Re: [R] Fisher test problem
As you say, I guess that's not possible. Meanwhile I've got R's results verified in Stata . tabi 17 6 \ 2 1, chi2 exact | col row | 1 2 | Total ---+--+-- 1 |17 6 |23 2 | 2 1 | 3 ---+--+-- Total |19 7 |26 Pearson chi2(1) = 0.0708 Pr = 0.790 Fisher's exact = 1.000 1-sided Fisher's exact = 0.627 The OpenEpi.com output was: (note the 2 tailed P value (fisher test)) Chi Square and Exact Measures of Association TestValue p-value(1-tail) p-value(2-tail) Uncorrected chi square 0.07083 0.3951 0.7901 Yates corrected chi square 0.1813 0.3351 0.6702 Mantel-Haenszel chi square 0.06811 0.3971 0.7941 Fisher exact0.6273 1.25 Mid-P exact 0.3971 0.7942 Thanks for pointing it out.. Had been under the notion that OpenEpi was very... strong (being associated with EpiInfo and WHO). Viju Tal Galili wrote: Viju Moses, Are you sure you got a P-value of 1.25 ? Since P-value could only be between 0 to 1... Tal On Sat, Mar 21, 2009 at 9:17 PM, Viju Moses vijumo...@gmail.com mailto:vijumo...@gmail.com wrote: Hi, I noted a discrepancy between R and openepi when I ran a fisher test with the same matrix. In R: a=matrix(c(1,2,6,17), nrow=2) a [,1] [,2] [1,]16 [2,]2 17 fisher.test(a, conf.int http://conf.int=T) Fisher's Exact Test for Count Data data: a p-value = 1 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.02061498 31.73691924 sample estimates: odds ratio 1.396646 But in openepi the P value is 1.25. (In another instance too for other sets of data, I had got a p value of 1 in 3 instances for a prop.test when I got 3 other answers on a friend's stata software with the same data. ) I'm using R on Ubuntu Intrepid. Is there anything I'm doing wrong? Any other packages I have to install? Thanks in advance Viju Moses __ 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 and provide commented, minimal, self-contained, reproducible code. -- -- My contact information: Tal Galili Phone number: 972-50-3373767 FaceBook: Tal Galili My Blogs: http://www.r-statistics.com/ http://www.talgalili.com http://www.biostatistics.co.il __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Fisher test problem
On Sat, 21 Mar 2009 markle...@verizon.net wrote: by definition, the one tailed p-value has to be = 0.5 so there is still something wrong with your OpenEpi calc. I think that's a little strong. Firstly, a one-tailed test in a pre-determined direction can have any p-value. Secondly, even the smaller one-sided tail probability can be 0.5. The problem is discreteness. It may be that 0.62 is the smaller of p(T=t) and p(T=t), and this turns out to be the case. The other tail probability is 0.83. OpenEpi should still be thresholding the p-value at 1, but that's just cosmetic. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Forestplot () box size question
On Sat, 21 Mar 2009, Gerard Smits wrote: I have tried several sites (2 in Ca and also Australia) and find only version 2.14. Which site did you pull 2.15 from? I have checked half a dozen sites in various countries, and they all have 2.15 (even in the Southern Hemisphere -- it's not that updates flow the wrong way south of the equator). The CRAN mirror status page says that no mirrors are more than a couple of days out of date. Mostly likely you have an old version of R. CRAN binaries are built only for the current version of R. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Estimating LC50 from a Weibull distribution
I am attempting to estimate LC50 (analogous to LD50, but uses exposure concentration rather than dose) by fitting a Weibull model; but I can't seem to get it to work. From what I can gather, I should be using survreg() from the survival package. The survreg() function relies on time-to-event data; my data result from 96 h exposures (i.e., dead or alive after a fixed period; 96 h). I've tried the following (doesn't work): conc - c(10.3, 10.8, 11.6, 13.2, 15.8, 20.1) # Exposure concentrations orign - c(76, 79, 77, 76, 78, 77) # Original number of subjects ndead - c(16, 22, 40, 69, 78, 77) # Number dead after 96 h d - data.frame(conc=conc, orign=orign, ndead=ndead) d$prop - d$ndead/d$orign # Calculate proportion dead after 96 h # Adjust for 100% mortalities d$prop[d$prop==1.00] - 1-(1/(2*d$orign[d$prop==1.00])) fit - survreg(Surv(d$prop) ~ d$conc, dist=weibull) summary(fit) Call: survreg(formula = Surv(d$prop) ~ d$conc, dist = weibull) Value Std. Error zp (Intercept) -2.254 0.9506 -2.37 0.017737 d$conc 0.135 0.0686 1.97 0.048532 Log(scale) -1.061 0.3203 -3.31 0.000927 Scale= 0.346 Weibull distribution Loglik(model)= 0.6 Loglik(intercept only)= -1.6 Chisq= 4.56 on 1 degrees of freedom, p= 0.033 Number of Newton-Raphson Iterations: 5 n= 6 Estimating the LC50 from these coefficients yields an unreasonable answer: LC50 = (0.5 + 2.254)/0.135 = 20.4; i.e., higher than the highest exposure concentration. I'm sure I'm doing something silly--I just don't know what. Essentially, I'm trying to do this, but with a Weibull model: library(MASS) resp - cbind(d$ndead, nalive = d$orign - d$ndead) mod - glm(resp ~ d$conc, family = binomial(link = probit)) result - dose.p(mod, p = 0.5) result DoseSE p = 0.5: 11.49053 0.1069564 Any help would be greatly appreciated. Sincerely, Greg. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. R
thx for ur fast responds. but sorry for asking stupid, i am a turn beginner of R (just trying it out 3 months, and i am taking my first course about it) so, to tackle this questions, i was told to use nested design method, could you actually show me how would u attempt this problem? (a) Determine if insulation in the house effects the average gas consumption. (b) How much extra gas is used when there is no insulation? Provide an interval estimate as well as a point estimate. i just got confused by the backgroud information. We are interested in looking at the effect of insulation on gas consumption. The average outside temperature (degrees celcius) was also measured. so how should my model looks like? i dont even know what should be my explanatory/response variables... thx in advance Gabor Grothendieck wrote: This works with the example. If the real data is different it may not work. To run the example below just copy and paste it into R. To run with the real data replace textConnection(Lines) with insulation.txt everywhere. Lines - Before insulAfter insul. tempgas tempgas -0.87.2-0.74.8 -0.76.90.84.6 0.46.41.04.7 2.56.01.44.0 2.95.81.54.2 3.25.81.64.2 3.65.62.34.1 3.94.72.54.0 4.25.82.53.5 4.35.23.13.2 5.44.93.93.9 6.04.94.03.5 6.04.34.03.7 6.04.44.23.5 6.24.54.33.5 6.34.64.63.7 6.93.74.73.5 7.03.94.93.4 7.44.24.93.7 7.54.04.94.0 7.53.95.03.6 7.63.55.33.7 8.04.06.22.8 8.53.67.13.0 9.13.17.22.8 10.2 2.67.52.6 8.02.7 8.72.8 8.81.3 9.71.5 nfld - count.fields(textConnection(Lines)) data.lines - readLines(textConnection(Lines)) data.lines - ifelse(nfld == 2, paste(NA NA, data.lines), data.lines) my.data - read.table(textConnection(data.lines), header = TRUE, skip = 1) On Sat, Mar 21, 2009 at 8:13 PM, UBC cheong0...@hotmail.com wrote: so i am having this question what should i do if the give data file (.txt) has 4 columns, but different lengths? how can i read them in R? any idea for the following problem? Gas consumption (1000 cubic feet) was measured before and after insulation was put into a house. We are interested in looking at the effect of insulation on gas consumption. The average outside temperature (degrees celcius) was also measured. The data are included in the file insulation.txt. (a) Determine if insulation in the house effects the average gas consumption. (b) How much extra gas is used when there is no insulation? Provide an interval estimate as well as a point estimate. heres the content in insulation.txt (u can just copy and paste it to the notepad so can be read in R) Before insul After insul. temp gas temp gas -0.8 7.2 -0.7 4.8 -0.7 6.9 0.8 4.6 0.4 6.4 1.0 4.7 2.5 6.0 1.4 4.0 2.9 5.8 1.5 4.2 3.2 5.8 1.6 4.2 3.6 5.6 2.3 4.1 3.9 4.7 2.5 4.0 4.2 5.8 2.5 3.5 4.3 5.2 3.1 3.2 5.4 4.9 3.9 3.9 6.0 4.9 4.0 3.5 6.0 4.3 4.0 3.7 6.0 4.4 4.2 3.5 6.2 4.5 4.3 3.5 6.3 4.6 4.6 3.7 6.9 3.7 4.7 3.5 7.0 3.9 4.9 3.4 7.4 4.2 4.9 3.7 7.5 4.0 4.9 4.0 7.5 3.9 5.0 3.6 7.6 3.5 5.3 3.7 8.0 4.0 6.2 2.8 8.5 3.6 7.1 3.0 9.1 3.1 7.2 2.8 10.2 2.6 7.5 2.6 8.0 2.7 8.7 2.8 8.8 1.3 9.7 1.5 thx and any ideas would help. -- View this message in context: http://www.nabble.com/data-analysis.-R-tp22641912p22641912.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. -- View this message in context: http://www.nabble.com/data-analysis.-R-tp22641912p22643290.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
[R] Peña, Rodríguez test
Hi, Anyone have done anything about this Peña, Rodríguez test b4. I have never seen it b4 in my life. Now i simply need to write some R code for this. any help would be appreciated. _ Chat with the whole group, and bring everyone together. [[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] variance/mean
At the risk of appearing ignorant why is the folowing true?
o - cbind(rep(1,3),rep(2,3),rep(3,3))
var(o)
[,1] [,2] [,3]
[1,]000
[2,]000
[3,]000
and
mean(o)
[1] 2
How do I get mean to return an array similar to var? I would expect in the
above example a vector of length 3 {1,2,3}.
Thank you for your help.
Kevin
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] variance/mean
On 22-Mar-09 08:17:29, rkevinbur...@charter.net wrote:
At the risk of appearing ignorant why is the folowing true?
o - cbind(rep(1,3),rep(2,3),rep(3,3))
var(o)
[,1] [,2] [,3]
[1,]000
[2,]000
[3,]000
and
mean(o)
[1] 2
How do I get mean to return an array similar to var? I would expect in
the above example a vector of length 3 {1,2,3}.
Thank you for your help.
Kevin
This is a consequence of (understandable) confusion about how var()
and mean() operate! It is not explicit, in ?var, that if you apply
var() to a matrix, as in your var(o) you get the covariance matrix
between the columns of 'o' -- except where it says (almost as an
aside) that 'var' is just another interface to 'cov'. Hence in
your example var(o) is equivalent to cov(o). Looked at in this
way, it is now straightforward to expect what you got.
This is, of course, different from what you would expect if you apply
var() to a vector, namely the variance of that series of numbers
(a single value).
On the other hand, mean() works differently. According to ?mean:
Arguments:
x: An R object. Currently there are methods for numeric
data frames, numeric vectors and dates.
[...]
Value:
For a data frame, a named vector with the appropriate method
being applied column by column.
which may have been what you expected. But a matrix is not a data
frame. Instead, it is an array, which (in effect) is a vector with
an attached dimensions attribute which tells R how to chop it up
into columns etc. -- whereas a data frame has its by-column
structure built in to it.
Now: ?mean says nothing about matrices. Nothing whatever.
So you have to find out the hard way that mean(o) treats the array
'o' as a vector, ignoring its dimensions attribute. Hence you
get a single number, which is the mean of all the values in the
matrix.
In order to get what you are apparently looking for (the means of
the columns of 'o'), you could:
a) (the smooth way) use the apply() function, causing mean() to be
applied to the second dimension (columns) of 'o':
apply(o,2,mean)
# [1] 1 2 3
b) (the heavy way) take a hint from ?mean and feed it a data frame:
mean(as.data.frame(o))
# V1 V2 V3
# 1 2 3
Hoping this helps to clarify things!
Ted.
E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
Fax-to-email: +44 (0)870 094 0861
Date: 22-Mar-09 Time: 09:01:40
-- XFMail --
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] variance/mean
rkevinbur...@charter.net wrote:
At the risk of appearing ignorant why is the folowing true?
o - cbind(rep(1,3),rep(2,3),rep(3,3))
var(o)
[,1] [,2] [,3]
[1,]000
[2,]000
[3,]000
and
mean(o)
[1] 2
How do I get mean to return an array similar to var? I would expect in the
above example a vector of length 3 {1,2,3}.
you may well be ignorant about how var works with matrices, but this
does not mean it's your fault. the documentation is typically cryptical.
when you apply var to a single matrix, it will compute covariances
between its columns rather than the overall variance:
set.seed(0)
x = matrix(rnorm(4), 2, 2)
var(x)
#[,1] [,2]
# [1,] 1.2629543 1.329799
# [2,] -0.3262334 1.272429
matrix(nrow=2, ncol=2, byrow=TRUE, c(
cov(x[,1], x[,1]), cov(x[,1], x[,2]),
cov(x[,2], x[,1]), cov(x[,2], x[,2])))
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] package sowas
There is a binary version for Windows in the homepage of it. http://www.cru.uea.ac.uk/~douglas/software/sowas_0.94.zip You can find more on http://tocsy.agnld.uni-potsdam.de/wavelets/ Best 2009/3/22 stephen sefick ssef...@gmail.com: You must go to his website because it is not on CRAN. I have built it on mac osx by installing it with R CMD install. This should work on other unix platforms. Stephen Sefick On Sat, Mar 21, 2009 at 9:26 PM, mau...@alice.it wrote: I cannot find sowas package by Douglas Mauran in CRAN packages list- On which platforms does sowas run ? Has anybody used such a package at all ? hHank you very much, Maura tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent Tel: (00852) 3442 3832 PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html A sociologist is someone who, when a beautiful women enters the room and everybody look at her, looks at everybody. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] limiting simulated animal movement
Umesh Srinivasan wrote:
Hi,
I am trying to simulate animal movement in a gridded landscape made up of
cells. At each time step (iteration), the animal moves from one cell to
another in a random fashion.
...
The problem is
1. I want to limit animals to a pre-defined circular home range
2. I want to limit animals to the borders of the landscape itself
I tried to limit animals to their home range using:
if (sqrt((x - a)^2 + (y - b)^2) radius.range) {
a - a[i-1]
b - b[i-1]
}
Where x and y are co-ordinates for the centre of the home range
But this is not working - giving NA values for x and y co-ordinates. Does
anyone know what to do?
Hi Umesh,
The reason the above is generating NAs may be that you have defined x
and y outside the function you are using to simulate movement. In
general, an animal would slow down near the edges of a range, so
something like this might give you a more realistic simulated exploration:
explore.circle-function(x,y,radius,nsteps) {
plot(0,xlim=c(x-radius-1,x+radius+1),ylim=c(y-radius-1,y+radius+1),type=n)
newx-newy-rep(NA,nsteps+1)
xyprob-matrix(NA,ncol=6,nrow=nsteps+1)
newx[1]-x
newy[1]-y
for(astep in 1:nsteps) {
xprob-c((newx[astep]-(x-radius))/radius,1,(x+radius-newx[astep])/radius)
newx[astep+1]-newx[astep]+sample(-1:1,1,prob=xprob)
yprob-c((newy[astep]-(y-radius))/radius,1,(y+radius-newy[astep])/radius)
newy[astep+1]-newy[astep]+sample(-1:1,1,prob=yprob)
text(newx[astep],newy[astep],astep)
xyprob[astep,]-c(xprob,yprob)
}
text(newx[nsteps+1],newy[nsteps+1],nsteps+1)
return(cbind(newx,newy,xyprob))
}
This is not guaranteed to stay inside the circle, so you might have to
add more constraints.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] limiting simulated animal movement
Wow, that was really quick, thanks a lot. Will try this and get back to you.
Cheers,
Umesh
On Sun, Mar 22, 2009 at 3:52 PM, Jim Lemon j...@bitwrit.com.au wrote:
Umesh Srinivasan wrote:
Hi,
I am trying to simulate animal movement in a gridded landscape made up of
cells. At each time step (iteration), the animal moves from one cell to
another in a random fashion.
...
The problem is
1. I want to limit animals to a pre-defined circular home range
2. I want to limit animals to the borders of the landscape itself
I tried to limit animals to their home range using:
if (sqrt((x - a)^2 + (y - b)^2) radius.range) {
a - a[i-1]
b - b[i-1]
}
Where x and y are co-ordinates for the centre of the home range
But this is not working - giving NA values for x and y co-ordinates. Does
anyone know what to do?
Hi Umesh,
The reason the above is generating NAs may be that you have defined x and y
outside the function you are using to simulate movement. In general, an
animal would slow down near the edges of a range, so something like this
might give you a more realistic simulated exploration:
explore.circle-function(x,y,radius,nsteps) {
plot(0,xlim=c(x-radius-1,x+radius+1),ylim=c(y-radius-1,y+radius+1),type=n)
newx-newy-rep(NA,nsteps+1)
xyprob-matrix(NA,ncol=6,nrow=nsteps+1)
newx[1]-x
newy[1]-y
for(astep in 1:nsteps) {
xprob-c((newx[astep]-(x-radius))/radius,1,(x+radius-newx[astep])/radius)
newx[astep+1]-newx[astep]+sample(-1:1,1,prob=xprob)
yprob-c((newy[astep]-(y-radius))/radius,1,(y+radius-newy[astep])/radius)
newy[astep+1]-newy[astep]+sample(-1:1,1,prob=yprob)
text(newx[astep],newy[astep],astep)
xyprob[astep,]-c(xprob,yprob)
}
text(newx[nsteps+1],newy[nsteps+1],nsteps+1)
return(cbind(newx,newy,xyprob))
}
This is not guaranteed to stay inside the circle, so you might have to add
more constraints.
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] bargraph.CI change se for sd
Thanks for the solution. And sorry about the not workable example (I actually edited the post a minute after posting it -too late I am afraid). library(sciplot) bargraph.CI(peptide, surface, group=adjunct,data = y) Error in eval(substitute(subset), envir = data) : object y not found #groan, ... why can't people offer a workable example? data(ToothGrowth) # se as default bargraph.CI(x.factor = dose, response = len, data = ToothGrowth) # create desired function bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) -- View this message in context: http://www.nabble.com/bargraph.CI-change-se-for-sd-tp22633770p22644429.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 plotting a histogram of X~Exp(1)
I want to plot a histogram of X~Exp(1) where X is the sum of Y + Z. To do this, should I simulate values of Y and Z using Y-runif(100) and Z-runif(100)? And where do I go from there? Many thanks. -- View this message in context: http://www.nabble.com/Help-plotting-a-histogram-of-X%7EExp%281%29-tp22645091p22645091.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] Plot and Boxplot in the same graph
I tried to do it with: y - rnorm(100) x - gl(2,50) boxplot(x,y) points(x,y) But the problem is, that the the y coordinates are shown for the boxplot and not for points(x,y) Is it possible to show the graph with the (x,y) coordinates with the points() function and the boxplots only for the x coordinates. A better solution could be to have a seperated y axis on the left side for the boxplots(). Is it possible to do it in this was ? Paul Johnson-11 wrote: On Fri, Mar 20, 2009 at 10:02 PM, johnhj jhar...@web.de wrote: Hii, Is it possible, to use the plot() funktion and the boxplot() funktion together ? I will plot a simple graph and additionally to the graph on certain places boxplots. I have imagined to plot the graph a little bit transparency and show in the same graph on certain places boxplots Is it possible to do it in this way ? greetings, johnh -- Run the boxplot first, then use points() or other subsidiary plot functions to add the points in the figure. y - rnorm(100) x - gl(2,50) boxplot(x,y) points(x,y) -- Paul E. Johnson Professor, Political Science 1541 Lilac Lane, Room 504 University of Kansas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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/Plot-and-Boxplot-in-the-same-graph-tp22632355p22645076.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] Estimating LC50 from a Weibull distribution
Hi Greg, you can use the extension package 'drc' from CRAN: conc - c(10.3, 10.8, 11.6, 13.2, 15.8, 20.1) # Exposure concentrations orign - c(76, 79, 77, 76, 78, 77) # Original number of subjects ndead - c(16, 22, 40, 69, 78, 77) # Number dead after 96 h d - data.frame(conc=conc, orign=orign, ndead=ndead) ## Loading 'drc' library(drc) ## Fitting model assuming 100% mortality for high concentrations ## and 0% for concentration 0 d.m1-drm(ndead/orign~conc, weight=orign, data=d, fct=W1.2(), type=binomial) plot(d.m1) ## Fitting model where mortality at conc=0 is estimated ## (ad hoc adjustments should be avoided) d.m2-update(d.m1, fct=W1.3u()) plot(d.m2, add=TRUE, type=none, lty=2) ## Calculating LC50 ED(d.m2, 50) Note that you don't really have time-to-event data as you only have observations reflecting the effect of exposure at one particular time point (96h). Time-to-event data occur if exposure is monitored repeatedly over time, in several intervals or at several time points, e.g. at 24h, 48h, and 96h, possibly subject to censoring. Therefore the above analysis is based on binomial distributions that are often appropriate for quantal data. Note also, that in ecotoxicology Weibull model may refer to one several related models depending on which guidelines, papers, or software you refer to. Christian __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help plotting a histogram of X~Exp(1)
If you run this you get sensible output, but whether it meets you needs depends on what you meant by Exp(1): Y - rnorm(100) Z - rnorm(100) X - exp(Y+Z) hist(X) If you are looking for methods to handle random variables in R, you may want to look at the series of packages all beginning with distr: distr, distrDoc, distrEx, distrMod, distrSim, distrTeach, distrTEst. -- David Winsemius On Mar 22, 2009, at 7:19 AM, BowlesMcCann wrote: I want to plot a histogram of X~Exp(1) where X is the sum of Y + Z. To do this, should I simulate values of Y and Z using Y-runif(100) and Z-runif(100)? And where do I go from there? Many thanks. -- View this message in context: http://www.nabble.com/Help-plotting-a-histogram-of-X%7EExp%281%29-tp22645091p22645091.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. 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] libRlapack.so not found
Johannes Huesing johan...@huesing.name [Sat, Mar 21, 2009 at 08:12:05AM CET]: Whenever I try to load the Matrix package, I get the following error message: libRlapack.so: cannot open shared object file: No such file or directory A file with that name is indeed not on the hard disk. http://lmgtfy.com/?q=libRlapack Thanks to Dirk who provided the answer about a year ago. sudo apt-get remove r-cran-matrix did it for me. -- Johannes Hüsing There is something fascinating about science. One gets such wholesale returns of conjecture mailto:johan...@huesing.name from such a trifling investment of fact. http://derwisch.wikidot.com (Mark Twain, Life on the Mississippi) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: package sowas
I downloaded it. But I could not find the installation executable file. How can I install it on Windows XP ? Thank you. Maura -Messaggio originale- Da: ronggui [mailto:ronggui.hu...@gmail.com] Inviato: dom 22/03/2009 10.47 A: mau...@alice.it; r-help@r-project.org Oggetto: Re: [R] package sowas There is a binary version for Windows in the homepage of it. http://www.cru.uea.ac.uk/~douglas/software/sowas_0.94.zip You can find more on http://tocsy.agnld.uni-potsdam.de/wavelets/ Best 2009/3/22 stephen sefick ssef...@gmail.com: You must go to his website because it is not on CRAN. I have built it on mac osx by installing it with R CMD install. This should work on other unix platforms. Stephen Sefick On Sat, Mar 21, 2009 at 9:26 PM, mau...@alice.it wrote: I cannot find sowas package by Douglas Mauran in CRAN packages list- On which platforms does sowas run ? Has anybody used such a package at all ? hHank you very much, Maura tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent Tel: (00852) 3442 3832 PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html A sociologist is someone who, when a beautiful women enters the room and everybody look at her, looks at everybody. tutti i telefonini TIM! [[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] R: package sowas
Executable file is not required. You must install R first, then you click packages--install package(s) from local zip file ... to install the sowas package. Best 2009/3/22 mau...@alice.it: I downloaded it. But I could not find the installation executable file. How can I install it on Windows XP ? Thank you. Maura -Messaggio originale- Da: ronggui [mailto:ronggui.hu...@gmail.com] Inviato: dom 22/03/2009 10.47 A: mau...@alice.it; r-help@r-project.org Oggetto: Re: [R] package sowas There is a binary version for Windows in the homepage of it. http://www.cru.uea.ac.uk/~douglas/software/sowas_0.94.zip You can find more on http://tocsy.agnld.uni-potsdam.de/wavelets/ Best 2009/3/22 stephen sefick ssef...@gmail.com: You must go to his website because it is not on CRAN. I have built it on mac osx by installing it with R CMD install. This should work on other unix platforms. Stephen Sefick On Sat, Mar 21, 2009 at 9:26 PM, mau...@alice.it wrote: I cannot find sowas package by Douglas Mauran in CRAN packages list- On which platforms does sowas run ? Has anybody used such a package at all ? hHank you very much, Maura tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent Tel: (00852) 3442 3832 PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html A sociologist is someone who, when a beautiful women enters the room and everybody look at her, looks at everybody. Alice Messenger ;-) chatti anche con gli amici di Windows Live Messenger e tutti i telefonini TIM! Vai su http://maileservizi.alice.it/alice_messenger/index.html?pmk=footer -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: R: package sowas
OK.Done. Tank you very much. Maura -Messaggio originale- Da: ronggui [mailto:ronggui.hu...@gmail.com] Inviato: dom 22/03/2009 15.40 A: mau...@alice.it Cc: r-help@r-project.org Oggetto: Re: R: [R] package sowas Executable file is not required. You must install R first, then you click packages--install package(s) from local zip file ... to install the sowas package. Best 2009/3/22 mau...@alice.it: I downloaded it. But I could not find the installation executable file. How can I install it on Windows XP ? Thank you. Maura -Messaggio originale- Da: ronggui [mailto:ronggui.hu...@gmail.com] Inviato: dom 22/03/2009 10.47 A: mau...@alice.it; r-help@r-project.org Oggetto: Re: [R] package sowas There is a binary version for Windows in the homepage of it. http://www.cru.uea.ac.uk/~douglas/software/sowas_0.94.zip You can find more on http://tocsy.agnld.uni-potsdam.de/wavelets/ Best 2009/3/22 stephen sefick ssef...@gmail.com: You must go to his website because it is not on CRAN. I have built it on mac osx by installing it with R CMD install. This should work on other unix platforms. Stephen Sefick On Sat, Mar 21, 2009 at 9:26 PM, mau...@alice.it wrote: I cannot find sowas package by Douglas Mauran in CRAN packages list- On which platforms does sowas run ? Has anybody used such a package at all ? hHank you very much, Maura tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent Tel: (00852) 3442 3832 PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html A sociologist is someone who, when a beautiful women enters the room and everybody look at her, looks at everybody. Alice Messenger ;-) chatti anche con gli amici di Windows Live Messenger e tutti i telefonini TIM! er -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html tutti i telefonini TIM! [[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] Multiple Comparisons for (multicomp - glht) for glm negative binomial (glm.nb)
Hi I have some experimental data where I have counts of the number of insects collected to different trap types rotated through 5 different location (variable -location), 4 different chemical attractants [A, B, C, D] were applied to the traps (variable - semio) and all were trialled at two different CO2 release rates [1, 2] (variable CO2) I also have a selection of continuous variables measuring meteorological conditions to account for any bias cause by changing weather conditions, the data is over dispersed so I have fitted a negative binomial glm (glm.nb) and simplified using stepAIC from the MASS package etc. There are significant differences in the number of insects attracted to the different chemical (semio) and to the two different CO2 (release rates) I have then used the glht function from the multcomp package to do multiple comparisons to see what the specific differences between the levels are for semio and CO2 using the code below which works great but what I would like to do is to do comparisons combining the factors e.g a comparison for semioA at CO at level 1 vs Semio A at CO2 level 2 etc to see which is the best combination, is this possible or should I have started of with my counts already split up into this e.g. a treatment variable(semioA at CO2 level 1 = A1, semioA at CO2 level 2 = A2 etc), I started with them this way as we have no prior knowledge that increasing co2 will have any effect. I have had a quick try with the data split into a treatment factor (instead of semio and CO2 level) but I can not get convergence with glm.nb I think this may be to do with to many zeros in the data set, do you know if glht or another multiple comparison will work or zeroinflated negative binomial regression(zeroinfl() from the pscl library)? Any help or ideas will be gratefully appreciated. Many thanks in advance. Lara semiochemical -read.csv(G:/semiochemical_data.csv, header=T) semiochemical$location2-factor(semiochemical$location) levels(semiochemical$location2)-c(1,2,3,4,5) semiochemical$semio2-factor(semiochemical$semio) levels(semiochemical$semio2)-c(A,B,C,D) semiochemical$CO22-factor(semiochemical$CO2) levels(semiochemical$CO22)-c(1,2) model1-glm.nb(total ~ semio + CO2 + location + temp + mean.wind.speed) model.glht.Semio - glht(model1, linfct=mcp(semio=Tukey)) model.glht.CO2 - glht(model1, linfct=mcp(CO2=Tukey)) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problems with combining plots
Hii, I will combine some plots. Like this example here http://www.statmethods.net/advgraphs/layout.html I tired to do it for 2 plots but without success. Here is my code: test-read.table(file=D:/file.txt) space-read.table(file=D:/space.txt) space$gruppe - 502*rep(1:6, each=7) x- c(test$V1) y- c(test$V2) par(mfrow=c(1,1)) png(filename = D:/example.png, width = 640, height = 480, pointsize = 12, bg = white, res = NA) boxplot(V2 ~ gruppe , data = space , col = lightgray,boxwex=0.2) plot(panel.first=grid(ny=NULL,nx=NULL),x,y, xlab = Zeit(sec), ylab =Datenrate(MBit(sec)),ylim=c(0,40), col =purple, type =l, main =combined plots,lwd=2) dev.off() What is the mistake in my code ? -- View this message in context: http://www.nabble.com/Problems-with-combining-plots-tp22646692p22646692.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] Help plotting a histogram of X~Exp(1)
Thank you very much! :) David Winsemius wrote: If you run this you get sensible output, but whether it meets you needs depends on what you meant by Exp(1): Y - rnorm(100) Z - rnorm(100) X - exp(Y+Z) hist(X) If you are looking for methods to handle random variables in R, you may want to look at the series of packages all beginning with distr: distr, distrDoc, distrEx, distrMod, distrSim, distrTeach, distrTEst. -- David Winsemius On Mar 22, 2009, at 7:19 AM, BowlesMcCann wrote: I want to plot a histogram of X~Exp(1) where X is the sum of Y + Z. To do this, should I simulate values of Y and Z using Y-runif(100) and Z-runif(100)? And where do I go from there? Many thanks. -- View this message in context: http://www.nabble.com/Help-plotting-a-histogram-of-X%7EExp%281%29-tp22645091p22645091.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. 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. :clap::clap::clap: -- View this message in context: http://www.nabble.com/Help-plotting-a-histogram-of-X%7EExp%281%29-tp22645091p22645942.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] Converting Matrix into List - problem (urgent)
Hi,
I'm trying to convert Matrix into a list format and have written the following
code:
path = (C:/2PL_Alpha_C_2PL_Loading)
setwd(path)
getwd()
congeneric = matrix(rep(NA,36),nrow=6,ncol=6)
conFirst = matrix(rep(NA,36),nrow=6,ncol=6)
conFirstTwenty = rep(NA, 20)
k = 1
#Reading all the Alpha and Congeneric 2PL values into variables
alpha - read.table(2PLAlphaGenPars_1.dat, header=FALSE)
congeneric - read.table(C_parameter_estimates_1.dat, header=FALSE)
for (i in 1:6)
{
for (j in 1:6)
{
conFirst[i,j] = as.matrix(congeneric[i,j])
if (k = 20)
{
conFirstTwenty[k] = as.list(conFirst[i,j])
k = k + 1
}
}
}
In the above program i'm picking up the first 20 values from the Matrix and
putting it in a list format. This is what i see now in conFirstTwenty:
[[1]]
[1] 0.520404D+00
[[2]]
[1] 0.601942D+00
[[3]]
[1] 0.603340D+00
[[4]]
[1] 0.655582D+00
[[5]]
[1] 0.490995D+00
.
..
..
...
[[20]]
[1] 0.627368D+00
I want to remove the Column name and Row name from the above output. Any help
on this will be greatly appreciated (I'm open to any other alternative way to
convert Matrix into List also)
P.S. I have tried using row.names and col.names in the read.table function
however getting error col.names object not found. I don't know what does this
mean
Regards,
Nidhi Kohli
***
Nidhi Kohli, M.Ed.
Doctoral Student
Department of Measurement, Statistics
and Evaluation
University of Maryland
1230 Benjamin Building
College Park, MD 20742-1115
e-mail: nid...@umd.edu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Accuracy of R and other platforms
Alejandro C. Frery:
@ARTICLE{AlmironSilvaMM:2009,
author = {Almiron, M. and Almeida, E. S. and Miranda, M.},
title = {The Reliability of Statistical Functions in Four Software
Packages Freely used in Numerical Computation},
journal = {Brazilian Journal of Probability and Statistics},
year = {in press},
volume = {Special Issue on Statistical Image and Signal Processing},
url = {http://www.imstat.org/bjps/}}
is freely available under the Future Papers link. It makes a nice
comparison of the numerical properties of R, Ox, Octave and Python.
Thanks for posting this. I’m happy to see that the results for R were
generally excellent, and almost always better than for the three other
software packages.
But there were a few cases where R did not turn out to be the winner.
Rather surprising that Ox was better than R for computing the
autocorrelation coefficient for two of the datasets, given its terrible
results for the standard deviation. Anybody have any ideas why?
--
Karl Ove Hufthammer
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Converting Matrix into List - problem (urgent)
Hi,
I'm trying to convert Matrix into a list format and have written the following
code:
path = (C:/2PL_Alpha_C_2PL_Loading)
setwd(path)
getwd()
congeneric = matrix(rep(NA,36),nrow=6,ncol=6)
conFirst = matrix(rep(NA,36),nrow=6,ncol=6)
conFirstTwenty = rep(NA, 20)
k = 1
#Reading all the Alpha and Congeneric 2PL values into variables
alpha - read.table(2PLAlphaGenPars_1.dat, header=FALSE)
congeneric - read.table(C_parameter_estimates_1.dat, header=FALSE)
for (i in 1:6)
{
for (j in 1:6)
{
conFirst[i,j] = as.matrix(congeneric[i,j])
if (k = 20)
{
conFirstTwenty[k] = as.list(conFirst[i,j])
k = k + 1
}
}
}
In the above program i'm picking up the first 20 values from the Matrix and
putting it in a list format. This is what i see now in conFirstTwenty:
[[1]]
[1] 0.520404D+00
[[2]]
[1] 0.601942D+00
[[3]]
[1] 0.603340D+00
[[4]]
[1] 0.655582D+00
[[5]]
[1] 0.490995D+00
.
..
..
...
[[20]]
[1] 0.627368D+00
I want to remove the Column name and Row name from the above output. Any help
on this will be greatly appreciated (I'm open to any other alternative way to
convert Matrix into List also)
P.S. I have tried using row.names and col.names in the read.table function
however getting error col.names object not found. I don't know what does this
mean
Regards,
Nidhi Kohli
***
Nidhi Kohli, M.Ed.
Doctoral Student
Department of Measurement, Statistics
and Evaluation
University of Maryland
1230 Benjamin Building
College Park, MD 20742-1115
e-mail: nid...@umd.edu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Forestplot () box size question
Yes, my R is a few versions old. I did not realize that the package version was dependent on the R version. Thanks. Gerard PS was able to apply the code suggested by David W. to the 2.14 version and got it to work. At 12:32 AM 3/22/2009, Thomas Lumley wrote: On Sat, 21 Mar 2009, Gerard Smits wrote: I have tried several sites (2 in Ca and also Australia) and find only version 2.14. Which site did you pull 2.15 from? I have checked half a dozen sites in various countries, and they all have 2.15 (even in the Southern Hemisphere -- it's not that updates flow the wrong way south of the equator). The CRAN mirror status page says that no mirrors are more than a couple of days out of date. Mostly likely you have an old version of R. CRAN binaries are built only for the current version of R. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle [[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 Matrix into List - problem (urgent)
Hi JiHo, Thank you so much for the quick reply. Let me explain you what I want. I have a data file in the following format: 0.610251D+00 0.615278D+00 0.583581D+00 0.560295D+00 0.501325D+00 0.639512D+00 0.701607D+00 0.544963D+00 0.589746D+00 0.648588D+00 0.608216D+00 0.582599D+00 0.625204D+00 0.523065D+00 0.627593D+00 0.621433D+00 0.733730D+00 0.498495D+00 0.748673D+00 0.591025D+00 0.578333D+00 0.564807D+00 0.652199D+00 0.579333D+00 I'm reading this file into a variable named congeneric (see my code) and now trying to pick up first 20 values and need these 20 values in a list format like 0.610251D+00 (row 1, col 1) 0.615278D+00 (row 1, col 2) 0.583581D+00 (row 1, col 3) ... Can you tell me how can i achieve this? I think i'm pretty close but don't know how to remove the Row Name and Col. name from the conFirstTwenty list (see my code) I really appreciate your help Regards Nidhi Original message Date: Sun, 22 Mar 2009 12:54:34 -0400 From: JiHO jo.li...@gmail.com Subject: Re: [R] Converting Matrix into List - problem (urgent) To: Nidhi Kohli nid...@umd.edu, R Help r-h...@stat.math.ethz.ch On 2009-March-22 , at 12:26 , Nidhi Kohli wrote: I want to remove the Column name and Row name from the above output. Any help on this will be greatly appreciated (I'm open to any other alternative way to convert Matrix into List also) What are you trying to achieve exactly? Do you just want to print a clean output on the screen? If yes, look at `cat`, or `print`. JiHO --- http://jo.irisson.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.
Re: [R] Hollander's test of bivariate symmetry
Couldn't find it anywhere, so for future users who stumble on this thread,
here is some code.
Note: If your data has missing values, delete those observations from the
data first before running this code.
Place these functions in the global environment (ie run the code below).
Then, to obtain the test statistic: H_bisym_stat(x,y) (x and y are the
two vectors you are comparing).
To obtain a p-value: H_bisym_stat(x,y). The p-value is calculated via a
Monte Carlo algorithm.
You can use the permutations argument to H_bisym_stat to increase the
number of Monte Carlo samples.
#Hollander's test of bivariate symmetry
#Calculating test statistic for Hollander's test of bivariate
symmetry
H_bisym_stat - function(x,y){
minXY - pmin(x,y);
sortX - x[order(minXY)];
sortY - y[order(minXY)];
minXY - pmin(sortX, sortY);
R - as.numeric(sortX = sortY);
maxXY - pmax(sortX,sortY);
D1 - matrix(0, length(x), length(x));
D2 - matrix(0, length(x), length(x));
for (i in 1:length(x)){
for (j in 1:length(x)){
D1[i,j] = as.numeric(minXY[j]
maxXY[i] maxXY[i] = maxXY[j]);
D2[i,j] = as.numeric(minXY[i] =
minXY[j]);
}
}
D = D1*D2;
S = 2*R - 1;
T = t( t(S) %*% D);
H_squared = (1/length(x)^2)*( t(T) %*% T );
H_squared};
#Create a permutation sample for paired data;
permsamp_paired - function(x,y){
new - rbinom(length(x),1,0.5);
x_new - x*new + y*(1 - new);
y_new - y*new + x*(1 - new);
list(x_new = x_new, y_new = y_new)};
#Calculate pvalue for Hollander statistic;
H_bisym_pv - function(x, y, permutations = 1000){
H_squared = H_bisym_stat(x=x, y=y);
numb_greater = 0;
for (i in 1:permutations) {
newvars - permsamp_paired(x=x, y=y);
x_new - newvars$x_new;
y_new - newvars$y_new;
newstat - H_bisym_stat(x=x_new, y=y_new);
numb_greater = numb_greater +
as.numeric(newstat = H_squared);
}
pv - numb_greater/permutations;
pv };
Joe Boyer
Statistical Sciences
GlaxoSmithKline
[[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 Matrix into List - problem (urgent)
On 2009-March-22 , at 13:06 , Nidhi Kohli wrote: I'm reading this file into a variable named congeneric (see my code) and now trying to pick up first 20 values and need these 20 values in a list format like 0.610251D+00 (row 1, col 1) 0.615278D+00 (row 1, col 2) 0.583581D+00 (row 1, col 3) ... Basically you want a list in which each element as only one value in it. And those elements are the first twenty of the previous matrix, reading by line, then column. I am not sure about: I want to remove the Column name and Row name from the above output. Any help on this will be greatly appreciated (I'm open to any other alternative way to convert Matrix into List also) I don't see any column or row name in [[1]] [1] 0.520404D+00 [[2]] [1] 0.601942D+00 [[3]] [1] 0.603340D+00 [[4]] [1] 0.655582D+00 [[5]] [1] 0.490995D+00 . .. .. ... [[20]] [1] 0.627368D+00 That's just a the way a list is printed. If you just want a clean visual output, then use cat in a for loop. So I guess my question is what are you trying to achieve with that? What do you need the list for? It would help to know (a bit) of the general context to help you. JiHO --- http://jo.irisson.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.
Re: [R] Converting Matrix into List - problem (urgent)
On 2009-March-22 , at 13:06 , Nidhi Kohli wrote:
Thank you so much for the quick reply. Let me explain you what I
want. I have a data file in the following format:
0.610251D+00 0.615278D+00 0.583581D+00 0.560295D+00
0.501325D+00 0.639512D+00 0.701607D+00 0.544963D+00
0.589746D+00 0.648588D+00 0.608216D+00 0.582599D+00
0.625204D+00 0.523065D+00 0.627593D+00 0.621433D+00
0.733730D+00 0.498495D+00 0.748673D+00 0.591025D+00
0.578333D+00 0.564807D+00 0.652199D+00 0.579333D+00
I'm reading this file into a variable named congeneric (see my code)
and now trying to pick up first 20 values and need these 20 values
in a list format like
0.610251D+00 (row 1, col 1)
0.615278D+00 (row 1, col 2)
0.583581D+00 (row 1, col 3)
...
Can you tell me how can i achieve this? I think i'm pretty close but
don't know how to remove the Row Name and Col. name from the
conFirstTwenty list (see my code)
Alternatives:
congeneric - scan(textConnection(0.610251 0.615278 0.583581 0.560295
0.501325 0.639512 0.701607 0.544963
0.589746 0.648588 0.608216 0.582599
0.625204 0.523065 0.627593 0.621433
0.733730 0.498495 0.748673 0.591025
0.578333 0.564807 0.652199 0.579333), sep= )
# you would use scan with the filename in which the data is, rather
than the textConnection. That's just for the purpose of the
demonstration here.
# possible outputs
congeneric[1:20]
as.list(congeneric[1:20])
for (i in 1:20) {
cat(congeneric[i],\n)
}
JiHO
---
http://jo.irisson.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.
Re: [R] Converting Matrix into List - problem (urgent)
JiHo,
Thanks for the reply, however i dont think it will help. Okay here is what i
will achieve once i get this into a list format. I have another file in the
following format:
0.63275433157105
0.686061949818395
0.786426681675948
0.954103894997388
0.600840965518728
0.949194842483848
0.972337634302676
0.830398896243423
0.81455702194944
0.530893135233782
0.602987287449650
0.588278376264498
0.843511423328891
0.692051859106869
0.88492070278
0.748849621042609
0.858809254132211
0.995953047415242
0.690017589717172
0.888722610659897
I need to compute correlation between these two sets. So i'm trying to convert
my other file also in the above format so that i can use the function to
compute correlation
Nidhi
Original message
Date: Sun, 22 Mar 2009 13:27:32 -0400
From: JiHO jo.li...@gmail.com
Subject: Re: [R] Converting Matrix into List - problem (urgent)
To: Nidhi Kohli nid...@umd.edu
Cc: R Help r-h...@stat.math.ethz.ch
On 2009-March-22 , at 13:06 , Nidhi Kohli wrote:
Thank you so much for the quick reply. Let me explain you what I
want. I have a data file in the following format:
0.610251D+00 0.615278D+00 0.583581D+00 0.560295D+00
0.501325D+00 0.639512D+00 0.701607D+00 0.544963D+00
0.589746D+00 0.648588D+00 0.608216D+00 0.582599D+00
0.625204D+00 0.523065D+00 0.627593D+00 0.621433D+00
0.733730D+00 0.498495D+00 0.748673D+00 0.591025D+00
0.578333D+00 0.564807D+00 0.652199D+00 0.579333D+00
I'm reading this file into a variable named congeneric (see my code)
and now trying to pick up first 20 values and need these 20 values
in a list format like
0.610251D+00 (row 1, col 1)
0.615278D+00 (row 1, col 2)
0.583581D+00 (row 1, col 3)
...
Can you tell me how can i achieve this? I think i'm pretty close but
don't know how to remove the Row Name and Col. name from the
conFirstTwenty list (see my code)
Alternatives:
congeneric - scan(textConnection(0.610251 0.615278 0.583581 0.560295
0.501325 0.639512 0.701607 0.544963
0.589746 0.648588 0.608216 0.582599
0.625204 0.523065 0.627593 0.621433
0.733730 0.498495 0.748673 0.591025
0.578333 0.564807 0.652199 0.579333), sep= )
# you would use scan with the filename in which the data is, rather
than the textConnection. That's just for the purpose of the
demonstration here.
# possible outputs
congeneric[1:20]
as.list(congeneric[1:20])
for (i in 1:20) {
cat(congeneric[i],\n)
}
JiHO
---
http://jo.irisson.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.
[R] Selecting closest values
Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you -- View this message in context: http://www.nabble.com/Selecting-closest-values-tp22647126p22647126.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] Problems with combining plots
par(mfrow=c(1,1)) will give you just one panel. Try par(mfrow=c(2,1)) or par(mfrow=c(1,2)). -Original Message- From: r-help-boun...@r-project.org on behalf of johnhj Sent: Sun 3/22/2009 10:50 AM To: r-help@r-project.org Subject: [R] Problems with combining plots Hii, I will combine some plots. Like this example here http://www.statmethods.net/advgraphs/layout.html I tired to do it for 2 plots but without success. Here is my code: test-read.table(file=D:/file.txt) space-read.table(file=D:/space.txt) space$gruppe - 502*rep(1:6, each=7) x- c(test$V1) y- c(test$V2) par(mfrow=c(1,1)) png(filename = D:/example.png, width = 640, height = 480, pointsize = 12, bg = white, res = NA) boxplot(V2 ~ gruppe , data = space , col = lightgray,boxwex=0.2) plot(panel.first=grid(ny=NULL,nx=NULL),x,y, xlab = Zeit(sec), ylab =Datenrate(MBit(sec)),ylim=c(0,40), col =purple, type =l, main =combined plots,lwd=2) dev.off() What is the mistake in my code ? -- View this message in context: http://www.nabble.com/Problems-with-combining-plots-tp22646692p22646692.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting closest values
It would be helpful is you supplied a subset of the data so that we could see what you expect. The 'outer' On Sun, Mar 22, 2009 at 1:48 PM, P_M pmart...@broadpark.no wrote: Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you -- View this message in context: http://www.nabble.com/Selecting-closest-values-tp22647126p22647126.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] Selecting closest values
Yes, of course. ID loc 1 144 2 144 3 140 4 126 5 120 6 112 7 100 8 99 9 91 11 90 12 90 13 89 15 86 16 85 17 85 18 80 19 79 20 79 21 78 22 78 23 76 24 74 25 73 26 72 27 71 28 68 29 68 30 68 . . . . 185 -22 P_M wrote: Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you -- View this message in context: http://www.nabble.com/Selecting-closest-values-tp22647126p22648754.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using wavelet transform to calculate mean frequency of a signal
Dear list,
in short: I would like to calculate the mean frequency
of a signal (e.g. time series) using the wavelet transform.
with details: I did not find a R function to calculate a
mean frequency using one of the cran packages.
My searches using R Site Search returned:
**(1)waveclock {waveclock}R Documentation
Reconstruction of the modal frequencies in a time series using continuous
wavelet transformation and the crazy climbers algorithm
(this look like what I wanted to do, but I can't figure it out)
**(2) tfmean {Rwave}R Documentation
Average frequency by frequency
(this seems not to help either)
I am a bit desperate at the moment
and would very, very much appreciate any help.
Regards,
Stephan
[[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] Pruning trees in a Random Forest
Jerry Friedman advocated a few mods to RF in his ISLE algorithm:
- draw smaller samples
- grow smaller trees
- post process the forest using something like PathSeeker (or LASSO)
Andy
From: Wensui Liu [mailto:liuwen...@gmail.com]
Sent: Fri 3/20/2009 4:46 PM
To: Liaw, Andy
Cc: Anirudh Kondaveeti; r-help@r-project.org
Subject: Re: [R] Pruning trees in a Random Forest
I don't think it necessary to prune trees in RF, per brieman's paper.
On 3/20/09, Liaw, Andy andy_l...@merck.com wrote:
The way the trees are structured in randomForest, there's no way to stop
tree growth by depth (what you called level).
(If anyone has ideas, I'm all ears.)
Andy
From: Anirudh Kondaveeti
Hi all!
The randomForest in R enables us to prune the trees using the nodesize
feature where we can stop splitting a node if it contains
less than the
specified no.of of records/entities at that node.
However is there a way to stop the tree growing after a
specified number of
levels. To be more clear on what I mean by a level. Level 0
is the parent
node, Level 1 has 2 daughter nodes, Level 2 has 4 daughter
nodes, Level 3
has 8 daughter nodes etc.
Thanks in advance!
Anirudh Kondaveeti
[[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.
Notice: This e-mail message, together with any attachme...{{dropped:12}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
==
WenSui Liu
Acquisition Risk, Chase
Blog : statcompute.spaces.live.com
Tough Times Never Last. But Tough People Do. - Robert Schuller
==
Notice: This e-mail message, together with any attachme...{{dropped:15}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting closest values
If we assume the distance is linear, this might work: x ID loc 1 1 144 2 2 144 3 3 140 4 4 126 5 5 120 6 6 112 7 7 100 8 8 99 9 9 91 10 11 90 11 12 90 12 13 89 13 15 86 14 16 85 15 17 85 16 18 80 17 19 79 18 20 79 19 21 78 20 22 78 21 23 76 22 24 74 23 25 73 24 26 72 25 27 71 26 28 68 27 29 68 28 30 68 x.outer - outer(x$loc, x$loc, function(a,b) abs(a - b)) # set diag to Inf so a point is not closest to itself diag(x.outer) - Inf # assume that you choose the 8th element, here are the closest IDs x$ID[order(x.outer[,8])] [1] 7 9 11 12 13 6 15 16 17 18 19 20 5 21 22 23 24 25 4 26 27 28 29 30 3 1 2 8 On Sun, Mar 22, 2009 at 2:04 PM, P_M pmart...@broadpark.no wrote: Yes, of course. ID loc 1 144 2 144 3 140 4 126 5 120 6 112 7 100 8 99 9 91 11 90 12 90 13 89 15 86 16 85 17 85 18 80 19 79 20 79 21 78 22 78 23 76 24 74 25 73 26 72 27 71 28 68 29 68 30 68 . . . . 185 -22 P_M wrote: Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you -- View this message in context: http://www.nabble.com/Selecting-closest-values-tp22647126p22648754.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] Converting Matrix into List - problem (urgent)
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Nidhi Kohli
Sent: Sunday, March 22, 2009 10:41 AM
To: JiHO
Cc: R Help
Subject: Re: [R] Converting Matrix into List - problem (urgent)
JiHo,
Thanks for the reply, however i dont think it will help. Okay
here is what i will achieve once i get this into a list
format. I have another file in the following format:
0.63275433157105
0.686061949818395
0.786426681675948
0.954103894997388
0.600840965518728
0.949194842483848
0.972337634302676
0.830398896243423
0.81455702194944
0.530893135233782
0.602987287449650
0.588278376264498
0.843511423328891
0.692051859106869
0.88492070278
0.748849621042609
0.858809254132211
0.995953047415242
0.690017589717172
0.888722610659897
I need to compute correlation between these two sets. So i'm
trying to convert my other file also in the above format so
that i can use the function to compute correlation
Nidhi
Original message
Date: Sun, 22 Mar 2009 13:27:32 -0400
From: JiHO jo.li...@gmail.com
Subject: Re: [R] Converting Matrix into List - problem (urgent)
To: Nidhi Kohli nid...@umd.edu
Cc: R Help r-h...@stat.math.ethz.ch
On 2009-March-22 , at 13:06 , Nidhi Kohli wrote:
Thank you so much for the quick reply. Let me explain you what I
want. I have a data file in the following format:
0.610251D+00 0.615278D+00 0.583581D+00 0.560295D+00
0.501325D+00 0.639512D+00 0.701607D+00 0.544963D+00
0.589746D+00 0.648588D+00 0.608216D+00 0.582599D+00
0.625204D+00 0.523065D+00 0.627593D+00 0.621433D+00
0.733730D+00 0.498495D+00 0.748673D+00 0.591025D+00
0.578333D+00 0.564807D+00 0.652199D+00 0.579333D+00
I'm reading this file into a variable named congeneric
(see my code)
and now trying to pick up first 20 values and need these
20 values
in a list format like
0.610251D+00 (row 1, col 1)
0.615278D+00 (row 1, col 2)
0.583581D+00 (row 1, col 3)
...
Can you tell me how can i achieve this? I think i'm pretty
close but
don't know how to remove the Row Name and Col. name from the
conFirstTwenty list (see my code)
Alternatives:
congeneric - scan(textConnection(0.610251 0.615278
0.583581 0.560295
0.501325 0.639512 0.701607 0.544963
0.589746 0.648588 0.608216 0.582599
0.625204 0.523065 0.627593 0.621433
0.733730 0.498495 0.748673 0.591025
0.578333 0.564807 0.652199 0.579333), sep= )
# you would use scan with the filename in which the data is, rather
than the textConnection. That's just for the purpose of the
demonstration here.
# possible outputs
congeneric[1:20]
as.list(congeneric[1:20])
for (i in 1:20) {
cat(congeneric[i],\n)
}
JiHO
---
You really haven't given us enough information. What stements did you
execute that produced the data and row/column labels shown above? What does
running str(congeneric) tell you about congeneric? I don't think you want
your data for the correlation in an R list structure. You want your data in
vectors, or the columns of a matrix or dataframe. Given the data in your
file, you could do something like this
#read data into a matrix
m - as.matrix(read.table(your_data_file,sep=' ', header=FALSE))
#get the 20 values you want
vector1 - as.vector(t(m))[1:20]
But without knowing the actual structure of your variable congeneric, it is
not possible to tell you how to manipulate it. Maybe someone else will be
better at guessing.
Dan
Daniel Nordlund
Bothell, WA USA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 Comparisons for (multicomp - glht) for glm negative binomial (glm.nb)
On Sun, 22 Mar 2009, lara harrup (IAH-P) wrote: Hi I have some experimental data where I have counts of the number of insects collected to different trap types rotated through 5 different location (variable -location), 4 different chemical attractants [A, B, C, D] were applied to the traps (variable - semio) and all were trialled at two different CO2 release rates [1, 2] (variable CO2) I also have a selection of continuous variables measuring meteorological conditions to account for any bias cause by changing weather conditions, the data is over dispersed so I have fitted a negative binomial glm (glm.nb) and simplified using stepAIC from the MASS package etc. There are significant differences in the number of insects attracted to the different chemical (semio) and to the two different CO2 (release rates) I have then used the glht function from the multcomp package to do multiple comparisons to see what the specific differences between the levels are for semio and CO2 using the code below which works great but what I would like to do is to do comparisons combining the factors e.g a comparison for semioA at CO at level 1 vs Semio A at CO2 level 2 etc to see which is the best combination, is this possible or should I have started of with my counts already split up into this e.g. a treatment variable(semioA at CO2 level 1 = A1, semioA at CO2 level 2 = A2 etc), I started with them this way as we have no prior knowledge that increasing co2 will have any effect. I have had a quick try with the data split into a treatment factor (instead of semio and CO2 level) but I can not get convergence with glm.nb I think this may be to do with to many zeros in the data set, do you know if glht or another multiple comparison will work or zeroinflated negative binomial regression(zeroinfl() from the pscl library)? Yes, it does, because zeroinfl() (and hurdle()) both return models with the usual extractor methods (coef, vcov, et al). The only nuisance is that you can't use the mcp() interface because it will be confused about the two parts of the model (count vs. zero-inflation part) which might both contain the same variables. What you have to do is set up your contrast matrix by hand. An example using the NMES1988 data from the AER package is included below. hth, Z ## packages library(pscl) library(multcomp) ## data data(NMES1988, package = AER) nmes - NMES1988[, c(1, 6:8)] ## models fm_pois - glm(visits ~ ., data = nmes, family = poisson) fm_nbin - glm.nb(visits ~ ., data = nmes) fm_zinb - zeroinfl(visits ~ . | ., data = nmes, dist = negbin) ## generalized linear hypotheses glht_pois - glht(fm_pois, linfct = mcp(health = Tukey)) glht_nbin - glht(fm_nbin, linfct = mcp(health = Tukey)) ## compute contrasts by hand for ZINB nr - length(levels(nmes$health)) contr - matrix(0, nrow = nr, ncol = length(coef(fm_zinb))) colnames(contr) - names(coef(fm_nbin)) rownames(contr) - paste(levels(nmes$health)[c(2, 3, 3)], levels(nmes$health)[c(1, 1, 2)], sep = - ) contr[,3:4] - contrMat(numeric(nrow(contr)), type = Tukey)[,-2] glht_zinb - glht(fm_zinb, linfct = contr) ## multiple comparisons summary(glht_pois) summary(glht_nbin) summary(glht_zinb) Any help or ideas will be gratefully appreciated. Many thanks in advance. Lara semiochemical -read.csv(G:/semiochemical_data.csv, header=T) semiochemical$location2-factor(semiochemical$location) levels(semiochemical$location2)-c(1,2,3,4,5) semiochemical$semio2-factor(semiochemical$semio) levels(semiochemical$semio2)-c(A,B,C,D) semiochemical$CO22-factor(semiochemical$CO2) levels(semiochemical$CO22)-c(1,2) model1-glm.nb(total ~ semio + CO2 + location + temp + mean.wind.speed) model.glht.Semio - glht(model1, linfct=mcp(semio=Tukey)) model.glht.CO2 - glht(model1, linfct=mcp(CO2=Tukey)) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] barplot2 x-axis
Dear R users, I am trying to build a barplot2 graph however I can't find a way of defining the scale for the x-axis. I would like to show in my x-axis only the numbers 0, 25, 50, 75 etc. (so far R is giving me a random scale hard to interpret and it doens't look nice...). Could anyone advise me on how to do this please, it would be a great help! Thank you. Below I show the code I have been using for my plots. my.plot- barplot2(mydata.y, names.arg=mydata.x, width=1, xlab=km, ylab=kg, main=plot.name, plot.ci=TRUE, ci.l=data.lci,ci.u=data.uci, ci.col = red, font=40, font.lab=60, xlim=c(0,250), xpd=FALSE) Thanking you in advance Mafalda -- Mafalda Viana [[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 Matrix into List - problem (urgent)
On 2009-March-22 , at 12:26 , Nidhi Kohli wrote: I want to remove the Column name and Row name from the above output. Any help on this will be greatly appreciated (I'm open to any other alternative way to convert Matrix into List also) What are you trying to achieve exactly? Do you just want to print a clean output on the screen? If yes, look at `cat`, or `print`. JiHO --- http://jo.irisson.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.
Re: [R] Selecting closest values
P_M pmart...@broadpark.no [Sun, Mar 22, 2009 at 06:48:36PM CET]: Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. In the general case, the nn function from the RANN package (not part of CRAN but found by rseek) does what you want, but for the one-dimensional case this might be overkill. -- Johannes Hüsing There is something fascinating about science. One gets such wholesale returns of conjecture mailto:johan...@huesing.name from such a trifling investment of fact. http://derwisch.wikidot.com (Mark Twain, Life on the Mississippi) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 'require' equivalent for local functions
Hello everyone,
I often create some local libraries of functions (.R files with only
functions in them) that I latter call. In scripts that call a function
from such library, I would like to be able to test whether the
function is already known in the namespace and, only if it is not,
source the library file. I.e. what `require` does for packages, I want
to do with my local functions.
Example:
lib.R
foo - function(x) { x*2 }
script.R
require.local(foo,lib.R)
# that searches for function foo and, if not found, executes
source(lib.R)
foo(2)
Obviously, I want the test to be quite efficient otherwise I might as
well source the local library every time. I am aware that it would
probably not be able to check for changes in lib.R (i.e. do
complicated things such as re-source lib.R if foo in the namespace and
foo in lib.R are different), but that I can handle manually.
This seems like a common enough workflow but I cannot find a pre-
existing solution. Does anyone have pointers?
Otherwise I tried to put that together:
require.local - function(fun, lib)
#
# Searches for function fun and sources lib in
# case it is not found
#
{
if (! (deparse(substitute(fun)) %in% ls(.GlobalEnv) class(fun)
== function) ) {
cat(Sourcing, lib,...\n)
source(lib)
}
}
but I am really not confident with all those deparse/substitute things
and the environment manipulation, so I guess there should be a better
way.
JiHO
---
http://jo.irisson.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.
Re: [R] 'require' equivalent for local functions
Try this:
if (!exists(myfun, mode = function)) source(myfile.R)
Also check the other arguments of exists in case you want to
restrict the search.
On Sun, Mar 22, 2009 at 5:05 PM, JiHO jo.li...@gmail.com wrote:
Hello everyone,
I often create some local libraries of functions (.R files with only
functions in them) that I latter call. In scripts that call a function from
such library, I would like to be able to test whether the function is
already known in the namespace and, only if it is not, source the library
file. I.e. what `require` does for packages, I want to do with my local
functions.
Example:
lib.R
foo - function(x) { x*2 }
script.R
require.local(foo,lib.R)
# that searches for function foo and, if not found, executes
source(lib.R)
foo(2)
Obviously, I want the test to be quite efficient otherwise I might as well
source the local library every time. I am aware that it would probably not
be able to check for changes in lib.R (i.e. do complicated things such as
re-source lib.R if foo in the namespace and foo in lib.R are different), but
that I can handle manually.
This seems like a common enough workflow but I cannot find a pre-existing
solution. Does anyone have pointers?
Otherwise I tried to put that together:
require.local - function(fun, lib)
#
# Searches for function fun and sources lib in
# case it is not found
#
{
if (! (deparse(substitute(fun)) %in% ls(.GlobalEnv) class(fun)
== function) ) {
cat(Sourcing, lib,...\n)
source(lib)
}
}
but I am really not confident with all those deparse/substitute things and
the environment manipulation, so I guess there should be a better way.
JiHO
---
http://jo.irisson.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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] 'require' equivalent for local functions
JiHO wrote:
Hello everyone,
I often create some local libraries of functions (.R files with only
functions in them) that I latter call. In scripts that call a function
from such library, I would like to be able to test whether the
function is already known in the namespace and, only if it is not,
source the library file. I.e. what `require` does for packages, I want
to do with my local functions.
Example:
lib.R
foo - function(x) { x*2 }
script.R
require.local(foo,lib.R)
# that searches for function foo and, if not found, executes
source(lib.R)
foo(2)
Obviously, I want the test to be quite efficient otherwise I might as
well source the local library every time. I am aware that it would
probably not be able to check for changes in lib.R (i.e. do
complicated things such as re-source lib.R if foo in the namespace and
foo in lib.R are different), but that I can handle manually.
This seems like a common enough workflow but I cannot find a
pre-existing solution. Does anyone have pointers?
Otherwise I tried to put that together:
require.local - function(fun, lib)
#
#Searches for function fun and sources lib in
#case it is not found
#
{
if (! (deparse(substitute(fun)) %in% ls(.GlobalEnv)
class(fun) == function) ) {
perhaps
find(deparse(substitute(fun)), mode='function')
but note that this will *not* tell you whether *the* function you want
to import is already known, but rather whether *some* function with the
specified name is already known.
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] barplot2 x-axis
On Mar 22, 2009, at 2:31 PM, Mafalda Viana wrote: Dear R users, I am trying to build a barplot2 graph however I can't find a way of defining the scale for the x-axis. I would like to show in my x-axis only the numbers 0, 25, 50, 75 etc. (so far R is giving me a random scale hard to interpret and it doens't look nice...). Could anyone advise me on how to do this please, it would be a great help! Thank you. Below I show the code I have been using for my plots. my.plot- barplot2(mydata.y, names.arg=mydata.x, width=1, xlab=km, ylab=kg, main=plot.name, plot.ci=TRUE, ci.l=data.lci,ci.u=data.uci, ci.col = red, font=40, font.lab=60, xlim=c(0,250), xpd=FALSE) It is not entirely clear what you are doing here an we cannot reproduce it lacking your data. Your x (horizontal) axis will have one tick mark below each bar, using the labels that you have indicated with mydata.x. Based upon your code, the x axis values would appear to be measures of distance. Your y (vertical) axis would appear to be a measure of weight, perhaps being some descriptive statistic (eg. mean) for weight at each distance. It is unclear what you want to do with the x axis in lieu of the default values. Modifying the tick marks on the y axis would seem to make more sense here. In general, with any R graphic, you would use the arguments 'xaxt = n' and/or 'yaxt = n' to disable the default drawing of the x and y axes, respectively. You would then call the axis() function passing the specific locations and values of the tick marks that you wish to draw. See ?par and ?axis for more information on these. BTW if you are not plotting counts or proportions/percentages, then a barplot is not the best approach for the display of summarized continuous data. I would use a point plot with error bars/confidence intervals. See the errbar() function in the Hmisc package or the plotCI() or plotmeans() functions in gplots. You could also create something manually by using plot() and then either segments() or arrows() for the CIs. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] If statement generates two outputs
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
Thanks,
James
--
View this message in context:
http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22650844.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] If statement generates two outputs
jimdare wrote:
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
almost there:
if (X5) {df1-c(4,5,6,7,8); df2-c(9,10,11,12,13)}
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] If statement generates two outputs
if (X5) {
df1-c(4,5,6,7,8)
df2-c(9,10,11,12,13)
}
hth,
Daniel Moreira, MD
jimdare jamesdar...@gmail.com
Sent by: r-help-boun...@r-project.org
03/22/2009 05:27 PM
To
r-help@r-project.org
cc
Subject
[R] If statement generates two outputs
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
Thanks,
James
--
View this message in context:
http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22650844.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.
[[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] If statement generates two outputs
Thanks very much
Wacek Kusnierczyk wrote:
jimdare wrote:
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
almost there:
if (X5) {df1-c(4,5,6,7,8); df2-c(9,10,11,12,13)}
vQ
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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/If-statement-generates-two-outputs-tp22650844p22650960.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] Following progress in a lapply() function
Dear all,
I am processing a very long and complicated list using lapply through a custom
function and I would like to generate some sort of progress report. For
instance, print a dot on the screen every time 1000 item have been process. Or
even better, reporting the percent of the list that have been process every
10%. However, I can't seem to figure out a way to achieve that.
For instance, I have a list of 50,000 slots:
aList - replicate(5,list(rnorm(50)))
That need to be process through the following custom function:
myFnc - function(x){
tTest - t.test(x)
return(list(p.value=tTest$p.value,t.stat=tTest$stat))
}
Using an lapply statement, as in:
myResults - lapply(aList, myFnc)
The goal would be to report on the progress of the lapply() function during
processing.
Any suggestion would be greatly appreciated.
Thanks
Marco
--
Marco Blanchette, Ph.D.
Assistant Investigator
Stowers Institute for Medical Research
1000 East 50th St.
Kansas City, MO 64110
Tel: 816-926-4071
Cell: 816-726-8419
Fax: 816-926-2018
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] barplot2 x-axis
Dear Mark, Thanks for your reply. I attach one of my plots as an example (e.g.Mafalda) to show what I want. If you notice in the plot the x-axis has a funny scale (0, 13, 28...) and what I would like to do is to make that scale from 0 and then increasing by 25 km for a better interpretation (0, 25, 50...). Hopefully my problem is more clear now, meanwhile I will have a look to your suggestions. Thank you very much Mafalda 2009/3/22 Marc Schwartz marc_schwa...@comcast.net On Mar 22, 2009, at 2:31 PM, Mafalda Viana wrote: Dear R users, I am trying to build a barplot2 graph however I can't find a way of defining the scale for the x-axis. I would like to show in my x-axis only the numbers 0, 25, 50, 75 etc. (so far R is giving me a random scale hard to interpret and it doens't look nice...). Could anyone advise me on how to do this please, it would be a great help! Thank you. Below I show the code I have been using for my plots. my.plot- barplot2(mydata.y, names.arg=mydata.x, width=1, xlab=km, ylab=kg, main=plot.name, plot.ci=TRUE, ci.l=data.lci,ci.u=data.uci, ci.col = red, font=40, font.lab=60, xlim=c(0,250), xpd=FALSE) It is not entirely clear what you are doing here an we cannot reproduce it lacking your data. Your x (horizontal) axis will have one tick mark below each bar, using the labels that you have indicated with mydata.x. Based upon your code, the x axis values would appear to be measures of distance. Your y (vertical) axis would appear to be a measure of weight, perhaps being some descriptive statistic (eg. mean) for weight at each distance. It is unclear what you want to do with the x axis in lieu of the default values. Modifying the tick marks on the y axis would seem to make more sense here. In general, with any R graphic, you would use the arguments 'xaxt = n' and/or 'yaxt = n' to disable the default drawing of the x and y axes, respectively. You would then call the axis() function passing the specific locations and values of the tick marks that you wish to draw. See ?par and ?axis for more information on these. BTW if you are not plotting counts or proportions/percentages, then a barplot is not the best approach for the display of summarized continuous data. I would use a point plot with error bars/confidence intervals. See the errbar() function in the Hmisc package or the plotCI() or plotmeans() functions in gplots. You could also create something manually by using plot() and then either segments() or arrows() for the CIs. HTH, Marc Schwartz -- Mafalda Viana __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 frame to array
Aloha all, I have a data frame with 4 columns. The first three are factors (f1, f2, f3) and the fourth is numeric. I'd like to explore these data using median polish. To do that I plan to use medpolish() on the matrix[f1,f2xf3], then medpolish on the resulting matrix[f2,f3]. This approach is described by Cook on page 141 of Exploring Data Tables, Trends, and Shapes. split() gets me close to where I want to be, but results in a list, rather than a matrix. How do I construct the matrix[f1,f2xf3] from my data frame? Also, any pointers to existing code that performs multi-way median polish will be appreciated. Sorry for the newbie-type query, but manipulating data prior to analysis is really hard for me in R. All the best, Tom Thomas S. Dye, Ph.D. T. S. Dye Colleagues, Archaeologists, Inc. Phone: (808) 529-0866 Fax: (808) 529-0884 http://www.tsdye.com [[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] Following progress in a lapply() function
There are several packages with progress bars:
RSiteSearch(progress)
On Sun, Mar 22, 2009 at 6:06 PM, Blanchette, Marco m...@stowers.org wrote:
Dear all,
I am processing a very long and complicated list using lapply through a
custom function and I would like to generate some sort of progress report.
For instance, print a dot on the screen every time 1000 item have been
process. Or even better, reporting the percent of the list that have been
process every 10%. However, I can't seem to figure out a way to achieve that.
For instance, I have a list of 50,000 slots:
aList - replicate(5,list(rnorm(50)))
That need to be process through the following custom function:
myFnc - function(x){
tTest - t.test(x)
return(list(p.value=tTest$p.value,t.stat=tTest$stat))
}
Using an lapply statement, as in:
myResults - lapply(aList, myFnc)
The goal would be to report on the progress of the lapply() function during
processing.
Any suggestion would be greatly appreciated.
Thanks
Marco
--
Marco Blanchette, Ph.D.
Assistant Investigator
Stowers Institute for Medical Research
1000 East 50th St.
Kansas City, MO 64110
Tel: 816-926-4071
Cell: 816-726-8419
Fax: 816-926-2018
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Following progress in a lapply() function
On Sun, Mar 22, 2009 at 5:06 PM, Blanchette, Marco m...@stowers.org wrote:
Dear all,
I am processing a very long and complicated list using lapply through a
custom function and I would like to generate some sort of progress report.
For instance, print a dot on the screen every time 1000 item have been
process. Or even better, reporting the percent of the list that have been
process every 10%. However, I can't seem to figure out a way to achieve that.
For instance, I have a list of 50,000 slots:
aList - replicate(5,list(rnorm(50)))
That need to be process through the following custom function:
myFnc - function(x){
tTest - t.test(x)
return(list(p.value=tTest$p.value,t.stat=tTest$stat))
}
Using an lapply statement, as in:
myResults - lapply(aList, myFnc)
The goal would be to report on the progress of the lapply() function during
processing.
install.packages(plyr)
library(plyr)
myResults - llply(aList, myFnc, .progress = text)
see http://had.co.nz/plyr for more info.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting closest values
Here is one way you might find useful id - 1:183 lo - sample((-22):144, 183, rep = TRUE) ds - outer(lo, lo, function(x,y) abs(x-y)) rs - t(apply(ds, 2, function(x) id[order(x)][2:11])) rownames(rs) - id Here's what I got head(rs) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] 1 118 23 125 182 1838 86 119 141 175 2 27 167 1764 126 45 113 81 144 153 3 437 18 60 148 11 12 150 85 163 4 126 27 167 1762 45 144 153 4855 5 15 46 86 141 175 23 125 1831 118 6 58 78 139 13 66 131 156 181 117 157 If you want anything more random in the case of ties than this, you will have to work a bit harder on the code. W. Bill Venables http://www.cmis.csiro.au/bill.venables/ -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of P_M Sent: Monday, 23 March 2009 1:32 AM To: r-help@r-project.org Subject: [R] Selecting closest values Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you -- View this message in context: http://www.nabble.com/Selecting-closest-values-tp22647126p22647126.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] I have a question about a programa in R
Hello, And what is exactly your problem? Best regards, Carlos J. Gil Bellosta http://www.datanalytics.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] data frame to array
Here is a possibility
Dat - cbind(expand.grid(f1 = letters[1:5], f2 = LETTERS[1:5],
f3 = as.character(1:5)), x = rnorm(125))
M - with(Dat, {
f23 - f2:f3
m - matrix(0, length(levels(f1)), length(levels(f23)))
i - match(f1, levels(f1))
j - match(f23, levels(f23))
m[cbind(i,j)] - x
dimnames(m) - list(levels(f1), levels(f23))
m
})
M[1:5, 1:5]
A:1A:2A:3A:4A:5
a 1.72686085 -2.0605242 1.0989119 0.8096139 1.0146972
b -0.34512446 -0.1709805 0.3401842 0.5815685 -1.4862872
c 1.14489491 -0.3959085 0.3222197 -1.1108793 0.3676764
d 0.02520386 -1.0018102 -0.7232067 -0.6142914 0.6694813
e -1.23366653 0.3826862 -0.6797035 0.6536055 0.8865669
This should work provided you have one entry per f1 x f2 x f3 cell. The rows
of the data frame may be in arbitrary order.
Bill Venables
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Thomas S. Dye
Sent: Monday, 23 March 2009 7:36 AM
To: r-help@r-project.org
Subject: [R] data frame to array
Aloha all,
I have a data frame with 4 columns. The first three are factors (f1,
f2, f3) and the fourth is numeric. I'd like to explore these data
using median polish. To do that I plan to use medpolish() on the
matrix[f1,f2xf3], then medpolish on the resulting matrix[f2,f3]. This
approach is described by Cook on page 141 of Exploring Data Tables,
Trends, and Shapes.
split() gets me close to where I want to be, but results in a list,
rather than a matrix. How do I construct the matrix[f1,f2xf3] from my
data frame?
Also, any pointers to existing code that performs multi-way median
polish will be appreciated.
Sorry for the newbie-type query, but manipulating data prior to
analysis is really hard for me in R.
All the best,
Tom
Thomas S. Dye, Ph.D.
T. S. Dye Colleagues, Archaeologists, Inc.
Phone: (808) 529-0866 Fax: (808) 529-0884
http://www.tsdye.com
[[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] barplot2 x-axis
Looking at the graphic (I got it here, but it was stripped for those getting it via the server), the CI's don't really add much to the interpretation of the data. They are small enough as to be a non-issue at least in this example and there is too much data to make them useful for visually comparing the means across the bars anyway. The values along the x axis that you are getting are the result of there being too many small vertical bars in the space and that the size of the font for the bars is too large to display all of the labels. The other consideration, is that the horizontal spacing of the bars will not be based upon their numeric value on a linear scale, but their physical sequence in your data. Thus, if there are sequence gaps in the distance measures along the x axis, those gaps will not be maintained when displayed, resulting in a compromised linear scale along the axis. For example, consider: x - c(1:5, 8, 15:18) names(x) = c(1:5, 8, 15:18) barplot(x) Note that there are gaps (6:7 and 9:14) where data is not present, but the bars (5/8 and 8/15) are nevertheless next to each other. You would need to insert NA's at the missing data points to maintain the proper bar locations along the x axis. Here is what it should look like: x - c(1:5, NA, NA, 8, rep(NA, 6), 15:18) names(x) - c(1:5, NA, NA, 8, rep(NA, 6), 15:18) barplot(x, cex.names = 0.75) Contrast how barplot() handles such data by default versus plot(): x - c(1:5, 8, 15:18) plot(x, x, type = h, lwd = 10) Recognizing that there may be community standards for you in how such data (presumably animal or vegetative abundance data) should be presented, my recommendation otherwise in a vacuum would be to plot the data using points and leave out the CIs: plot(mydata.x, mydata.y, xlab = km, ylab = kg, main = plot.name, xlim = c(0, 250), xaxt = n, pch = 19) axis(1, at = seq(0, 250, 25)) or if you need bars (use type = h as above), if that is the preferred presentation format: plot(mydata.x, mydata.y, xlab = km, ylab = kg, main = plot.name, xlim = c(0, 250), xaxt = n, type = h, lwd = 5) axis(1, at = seq(0, 250, 25)) The bottom line is the key take away message that you are trying to convey, without that message being lost in too much ink. HTH, Marc On Mar 22, 2009, at 4:45 PM, Mafalda Viana wrote: Dear Mark, Thanks for your reply. I attach one of my plots as an example (e.g.Mafalda) to show what I want. If you notice in the plot the x-axis has a funny scale (0, 13, 28...) and what I would like to do is to make that scale from 0 and then increasing by 25 km for a better interpretation (0, 25, 50...). Hopefully my problem is more clear now, meanwhile I will have a look to your suggestions. Thank you very much Mafalda 2009/3/22 Marc Schwartz marc_schwa...@comcast.net On Mar 22, 2009, at 2:31 PM, Mafalda Viana wrote: Dear R users, I am trying to build a barplot2 graph however I can't find a way of defining the scale for the x-axis. I would like to show in my x-axis only the numbers 0, 25, 50, 75 etc. (so far R is giving me a random scale hard to interpret and it doens't look nice...). Could anyone advise me on how to do this please, it would be a great help! Thank you. Below I show the code I have been using for my plots. my.plot- barplot2(mydata.y, names.arg=mydata.x, width=1, xlab=km, ylab=kg, main=plot.name, plot.ci=TRUE, ci.l=data.lci,ci.u=data.uci, ci.col = red, font=40, font.lab=60, xlim=c(0,250), xpd=FALSE) It is not entirely clear what you are doing here an we cannot reproduce it lacking your data. Your x (horizontal) axis will have one tick mark below each bar, using the labels that you have indicated with mydata.x. Based upon your code, the x axis values would appear to be measures of distance. Your y (vertical) axis would appear to be a measure of weight, perhaps being some descriptive statistic (eg. mean) for weight at each distance. It is unclear what you want to do with the x axis in lieu of the default values. Modifying the tick marks on the y axis would seem to make more sense here. In general, with any R graphic, you would use the arguments 'xaxt = n' and/or 'yaxt = n' to disable the default drawing of the x and y axes, respectively. You would then call the axis() function passing the specific locations and values of the tick marks that you wish to draw. See ? par and ?axis for more information on these. BTW if you are not plotting counts or proportions/percentages, then a barplot is not the best approach for the display of summarized continuous data. I would use a point plot with error bars/confidence intervals. See the errbar() function in the Hmisc package or the plotCI() or plotmeans() functions in gplots. You could also create something manually by using plot() and then either segments() or arrows() for the CIs. HTH, Marc Schwartz -- Mafalda Viana
Re: [R] If statement generates two outputs
Hi,
I tried to create the following if / else statement but I keep getting the
error Error: unexpected '}' in size=large,center=none)
} (I have highlighted the } in bold where the error is occuring). I can't
seem to find a reason for this, does anyone know how I can fix it?
Thanks,
James
if (nostocks=3)
{data1-test[,data1stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,(nostocks+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none)
}else)
{data1-test[,1:9];data2-test[,data2stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,4)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none);
tex2-latex(data2, file=paste(i$Species[1], 2.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,((nostocks-4)+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),size=large,center=none)
}
jimdare wrote:
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
Thanks,
James
--
View this message in context:
http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22652363.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] using xyplot
Dear R-sians!
I am trying generate a bunch of xyplots
library(lattice)
myPanel - function(x,y,xl=range(x),yl=range(y),...) {
panel.xyplot(x,y,pch=20,col='blue',cex=0.7,xlim=xl,ylim=yl,...)
panel.abline(v=0, col='gray30',lty=2,lwd=1.5,...)
panel.loess(x,y,
span=2/3,family='gaussian',normalize=T,col='red',lwd=1.5,...)
# panel.lmline(x,y, col='red',lwd=1.5,...)
}
nms - names(tmp7)[c(3:7,9)]
for (i in nms) {
tmp1 - xyplot(i~pk, data=tmp7, # different y's in the loop; x is same for
all
as.table=T,panel=myPanel,main=paste('PK vs',i,'- overall'))
pdf(paste(pkvs,i,a.pdf,sep=),w=10,h=8)
print(tmp1)
dev.off()
}
The above code does not seem to work. I was not able to use sapply with a
FUN=xyplot, either.
I was, however, able to generate the plots (with xyplot) when I hard-code
the index value!
Could you please help me troubleshooting the above code? If you can also
help with an alternate code using sapply, it would be helpful!
Thanks much in advance!
Regards,
Santosh
[[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] 'require' equivalent for local functions
On 22/03/2009 5:05 PM, JiHO wrote:
Hello everyone,
I often create some local libraries of functions (.R files with only
functions in them) that I latter call. In scripts that call a function
from such library, I would like to be able to test whether the
function is already known in the namespace and, only if it is not,
source the library file. I.e. what `require` does for packages, I want
to do with my local functions.
That's pretty hard to make bulletproof. Why not just put those
functions in a package, and use that package? If the functions are all
written in R, creating the package is very easy: see package.skeleton.
(And if you have a perfect memory and don't plan to distribute the
package to anyone, you can skip documenting the functions: then it's
almost no work at all.)
Duncan Murdoch
Example:
lib.R
foo - function(x) { x*2 }
script.R
require.local(foo,lib.R)
# that searches for function foo and, if not found, executes
source(lib.R)
foo(2)
Obviously, I want the test to be quite efficient otherwise I might as
well source the local library every time. I am aware that it would
probably not be able to check for changes in lib.R (i.e. do
complicated things such as re-source lib.R if foo in the namespace and
foo in lib.R are different), but that I can handle manually.
This seems like a common enough workflow but I cannot find a pre-
existing solution. Does anyone have pointers?
Otherwise I tried to put that together:
require.local - function(fun, lib)
#
# Searches for function fun and sources lib in
# case it is not found
#
{
if (! (deparse(substitute(fun)) %in% ls(.GlobalEnv) class(fun)
== function) ) {
cat(Sourcing, lib,...\n)
source(lib)
}
}
but I am really not confident with all those deparse/substitute things
and the environment manipulation, so I guess there should be a better
way.
JiHO
---
http://jo.irisson.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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Axes in 3D plots
I am wondering if anyone knows how to change or rotate the default axes on a 3D scatterplot. I would like to change which sides of the cube the 3 axes are displayed on. Many thanks, Leah Gerber [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Axes in 3D plots
To rotate the graph you can change the viewpoint. Try: ?rgl.viewpoint
library(rgl)
x - sort(rnorm(1000))
y - rnorm(1000)
z - rnorm(1000) + atan2(x,y)
plot3d(x, y, z, col=rainbow(1000), size=2)
start - proc.time()[3]
while ((i - 36*(proc.time()[3]-start)) 360) {
rgl.viewpoint(i,i/4);
}
To define the side which the axes will be displayed, try: ?axes3d
library(rgl)
x - sort(rnorm(1000))
y - rnorm(1000)
z - rnorm(1000) + atan2(x,y)
plot3d(x, y, z, col=rainbow(1000), size=2,axes=F)
axes3d(c('x--','x+-','x++','x-+','z--','y++'))
hth,
Daniel Moreira, MD
Leah Gerber leahger...@yahoo.com
Sent by: r-help-boun...@r-project.org
03/22/2009 09:07 PM
To
r-help@r-project.org
cc
Subject
[R] Axes in 3D plots
I am wondering if anyone knows how to change or rotate the default axes on
a 3D scatterplot. I would like to change which sides of the cube the 3
axes are displayed on.
Many thanks,
Leah Gerber
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] If statement generates two outputs
You appear to have a number of unbalanced parentheses in your
statements. Here is one that seems to work:
if (nostocks=3)
{data1-test[,data1stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,(nostocks+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none))
}else
{data1-test[,1:9];data2-test[,data2stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,4)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none);
tex2-latex(data2, file=paste(i$Species[1], 2.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,((nostocks-4)+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),size=large,center=none))
}
On Sun, Mar 22, 2009 at 7:59 PM, jimdare jamesdar...@gmail.com wrote:
Hi,
I tried to create the following if / else statement but I keep getting the
error Error: unexpected '}' in size=large,center=none)
} (I have highlighted the } in bold where the error is occuring). I can't
seem to find a reason for this, does anyone know how I can fix it?
Thanks,
James
if (nostocks=3)
{data1-test[,data1stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,(nostocks+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none)
}else)
{data1-test[,1:9];data2-test[,data2stocks];
tex1-latex(data1, file=paste(i$Species[1], 1.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,4)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),
size=large,center=none);
tex2-latex(data2, file=paste(i$Species[1], 2.tex, sep=),
rowname = NULL,
cgroup = c(Fishstock, stocknames,Total),
n.cgroup = c(1, rep(2,((nostocks-4)+1)),
colheads = c(Year, rep(c(Catch, TACC),
nostocks+1)),size=large,center=none)
}
jimdare wrote:
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X5 I want to create df1 and df2:
if (X5) {df1-c(4,5,6,7,8) AND df2-c(9,10,11,12,13)}
Thanks,
James
--
View this message in context:
http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22652363.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.
[R] mvpart error
Hello all, When attempting a classification tree using mvpart, I get the following error: thesis2.mvp=mvpart(bat_sp~., data=alltrees.df) Error in all(keep) : unused argument(s) (c(TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE)) bat_sp is an unordered categorical variable, and the predictors are a mixture of categorical and continuous variables. I had run the algorithm successfully in the past (about 1.5 years ago) on a preliminary subset of the same dataset using a previous version of R The current system is R 2.8.1 running under Ubuntu Jaunty. Any ideas as to the cause? Josh Stumpf Eastern Michigan University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with 'boot'
Hi, I'm wanting to test for a difference in medians between 2 groups using resampling methods. I found the boot package, but don't really understand how to write the 'statistic' function required as the 2nd argument for the bootstrap test. Thanks if you can help, Paul -- View this message in context: http://www.nabble.com/Help-with-%27boot%27-tp22653487p22653487.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] newton method
Hi R-users, Does R has a topic on newton's method? Thank you for the info. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting closest values
bill.venab...@csiro.au wrote: Here is one way you might find useful id - 1:183 lo - sample((-22):144, 183, rep = TRUE) ds - outer(lo, lo, function(x,y) abs(x-y)) rs - t(apply(ds, 2, function(x) id[order(x)][2:11])) rownames(rs) - id Here's what I got head(rs) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] 1 118 23 125 182 1838 86 119 141 175 2 27 167 1764 126 45 113 81 144 153 3 437 18 60 148 11 12 150 85 163 4 126 27 167 1762 45 144 153 4855 5 15 46 86 141 175 23 125 1831 118 6 58 78 139 13 66 131 156 181 117 157 If you want anything more random in the case of ties than this, you will have to work a bit harder on the code. W. Bill Venables http://www.cmis.csiro.au/bill.venables/ -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of P_M Sent: Monday, 23 March 2009 1:32 AM To: r-help@r-project.org Subject: [R] Selecting closest values Hi I have a table with ID (1 to 183) and Location (144 to -22). My problem is that I want to select the 10 ID's that are closest in Location to ID 1, ID 2 and so on. Also, some ID have the same Location. Say, if 11 ID's are closest to ID 100 I want to randomly choose one of the ID's to select 10 ID's total. Thank you Also take a look at the whichClosest function in the Hmisc package. But whichClosest cheats by calling Fortran. Cheers Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] newton method
Take a look at the functionsnlm(), optim() in the stats package and maxNR() in the maxLik package. On Mar 22, 2009, at 11:15 PM, Roslina Zakaria wrote: Does R has a topic on newton's method? _ Professor Michael Kubovy University of Virginia Department of Psychology Postal Address: P.O.Box 400400, Charlottesville, VA 22904-4400 Express Parcels Address: Gilmer Hall, Room 102, McCormick Road, Charlottesville, VA 22903 Office:B011;Phone: +1-434-982-4729 Lab:B019; Phone: +1-434-982-4751 WWW:http://www.people.virginia.edu/~mk9y/ Skype name: polyurinsane [[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] 'require' equivalent for local functions
I agree with Duncan. I used to do exactly what you did - source()ing data files inside a wrapper not unlike C #define wrappers, but it became a headache with more files and the files began looking more cluttered. It has taken me several days to learn about how create a package properly, along with package RUnit for unit-testing, and with documentation. The R Extensions file is often a good source of information. Be sure you find information about Rcmd install and Rcmd check, which are also very useful. prompt() can help you build your .Rd (help files). Alternatively, you may use Rdoc$compile() (from package R.oo) if you intend to embed your Rdoc-style comments inside your R code, as I do. I also use R.oo as a more traditional object-oriented alternative to S3/S4. Once set-up, you can automagically generate .pdf files and .chm (windows-based help) for your package. Help for my own package has helped me keep my code consistent, clean, and re-factorable. Best of all, you can use put require( my.package ) or data( my.data) and voila. It has been a bit of a learning curve, but the packaging facilities in R are actually very well developed. Once set-up, maintenance becomes less of a chore. Good luck. Duncan Murdoch-2 wrote: On 22/03/2009 5:05 PM, JiHO wrote: Hello everyone, I often create some local libraries of functions (.R files with only functions in them) that I latter call. In scripts that call a function from such library, I would like to be able to test whether the function is already known in the namespace and, only if it is not, source the library file. I.e. what `require` does for packages, I want to do with my local functions. That's pretty hard to make bulletproof. Why not just put those functions in a package, and use that package? If the functions are all written in R, creating the package is very easy: see package.skeleton. (And if you have a perfect memory and don't plan to distribute the package to anyone, you can skip documenting the functions: then it's almost no work at all.) -- View this message in context: http://www.nabble.com/%27require%27-equivalent-for-local-functions-tp22650626p22653884.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.
