plot
generated
has the first label repeated in all panels!
How can I get the labels to be different in each panel?
library(lattice)
x - rnorm(400)
y - rnorm(400)
a - gl(4, 100)
xyplot(y~x|a,
panel=function(...){
panel.loess(...)
panel.text(0,2,label=c('best','better','bad','worst'))})
Thanks
Osman
Osman O. Al-Radi, MD, MSc, FRCSC
Staff Cardiovascular Surgeon
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Re: [R] Adding variables
Jim, Both my emails contained reproducible code (the first one wasn't completely reproducible - it required one to know that attenu is a base R dataset). Anyway, thanks for your help. On Sat, Sep 26, 2009 at 8:11 PM, jim holtman jholt...@gmail.com wrote: I assumed (since you did not provide reproducible code) that 'mag' was a matrix. If 'station' is a matrix, then mag + rowSums(station) will work. If that does not work, then you need to tell us what your data objects are. On Sat, Sep 26, 2009 at 11:39 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi Jim, I might be missing something but your command gives the error: Error in rowSums(mag) : 'x' must be an array of at least two dimensions # data(attenu) attach(attenu) rowSums(mag) + rowSums(station) attenu$new-rowSums(cbind(mag, station)) # Thanks On Sat, Sep 26, 2009 at 4:30 PM, jim holtman jholt...@gmail.com wrote: Probably more efficient if you remove the 'cbind' which would create a combined matrix. Use the following: rowSums(mag) + rowSums(station) On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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? -- 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] QQ plotting of various distributions...
Eric Thompson wrote: The supposed example of a Q-Q plot is most certainly not how to make a Q-Q plot. I don't even know where to start First off, the two Q:s in the title of the plot stand for quantile, not random. The answer supplied simply plots two sorted samples of a distribution against each other. While this may resemble the general shape of a QQ plot, that is where the similarities end. The empirical quantiles of a sample are simply the sorted values. You can plot empirical quantiles of one sample versus some version of quantiles from a distribution (what qqnorm does) or versus empirical quantiles of another sample (what Sunil did). The randomness in his demonstration did two things: it generated some data, and it showed the variability of the plot under repeated sampling. Some general advice: be careful who you take advice from on the internet. That's good advice. Duncan Murdoch The Wikipedia entry for Q-Q plot may be a good start if you don't know what a Q-Q plot is, although you should also use it with caution. Lets say you have some samples that may be normally distributed: set.seed(1) x - rnorm(30) # now try with R's built in function qqnorm(x, xlim = c(-3, 3), ylim = c(-3, 3)) # Now try Sunil's Q-Q plot method, but for rnorm # rather than rgamma some_data - x test_data - rnorm(30) points(sort(some_data),sort(test_data), col = blue) # Note that the points are NOT the same! This should have been obvious for the simple reason that the QQ plot should not be influenced by the random number generator that you are using! A QQ plot is uniquely reproducible. The more general (and correct) way to get the QQ plot involves choosing a plotting position and the quantile function (e.g. qnorm or qgamma functions in R) of the pertinent distribution: # Sort the data: x.s - sort(x) n - length(x) # Plotting position (must be careful here in general!) p - ppoints(n) # Compute the quantile x.q - qnorm(p) points(x.q, x.s, col = red) # and they fall exactly on the points generated by qqnorm(). Now, you should be able to generalize this for any distribution. Hope this helps. Eric Thompson 2009/9/27 Petar Milin pmi...@ff.uns.ac.rs: Thanks for the answer. Now, only problem is to to get parameter(s) of a given function. For gamma, I shall try with gammafit() from mhsmm package. Also, I shall look for others appropriate parameter estimates. Will use SuppDists too. Best, PM Sunil Suchindran wrote: #same shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=6,scale=2) plot(sort(some_data),sort(test_data)) # You can also use qqplot(some_data,test_data) abline(0,1) # different shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=4,scale=2) plot(sort(some_data),sort(test_data)) abline(0,1) It is helpful to assess the sampling variability, by creating repeated sets of test_data, and plotting all of these along with your observations to create a confidence envelope. The SuppDists provides Inverse Gauss. On Thu, Sep 17, 2009 at 11:46 AM, Petar Milin pmi...@ff.uns.ac.rs wrote: Hello! I am trying with this question again: I would like to test few distributional assumptions for some behavioral response data. There are few theories about true distribution of those data, like: normal, lognormal, gamma, ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would be via qq-plot, to show to students differences. First two are trivial: qqnorm(dat$X) qqnorm(log(dat$X)) Then, things are getting more hairy. I am not sure how to make plots for the rest. I tried gamma with: qqmath(~ X, data=dat, distribution=function(X) � qgamma(X, shape, scale)) Which should be the same as: plot(qgamma(ppoints(dat$X), shape, scale), sort(dat$X)) Shape and scale parameters I got via mhsmm package that has gammafit() for shape and scale parameters estimation. Am I on right track? Does anyone know how to plot the rest: ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian)? Thanks, PM __ R-help@r-project.org mailto: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 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-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,
Re: [R] Re ad in multiple datasets
Try this: filenames - sprintf(data%d.csv, 1:20) DFs - sapply(filenames, read.csv, simplify = FALSE) which will return a list of data frames, DFs, each named by its filename so that DFs[[1]] or DFs[[data1.csv]] give the data frame read from data1.csv, etc. and names(DFs) gives a vector of their names as does filenames. On Sat, Sep 26, 2009 at 11:47 PM, legen lege...@gmail.com wrote: Hello, all: I have twenty datasets named as: data1.csv, data2.csv, …, data20.csv. I am trying to read all of them into R by using loop and function read.table(), but I don't know how to handle the name of datasets. Has anybody have encountered a similar problem? Or do you have any suggestions? Your help would be greatly appreciated. Legen -- View this message in context: http://www.nabble.com/Read-in-multiple-datasets-tp25630688p25630688.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Gini importance measure in RF package
Dear all, I am working with randomForest package and I am interested in examining the Gini importance measures that are used as a general indicator of feature relevance. Is there a possibility of getting the Gini measure that is being estimated in each tree by the output of the getTree() function? Thanks a lot, Chrysanthi [[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] Adding variables
Well, I had never seen any help pages use attenu as an example. Like Jim I assumed that you were offering code that was operating on some private copy of data, If you look at the number of datasets, I think it unreasonable to expect the great majority of potentially helpful persons to know all of them either. So why were you trying to add a factor variable to a numeric, anyway? If your hope was to convert those codes to numeric then it's a FAQ: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f If the problem were more basic, and you did not know what was in that dataset then the answer might have bee: str(attenu) -- David On Sep 27, 2009, at 7:47 AM, tzygmund mcfarlane wrote: Jim, Both my emails contained reproducible code (the first one wasn't completely reproducible - it required one to know that attenu is a base R dataset). Anyway, thanks for your help. On Sat, Sep 26, 2009 at 8:11 PM, jim holtman jholt...@gmail.com wrote: I assumed (since you did not provide reproducible code) that 'mag' was a matrix. If 'station' is a matrix, then mag + rowSums(station) will work. If that does not work, then you need to tell us what your data objects are. On Sat, Sep 26, 2009 at 11:39 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi Jim, I might be missing something but your command gives the error: Error in rowSums(mag) : 'x' must be an array of at least two dimensions # data(attenu) attach(attenu) rowSums(mag) + rowSums(station) attenu$new-rowSums(cbind(mag, station)) # Thanks On Sat, Sep 26, 2009 at 4:30 PM, jim holtman jholt...@gmail.com wrote: Probably more efficient if you remove the 'cbind' which would create a combined matrix. Use the following: rowSums(mag) + rowSums(station) On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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? -- 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding variables
The station column has non numeric values, so you need convert to numeric before: with(attenu, as.numeric(as.character(station)) + mag) On Sun, Sep 27, 2009 at 8:47 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Jim, Both my emails contained reproducible code (the first one wasn't completely reproducible - it required one to know that attenu is a base R dataset). Anyway, thanks for your help. On Sat, Sep 26, 2009 at 8:11 PM, jim holtman jholt...@gmail.com wrote: I assumed (since you did not provide reproducible code) that 'mag' was a matrix. If 'station' is a matrix, then mag + rowSums(station) will work. If that does not work, then you need to tell us what your data objects are. On Sat, Sep 26, 2009 at 11:39 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi Jim, I might be missing something but your command gives the error: Error in rowSums(mag) : 'x' must be an array of at least two dimensions # data(attenu) attach(attenu) rowSums(mag) + rowSums(station) attenu$new-rowSums(cbind(mag, station)) # Thanks On Sat, Sep 26, 2009 at 4:30 PM, jim holtman jholt...@gmail.com wrote: Probably more efficient if you remove the 'cbind' which would create a combined matrix. Use the following: rowSums(mag) + rowSums(station) On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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? -- 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. -- 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] Adding variables
So why were you trying to add a factor variable to a numeric, anyway? For no other reason than to illustrate the task of addition. It is, admittedly, meaningless. Well, I had never seen any help pages use attenu as an example. I literally went to: http://stat.ethz.ch/R-manual/R-patched/library/datasets/html/00Index.html and picked a random dataset. I haven't enough R experience to know which datasets are more commonly used to illustrate issues than others. My original question was about the efficiency of simple arithmetic for large datasets and managing them. While the advice I have received so far is very good for my R style, I was wondering if you might add something regarding this. PS. Thanks Henrique, but again the example was chosen completely at random to provide reproducible code I wasn't thinking about numeric and factor variables. On Sun, Sep 27, 2009 at 3:14 PM, David Winsemius dwinsem...@comcast.net wrote: Well, I had never seen any help pages use attenu as an example. Like Jim I assumed that you were offering code that was operating on some private copy of data, If you look at the number of datasets, I think it unreasonable to expect the great majority of potentially helpful persons to know all of them either. So why were you trying to add a factor variable to a numeric, anyway? If your hope was to convert those codes to numeric then it's a FAQ: http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f If the problem were more basic, and you did not know what was in that dataset then the answer might have bee: str(attenu) -- David On Sep 27, 2009, at 7:47 AM, tzygmund mcfarlane wrote: Jim, Both my emails contained reproducible code (the first one wasn't completely reproducible - it required one to know that attenu is a base R dataset). Anyway, thanks for your help. On Sat, Sep 26, 2009 at 8:11 PM, jim holtman jholt...@gmail.com wrote: I assumed (since you did not provide reproducible code) that 'mag' was a matrix. If 'station' is a matrix, then mag + rowSums(station) will work. If that does not work, then you need to tell us what your data objects are. On Sat, Sep 26, 2009 at 11:39 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi Jim, I might be missing something but your command gives the error: Error in rowSums(mag) : 'x' must be an array of at least two dimensions # data(attenu) attach(attenu) rowSums(mag) + rowSums(station) attenu$new-rowSums(cbind(mag, station)) # Thanks On Sat, Sep 26, 2009 at 4:30 PM, jim holtman jholt...@gmail.com wrote: Probably more efficient if you remove the 'cbind' which would create a combined matrix. Use the following: rowSums(mag) + rowSums(station) On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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? -- 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding variables
with(attenu, mag + as.numeric(station)) is nearly twice as fast: system.time(for(i in 1:1000) with(attenu, mag + as.numeric(station))) user system elapsed 0.050.020.06 system.time(for(i in 1:1000) rowSums(cbind(mag, station))) user system elapsed 0.090.000.10 See ?system.time, ?Rprof and http://code.google.com/p/rbenchmark/ for timing commands. On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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] Adding variables
Thank you Gabor ( Henrique)! On Sun, Sep 27, 2009 at 3:26 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: with(attenu, mag + as.numeric(station)) is nearly twice as fast: system.time(for(i in 1:1000) with(attenu, mag + as.numeric(station))) user system elapsed 0.05 0.02 0.06 system.time(for(i in 1:1000) rowSums(cbind(mag, station))) user system elapsed 0.09 0.00 0.10 See ?system.time, ?Rprof and http://code.google.com/p/rbenchmark/ for timing commands. On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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] QQ plotting of various distributions...
I think it's helpful to show the sampling variability in a QQ plot under repeated sampling. An example is given in Venables, Ripley pg 86. The variance is higher at the tails. Even when the distributions are the same, the QQ plot does not have to resemble a straight line because of sampling. I don't think you can think of any one of these as the correct plot. Also, if the two data sets have an equal number of points, the empirical qq plot is simply a plot of one sorted data set against the other. (Kundu, Statistical Computing, pg 42). On Sun, Sep 27, 2009 at 9:06 AM, Duncan Murdoch murd...@stats.uwo.ca wrote: Eric Thompson wrote: The supposed example of a Q-Q plot is most certainly not how to make a Q-Q plot. I don't even know where to start First off, the two Q:s in the title of the plot stand for quantile, not random. The answer supplied simply plots two sorted samples of a distribution against each other. While this may resemble the general shape of a QQ plot, that is where the similarities end. The empirical quantiles of a sample are simply the sorted values. You can plot empirical quantiles of one sample versus some version of quantiles from a distribution (what qqnorm does) or versus empirical quantiles of another sample (what Sunil did). The randomness in his demonstration did two things: it generated some data, and it showed the variability of the plot under repeated sampling. __ 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] Adding variables
Note that Henrique's code does not give the same result as the expression you posted although its possible that his is what you really intended. On Sun, Sep 27, 2009 at 10:27 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Thank you Gabor ( Henrique)! On Sun, Sep 27, 2009 at 3:26 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: with(attenu, mag + as.numeric(station)) is nearly twice as fast: system.time(for(i in 1:1000) with(attenu, mag + as.numeric(station))) user system elapsed 0.05 0.02 0.06 system.time(for(i in 1:1000) rowSums(cbind(mag, station))) user system elapsed 0.09 0.00 0.10 See ?system.time, ?Rprof and http://code.google.com/p/rbenchmark/ for timing commands. On Sat, Sep 26, 2009 at 11:16 AM, tzygmund mcfarlane tzygm...@googlemail.com wrote: Hi, For very large matrices, is this the most efficient way to add two variables together? # attach(attenu) new-rowSums(cbind(mag, station)) # Also, could I be directed to some resources for working with very large datasets? Thanks __ 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] zoo: merging aggregated zoo-objects fails
Dear all,
I have several text files looking like this:
9063032 19700201 22:00 174.067
9063032 19700201 23:00 174.076
9063032 19700202 00:00 174.085
9063032 19700202 01:00 174.091
9063032 19700202 02:00 174.094
9063032 19700202 03:00 174.091
9063032 19700202 04:00 174.082
9063032 19700202 05:00 174.079
And I run this loop:
for (j in 1:nr.of.files)
{
#Import:
DF - read.table(path,header=FALSE,na.string='-',
colClasses=c(NULL,character,character,numeric))
z - zoo(DF$V4,chron(dates=DF$V2,times=paste(DF$V3,0,sep=:),
format=c(dates=ymd,times=h:m:s)))
#getting daily values:
x - aggregate(z,trunc,mean)
#summing everything up in one nr.of.files-dimensional zoo-object
if (j 1) final - merge(final,x) else final - x
}
Unfortunately I get the following error message:
Fehler in matrix(unlist(lapply(dots, origin)), nrow = 3) :
Versuch ein Attribut von NULL zu setzen
(Error ... Trying to set an attribute NULL)
There is no error when I first run merge() and then aggregate(), but this is
not possible due to the rest of the program.
Does any one have a clue?
regards, gunnar
(R vers.2.9.1 (Debian Lenny, XFCE 4.4, EMACS 22.2.1, ESS 5.3.8))
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Re: [R] for cycle with uncontinuous numbers
Hi Tobias, thanks for the help,
the code I am using is quite long, but basically what I tried to do was
test - matrix(0,6,1)
x - matrix( c(50,100,200,300,900,2343) ,ncol = 1)
for (i in x){
test [i] - (i)
}
but this code returns NA for all the elements which are not x
Tobias Verbeke-2 wrote:
Hi nice people,
:-)
I would like to do a for cycle but i wish it to assume only the numers
50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
If you would use a reproducible code example we
could point out where the error comes from;
otherwise we can only tell this should work;
try
x - c(50,100,200,300,900,2343)
for (i in x){
cat(i^2, \n)
}
# 2500
# 1
# 4
# 9
# 81
# 5489649
HTH,
Tobias
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[R] Gini importance measure in RF package
Dear all, I am working with randomForest package and I am interested in examining the Gini importance measures that are used as a general indicator of feature relevance. Is there a possibility of getting the Gini measure that is being estimated in each tree by the output of the getTree() function? Thanks a lot, Chrysanthi [[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] Problem with Random Forest predict
Dear all, I am working with randomForest package and I am interested in examining the Gini importance measures that are used as a general indicator of feature relevance. Is there a possibility of getting the Gini measure that is being estimated in each tree by the output of the getTree() function? Thanks a lot, Chrysanthi [[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] Xterm escape sequences in Prompt
Dear list, I would like to know if there is any way to include xterm escape sequences in R's prompt using options( prompt= XXX, continue= XXX ), where XXX can contain, for example, \033[1;31mRed for chancing text color to red (see http://www.frexx.de/xterm-256-notes/). If you do this, the color escape sequences work (with an appropriate terminal emulator, e.g., xterm under Linux), providing you with a colored R prompt. However, when using the command history (arrow up/down) the edit line gets messed up because the algorithm counts the escape sequence not as a single char but as multiple. Thus, I was wondering if there would be any way to tell options(prompt) not only the string but also the number of characters in the string (of course, defaulting to strlen). The specific reason I ask this question is because I have recently written a gedit plug-in for R, allowing it to become a lightweight IDE for R (http://sourceforge.net/projects/rgedit/) and one feature I would like to have is colored prompts (to make the R console easier to navigate through). Alas, as I said before, using options() with escape sequences kind of works as long as you don't use the command history (implemented in my rgedit but disabled by default). Thanks you in advance, Dan _ [[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] Optional libraries (libtiff, etc) not found
I installed the (binary) biOps package, which can use libtiff and libfftw. Then I used macports to install the libraries (in /opt/local/lib). But I restart R and biOps still does not see the libraries. I've tried adding symbolic links from /opt/local/lib to /usr/local/lib, I've added /opt/local/lib to LIBRARY_PATH, LD_LIBRARY_PATH, and DYLD_LIBRARY_PATH, and it doesn't work. Do you need to install biOps source in order for it to realize that the libraries are there? Or something else? -- View this message in context: http://www.nabble.com/Optional-libraries-%28libtiff%2C-etc%29-not-found-tp25634024p25634024.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] zoo: merging aggregated zoo-objects fails
Please read the last line to every message on r-help. In particular
make it reproducible and minimal. The code you post should look like
this where you have cut down DF1, DF2 and DF3 to the smallest number
of rows that still exhibits the error.
DF1 - ...output from dput(DF1)
DF2 - ...output from dput(DF2)...
DF3 - ...output from dput(DF3)...
z1 - zoo(...)
z2 - zoo(...)
z3 - zoo(...)
m - merge(z1, z2)
m2 - merge(z3, m) # error
On Sun, Sep 27, 2009 at 8:42 AM, gunnar.p pr...@uni-potsdam.de wrote:
Dear all,
I have several text files looking like this:
9063032 19700201 22:00 174.067
9063032 19700201 23:00 174.076
9063032 19700202 00:00 174.085
9063032 19700202 01:00 174.091
9063032 19700202 02:00 174.094
9063032 19700202 03:00 174.091
9063032 19700202 04:00 174.082
9063032 19700202 05:00 174.079
And I run this loop:
for (j in 1:nr.of.files)
{
#Import:
DF - read.table(path,header=FALSE,na.string='-',
colClasses=c(NULL,character,character,numeric))
z - zoo(DF$V4,chron(dates=DF$V2,times=paste(DF$V3,0,sep=:),
format=c(dates=ymd,times=h:m:s)))
#getting daily values:
x - aggregate(z,trunc,mean)
#summing everything up in one nr.of.files-dimensional zoo-object
if (j 1) final - merge(final,x) else final - x
}
Unfortunately I get the following error message:
Fehler in matrix(unlist(lapply(dots, origin)), nrow = 3) :
Versuch ein Attribut von NULL zu setzen
(Error ... Trying to set an attribute NULL)
There is no error when I first run merge() and then aggregate(), but this is
not possible due to the rest of the program.
Does any one have a clue?
regards, gunnar
(R vers.2.9.1 (Debian Lenny, XFCE 4.4, EMACS 22.2.1, ESS 5.3.8))
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Re: [R] for cycle with uncontinuous numbers
It is unclear exactly what you are trying to do. Since your 'test'
matrix is only 6 rows in length, then when you extend it by trying to
store into 50 for example, it will fill in the rest of the new values
with NA. Can you explain what you think the code is supposed to do?
It is doing exactly what you are asking it to do with the script you
provided.
On Sun, Sep 27, 2009 at 10:09 AM, Marcio Resende
mresende...@yahoo.com.br wrote:
Hi Tobias, thanks for the help,
the code I am using is quite long, but basically what I tried to do was
test - matrix(0,6,1)
x - matrix( c(50,100,200,300,900,2343) ,ncol = 1)
for (i in x){
test [i] - (i)
}
but this code returns NA for all the elements which are not x
Tobias Verbeke-2 wrote:
Hi nice people,
:-)
I would like to do a for cycle but i wish it to assume only the numers
50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
If you would use a reproducible code example we
could point out where the error comes from;
otherwise we can only tell this should work;
try
x - c(50,100,200,300,900,2343)
for (i in x){
cat(i^2, \n)
}
# 2500
# 1
# 4
# 9
# 81
# 5489649
HTH,
Tobias
__
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and provide commented, minimal, self-contained, reproducible code.
--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Generation of Multariate Gumbel Random Numbers VGAM or GUMBEL
Dear R users: I'm trying to generate multivariate random numbers where some variables are correlated. I tried VGAM and GUMBEL, but I really don't understand from the documentation, how to move from the single random to the multivariate. Does anybody have any insight about about this? Thanks in advance Angelo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting SAS Data code to R.
On Sat, Sep 26, 2009 at 11:33 PM, David Winsemius dwinsem...@comcast.net wrote: I am contemplating bringing in and merging three NHANES-III datasets from the National Center for Health Statistics that are fixed format with record length=3348, line counts around 20,000 and described by SAS DATA steps. I have downloaded and linked similar datasets from the Continuous NHANES public data releases, but never ones with this many variables at once. In the prior effort I managed the task by some cut-paste-editing from the SAS code file into a corresponding read.fwf R call, but the earlier NHANES-III data is far more voluminous than the more recent Continuous version. I am wondering if anyone has experience with such a process and would be willing to share some advice? The SAS code can be seen here: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.sas The main code file Data step starts out... FILENAME ADULT D:\Questionnaire\DAT\ADULT.DAT LRECL=3348; *** LRECL includes 2 positions for CRLF, assuming use of PC SAS; DATA WORK; INFILE ADULT MISSOVER; LENGTH SEQN 7 DMPFSEQ 5 DMPSTAT 3 DMARETHN 3 DMARACER 3 DMAETHNR 3 HSSEX 3 The corresponding positions in the INPUT section are INPUT SEQN 1-5 DMPFSEQ 6-10 DMPSTAT 11 DMARETHN 12 DMARACER 13 DMAETHNR 14 HSSEX 15 The note about CRLF appears to be implying that those characters are being counted as part of the length of the first variable, SEQN, but that there are only 5 meaningful positions. I suppose I can find out by trial and error how to read such files, but it would save me some time if anyone in the audience has worked through this on this data before. One thought would be to import the data with the SAS work-alike program, WKS, (which I have not used before) and then to read in with read.xport from the foreign library. That would obviate the need to understand the character position issue, but probably has a time commitment to get it up and running and learn how to use it. Another thought would be to parse the fixed width SAS Data step code into pieces and build a data.frame from which I then extract the row.names, col.names, and colClasses from that centralized structure. Are the data available to the public somewhere or could just a few records be made available? The reason I ask is because I imagine there are a lot of missing data in each record (the data are arranged in the wide format for longitudinal data and includes follow-up questions that will not apply to most respondents). The missing data indicator, if any, and the format of the other fields will be important in deciding how to split the data. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] for cycle with uncontinuous numbers
Dear John,
What I am trying to do is a genetic analysis and the i in my cycle are the
numbers of markers I am testing.
I know the code is not right, but what i wanted to do was to fill this 6 row
matrix with the numbers in x
So basically I wanted the cycle to loop only with the values on x
(50,100,200,300,900,2343) and not the the rest of the values
Do you know how to do it?
In the example I wanted my final test matrix to be
[50
100
200
300
900
2343]
dim(test) - still 6x1
Thanks for the help
jholtman wrote:
It is unclear exactly what you are trying to do. Since your 'test'
matrix is only 6 rows in length, then when you extend it by trying to
store into 50 for example, it will fill in the rest of the new values
with NA. Can you explain what you think the code is supposed to do?
It is doing exactly what you are asking it to do with the script you
provided.
On Sun, Sep 27, 2009 at 10:09 AM, Marcio Resende
mresende...@yahoo.com.br wrote:
Hi Tobias, thanks for the help,
the code I am using is quite long, but basically what I tried to do was
test - matrix(0,6,1)
x - matrix( c(50,100,200,300,900,2343) ,ncol = 1)
for (i in x){
test [i] - (i)
}
but this code returns NA for all the elements which are not x
Tobias Verbeke-2 wrote:
Hi nice people,
:-)
I would like to do a for cycle but i wish it to assume only the numers
50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
If you would use a reproducible code example we
could point out where the error comes from;
otherwise we can only tell this should work;
try
x - c(50,100,200,300,900,2343)
for (i in x){
cat(i^2, \n)
}
# 2500
# 1
# 4
# 9
# 81
# 5489649
HTH,
Tobias
__
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.
--
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__
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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?
__
R-help@r-project.org mailing list
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Optional libraries (libtiff, etc) not found
There is sufficient circumstantial evidence to suggest you are posting in the wrong mailing list. Follow-up should go there. You also have not posted sessionInfo() output or the error messages you are getting when what-ever-it-is-that-you-have-done that didn't work. You should post in R-SIG-Mac and include the appropriate information requested in the Posting Guide (sessionInfo(), console output prior to error message, and error messages). On Sep 27, 2009, at 10:23 AM, Wayne F wrote: I installed the (binary) biOps package, which can use libtiff and libfftw. Then I used macports to install the libraries (in /opt/local/lib). But I restart R and biOps still does not see the libraries. There is a binary package called devpack4 at: http://cran.r-project.org/bin/macosx/tools/ I don't see libtiff in there, but libjpeg which biOps requires (according to the docs) is in there. If you are going to compile from sources you should read that information carefully. I've tried adding symbolic links from /opt/local/lib to /usr/local/ lib, I've added /opt/local/lib to LIBRARY_PATH, LD_LIBRARY_PATH, and DYLD_LIBRARY_PATH, and it doesn't work. Do you need to install biOps source in order for it to realize that the libraries are there? Or something else? I got this error when I tried to load the binary installed package: library(biOps) Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/Library/Frameworks/R.framework/ Resources/library/biOps/libs/x86_64/biOps.so': dlopen(/Library/Frameworks/R.framework/Resources/library/biOps/libs/ x86_64/biOps.so, 6): Symbol not found: _TIFFClose Referenced from: /Library/Frameworks/R.framework/Resources/library/ biOps/libs/x86_64/biOps.so Expected in: dynamic lookup Compiling from sources using the GUI Package Installer succeeded and the package loaded without error. I did encounter some warnings in the check phase: checking fftw3.h usability... no checking fftw3.h presence... no checking for fftw3.h... no checking jpeglib.h usability... configure: WARNING: Can't find fftw3 header yes checking jpeglib.h presence... yes checking for jpeglib.h... yes checking tiff.h usability... no checking tiff.h presence... no checking for tiff.h... no configure: WARNING: Can't find libtiff header My scan of the compilation messages suggested that the lack of tiff support was not remedied in the process but I cannot be sure about that (see below). In the past (say a year ago) I generally needed to compile from sources if I were using the 64-bit GUI, but that is an unusual occurrence more recently. Being a UNIX-noob, I cannot tell whether this error was due to my being in the 64 bit configuration. Whether the package will do anything useful now, I cannot say: capabilities() still reported FALSE for tiff The first example in hte documentation produces a black square jpeg() p - q - seq(-1, 1, length=20) r - 1 - outer(p^2, q^2, +) / 2 plot(imagedata(r)) dev.off() null device 1 The second example I tried produced a very pretty picture of a flower: x - readJpeg(system.file(samples, violet.jpg, package=biOps)) cat(Image Type, imageType(x)) Image Type rgb x size: 499 x 333 type: rgb plot(x) (And plotting to the tiff device produces a nice flower, despite what capabilities says.) Best of luck; -- David Winsemius, MD Heritage Laboratories West Hartford, CT sessionInfo() R version 2.9.2 (2009-08-24) x86_64-apple-darwin9.8.0 locale: en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] biOps_0.2.1 gdata_2.6.1 Design_2.3-0foreign_0.8-37 Hmisc_3.7-0 [6] survival_2.35-7 loaded via a namespace (and not attached): [1] cluster_1.12.0 grid_2.9.2 gtools_2.6.1lattice_0.17-25 tools_2.9.2 __ 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] QQ plotting of various distributions...
Thanks all! I did not want to cause any trouble and, God forbid, offense. I thought, I asked a simple question to improve my understanding and R-skills. It seems that there ain't single gospel truth about QQs. :-) Thanks, again! Best, PM Juliet Hannah wrote: I think it's helpful to show the sampling variability in a QQ plot under repeated sampling. An example is given in Venables, Ripley pg 86. The variance is higher at the tails. Even when the distributions are the same, the QQ plot does not have to resemble a straight line because of sampling. I don't think you can think of any one of these as the correct plot. Also, if the two data sets have an equal number of points, the empirical qq plot is simply a plot of one sorted data set against the other. (Kundu, Statistical Computing, pg 42). On Sun, Sep 27, 2009 at 9:06 AM, Duncan Murdoch murd...@stats.uwo.ca wrote: Eric Thompson wrote: The supposed example of a Q-Q plot is most certainly not how to make a Q-Q plot. I don't even know where to start First off, the two Q:s in the title of the plot stand for quantile, not random. The answer supplied simply plots two sorted samples of a distribution against each other. While this may resemble the general shape of a QQ plot, that is where the similarities end. The empirical quantiles of a sample are simply the sorted values. �You can plot empirical quantiles of one sample versus some version of quantiles from a distribution (what qqnorm does) or versus empirical quantiles of another sample (what Sunil did). �The randomness in his demonstration did two things: it generated some data, and it showed the variability of the plot under repeated sampling. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with downloading workspace file from a web address
Paul Smith wrote: Dear All, To load a previously saved workspace, one can do the following: load(/path/to/the/saved/workspace/file) However, if the path to the saved workspace file is a web address, one gets the following error: «Error in readChar(con, 5L, useBytes = TRUE) : cannot open the connection In addition: Warning message: In readChar(con, 5L, useBytes = TRUE) : cannot open compressed file 'http://phhs80.googlepages.com/workspace20090922', probable reason 'No such file or directory'» To circumvent this problem, one can download the saved workspace file to a local folder with download.file() and the option mode=wb active. My question is: Should not load() have the same mode option so that everything could be done only with load() (and not with two instructions: downaload.file() and load())? How did you try? Does load(url(http://phhs80.googlepages.com/workspace20090922;)) work for you? Uwe Ligges Thanks in advance, Paul __ 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] Converting SAS Data code to R.
On Sep 27, 2009, at 11:49 AM, Douglas Bates wrote: On Sat, Sep 26, 2009 at 11:33 PM, David Winsemius dwinsem...@comcast.net wrote: I am contemplating bringing in and merging three NHANES-III datasets from the National Center for Health Statistics that are fixed format with record length=3348, line counts around 20,000 and described by SAS DATA steps. I have downloaded and linked similar datasets from the Continuous NHANES public data releases, but never ones with this many variables at once. In the prior effort I managed the task by some cut-paste-editing from the SAS code file into a corresponding read.fwf R call, but the earlier NHANES-III data is far more voluminous than the more recent Continuous version. I am wondering if anyone has experience with such a process and would be willing to share some advice? The SAS code can be seen here: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.sas The main code file Data step starts out... FILENAME ADULT D:\Questionnaire\DAT\ADULT.DAT LRECL=3348; *** LRECL includes 2 positions for CRLF, assuming use of PC SAS; DATA WORK; INFILE ADULT MISSOVER; LENGTH SEQN 7 DMPFSEQ 5 DMPSTAT 3 DMARETHN 3 DMARACER 3 DMAETHNR 3 HSSEX 3 The corresponding positions in the INPUT section are INPUT SEQN 1-5 DMPFSEQ 6-10 DMPSTAT 11 DMARETHN 12 DMARACER 13 DMAETHNR 14 HSSEX15 The note about CRLF appears to be implying that those characters are being counted as part of the length of the first variable, SEQN, but that there are only 5 meaningful positions. I suppose I can find out by trial and error how to read such files, but it would save me some time if anyone in the audience has worked through this on this data before. One thought would be to import the data with the SAS work-alike program, WKS, (which I have not used before) and then to read in with read.xport from the foreign library. That would obviate the need to understand the character position issue, but probably has a time commitment to get it up and running and learn how to use it. Another thought would be to parse the fixed width SAS Data step code into pieces and build a data.frame from which I then extract the row.names, col.names, and colClasses from that centralized structure. Are the data available to the public somewhere or could just a few records be made available? Yes. Just trim the file name and the CDC ftp server accepts the path specification: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/ The file that goes with that SAS code is adult.dat ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.dat The reason I ask is because I imagine there are a lot of missing data in each record (the data are arranged in the wide format for longitudinal data and includes follow-up questions that will not apply to most respondents). The missing data indicator, if any, and the format of the other fields will be important in deciding how to split the data. Thanks for that. It was not designed as a longitudinal study, but rather as cross-sectional study that was spaced over several years. They did a re-exam of some sort, but that was not the primary purpose, nor will it be my particular interest. I have tried to determine by examination whether . or is the missing value indicator and it appears that both may used although there are many more spaces. Most of the input suggests to my 15-year-old memories of SAS that the data is numeric but there are 17 variables where input spec is $nn varLines[grep([[:punct:]], varLines)] [1] HAX11AG $6 HAX11AH $6 HAX11AI $6 [4] HAX11AJ $6 HAX11AK $6 HAX11AL $6 [7] HAX11AM $6 HAX11AN $6 HAX11AO $6 [10] HAX11AP $6 HAX11AQ $6 HAX11AR $6 [13] HAX11AS $6 HAX11AT $6 HAX11AU $6 [16] HAX11AV $6 HAZA1CC $30 -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] for cycle with uncontinuous numbers
You did not make test large enough to accept an index of 50 or any
of the other numbers in x for that matter. On the first pass through
that loop you to assigned to the 50 to the 50th element in test. If
you wanted to assign 50 to the first element of test the you should
look at the seq_along function:
x - matrix( c(50,100,200,300,900,2343) ,ncol = 1)
for (i in seq_along(x) ){
+ test [i] - x[i]
+ }
test
[,1]
[1,] 50
[2,] 100
[3,] 200
[4,] 300
[5,] 900
[6,] 2343
On Sep 27, 2009, at 11:45 AM, Marcio Resende wrote:
Dear John,
What I am trying to do is a genetic analysis and the i in my cycle
are the
numbers of markers I am testing.
I know the code is not right, but what i wanted to do was to fill
this 6 row
matrix with the numbers in x
So basically I wanted the cycle to loop only with the values on x
(50,100,200,300,900,2343) and not the the rest of the values
Do you know how to do it?
In the example I wanted my final test matrix to be
[50
100
200
300
900
2343]
dim(test) - still 6x1
And probably all zeroes, right?
Thanks for the help
jholtman wrote:
It is unclear exactly what you are trying to do. Since your 'test'
matrix is only 6 rows in length, then when you extend it by trying to
store into 50 for example, it will fill in the rest of the new values
with NA. Can you explain what you think the code is supposed to do?
It is doing exactly what you are asking it to do with the script you
provided.
On Sun, Sep 27, 2009 at 10:09 AM, Marcio Resende
mresende...@yahoo.com.br wrote:
Hi Tobias, thanks for the help,
the code I am using is quite long, but basically what I tried to
do was
test - matrix(0,6,1)
x - matrix( c(50,100,200,300,900,2343) ,ncol = 1)
for (i in x){
test [i] - (i)
}
but this code returns NA for all the elements which are not x
Tobias Verbeke-2 wrote:
Hi nice people,
:-)
I would like to do a for cycle but i wish it to assume only the
numers
50,
100, 200, 300, 900 and 2343
I tried to do something like
x - c(50,100,200,300,900,2343)
for (i in x){
#.
}
But it didn´t work
If you would use a reproducible code example we
could point out where the error comes from;
otherwise we can only tell this should work;
try
x - c(50,100,200,300,900,2343)
for (i in x){
cat(i^2, \n)
}
# 2500
# 1
# 4
# 9
# 81
# 5489649
HTH,
Tobias
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What is the problem that you are trying to solve?
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Re: [R] Select.spatial on spplots
Julius Tesoro wrote: Hi everyone. I posted this on R-sig-geo but got no response. Can select.spatial() be used in an existing spplot? I have tried selecting points (eq) from a plot generated from sp. However, when I invoke select.spatial(eq). It generates only the points without the background containing the faults. I need the background to select which earthquakes coalesce on which fault. Is there an alternative? eq.pts-list(sp.points,eq, col=blue, lwd=0.5, pch = 4) spplot(faults,Dip, xlim = c(11,12), ylim = c(376,389), sp.layout=list(eq.pts), col = heat.colors(3)) select.spatial(eq) Cheers, Julius Tesoro Unfortunately, it looks like select.spatial() wipes the plotting region by executing it's own call to plot(). However, the function it's self is very, very simple, just type: select.spatial To see what goes on inside. It looks like you could obtain point-wise selection by just calling identify(): identify( coordinates( eq )[,1], coordinates( eq )[,2] To use areas, call locator() and then process the results using point.in.polygon(). As for whether this will work using a spplot-- I really couldn't say. I was not able to reproduce a spplot using the information you provided-- lack of a working example may be a reason you got nothing but silence on R-sig-geo. Good luck! -Charlie - Charlie Sharpsteen Undergraduate Environmental Resources Engineering Humboldt State University -- View this message in context: http://www.nabble.com/Select.spatial-on-spplots-tp25632668p25635166.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] implementation of matrix logarithm (inverse of matrix exponential)
On Sat, 26 Sep 2009, spencerg wrote:
Sylvester's formula (http://en.wikipedia.org/wiki/Sylvester%27s_formula)
applies to a square matrix A = S L solve(S), where L = a diagonal matrix and
S = matrix of eigenvectors. Let f be an analytic function [for which f(A)
is well defined]. Then f(A) = S f(L) solve(S).
We can code this as follows:
sylvester - function(x, f){
n - nrow(x)
eig - eigen(x)
vi - solve(eig$vectors)
with(eig, (vectors * rep(f(values), each=n)) %*% vi)
}
logm - function(x)sylvester(x, log)
Example:
A - matrix(1:4, 2)
eA - expm(A)
logm(eA)
With Chuck Berry's example, we get the following:
M - matrix( c(0,1,0,0), 2 )
sylvester(M, log)
The case I gave would be
sylvester( as.matrix( expm( M ) ), log )
for which the perfectly sensible answer is M, not what appears here:
Error in solve.default(eig$vectors) :
system is computationally singular: reciprocal condition number =
1.00208e-292
This is a perfectly sensible answer in this case. We get the same
result from sylvester(M, exp), though expm(M) works fine.
A better algorithm for this could be obtains by studying the code
for expm in the Matrix package and the references in the associated help
page.
I doubt that code already in R will handle cases requiring Jordan blocks
for evaluation of the matrix logarithm (which cases arise in the context
of discrete state, continuous time Markov chains) without requiring one to
built that code more or less from scratch.
I'd be happy to hear that this is not so.
HTH,
Chuck
Hope this helps. Spencer
Gabor Grothendieck wrote:
Often one uses matrix logarithms on symmetric positive definite
matrices so the assumption of being symmetric is sufficient in many
cases.
On Sat, Sep 26, 2009 at 7:28 PM, Charles C. Berry cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
OK. Try this:
library(Matrix)
M - matrix(c(2, 1, 1, 2), 2); M
[,1] [,2]
[1,]21
[2,]12
Right. expm( M ) is diagonalizable.
But for
M - matrix( c(0,1,0,0), 2 )
you get the wrong result.
Maybe I should have added that I do not see the machinery in R for
dealing
with Jordan blocks.
HTH,
Chuck
# log of expm(M) is original matrix M
with(eigen(expm(M)), vectors %*% diag(log(values)) %*% t(vectors))
[,1] [,2]
[1,]21
[2,]12
On Sat, Sep 26, 2009 at 6:24 PM, Charles C. Berry
cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
Try:
expm( - M)
Mimosa probably meant say 'the inverse function'.
I do not see one in R.
Chuck
On Sat, Sep 26, 2009 at 5:06 PM, Mimosa Zeus mimosa1...@yahoo.fr
wrote:
Dear R users,
Does anyone has implemented the inverse of the matrix
exponential (expm
in the package Matrix)?
In Matlab, there're logm and expm, there's only expm in R.
Cheers
Mimosa
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
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Dept of Family/Preventive
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E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
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[R] Fwd: generic methods - in particular the summary function
Nobody wants to answer my question... is there something stupid in it? Début du message réexpédié : De : Christophe Dutang duta...@gmail.com Date : 19 septembre 2009 11:26:51 HAEC À : r-help@r-project.org Objet : generic methods - in particular the summary function Hi all, I'm currently working on the fitdistrplus package (that basically fit distributions). There is something I do not understand about the generic function summary. In the current version on CRAN, there is no NAMESPACE saying S3method(summary, fitdist) . However if we use summary on an object send by fitdist function it works fine... According to R-lang, we have The most common use of generic functions is to provide print and summary methods for statistical ob jects, generally the output of some model fitting process. To do this, each model attaches a class attribute to its output and then provides a special method that takes that output and provides a nice readable version of it. The user then needs only remember that print or summary will provide nice output for the results of any analysis. I would like to be sure, that if the summary.fitdist is not exported in the NAMESPACE, then we must use declare it with S3method. Thanks in advance Christophe -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr [[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] Fwd: generic methods - in particular the summary function
Christophe Dutang wrote: Nobody wants to answer my question... is there something stupid in it? Maybe, or maybe just too much traffic on the lists these days. Début du message réexpédié : De : Christophe Dutang duta...@gmail.com Date : 19 septembre 2009 11:26:51 HAEC À : r-help@r-project.org Objet : generic methods - in particular the summary function Hi all, I'm currently working on the fitdistrplus package (that basically fit distributions). There is something I do not understand about the generic function summary. In the current version on CRAN, there is no NAMESPACE saying S3method(summary, fitdist) . However if we use summary on an object send by fitdist function it works fine... Yes, because your package's CRAN version does not have any namespace at all, hence nothing can be hidden in it. According to R-lang, we have The most common use of generic functions is to provide print and summary methods for statistical ob jects, generally the output of some model fitting process. To do this, each model attaches a class attribute to its output and then provides a special method that takes that output and provides a nice readable version of it. The user then needs only remember that print or summary will provide nice output for the results of any analysis. I would like to be sure, that if the summary.fitdist is not exported in the NAMESPACE, then we must use declare it with S3method. Right, and that's the way you should do it if you have a namespace (do not export the function but declare it as an S3 method). And yes, it is a good idea to add a namespace. Best, Uwe Ligges Thanks in advance Christophe -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr [[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] NLM
Andrew Wang wrote:
I am trying to understand NLM package, so I generated this data set consisting y and x using
y= a + b*x +c*x^2 + N(0,10), with a=3.5,b=4.5,c=5.5
Given y and x, I am trying to use NLM to have estimates of parameters a, b and c that minimize the least square error
my code looks like
f- function(y,x,a,b,c) {sum((y-(a+b*x+c*x^2))^2)}
nlm(f, y, x, a=3, b=4,c=5)
But it comes up with rediculous result. My understanding of this NLM must be
wrong.
See ?nlm for the arguments you have to pass to nlm() and how the
function needs to be written. There are also some examples on that help
page.
Uwe Ligges
Please help!
Andy
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Re: [R] How to open only one file in a .gz file?
Peng Yu wrote: Hi, Suppose that there are multiple files in a .gz file. Well, the extension .gz regularly stand for a (single) gzip compressed file. Tyoically you will find more tan one file in a tar archive or a zip compressed archive. For the latter see ?unz. Uwe Ligges How to open only one file in it? I don't find such options in the help. Regards, Peng __ 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] Fwd: generic methods - in particular the summary function
On Sep 27, 2009, at 12:52 PM, Christophe Dutang wrote: Nobody wants to answer my question... is there something stupid in it? I couldn't really say. Not being a user of fitdistrplus I don't have much baseline experience. Had you posted code that produced something to work on, I might have made an effort at applying the methods I typically use, such as str(object), methods(function), and printing out the function code. But I am a rather low-level R-user, and as soon as you strart throwing around questions about NAMESPACE, my eyes glaze over and I move to the next question. When I do install and load fitdistrplus and then execute: methods(summary) .. I see a new function which is not even invisible, called summary.fitdist. So it seems to be in my NAMESPACE ... or at least what I dimly understand about such metaphysical entities. I get an error when I execute your suggestion: S3method(summary, fitdist) Error: could not find function S3method So I would have assumed that you had a package with that S3method function about which you had not informed us. Not sure this helps because, as I said, your question appeared more complex that I considered myself competent to answer. -- David Début du message réexpédié : De : Christophe Dutang duta...@gmail.com Date : 19 septembre 2009 11:26:51 HAEC À : r-help@r-project.org Objet : generic methods - in particular the summary function Hi all, I'm currently working on the fitdistrplus package (that basically fit distributions). There is something I do not understand about the generic function summary. In the current version on CRAN, there is no NAMESPACE saying S3method(summary, fitdist) . However if we use summary on an object send by fitdist function it works fine... According to R-lang, we have The most common use of generic functions is to provide print and summary methods for statistical ob jects, generally the output of some model fitting process. To do this, each model attaches a class attribute to its output and then provides a special method that takes that output and provides a nice readable version of it. The user then needs only remember that print or summary will provide nice output for the results of any analysis. I would like to be sure, that if the summary.fitdist is not exported in the NAMESPACE, then we must use declare it with S3method. Thanks in advance Christophe -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr [[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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error while plotting
Nair, Murlidharan T wrote: I am getting the following errors when I am trying to plot the data below. I cannot figure out the error. Error in plot.window(...) : need finite 'xlim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf #I am using the following code #== library(multcomp) sig.data-structure(list(X = 1:10, Cell.lines = structure(c(2L, 5L, 8L, 9L, 3L, 6L, 10L, 1L, 4L, 7L), .Label = c(T(70%)a-N(0%)c, T(70%)a-N(0%)f, T(70%)a-N(0%)i, T(70%)c-N(0%)c, T(70%)c-N(0%)f, T(70%)c-N(0%)i, T(80%)a-N(0%)c, T(80%)a-N(0%)f, T(90%)-N(0%)f, T(90%)-N(0%)i ), class = factor), estimate = c(9859.74333, -5553.64802, 6227.17947, 8063.6472, 6548.86032, -8864.53103, 4752.7642, 9057.72021, -6355.67115, 5425.15635), lower = c(5560.57875, -9852.8126, 1928.01489, 3764.48262, 2249.69575, -13163.69561, 453.59962, 4758.55563, -10654.83573, 1125.99177), upper = c(14158.90791, -1254.48344, 10526.34405, 12362.81178, 10848.0249, -4565.36645, 9051.92877, 13356.88479, -2056.50657, 9724.32092), p.val.raw = c(1.15e-08, 5.78e-05, 1.36e-05, 3.21e-07, 6.91e-06, 6.97e-08, 0.000331, 4.87e-08, 1.04e-05, 7.63e-05 ), p.val.bon = c(2.66e-06, 0.0133, 0.00315, 7.41e-05, 0.0016, 1.61e-05, 0.0764, 1.13e-05, 0.0024, 0.0176), p.val.adj = c(2.65e-13, 0.000592, 2.82e-05, 9.72e-08, 6.56e-05, 8.76e-09, 0.0117, 6.22e-09, 6.44e-06, 0.000334)), .Names = c(X, Cell.lines, estimate, lower, upper, p.val.raw, p.val.bon, p.val.adj), class = data.frame, row.names = c(T(70%)a-N(0%)f, T(70%)c-N(0%)f, T(80%)a-N(0%)f, T(90%)-N(0%)f, T(70%)a-N(0%)i, T(70%)c-N(0%)i, T(90%)-N(0%)i, T(70%)a-N(0%)c, T(70%)c-N(0%)c, T(80%)a-N(0%)c)) rownames(sig.data)-sig.data[,2] my.hmtest - structure(list( estimate = t(t(structure(sig.data[,estimate], .Names = rownames(sig.data, conf.int = sig.data[,4:5], ctype = ABCC4-2007), class = hmtest) par(mex=0.5) #This helps to accomodate the margins when text is getting cut off plot(my.hmtest, cex.axis=0.7) There is not method plot.hmtest defined anywhere. Hence plot.default is used and that one does not know hoe to handle an object like the one you just defined. Uwe Ligges __ 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] Check value interval in a if loop
Hi R community I have a little problem, and I tried to solve it by myself but I couldn't. I building an if loop, and I want to check a value inside an interval. This would be the case: pvalue=0,2999 if(pvalue0.05 or pvalue0.1) as you can see I would like to check in that if loop if my pvalue is inside of that interval(from 0.05 to 0.1), and I tried these options: if(pvalue0.05 or pvalue0.1) (not possible because R don't recognise OR as operator) if(pvalue0.05 || pvalue0.1) (this one is not good enough, cause it fulfills one condition pvalue0.05 but it doesn't get other contidion pvalue0.1) does anyone know a way to stablish a inteval as an statement for my if loop. Thanks in advance Lucas _ [[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] xtable - how to add a sum of values in a row column?
Hi, I saw this example for 2.10 Time series in the xtable gallery documentation. http://cran.r-project.org/web/packages/xtable/vignettes/xtableGallery.pdf How would I add a column at the end Total which sums the row, with minimal changes to the code below? Thanks in advance. - Ken 2.10 Time series temp.ts - ts(cumsum(1 + round(rnorm(100), 0)), start = c(1954, + 7), frequency = 12) temp.table - xtable(temp.ts, digits = 0) caption(temp.table) - Time series example print(temp.table, floating = FALSE) Time series example Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1954 2 3 6 8 11 10 1955 11 13 15 16 16 18 20 22 21 22 24 24 1956 25 26 28 28 28 28 29 31 31 32 33 34 1957 35 36 38 39 39 41 42 42 41 42 43 45 1958 46 46 47 47 49 51 54 56 58 59 61 61 1959 62 61 62 62 62 63 62 64 64 66 67 68 1960 67 67 69 71 74 75 77 78 79 80 82 81 1961 84 86 87 88 89 91 94 94 94 94 96 97 1962 98 99 101 102 104 105 108 107 106 107 -- View this message in context: http://www.nabble.com/xtable---how-to-add-a-%22sum-of-values-in-a-row%22-column--tp25635552p25635552.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to open only one file in a .gz file?
gzfile() can handle a single file in *.gz. If you want to read one of the files in a bunch of files, you may as well extract them first using tar -zxvf yourarchive.tar.gz (sounds like a stupid solution), or if you only want to use R, system() could help, e.g. system(tar -zxvf yourarchive.tar.gz). You didn't tell us sessionInfo(), so I'm not sure whether you are able to use the tar command. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: 515-294-6609 Web: http://yihui.name Department of Statistics, Iowa State University 3211 Snedecor Hall, Ames, IA 2009/9/27 Uwe Ligges lig...@statistik.tu-dortmund.de: Peng Yu wrote: Hi, Suppose that there are multiple files in a .gz file. Well, the extension .gz regularly stand for a (single) gzip compressed file. Tyoically you will find more tan one file in a tar archive or a zip compressed archive. For the latter see ?unz. Uwe Ligges How to open only one file in it? I don't find such options in the help. Regards, Peng __ 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] constraining covariance structure for random effects in lme
I would like to impose a unique structure on the random effects covariance matrix in a multilevel model (i.e., a structure that is not included in the standard pdMat classes listed in Table 4.3 of Pinheiro and Bates' lme book). Is there any way to manually impose equality constraints on the elements of the random effects covariance matrix? Thank you. __ 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] Check value interval in a if loop
Lucas Sevilla García wrote: Hi R community I have a little problem, and I tried to solve it by myself but I couldn't. I building an if loop, and I want to check a value inside an interval. This would be the case: pvalue=0,2999 if(pvalue0.05 or pvalue0.1) as you can see I would like to check in that if loop if my pvalue is inside of that interval(from 0.05 to 0.1), and I tried these options: if(pvalue0.05 or pvalue0.1) (not possible because R don't recognise OR as operator) if(pvalue0.05 || pvalue0.1) (this one is not good enough, cause it fulfills one condition pvalue0.05 but it doesn't get other contidion pvalue0.1) does anyone know a way to stablish a inteval as an statement for my if loop. 1. if() is not used for constructing loops (and I think you know that) 2. You need an and operator rather than or, such as (for vectors) or (for scalar values). Uwe Ligges Thanks in advance Lucas _ [[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 CMD INSTALL --build: Folders /inst and /etc not in zip-file and WindowsXP locks /library/[package]/etc/
Tobias Schoch wrote: Dear R users, My set-up: OS=Windows XP, R-2.9.2, Rtools210 I faced the follwing problem with the package compilation: There is no /inst or /etc subdirectory in the package-zip-file. And the content of the /etc subdirectory is lost, too. I tried a simplified test package. The test package has the following structure (see also attachement: test package as source file): /test |---/inst | |---/etc | |---menus.txt | |---/man | |---mymean.Rd | |---test-package.Rd | |---/R | |---mymean.R | |---NAMESPACE |---DESCRIPTION The file menus.txt (inspired by the Rcmdr menu structure) contains one single comment line. The file mymean.R contains a simple function that computes the mean. Situation A) [R CMD BUILD test] works fine and the tar-ball contains the subdirectory /inst/etc/ and the file menus.txt. Situation B) [R CMD INSTALL --build test] generates the test_1.0.zip file without any error message. But: 1) This zip-file does neither contain the /etc folder nor the menus.txt file. 2) The installation created the folder /test in the library path /R-2.9.2/library/test/ , but this folder is locked by WindowsXP. That is, the access is denied and can only be resolved by re-defining the owner and the permission rights. 3) Having removed the package from the library, and trying to install the zip-package using install from local zip files... does the installation. But, the /etc folder is empty; the file menus.txt is missing. Probably the installation process in your Situation B) is never finished and hangs up due to misconfiguration of your tools or due to wrong permissions somewhere. I do not have any problem and explicitly tested with ./inst/etc/menus.txt (only difference is that I do not have exactly Rtools10 installed, but something that is extremely close). Try to make a clean setup of all your tools and double check your PATH variable. Best, Uwe Ligges Is there a known problem/bug in the R CMD INSTALL --build process dealing with the subdirectories /inst and /etc and the contents of the these folders? How to resolve the problem? Thanks Tobias http://www.nabble.com/file/p25609569/source_test.zip source_test.zip __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with downloading workspace file from a web address
2009/9/27 Uwe Ligges lig...@statistik.tu-dortmund.de: To load a previously saved workspace, one can do the following: load(/path/to/the/saved/workspace/file) However, if the path to the saved workspace file is a web address, one gets the following error: «Error in readChar(con, 5L, useBytes = TRUE) : cannot open the connection In addition: Warning message: In readChar(con, 5L, useBytes = TRUE) : cannot open compressed file 'http://phhs80.googlepages.com/workspace20090922', probable reason 'No such file or directory'» To circumvent this problem, one can download the saved workspace file to a local folder with download.file() and the option mode=wb active. My question is: Should not load() have the same mode option so that everything could be done only with load() (and not with two instructions: downaload.file() and load())? How did you try? Does load(url(http://phhs80.googlepages.com/workspace20090922;)) work for you? Thanks, Uwe, that works pretty fine! Paul __ 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] Check value interval in a if loop
Hi Lucas, try: if(pvalue0.05 pvalue0.1) HTH Marcio Lucas Sevilla García wrote: Hi R community I have a little problem, and I tried to solve it by myself but I couldn't. I building an if loop, and I want to check a value inside an interval. This would be the case: pvalue=0,2999 if(pvalue0.05 or pvalue0.1) as you can see I would like to check in that if loop if my pvalue is inside of that interval(from 0.05 to 0.1), and I tried these options: if(pvalue0.05 or pvalue0.1) (not possible because R don't recognise OR as operator) if(pvalue0.05 || pvalue0.1) (this one is not good enough, cause it fulfills one condition pvalue0.05 but it doesn't get other contidion pvalue0.1) does anyone know a way to stablish a inteval as an statement for my if loop. Thanks in advance Lucas _ [[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. -- View this message in context: http://www.nabble.com/Check-value-interval-in-a-if-loop-tp25635562p25635651.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] Clustering with R - efficient processing of large sparse data sets (text data)
I checked the R procedure HCLUST (hierarchical clustering) but it looks like it requires a full triangular n x n similarity matrix as input, where n = number of observations. The number of variables is 200. My data set has n = 50,000 observations (keywords), and I use ad-hoc similarity measures, not available in R, to measure keyword similarity. Here, the vast majority of the n x n similarities are equal to zero. So I am looking for a clustering procedure that would accept the following alternate input: x1, y1, s1 x2, y2, s2 ... xk, yk, sk where xi, yi are 2 keywords with similarity si 0 (1 = i = k). This input would contain k = 10,000 rows, which is much smaller than n x n = 50,000 x 50,000 elements when using the similarity matrix. The HCLUST function would crash if it used the dissimilarity matrix as input. Do you know how to use my small data input in R, instead of a very large sparse similarity matrix? Or in SAS? I need a simple solution, otherwise I'll just write myself the code that does hierarchical clustering, in C or Perl, or use a library. It would take me 2 hours to write the hierarchical clustering code from scratch, so I'm looking for a simple solution that will take less than 2 hours to implement. Follow up at http://www.analyticbridge.com/group/R_Packages/forum/topics/clustering-with-r-efficient __ 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] Clustering with R - efficient processing of large sparse data sets (text data)
I checked the R procedure HCLUST (hierarchical clustering) but it looks like it requires a full triangular n x n similarity matrix as input, where n = number of observations. The number of variables is 200. My data set has n = 50,000 observations (keywords), and I use ad-hoc similarity measures, not available in R, to measure keyword similarity. Here, the vast majority of the n x n similarities are equal to zero. So I am looking for a clustering procedure that would accept the following alternate input: x1, y1, s1 x2, y2, s2 ... xk, yk, sk where xi, yi are 2 keywords with similarity si 0 (1 = i = k). This input would contain k = 10,000 rows, which is much smaller than n x n = 50,000 x 50,000 elements when using the similarity matrix. The HCLUST function would crash if it used the dissimilarity matrix as input. Do you know how to use my small data input in R, instead of a very large sparse similarity matrix? Or in SAS? I need a simple solution, otherwise I'll just write myself the code that does hierarchical clustering, in C or Perl, or use a library. It would take me 2 hours to write the hierarchical clustering code from scratch, so I'm looking for a simple solution that will take less than 2 hours to implement. Follow up at: http://www.analyticbridge.com/group/R_Packages/forum/topics/clustering-with-r-efficient __ 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] View this page R-Packages (Statistical Computing)
Click on http://groups.google.com/group/r-help-archive/web/r-packages-statistical-computing - or copy paste it into your browser's address bar if that doesn't work. __ 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] View this page R-Packages (Statistical Computing)
Click on http://groups.google.com/group/r-help-archive/web/r-packages-statistical-computing - or copy paste it into your browser's address bar if that doesn't work. __ 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] Check value interval in a if loop
Lucas Sevilla García wrote:
Hi R community
I have a little problem, and I tried to solve it by myself but I
couldn't. I building an if loop, and I want to check a value inside
an interval. This would be the case:
pvalue=0,2999
if(pvalue0.05 or pvalue0.1)
as you can see I would like to check in that if loop if my pvalue is
inside of that interval(from 0.05 to 0.1), and I tried these options:
Hi Lucas,
Your if statement is probably embedded in a loop like this:
for(mytest in 1:100) {
pvalue-t.test(rnorm(50),rnorm(50))$p.value
if(pvalue 0.05 pvalue 0.1) cat(Just missed!\n)
else cat(I don't care\n)
}
so you only want it to test the current p value. If you want to get all
the p values and test them later:
pvalue-rep(NA,100)
for(mytest in 1:100)
pvalue[mytest]-t.test(rnorm(50),rnorm(50))$p.value
cat(c(Just missed\n,I don't care\n)
[(pvalue 0.05 pvalue 0.1)+1])
Jim
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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] Count number of zeros in a collumn
I have a matrix 700x2000 which is sampled in each cycle from another matrix 788x2000 with the numbers 0,1 and 9 There is one specific collumn of this matrix, dart[,1977], that usually, after the samplimg procedure has only 1 and 9 (because the zero frequency in this collumn is low). However, when this happens, I want to include an IF conditional in my code. so basically what i wanted to do was to count the number of zeros in this collumn to use this information in my conditional: something like: if (the number of zeros in collumn [,1977] is zero) ... else ... I tried to find out indirectly by the rowsum but because of the sampling procedure the rowsum is not always the same Thank you very much -- View this message in context: http://www.nabble.com/Count-number-of-zeros-in-a-collumn-tp25637516p25637516.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] QQ plotting of various distributions...
It seems I misunderstood Sunil's response and somewhat freaked out because it appeared that he was giving the wrong method for making a QQ plot, but was actually demonstrating the sampling variability. My apologies to Sunil. 2009/9/27 Duncan Murdoch murd...@stats.uwo.ca: Eric Thompson wrote: The supposed example of a Q-Q plot is most certainly not how to make a Q-Q plot. I don't even know where to start First off, the two Q:s in the title of the plot stand for quantile, not random. The answer supplied simply plots two sorted samples of a distribution against each other. While this may resemble the general shape of a QQ plot, that is where the similarities end. The empirical quantiles of a sample are simply the sorted values. You can plot empirical quantiles of one sample versus some version of quantiles from a distribution (what qqnorm does) or versus empirical quantiles of another sample (what Sunil did). The randomness in his demonstration did two things: it generated some data, and it showed the variability of the plot under repeated sampling. Some general advice: be careful who you take advice from on the internet. That's good advice. Duncan Murdoch The Wikipedia entry for Q-Q plot may be a good start if you don't know what a Q-Q plot is, although you should also use it with caution. Lets say you have some samples that may be normally distributed: set.seed(1) x - rnorm(30) # now try with R's built in function qqnorm(x, xlim = c(-3, 3), ylim = c(-3, 3)) # Now try Sunil's Q-Q plot method, but for rnorm # rather than rgamma some_data - x test_data - rnorm(30) points(sort(some_data),sort(test_data), col = blue) # Note that the points are NOT the same! This should have been obvious for the simple reason that the QQ plot should not be influenced by the random number generator that you are using! A QQ plot is uniquely reproducible. The more general (and correct) way to get the QQ plot involves choosing a plotting position and the quantile function (e.g. qnorm or qgamma functions in R) of the pertinent distribution: # Sort the data: x.s - sort(x) n - length(x) # Plotting position (must be careful here in general!) p - ppoints(n) # Compute the quantile x.q - qnorm(p) points(x.q, x.s, col = red) # and they fall exactly on the points generated by qqnorm(). Now, you should be able to generalize this for any distribution. Hope this helps. Eric Thompson 2009/9/27 Petar Milin pmi...@ff.uns.ac.rs: Thanks for the answer. Now, only problem is to to get parameter(s) of a given function. For gamma, I shall try with gammafit() from mhsmm package. Also, I shall look for others appropriate parameter estimates. Will use SuppDists too. Best, PM Sunil Suchindran wrote: #same shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=6,scale=2) plot(sort(some_data),sort(test_data)) # You can also use qqplot(some_data,test_data) abline(0,1) # different shape some_data - rgamma(500,shape=6,scale=2) test_data - rgamma(500,shape=4,scale=2) plot(sort(some_data),sort(test_data)) abline(0,1) It is helpful to assess the sampling variability, by creating repeated sets of test_data, and plotting all of these along with your observations to create a confidence envelope. The SuppDists provides Inverse Gauss. On Thu, Sep 17, 2009 at 11:46 AM, Petar Milin pmi...@ff.uns.ac.rs wrote: Hello! I am trying with this question again: I would like to test few distributional assumptions for some behavioral response data. There are few theories about true distribution of those data, like: normal, lognormal, gamma, ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would be via qq-plot, to show to students differences. First two are trivial: qqnorm(dat$X) qqnorm(log(dat$X)) Then, things are getting more hairy. I am not sure how to make plots for the rest. I tried gamma with: qqmath(~ X, data=dat, distribution=function(X) � qgamma(X, shape, scale)) Which should be the same as: plot(qgamma(ppoints(dat$X), shape, scale), sort(dat$X)) Shape and scale parameters I got via mhsmm package that has gammafit() for shape and scale parameters estimation. Am I on right track? Does anyone know how to plot the rest: ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian)? Thanks, PM __ R-help@r-project.org mailto: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 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
Re: [R] Count number of zeros in a collumn
Hi Marcio, How about if (min(dart[,1977])==0)... cheers milton On Sun, Sep 27, 2009 at 4:54 PM, Marcio Resende mresende...@yahoo.com.brwrote: I have a matrix 700x2000 which is sampled in each cycle from another matrix 788x2000 with the numbers 0,1 and 9 There is one specific collumn of this matrix, dart[,1977], that usually, after the samplimg procedure has only 1 and 9 (because the zero frequency in this collumn is low). However, when this happens, I want to include an IF conditional in my code. so basically what i wanted to do was to count the number of zeros in this collumn to use this information in my conditional: something like: if (the number of zeros in collumn [,1977] is zero) ... else ... I tried to find out indirectly by the rowsum but because of the sampling procedure the rowsum is not always the same Thank you very much -- View this message in context: http://www.nabble.com/Count-number-of-zeros-in-a-collumn-tp25637516p25637516.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.htmlhttp://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] RSQLite column names underscores
Hi, When I use RSQLite's dbWriteTable(...) function, the columns in the db table frequently end up having __1 (two underscores and a one) added to them. The names are unique within the table and contain no weird characters. For example a position column from a data.frame was written to the DB as position__1. Does anyone understand how this name munging happens and how I can prevent or work around it? Having these suffixes added seemingly arbitrarily makes it hard to construct the next query. Thanks for any hints, -chris __ 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] rsolnp- Error (Help!)
Hi I am relatively new to R and was trying to run an optimization problem using rsolnp. I am getting an error which seems to be not related to my construct of the optimization equations. Error in if (max(tt[2] - 10 * tol, nineq) = 0) rho = 0 : missing value where TRUE/FALSE needed I have attached the file code. I would greatly appreciate any help. Many thanks http://www.nabble.com/file/p25637806/OptTS.txt OptTS.txt -- View this message in context: http://www.nabble.com/rsolnp--Error-%28Help%21%29-tp25637806p25637806.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] RSQLite column names underscores
You are probably trying to use SQL reserved keywords as column names. Try entering this at the R prompt: library(RSQLite) .SQL92Keywords On Sun, Sep 27, 2009 at 5:25 PM, Christopher Bare cb...@systemsbiology.org wrote: Hi, When I use RSQLite's dbWriteTable(...) function, the columns in the db table frequently end up having __1 (two underscores and a one) added to them. The names are unique within the table and contain no weird characters. For example a position column from a data.frame was written to the DB as position__1. Does anyone understand how this name munging happens and how I can prevent or work around it? Having these suffixes added seemingly arbitrarily makes it hard to construct the next query. Thanks for any hints, -chris __ 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] Converting SAS Data code to R.
On Sep 27, 2009, at 12:10 PM, David Winsemius wrote: On Sep 27, 2009, at 11:49 AM, Douglas Bates wrote: On Sat, Sep 26, 2009 at 11:33 PM, David Winsemius dwinsem...@comcast.net wrote: I am contemplating bringing in and merging three NHANES-III datasets from the National Center for Health Statistics that are fixed format with record length=3348, line counts around 20,000 and described by SAS DATA steps. I have downloaded and linked similar datasets from the Continuous NHANES public data releases, but never ones with this many variables at once. In the prior effort I managed the task by some cut-paste-editing from the SAS code file into a corresponding read.fwf R call, but the earlier NHANES-III data is far more voluminous than the more recent Continuous version. I am wondering if anyone has experience with such a process and would be willing to share some advice? The SAS code can be seen here: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.sas The main code file Data step starts out... FILENAME ADULT D:\Questionnaire\DAT\ADULT.DAT LRECL=3348; *** LRECL includes 2 positions for CRLF, assuming use of PC SAS; DATA WORK; INFILE ADULT MISSOVER; LENGTH SEQN 7 DMPFSEQ 5 DMPSTAT 3 DMARETHN 3 DMARACER 3 DMAETHNR 3 HSSEX 3 The corresponding positions in the INPUT section are INPUT SEQN 1-5 DMPFSEQ 6-10 DMPSTAT 11 DMARETHN 12 DMARACER 13 DMAETHNR 14 HSSEX15 The note about CRLF appears to be implying that those characters are being counted as part of the length of the first variable, SEQN, but that there are only 5 meaningful positions. I suppose I can find out by trial and error how to read such files, but it would save me some time if anyone in the audience has worked through this on this data before. One thought would be to import the data with the SAS work-alike program, WKS, (which I have not used before) and then to read in with read.xport from the foreign library. That would obviate the need to understand the character position issue, but probably has a time commitment to get it up and running and learn how to use it. Another thought would be to parse the fixed width SAS Data step code into pieces and build a data.frame from which I then extract the row.names, col.names, and colClasses from that centralized structure. Are the data available to the public somewhere or could just a few records be made available? Yes. Just trim the file name and the CDC ftp server accepts the path specification: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/ The file that goes with that SAS code is adult.dat ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.dat The reason I ask is because I imagine there are a lot of missing data in each record (the data are arranged in the wide format for longitudinal data and includes follow-up questions that will not apply to most respondents). The missing data indicator, if any, and the format of the other fields will be important in deciding how to split the data. Thanks for that. It was not designed as a longitudinal study, but rather as cross-sectional study that was spaced over several years. They did a re-exam of some sort, but that was not the primary purpose, nor will it be my particular interest. I have tried to determine by examination whether . or is the missing value indicator and it appears that both may used although there are many more spaces. Most of the input suggests to my 15-year-old memories of SAS that the data is numeric but there are 17 variables where input spec is $nn varLines[grep([[:punct:]], varLines)] [1] HAX11AG $6 HAX11AH $6 HAX11AI $6 [4] HAX11AJ $6 HAX11AK $6 HAX11AL $6 [7] HAX11AM $6 HAX11AN $6 HAX11AO $6 [10] HAX11AP $6 HAX11AQ $6 HAX11AR $6 [13] HAX11AS $6 HAX11AT $6 HAX11AU $6 [16] HAX11AV $6 HAZA1CC $30 My progress on this effort so far consists of having figured out how to extract the variable names and their associated lengths so I can set up a call to read.fwf(). This is waht I did on hte section of the SAS code following INPUT that contains those elements: trim.ws - function(x) gsub(^[[:space:]]+|[[:space:]]+$, ,x) # courtesy of a Grothendieck r-help posting of a couple or three years ago. adult.var - data.frame(varnames = sapply( strsplit(trim.ws(varLines) , +) , [, 1:2)[1,], varlen= sapply( strsplit(trim.ws(varLines) , +) , [, 1:2)[2,]) #so that I can split the trimmed strings on an arbitrary number of spaces. adult.var[,][1:5,] varnames varlen 1 SEQN 7 2 DMPFSEQ 5 3 DMPSTAT 3 4 DMARETHN 3 5 DMARACER 3
Re: [R] implementation of matrix logarithm (inverse of matrix exponential)
There is a logm function in the expm package in the expm project on
R-forge. See http://expm.R-forge.R-project.org/
Martin was the person who added that function so I will defer to his
explanations of what it does. I know he has been traveling and it may
be a day or two before he can get to this thread.
On Sun, Sep 27, 2009 at 11:44 AM, Charles C. Berry cbe...@tajo.ucsd.edu wrote:
On Sat, 26 Sep 2009, spencerg wrote:
Sylvester's formula
(http://en.wikipedia.org/wiki/Sylvester%27s_formula) applies to a square
matrix A = S L solve(S), where L = a diagonal matrix and S = matrix of
eigenvectors. Let f be an analytic function [for which f(A) is well
defined]. Then f(A) = S f(L) solve(S).
We can code this as follows:
sylvester - function(x, f){
n - nrow(x)
eig - eigen(x)
vi - solve(eig$vectors)
with(eig, (vectors * rep(f(values), each=n)) %*% vi)
}
logm - function(x)sylvester(x, log)
Example:
A - matrix(1:4, 2)
eA - expm(A)
logm(eA)
With Chuck Berry's example, we get the following:
M - matrix( c(0,1,0,0), 2 )
sylvester(M, log)
The case I gave would be
sylvester( as.matrix( expm( M ) ), log )
for which the perfectly sensible answer is M, not what appears here:
Error in solve.default(eig$vectors) :
system is computationally singular: reciprocal condition number =
1.00208e-292
This is a perfectly sensible answer in this case. We get the
same result from sylvester(M, exp), though expm(M) works fine.
A better algorithm for this could be obtains by studying the code
for expm in the Matrix package and the references in the associated help
page.
I doubt that code already in R will handle cases requiring Jordan blocks for
evaluation of the matrix logarithm (which cases arise in the context of
discrete state, continuous time Markov chains) without requiring one to
built that code more or less from scratch.
I'd be happy to hear that this is not so.
HTH,
Chuck
Hope this helps. Spencer
Gabor Grothendieck wrote:
Often one uses matrix logarithms on symmetric positive definite
matrices so the assumption of being symmetric is sufficient in many
cases.
On Sat, Sep 26, 2009 at 7:28 PM, Charles C. Berry cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
OK. Try this:
library(Matrix)
M - matrix(c(2, 1, 1, 2), 2); M
[,1] [,2]
[1,] 2 1
[2,] 1 2
Right. expm( M ) is diagonalizable.
But for
M - matrix( c(0,1,0,0), 2 )
you get the wrong result.
Maybe I should have added that I do not see the machinery in R for
dealing
with Jordan blocks.
HTH,
Chuck
# log of expm(M) is original matrix M
with(eigen(expm(M)), vectors %*% diag(log(values)) %*% t(vectors))
[,1] [,2]
[1,] 2 1
[2,] 1 2
On Sat, Sep 26, 2009 at 6:24 PM, Charles C. Berry
cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
Try:
expm( - M)
Mimosa probably meant say 'the inverse function'.
I do not see one in R.
Chuck
On Sat, Sep 26, 2009 at 5:06 PM, Mimosa Zeus
mimosa1...@yahoo.fr
wrote:
Dear R users,
Does anyone has implemented the inverse of the
matrix exponential (expm
in the package Matrix)?
In Matlab, there're logm and expm, there's only expm
in R.
Cheers
Mimosa
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Re: [R] array to matrix or data.frame
On Sep 27, 2009, at 6:33 PM, Daniel Malter wrote: Hi, I have an array of dimension 6:6:16. I want to stack the 16 slices of the array into a matrix or a data frame of dimension 6:(6*16=96) in order to write it to a csv file. It seems I cannot get the matrix or data.frame functions to put the values from the array in the same order as they appear in the array easily, but I am almost sure there is an easy way to do this. #Here is code for a small 6:6:2 array. x=rep(rep(1:6,each=6),2) dim(x)=c(6,6,2) x #Matrix approach 1: matrix(x,nrow=12,ncol=6,byrow=F) #Matrix approach 2: matrix(x,nrow=12,ncol=6,byrow=T) #Data frame approach (cbinds rather than rbinds): data.frame(x) None of the above works as I would like to. The correct approach should stack the same values all in the same column of the matrix or data frame. I also could not get stack to work correctly and would appreciate your help. apply(x, 2 , I) [,1] [,2] [,3] [,4] [,5] [,6] [1,]123456 [2,]123456 [3,]123456 [4,]123456 [5,]123456 [6,]123456 [7,]123456 [8,]123456 [9,]123456 [10,]123456 [11,]123456 [12,]123456 -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] array to matrix or data.frame
Try this: matrix(aperm(x, c(1,3,2)), nc = 6) On Sun, Sep 27, 2009 at 6:33 PM, Daniel Malter dan...@umd.edu wrote: Hi, I have an array of dimension 6:6:16. I want to stack the 16 slices of the array into a matrix or a data frame of dimension 6:(6*16=96) in order to write it to a csv file. It seems I cannot get the matrix or data.frame functions to put the values from the array in the same order as they appear in the array easily, but I am almost sure there is an easy way to do this. #Here is code for a small 6:6:2 array. x=rep(rep(1:6,each=6),2) dim(x)=c(6,6,2) x #Matrix approach 1: matrix(x,nrow=12,ncol=6,byrow=F) #Matrix approach 2: matrix(x,nrow=12,ncol=6,byrow=T) #Data frame approach (cbinds rather than rbinds): data.frame(x) None of the above works as I would like to. The correct approach should stack the same values all in the same column of the matrix or data frame. I also could not get stack to work correctly and would appreciate your help. Daniel -- View this message in context: http://www.nabble.com/array-to-matrix-or-data.frame-tp25638310p25638310.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] dimension-preserving matrix coersion
i've written a function to coerce a matrix (e.g. from numeric to logical), but i'd like to know if someone has a more elegant method for this: m - matrix(c(0, 1, 1, 0), ncol = 2) m - as.logical(m) m [1] FALSE TRUE TRUE FALSE i'd like 'm' to still be a matrix with the original dimensions. in my function to do this, i coerce 'm' to a logical, then re-form it as a matrix, which seems like an extra (possibly bug-introducing) step that might be avoided if i knew of some hidden feature that might permit this in one fell swoop. any ideas? __ 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] array to matrix or data.frame
On 28/09/2009, at 11:51 AM, David Winsemius wrote:
snip
apply(x, 2 , I)
snip
***Much*** sexier than my solution!
cheers,
Rolf Turner
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Re: [R] rsolnp- Error (Help!)
It means that your expression max(tt[2] - 10 * tol, nineq) is returning NA: Notice I get the same error: if (1==1)1 [1] 1 if (NA == 1) 1 Error in if (NA == 1) 1 : missing value where TRUE/FALSE needed Check your script and see why it is NA. you might need: max(tt[2] - 10 * tol, nineq, na.rm=TRUE) If your data has NAs. On Sun, Sep 27, 2009 at 5:29 PM, tushar_kul tus...@gmail.com wrote: Hi I am relatively new to R and was trying to run an optimization problem using rsolnp. I am getting an error which seems to be not related to my construct of the optimization equations. Error in if (max(tt[2] - 10 * tol, nineq) = 0) rho = 0 : missing value where TRUE/FALSE needed I have attached the file code. I would greatly appreciate any help. Many thanks http://www.nabble.com/file/p25637806/OptTS.txt OptTS.txt -- View this message in context: http://www.nabble.com/rsolnp--Error-%28Help%21%29-tp25637806p25637806.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. -- 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] dimension-preserving matrix coersion
How about this: m - matrix(c(0, 1, 1, 0), ncol = 2) mode(m) - 'logical' m [,1] [,2] [1,] FALSE TRUE [2,] TRUE FALSE On Sun, Sep 27, 2009 at 6:59 PM, Murat Tasan mmu...@gmail.com wrote: i've written a function to coerce a matrix (e.g. from numeric to logical), but i'd like to know if someone has a more elegant method for this: m - matrix(c(0, 1, 1, 0), ncol = 2) m - as.logical(m) m [1] FALSE TRUE TRUE FALSE i'd like 'm' to still be a matrix with the original dimensions. in my function to do this, i coerce 'm' to a logical, then re-form it as a matrix, which seems like an extra (possibly bug-introducing) step that might be avoided if i knew of some hidden feature that might permit this in one fell swoop. any ideas? __ 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] dimension-preserving matrix coersion
HA! yeah, that'll do it! forgot that mode() can be used to set modes as well as get them. thanks much! -murat On Sep 27, 7:10 pm, jim holtman jholt...@gmail.com wrote: How about this: m - matrix(c(0, 1, 1, 0), ncol = 2) mode(m) - 'logical' m [,1] [,2] [1,] FALSE TRUE [2,] TRUE FALSE On Sun, Sep 27, 2009 at 6:59 PM, Murat Tasan mmu...@gmail.com wrote: i've written a function to coerce a matrix (e.g. from numeric to logical), but i'd like to know if someone has a more elegant method for this: m - matrix(c(0, 1, 1, 0), ncol = 2) m - as.logical(m) m [1] FALSE TRUE TRUE FALSE i'd like 'm' to still be a matrix with the original dimensions. in my function to do this, i coerce 'm' to a logical, then re-form it as a matrix, which seems like an extra (possibly bug-introducing) step that might be avoided if i knew of some hidden feature that might permit this in one fell swoop. any ideas? __ r-h...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://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-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://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] Determining name of calling function.
I have vague recollections of seeing this question discussed on r-help
previously, but I can't find the relevant postings.
I want to determine (from within a given function) the name of the
function
calling that given function.
E.g. if I have a function foo() which calls a function bar(), and also
a function clyde() which calls bar(), I want to have, in the code of
bar(),
an instruction which will return the character string foo if bar() was
called from foo() and the string clyde if bar() was called from
clyde().
Without really understanding what I'm doing I cobbled together the
following:
fname - as.character(sys.call(-1))[1]
This ***seems*** to work, at least in simple test cases.
But is it reliably robust? Are there traps for young players that I am
not seeing?
My ``solution'' returns NA as the value of fname if bar() is called
from the
command line, rather than being called by foo() or clyde(). This is
acceptable.
I think
Any avuncular advice from those younger and wiser than myself? :-)
cheers,
Rolf Turner
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Re: [R] array to matrix or data.frame
Works a charme; thanks much everybody. Daniel. David Winsemius wrote: On Sep 27, 2009, at 6:33 PM, Daniel Malter wrote: Hi, I have an array of dimension 6:6:16. I want to stack the 16 slices of the array into a matrix or a data frame of dimension 6:(6*16=96) in order to write it to a csv file. It seems I cannot get the matrix or data.frame functions to put the values from the array in the same order as they appear in the array easily, but I am almost sure there is an easy way to do this. #Here is code for a small 6:6:2 array. x=rep(rep(1:6,each=6),2) dim(x)=c(6,6,2) x #Matrix approach 1: matrix(x,nrow=12,ncol=6,byrow=F) #Matrix approach 2: matrix(x,nrow=12,ncol=6,byrow=T) #Data frame approach (cbinds rather than rbinds): data.frame(x) None of the above works as I would like to. The correct approach should stack the same values all in the same column of the matrix or data frame. I also could not get stack to work correctly and would appreciate your help. apply(x, 2 , I) [,1] [,2] [,3] [,4] [,5] [,6] [1,]123456 [2,]123456 [3,]123456 [4,]123456 [5,]123456 [6,]123456 [7,]123456 [8,]123456 [9,]123456 [10,]123456 [11,]123456 [12,]123456 -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/array-to-matrix-or-data.frame-tp25638310p25638689.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] dimension-preserving matrix coersion
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Murat Tasan Sent: Sunday, September 27, 2009 3:59 PM To: r-help@r-project.org Subject: [R] dimension-preserving matrix coersion i've written a function to coerce a matrix (e.g. from numeric to logical), but i'd like to know if someone has a more elegant method for this: m - matrix(c(0, 1, 1, 0), ncol = 2) m - as.logical(m) m [1] FALSE TRUE TRUE FALSE i'd like 'm' to still be a matrix with the original dimensions. in my function to do this, i coerce 'm' to a logical, then re-form it as a matrix, which seems like an extra (possibly bug-introducing) step that might be avoided if i knew of some hidden feature that might permit this in one fell swoop. m - m != 0 will do it, as will mode(m) - logical The former has the advantage that you don't have to use the canonical mapping of 0-FALSE,anything else-TRUE, but can use any standard comparison or vectorized logical operator. It also seems cleaner in that it can be used inside a larger expression, like m[m!=0]. Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com any ideas? __ 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] Count number of zeros in a collumn
more generally, if you want to test for some minimum threshold T on
the number of zeros, try:
if(length(which(dart[,1977] == 0)) T) { # some code to handle the
too-few-zeros-case }
On Sep 27, 4:54 pm, Marcio Resende mresende...@yahoo.com.br wrote:
I have a matrix 700x2000 which is sampled in each cycle from another matrix
788x2000 with the numbers 0,1 and 9
There is one specific collumn of this matrix, dart[,1977], that usually,
after the samplimg procedure has only 1 and 9 (because the zero frequency in
this collumn is low).
However, when this happens, I want to include an IF conditional in my code.
so basically what i wanted to do was to count the number of zeros in this
collumn to use this information in my conditional:
something like:
if (the number of zeros in collumn [,1977] is zero) ... else ...
I tried to find out indirectly by the rowsum but because of the sampling
procedure the rowsum is not always the same
Thank you very much
--
View this message in
context:http://www.nabble.com/Count-number-of-zeros-in-a-collumn-tp25637516p2...
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Determining name of calling function.
Not sure if this is important to you but R functions don't have to
have names so what you get back won't be a name if the function was
anonymous. In the example below an anonymous function calls fname and
the returned string is the calling sequence but that's not its name
since it has no name. In fact, in a sense no R functions have names.
You can store them in variables and call that variable its name but
that is not an intrinsic part of the function itself. A function is
just an environment, an argument list and a body -- no name.
fname - function() as.character(sys.call(-1))[1]
(function() fname())()
[1] (function() fname())
On Sun, Sep 27, 2009 at 7:24 PM, Rolf Turner r.tur...@auckland.ac.nz wrote:
I have vague recollections of seeing this question discussed on r-help
previously, but I can't find the relevant postings.
I want to determine (from within a given function) the name of the function
calling that given function.
E.g. if I have a function foo() which calls a function bar(), and also
a function clyde() which calls bar(), I want to have, in the code of bar(),
an instruction which will return the character string foo if bar() was
called from foo() and the string clyde if bar() was called from clyde().
Without really understanding what I'm doing I cobbled together the
following:
fname - as.character(sys.call(-1))[1]
This ***seems*** to work, at least in simple test cases.
But is it reliably robust? Are there traps for young players that I am
not seeing?
My ``solution'' returns NA as the value of fname if bar() is called from the
command line, rather than being called by foo() or clyde(). This is
acceptable.
I think
Any avuncular advice from those younger and wiser than myself? :-)
cheers,
Rolf Turner
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[R] CRAN (and crantastic) updates this week
CRAN (and crantastic) updates this week New packages * bdoc (1.0) Michael Anderson http://crantastic.org/packages/bdoc This package contains a function that will classify DNA barcodes as well as a few test and reference data sets. * bdsmatrix (1.0) Terry Therneau http://crantastic.org/packages/bdsmatrix This is a special case of sparse matrices, used by coxme and kinship * CircSpatial (1.0) Bill Morphet http://crantastic.org/packages/CircSpatial The package is a collection of functions for color continuous high resolution images of circular spatial data, circular kriging, and simulation of circular random fields. * cond (1.1-3) Alessandra R. Brazzale http://crantastic.org/packages/cond Higher order likelihood-based inference for logistic and loglinear models * coxme (2.0) Terry Therneau http://crantastic.org/packages/coxme Cox proportional hazards models containing Gaussian random effects, also known as frailty models. * csampling (1.1-3) Alessandra R. Brazzale http://crantastic.org/packages/csampling Monte Carlo conditional inference for the parameters of a linear nonnormal regression model * desire (1.0.5) Olaf Mersmann http://crantastic.org/packages/desire Harrington and Derringer-Suich type desirability functions * difR (1.0) Sebastien Beland http://crantastic.org/packages/difR The difR package contains several traditional methods to detect DIF in dichotomously scored items. Both uniform and non-uniform DIF effects can be detected, with methods relying upon item response models or not. Some methods deal with more than one focal group. * DTDA (1.1) Carla Moreira http://crantastic.org/packages/DTDA This package implements different algorithms for analyzing randomly truncated data, one-sided and two-sided (i.e. doubly) truncated data. * gRapHD (0.1.0) Gabriel Coelho Goncalves de Abreu http://crantastic.org/packages/gRapHD gRapHD is designed for efficient selection of high-dimensional undirected graphical models. The package provides tools for selecting trees, forests and decomposable models minimizing information criteria such as AIC or BIC, and for displaying the independence graphs of the models. It has also some useful tools for analysing graphical structures. It supports the use of discrete, continuous, or both types of variables. * IQCC (1.0) Emanuel P. Barbosa http://crantastic.org/packages/IQCC Builds statistical control charts with exact limits for univariate and multivariate cases. * marg (1.1-3) Alessandra R. Brazzale http://crantastic.org/packages/marg Likelihood inference based on higher order approximations for linear nonnormal regression models * MCMChybridGP (2.1) Mark J. Fielding http://crantastic.org/packages/MCMChybridGP Hybrid Markov chain Monte Carlo (MCMC) to simulate from a multimodal target distribution. A Gaussian process approximation makes this possible when derivatives are unknown. The Package serves to minimize the number of function evaluations in Bayesian calibration of computer models using parallel tempering. It allows replacement of the true target distribution in high temperature chains, or complete replacement of the target. Methods used are described in, Efficient MCMC schemes for Bayesian calibration of computer models, Fielding, Mark, Nott, David J. and Liong Shie-Yui (2009), in preparation. The authors gratefully acknowledge the support contributions of the Singapore-Delft Water Alliance (SDWA). The research presented in this work was carried out as part of the SDWA's Multi-Objective Multi-Reservoir Management research programme (R-264-001-272). * nlreg (1.1-3) Alessandra R. Brazzale http://crantastic.org/packages/nlreg Likelihood inference based on higher order approximations for nonlinear models with possibly non constant variance * nnclust (2.1) Thomas Lumley http://crantastic.org/packages/nnclust Finds nearest neighours and the minimum spanning tree for large data sets, does clustering using the minimum spanning tree. * phybase (1.1) Liang Liu http://crantastic.org/packages/phybase This package provides functions to read, write, manipulate, estimate, and summarize phylogenetic trees including species trees which contain not only the topology and branch lengths but also population sizes. The input/output functions can read tree files in which trees are presented in parenthetic format. The trees are read in as a string and then transformed to a matrix which describes the relationship of nodes and branch lengths. The nodes matrix provides an easy access for developers to further manipulate the tree, while the tree string provides interface with other phylogenetic R packages such as ape. The input/output functions can also be used to change the format of tree files between NEXUS and PHYLIP. Some basic functions have already been
Re: [R] Determining name of calling function.
On 28/09/2009, at 12:34 PM, Gabor Grothendieck wrote:
Not sure if this is important to you but R functions don't have to
have names so what you get back won't be a name if the function was
anonymous. In the example below an anonymous function calls fname and
the returned string is the calling sequence but that's not its name
since it has no name. In fact, in a sense no R functions have names.
You can store them in variables and call that variable its name but
that is not an intrinsic part of the function itself. A function is
just an environment, an argument list and a body -- no name.
fname - function() as.character(sys.call(-1))[1]
(function() fname())()
[1] (function() fname())
snip
Good point. Thanks. I don't ***think*** that this issue will call
problems for me. In my real application ``bar()'' will always be
called by a named function. Still, it's something to keep in mind.
cheers,
Rolf Turner
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Count number of zeros in a collumn
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Murat Tasan
Sent: Sunday, September 27, 2009 4:27 PM
To: r-help@r-project.org
Subject: Re: [R] Count number of zeros in a collumn
more generally, if you want to test for some minimum threshold T on
the number of zeros, try:
if(length(which(dart[,1977] == 0)) T) { # some code to
handle the too-few-zeros-case }
Note that
sum( dart[,1977]==0 )
gives the same result as
length(which( dart[,1977]==0 ))
with less typing and using less memory and time.
Arithmetic on logicals treats TRUE as 1 and FALSE
as 0.
(sum() forces you to decide what to do with NA's
in the data; which() always discards them.)
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
On Sep 27, 4:54 pm, Marcio Resende mresende...@yahoo.com.br wrote:
I have a matrix 700x2000 which is sampled in each cycle
from another matrix
788x2000 with the numbers 0,1 and 9
There is one specific collumn of this matrix, dart[,1977],
that usually,
after the samplimg procedure has only 1 and 9 (because the
zero frequency in
this collumn is low).
However, when this happens, I want to include an IF
conditional in my code.
so basically what i wanted to do was to count the number of
zeros in this
collumn to use this information in my conditional:
something like:
if (the number of zeros in collumn [,1977] is zero) ... else ...
I tried to find out indirectly by the rowsum but because of
the sampling
procedure the rowsum is not always the same
Thank you very much
--
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[R] Bubble Plot
Hello, I am using the bubble plot and have been able to overlay two different data sets on the same graphic successfully. I would like to do the following and cannot: 1) suppress the zero values such that there is no representation of them on my plot (i.e., the zeroes show up as the smallest dot size, and I can't change this) 2) Give values to y or x axes with values, and labels My script looks as such: coordinates(data) = ~y + x a1 = bubble(data, Alive, zero.print = .,maxsize = 5.0, key.entries = 4*(1:6),col=c(0,3)) a2 = bubble(handcore, Dead, maxsize = 5.0, main = , key.entries = 5*(0:10),col=c(0,4)) print(b1, more = TRUE) print(b2, more = FALSE) Thanks in advance for your help. mw Marion Wittmann, Ph.D. Tahoe Environmental Research Center University of California Davis __ 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] Strata formation using R
Hi Everybody Its me again. I have been able to sample using SRS and stratified sampling when i have predefined groups as follows: Hypermarket - matrix(rnorm(100, mean=5, sd=5000)) Supermarket - matrix(rnorm(400, mean=34000, sd=3000)) Minimarket - matrix(rnorm(1000, mean=1,sd=2000)) Cornershop - matrix(rnorm(1500, mean=2500, sd=500)) Spazashop - matrix(rnorm(2000, mean=1000, sd=250)) dat=data.frame(type=c(rep(Hypermarket,100), rep(Supermarket,400), rep(Minimarket,1000),rep(Cornershop,1500), rep(Spazashop,2000)), value=c(Hypermarket, Supermarket, Minimarket, Cornershop,Spazashop)) dat names(dat)=c(type,value) #Number of observations for each outlet type in a population dat$type - factor(dat$type, levels = c(Hypermarket,Supermarket,Minimarket,Cornershop,Spazashop)) (stratas-data.frame(table(dat$type))) ##sampling and proportional allocation as.matrix(table(dat$type)) s=strata(dat,c(type),size=c(20,80,200,300,400), method=srswor) dat.strat-getdata(dat,s) dat.strat Now i want to use the rectangular method to determine strata boundaries, thenfor the allocation use Neyman, so please help me achieve this. [[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] Data analysis package for positively skewed data
R-helpers A curious question: Can you make suggestions as to what to use in R for the data from a sample of the following: Hypermarket - matrix(rnorm(100, mean=5, sd=5000)) Supermarket - matrix(rnorm(400, mean=34000, sd=3000)) Minimarket - matrix(rnorm(1000, mean=1,sd=2000)) Cornershop - matrix(rnorm(1500, mean=2500, sd=500)) Spazashop - matrix(rnorm(2000, mean=1000, sd=250)) dat=data.frame(type=c(rep(Hypermarket,100), rep(Supermarket,400), rep(Minimarket,1000),rep(Cornershop,1500), rep(Spazashop,2000)), value=c(Hypermarket, Supermarket, Minimarket, Cornershop,Spazashop)) #Sampling with without replacement n-1000 dat.srs-dat[sample(1:dim(dat)[1], size=n,replace=F),] dat.srs #Number of observations for each outlet type in a sample dat.srs$type - factor(dat.srs$type, levels = c(Hypermarket,Supermarket,Minimarket,Cornershop,Spazashop)) (numoutlets-data.frame(table(dat.srs$type))) suggest a package that can help me get all the analysis info such as meean,var,std,dev,cv,ci,proportions,... [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting SAS Data code to R.
On Sep 27, 2009, at 6:01 PM, David Winsemius wrote: On Sep 27, 2009, at 12:10 PM, David Winsemius wrote: On Sep 27, 2009, at 11:49 AM, Douglas Bates wrote: On Sat, Sep 26, 2009 at 11:33 PM, David Winsemius dwinsem...@comcast.net wrote: I am contemplating bringing in and merging three NHANES-III datasets from the National Center for Health Statistics that are fixed format with record length=3348, line counts around 20,000 and described by SAS DATA steps. I have downloaded and linked similar datasets from the Continuous NHANES public data releases, but never ones with this many variables at once. In the prior effort I managed the task by some cut-paste-editing from the SAS code file into a corresponding read.fwf R call, but the earlier NHANES-III data is far more voluminous than the more recent Continuous version. I am wondering if anyone has experience with such a process and would be willing to share some advice? The SAS code can be seen here: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.sas The main code file Data step starts out... FILENAME ADULT D:\Questionnaire\DAT\ADULT.DAT LRECL=3348; *** LRECL includes 2 positions for CRLF, assuming use of PC SAS; DATA WORK; INFILE ADULT MISSOVER; LENGTH SEQN 7 DMPFSEQ 5 DMPSTAT 3 DMARETHN 3 DMARACER 3 DMAETHNR 3 HSSEX 3 The corresponding positions in the INPUT section are INPUT SEQN 1-5 DMPFSEQ 6-10 DMPSTAT 11 DMARETHN 12 DMARACER 13 DMAETHNR 14 HSSEX15 The note about CRLF appears to be implying that those characters are being counted as part of the length of the first variable, SEQN, but that there are only 5 meaningful positions. I suppose I can find out by trial and error how to read such files, but it would save me some time if anyone in the audience has worked through this on this data before. One thought would be to import the data with the SAS work-alike program, WKS, (which I have not used before) and then to read in with read.xport from the foreign library. That would obviate the need to understand the character position issue, but probably has a time commitment to get it up and running and learn how to use it. Another thought would be to parse the fixed width SAS Data step code into pieces and build a data.frame from which I then extract the row.names, col.names, and colClasses from that centralized structure. Are the data available to the public somewhere or could just a few records be made available? Yes. Just trim the file name and the CDC ftp server accepts the path specification: ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/ The file that goes with that SAS code is adult.dat ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/1A/adult.dat The reason I ask is because I imagine there are a lot of missing data in each record (the data are arranged in the wide format for longitudinal data and includes follow-up questions that will not apply to most respondents). The missing data indicator, if any, and the format of the other fields will be important in deciding how to split the data. Thanks for that. It was not designed as a longitudinal study, but rather as cross-sectional study that was spaced over several years. They did a re-exam of some sort, but that was not the primary purpose, nor will it be my particular interest. I have tried to determine by examination whether . or is the missing value indicator and it appears that both may used although there are many more spaces. Most of the input suggests to my 15-year-old memories of SAS that the data is numeric but there are 17 variables where input spec is $nn varLines[grep([[:punct:]], varLines)] [1] HAX11AG $6 HAX11AH $6 HAX11AI $6 [4] HAX11AJ $6 HAX11AK $6 HAX11AL $6 [7] HAX11AM $6 HAX11AN $6 HAX11AO $6 [10] HAX11AP $6 HAX11AQ $6 HAX11AR $6 [13] HAX11AS $6 HAX11AT $6 HAX11AU $6 [16] HAX11AV $6 HAZA1CC $30 My progress on this effort so far consists of having figured out how to extract the variable names and their associated lengths so I can set up a call to read.fwf(). This is waht I did on hte section of the SAS code following INPUT that contains those elements: trim.ws - function(x) gsub(^[[:space:]]+|[[:space:]]+$, ,x) # courtesy of a Grothendieck r-help posting of a couple or three years ago. adult.var - data.frame(varnames = sapply( strsplit(trim.ws(varLines) , +) , [, 1:2)[1,], varlen= sapply( strsplit(trim.ws(varLines) , +) , [, 1:2)[2,]) #so that I can split the trimmed strings on an arbitrary number of spaces. adult.var[,][1:5,] varnames varlen 1 SEQN 7 2 DMPFSEQ 5 3 DMPSTAT 3 4
[R] Re ading Functions that are in a Vector
I am trying to write a function that will have an input of a vector of functions. Here is a simplistic example. sumstats - c(mean,sd) sumstats[1] #Gives this error # sumstats[1] #[[1]] #function (x, ...) #UseMethod(mean) #environment: namespace:base I thought about restricting the input to character variables such as the following sumstats2 - c(mean,sd) Is there a way to change mean to the function mean? -- View this message in context: http://www.nabble.com/Reading-Functions-that-are-in-a-Vector-tp25639720p25639720.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] Data formatting for matplot
Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ 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] Data formatting for matplot
You can try this: matplot(do.call(cbind, split.dat)) On Sun, Sep 27, 2009 at 11:42 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ 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] implementation of matrix logarithm (inverse of matrix exponential)
On Sun, 27 Sep 2009, Douglas Bates wrote:
There is a logm function in the expm package in the expm project on
R-forge. See http://expm.R-forge.R-project.org/
Martin was the person who added that function so I will defer to his
explanations of what it does. I know he has been traveling and it may
be a day or two before he can get to this thread.
Douglas,
Thanks for this note.
The logm() function in the expm package does the trick.
It seems to properly handle block structures for non-diagonalizable
matrices:
require(expm)
M - matrix(c(0,1,0,0),2)
expm(M)
[,1] [,2]
[1,]10
[2,]11
logm(expm(M))
[,1] [,2]
[1,]00
[2,]10
And thanks to Vincent, Martin, et al for this package!
:-)
Chuck
On Sun, Sep 27, 2009 at 11:44 AM, Charles C. Berry cbe...@tajo.ucsd.edu wrote:
On Sat, 26 Sep 2009, spencerg wrote:
Sylvester's formula
(http://en.wikipedia.org/wiki/Sylvester%27s_formula) applies to a square
matrix A = S L solve(S), where L = a diagonal matrix and S = matrix of
eigenvectors. Let f be an analytic function [for which f(A) is well
defined]. Then f(A) = S f(L) solve(S).
We can code this as follows:
sylvester - function(x, f){
n - nrow(x)
eig - eigen(x)
vi - solve(eig$vectors)
with(eig, (vectors * rep(f(values), each=n)) %*% vi)
}
logm - function(x)sylvester(x, log)
Example:
A - matrix(1:4, 2)
eA - expm(A)
logm(eA)
With Chuck Berry's example, we get the following:
M - matrix( c(0,1,0,0), 2 )
sylvester(M, log)
The case I gave would be
sylvester( as.matrix( expm( M ) ), log )
for which the perfectly sensible answer is M, not what appears here:
Error in solve.default(eig$vectors) :
system is computationally singular: reciprocal condition number =
1.00208e-292
This is a perfectly sensible answer in this case. We get the
same result from sylvester(M, exp), though expm(M) works fine.
A better algorithm for this could be obtains by studying the code
for expm in the Matrix package and the references in the associated help
page.
I doubt that code already in R will handle cases requiring Jordan blocks for
evaluation of the matrix logarithm (which cases arise in the context of
discrete state, continuous time Markov chains) without requiring one to
built that code more or less from scratch.
I'd be happy to hear that this is not so.
HTH,
Chuck
Hope this helps. Spencer
Gabor Grothendieck wrote:
Often one uses matrix logarithms on symmetric positive definite
matrices so the assumption of being symmetric is sufficient in many
cases.
On Sat, Sep 26, 2009 at 7:28 PM, Charles C. Berry cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
OK. Try this:
library(Matrix)
M - matrix(c(2, 1, 1, 2), 2); M
[,1] [,2]
[1,] 2 1
[2,] 1 2
Right. expm( M ) is diagonalizable.
But for
M - matrix( c(0,1,0,0), 2 )
you get the wrong result.
Maybe I should have added that I do not see the machinery in R for
dealing
with Jordan blocks.
HTH,
Chuck
# log of expm(M) is original matrix M
with(eigen(expm(M)), vectors %*% diag(log(values)) %*% t(vectors))
[,1] [,2]
[1,] 2 1
[2,] 1 2
On Sat, Sep 26, 2009 at 6:24 PM, Charles C. Berry
cbe...@tajo.ucsd.edu
wrote:
On Sat, 26 Sep 2009, Gabor Grothendieck wrote:
Try:
expm( - M)
Mimosa probably meant say 'the inverse function'.
I do not see one in R.
Chuck
On Sat, Sep 26, 2009 at 5:06 PM, Mimosa Zeus
mimosa1...@yahoo.fr
wrote:
Dear R users,
Does anyone has implemented the inverse of the
matrix exponential (expm
in the package Matrix)?
In Matlab, there're logm and expm, there's only expm
in R.
Cheers
Mimosa
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Charles C. Berry (858)
534-2098
Dept of
Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
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Re: [R] Data formatting for matplot
Try this: library(lattice) xyplot(y ~ x, mydat, groups = id) On Sun, Sep 27, 2009 at 10:42 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I am wanting to produce a multiple line plot, and know I can do it with matplot but can't get my data in the format I need. I have a dataframe with three columns; individuals ID, x, and y. I have tried split() but it gives me a list of matrices, which is closer but not quite what I need. For example: id-rep(seq(1,5,1),length.out=100) x-rnorm(100,5,1) y-rnorm(100,20,5) mydat-data.frame(id,x,y) split.dat-split(mydat[,2:3],mydat[,1]) I would appreciate your help in either how to get this into a format acceptable to matplot or other options for creating a multiple line plot. Thanks, Tim Tim Clark Department of Zoology University of Hawaii __ 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] Re ading Functions that are in a Vector
On Sun, Sep 27, 2009 at 10:36 PM, trumpetsaz stephaniezim...@gmail.com wrote: I am trying to write a function that will have an input of a vector of functions. Here is a simplistic example. sumstats - c(mean,sd) sumstats[1] #Gives this error # sumstats[1] #[[1]] #function (x, ...) #UseMethod(mean) #environment: namespace:base That's not an error. The code asked it to return the first function so that's what it did. I thought about restricting the input to character variables such as the following sumstats2 - c(mean,sd) Is there a way to change mean to the function mean? This gives a vector of functions given a vector of their names: sumstat1 - lapply(sumstats2, get) __ 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] Re ading Functions that are in a Vector
On 28/09/2009, at 3:36 PM, trumpetsaz wrote:
I am trying to write a function that will have an input of a vector of
functions. Here is a simplistic example.
sumstats - c(mean,sd)
sumstats[1]
#Gives this error
# sumstats[1]
#[[1]]
#function (x, ...)
#UseMethod(mean)
#environment: namespace:base
I thought about restricting the input to character variables such
as the
following
sumstats2 - c(mean,sd)
Is there a way to change mean to the function mean?
You *can't* have a vector of functions, as far I know. The entries
of a vector must be (numeric, character, or logical) scalars.
You *can* have a *list* of functions; this might be the way you
want to go.
To ``change mean to the function mean'' --- set ?get.
cheers,
Rolf Turner
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Re: [R] Re ading Functions that are in a Vector
On Sep 27, 2009, at 11:07 PM, Rolf Turner wrote: On 28/09/2009, at 3:36 PM, trumpetsaz wrote: I am trying to write a function that will have an input of a vector of functions. Here is a simplistic example. sumstats - c(mean,sd) sumstats[1] #Gives this error # sumstats[1] #[[1]] #function (x, ...) #UseMethod(mean) #environment: namespace:base I thought about restricting the input to character variables such as the following sumstats2 - c(mean,sd) Is there a way to change mean to the function mean? You *can't* have a vector of functions, as far I know. The entries of a vector must be (numeric, character, or logical) scalars. You *can* have a *list* of functions; this might be the way you want to go. To ``change mean to the function mean'' --- set ?get. That is prezactly what he created: umstats - c(mean,sd) str(umstats) List of 2 $ :function (x, ...) $ :function (x, na.rm = FALSE) And what's more it works if it is accessed with [[: umstats[[1]](c(1,2,3,4,5)) [1] 3 David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re ading Functions that are in a Vector
On 28/09/2009, at 4:30 PM, David Winsemius wrote:
On Sep 27, 2009, at 11:07 PM, Rolf Turner wrote:
On 28/09/2009, at 3:36 PM, trumpetsaz wrote:
I am trying to write a function that will have an input of a vector
of
functions. Here is a simplistic example.
sumstats - c(mean,sd)
sumstats[1]
#Gives this error
# sumstats[1]
#[[1]]
#function (x, ...)
#UseMethod(mean)
#environment: namespace:base
I thought about restricting the input to character variables such
as the
following
sumstats2 - c(mean,sd)
Is there a way to change mean to the function mean?
You *can't* have a vector of functions, as far I know. The entries
of a vector must be (numeric, character, or logical) scalars.
You *can* have a *list* of functions; this might be the way you
want to go.
To ``change mean to the function mean'' --- set ?get.
That is prezactly what he created:
umstats - c(mean,sd)
str(umstats)
List of 2
$ :function (x, ...)
$ :function (x, na.rm = FALSE)
And what's more it works if it is accessed with [[:
umstats[[1]](c(1,2,3,4,5))
[1] 3
Never ceases to amaze me how R can get it right in settings and
with syntaxes that you would think can't ***possibly*** work!
cheers,
Rolf Turner
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: generic methods - in particular the summary function
Thanks for your answer. Yes, I use now a proper NAMESPACE with function declared correctly. Christophe Le 27 sept. 2009 à 19:16, David Winsemius a écrit : On Sep 27, 2009, at 12:52 PM, Christophe Dutang wrote: Nobody wants to answer my question... is there something stupid in it? I couldn't really say. Not being a user of fitdistrplus I don't have much baseline experience. Had you posted code that produced something to work on, I might have made an effort at applying the methods I typically use, such as str(object), methods(function), and printing out the function code. But I am a rather low-level R- user, and as soon as you strart throwing around questions about NAMESPACE, my eyes glaze over and I move to the next question. When I do install and load fitdistrplus and then execute: methods(summary) .. I see a new function which is not even invisible, called summary.fitdist. So it seems to be in my NAMESPACE ... or at least what I dimly understand about such metaphysical entities. I get an error when I execute your suggestion: S3method(summary, fitdist) Error: could not find function S3method So I would have assumed that you had a package with that S3method function about which you had not informed us. Not sure this helps because, as I said, your question appeared more complex that I considered myself competent to answer. -- David Début du message réexpédié : De : Christophe Dutang duta...@gmail.com Date : 19 septembre 2009 11:26:51 HAEC À : r-help@r-project.org Objet : generic methods - in particular the summary function Hi all, I'm currently working on the fitdistrplus package (that basically fit distributions). There is something I do not understand about the generic function summary. In the current version on CRAN, there is no NAMESPACE saying S3method(summary, fitdist) . However if we use summary on an object send by fitdist function it works fine... According to R-lang, we have The most common use of generic functions is to provide print and summary methods for statistical ob jects, generally the output of some model fitting process. To do this, each model attaches a class attribute to its output and then provides a special method that takes that output and provides a nice readable version of it. The user then needs only remember that print or summary will provide nice output for the results of any analysis. I would like to be sure, that if the summary.fitdist is not exported in the NAMESPACE, then we must use declare it with S3method. Thanks in advance Christophe -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr [[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. David Winsemius, MD Heritage Laboratories West Hartford, CT -- Christophe Dutang Ph.D. student at ISFA, Lyon, France website: http://dutangc.free.fr __ 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.
