Re: [R] Frequencies from a matrix - spider from frequencies
Jim Lemon wrote: Yes, I realized that I had forgotten to require(plotrix) after I sent the message. From your example, you might also want to look at the diamondplot function, also in plotrix. Jim, thanks for the hint to diamondplot. It is much closer natively to what I wanted to do, and simple to use. Hats off!: Just entering the data frame produces a quick print. However, it fails to make sense w.r.t. units and values here. I use the example data and code given in ?diamondplot: data(mtcars) mysubset-mtcars[substr(dimnames(mtcars)[[1]],1,1)==M,c(mpg,hp,wt,disp)] diamondplot(mysubset) and get a plot (I think I can't attach it here?), with, e.g. hp and disp crossing the Maserati radial axis at 17, wt at 15 and mpg at 10. The actual data row, though, is mpg hpwt disp Maserati Bora 15.0 335 3.570 301.0 Looking closer, the plot seems to arbitrarily scale all values (columns) to a(n arbitrary?) maximum of '17'. And when I print my data (submitted earlier), the same happens: all responses are scaled to 17 as the highest in each category. From that point of view, diamondplot is not that useful. How can I force it not to scale arbitrarily, but print the actual numbers? Uwe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] different forms of nls recommendations
emorway wrote: So I wanted to try a different equation of the general form a/(b+c*x^d) US.nls.2-nls(US.final.values$ECe~(a/(b+c*US.final.values$WTD^d)),data=US.final.values,start=list(a=100.81,b=73.7299,c=0.0565,d=-6.043),trace=TRUE,algorithm=port) but that ended with Convergence failure: false convergence (8). I tried relaxing the convergence You want 4 parameters from a set of data that has a very large variance. Try to fix d and c to a reasonable constant; it probably will converge. From looking at your data and the variances, I would suggest to log-transform first. If you feel bad about it, you can always fit the original data later after you have found a good model. Dieter -- View this message in context: http://n4.nabble.com/different-forms-of-nls-recommendations-tp1676330p1676547.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] Converting a character string into a data frame name and performing assignments to that data frame
Kavitha Venkatesan-2 wrote: variable.df is a character string that contains the name of the data frame that I want to do the following operations on: variable.df - data.frame(); # I can do the above command using assign( variable.df, data.frame() ) How can I perform the assignment statements below ? colnames(variable.df) = colnames(some.other.df) Try to avoid assign and eval when you are not fiRm; the first is rarely used, and the second is a bit to powerful for safeR. data(iris) head(iris) names(iris) = c(A,B,C,D,Spec) head(iris) Dieter -- View this message in context: http://n4.nabble.com/Converting-a-character-string-into-a-data-frame-name-and-performing-assignments-to-that-data-frame-tp1676236p1676552.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] using a condition given as string in subset function - how?
Hi, Is eval always used in conjunction with parse? Based on other languages, I'd expect the expression already to work without the use of parse(), but indeed it doesn't, or at least not as intended. Just a newbie question.. Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ --- On Sun, 3/21/10, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] using a condition given as string in subset function - how? To: Mark Heckmann mark.heckm...@gmx.de Cc: r-help@r-project.org Date: Sunday, March 21, 2010, 2:33 AM I know that if you have to resort to 'parse(text=...)', you should look for another way (it is a 'fortune'), but it is getting late, and at least it works: eval(parse(text=subset(df, A==1 B==1)))  A B 1 1 1 On Sat, Mar 20, 2010 at 9:09 PM, Mark Heckmann mark.heckm...@gmx.de wrote: df - data.frame(A=c(1,2), B=c(1,1)) I have a string containing a condition for a subset function, like: conditionAsString - paste(names(df), df[1,], sep===, collapse= )  conditionAsString  A==1 B==1 Now I want to use this string in the subset call, like subset(df, conditionAsString) I do not exactly now how to combine substitute, expression, parse and so on to get what I want, which would be: subset(df, A==1 B==1) but using the string conditionAsString. Thanks, Mark âââââââââââââââââââââââââââââââââââââââ Mark Heckmann Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.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.htmlhttp://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?    [[alternative HTML version deleted]] -Inline Attachment Follows- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Find a rectangle of maximal area
For an application in image processing -- using R for statistical purposes -- I need to solve the following task: Given n (e.g. n = 100 or 200) points in the unit square, more or less randomly distributed. Find a rectangle of maximal area within the square that does not contain any of these points in its interior. If a, b are height and width of the rectangel, other constraints may have to be imposed such as a, b = 0.5 and/or 0.5 = a/b = 2.0 . The rectangle is allowed to touch the border of the square. For each new image the points will be identified by the application, like all stars of a certain brightness on an astronomical picture. So the task will have to be performed several times. I assume this problem is computationally hard. I would like to find a solution that is reasonably fast for n = 100..200 points. Exhaustive search along the x, y coordinates of the points will not be fast enough. I know this request is not about R syntax and does not contain 'repro-code'. But perhaps a somehow similar problem has been solved before. Thanks in advance for any suggestions, Hans Werner P.S.: Example Is this rectangle of maximal area? n - 100; set.seed(832) x - runif(n); y - runif(n) plot(c(0,1), c(0,1), type=n, axes=FALSE, asp=1, xlab=, ylab=, main=Rectangle Problem, sub=) rect(0,0,1,1, col=darkblue) xl-100; yu-43; xr-40; yo-3 area - (x[xr]-x[xl])*(y[yo]-y[yu]) rect(x[xl], y[yu], x[xr], y[yo], lty=2, col=darkblue, border=red) text((x[xl]+x[xr])/2, (y[yu]+y[yo])/2, paste(area =, round(area, 4)), cex=0.75, col=red) points(x, y, pch=20, col=white) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Levene's Test for Homogeneity of Variance
Hi, All! To calculate Levene's Test for Homogeneity of Variance I use R Commander, and this is the output: levene.test(Dataset$age, Dataset$sex) Levene's Test for Homogeneity of Variance Df F value Pr(F) group 1 0.8739 0.3567 33 I am not sure what means Pr(F)? Can anyone explain/translate this? Regards, Iurie Malai Department of Psychology and Special Education Moldova Pedagogical State University iurie.ma...@gmail.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] Levene's Test for Homogeneity of Variance
On 21 March 2010 13:13, Iurie Malai iurie.ma...@gmail.com wrote: To calculate Levene's Test for Homogeneity of Variance I use R Commander, and this is the output: levene.test(Dataset$age, Dataset$sex) Levene's Test for Homogeneity of Variance Df F value Pr(F) group 1 0.8739 0.3567 33 I am not sure what means Pr(F)? Can anyone explain/translate this? this is shorthand for: 1 - pf(0.8739, 1 ,33) [1] 0.3567 see: ?pf Philippe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Levene's Test for Homogeneity of Variance
Dear Iurie, Pr(F) is the p-value for the test of the null hypotheses that the population variances are equal. This is the typical format for labelling a p-value in R output. I hope this helps, John John Fox Senator William McMaster Professor of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Iurie Malai Sent: March-21-10 8:14 AM To: r-help@r-project.org Subject: [R] Levene's Test for Homogeneity of Variance Hi, All! To calculate Levene's Test for Homogeneity of Variance I use R Commander, and this is the output: levene.test(Dataset$age, Dataset$sex) Levene's Test for Homogeneity of Variance Df F value Pr(F) group 1 0.8739 0.3567 33 I am not sure what means Pr(F)? Can anyone explain/translate this? Regards, Iurie Malai Department of Psychology and Special Education Moldova Pedagogical State University iurie.ma...@gmail.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] Problem specifying Gamma distribution in lme4/glmer
Matthew Giovanni matthewgiovanni at gmail.com writes: Dear R and lme4 users- I am trying to fit a mixed-effects model, with the glmer function in lme4, to right-skewed, zero-inflated, non-normal data representing understory grass and forb biomass (continuous) as a function of tree density (indicated by leaf-area). Thus, I have tried to specify a Gamma distribution with a log-link function but consistently receive an error as follows: total=glmer(total~gla4+(1|plot)+ (1|year/month),data=veg,family=Gamma(link=log)) summary(total) Error in asMethod(object) : matrix is not symmetric [1,2] I have also tried fitting glmm's with lme4 and glmer to other Gamma-distributed data but receive the same error. Has anyone had similar problems and found any solutions? 1. probably best to post questions like this to r-sig-mixed-mod...@r-project.org 2. haven't seen this particular problem. Can you please provide a reproducible example (post your data, or a small subset of your data, or a simulated example that displays the same problem), and give the results of the sessionInfo() function? f - factor(rep(1:10,each=10)) x - runif(100) dat - data.frame(x,f) library(lme4) [snip messages] g1 - glmer(x~1+(1|f),data=dat,family=Gamma(link=log)) Generalized linear mixed model fit by the Laplace approximation [...] summary(g1) [ works fine] sessionInfo() R version 2.10.1 (2009-12-14) i486-pc-linux-gnu [snip] other attached packages: [1] lme4_0.999375-32-2 Matrix_0.999375-38 lattice_0.18-3 3. zero-inflated data may not be particularly well-represented by a Gamma distribution: if you actually have a significant number of exactly-zero values, you may want to analyze your data in two stages, first as a presence-absence problem and then as a conditional density (i.e., what is the distribution of the non-zero values)? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem specifying Gamma distribution in lme4/glmer
Ben Bolker wrote: 3. zero-inflated data may not be particularly well-represented by a Gamma distribution: if you actually have a significant number of exactly-zero values, you may want to analyze your data in two stages, first as a presence-absence problem and then as a conditional density (i.e., what is the distribution of the non-zero values)? Interesting idea. Do you know of a example where this was done (independent of lmer)? We have similar data, were people are either symptom free (50% with score 0), and the rest is smoothly distributed. Dieter -- View this message in context: http://n4.nabble.com/Problem-specifying-Gamma-distribution-in-lme4-glmer-tp1676344p1676746.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] EM algorithm in R
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of tj
Sent: Saturday, March 20, 2010 10:35 AM
To: r-help@r-project.org
Subject: Re: [R] EM algorithm in R
here is my program... Im trying to fit 1 component to 6 components in each
of
the 300 generated samples. Each sample has size=200. For each of the 300
generated samples and for each modeled component (v=1,2,3,4,5,6), I will
get
estimates of the parameters that will maximize the likelihood, AND then I
will determine when AIC obtained its minimum value - - is it in the 1
component or 2 components or .. 6 components? Then I will count the
times that AIC has its minimum at component=2. I will do the same for BIC
and AIC.
MY program right now:
h=5
tol=0.0001
size=200
B=300
mean1=-0.8
mean2=0.8
q=c(0.25,0.75,0,0,0,0) #i will use this later for the qth
quantile
mu=0
a1=0; a2=0; a3=0; a4=0; a5=0; a6=0
mu1s=0 ; mu2s=0 ; mu3s=0 ; mu4s=0 ; mu5s=0 ; mu6s=0
as1=rep(NA,B) ; as2=rep(NA,B); as3=rep(NA,B); as4=rep(NA,B);
as5=rep(NA,B);
as6=rep(NA,B)
vs=0
itrs=0
AIC=0
BIC=0
CAIC=0
Aicno=0
Bicno=0
Caicno=0
mylog=function(y,mu1,mu2,mu3,mu4,mu5,mu6,var,a1,a2,a3,a4,a5,a6){
sum(log(
(a1/sqrt(2*pi*var))*(exp(-((y-mu1)^2)/(2*var))) +
(a2/sqrt(2*pi*var))*(exp(-((y-mu2)^2)/(2*var))) +
(a3/sqrt(2*pi*var))*(exp(-((y-mu3)^2)/(2*var))) +
(a4/sqrt(2*pi*var))*(exp(-((y-mu4)^2)/(2*var))) +
(a5/sqrt(2*pi*var))*(exp(-((y-mu5)^2)/(2*var))) +
(a6/sqrt(2*pi*var))*(exp(-((y-mu6)^2)/(2*var)))
),na.rm=TRUE)
}
set.seed(h)
for (j in 1:B){#generating my sample
t=rbinom(1,200,0.5)
y1=rnorm(mean=mean1,sd=1,n=t)
y2=rnorm(mean=mean2,sd=1,n=200-t)
y=c(y1,y2)
var=var(y)
mu1=as.numeric(quantile(y,q[1])) # setting my starting values for the
means
mu2=as.numeric(quantile(y,q[2]))
mu3=as.numeric(quantile(y,q[3]))
mu4=as.numeric(quantile(y,q[4]))
mu5=as.numeric(quantile(y,q[5]))
mu6=as.numeric(quantile(y,q[6]))
for (v in 1:6)
{
for (i in 2:600){#my maximum number of
iteration
to get the maximum likelihood estimates for my parameters is 600
a1[v]=ifelse(1v,NA,1/v)#here, i set my starting values
for
proportions as 1/v, where v is the number of components. Example, When
the
number
a2[v]=ifelse(2v,NA,1/v) #of components is 2, the program
will only give two starting values for proportions. I'm having a problem
when the
a3[v]=ifelse(3v,NA,1/v) #program gives NA... there are
problems encountered in the preceding procedures.
a4[v]=ifelse(4v,NA,1/v)
a5[v]=ifelse(5v,NA,1/v)
a6[v]=ifelse(6v,NA,1/v)
ms=
c(a1[i-1]*dnorm(y,mu1[i-1],sqrt(var[i-1])),a2[i-1]*dnorm(y,mu2[i-
1],sqrt(var[i-1])),
a3[i-1]*dnorm(y,mu3[i-1],sqrt(var[i-1])),
a4[i-1]*dnorm(y,mu4[i-1],sqrt(var[i-1])),
a5[i-1]*dnorm(y,mu5[i-1],sqrt(var[i-1])),
a6[i-1]*dnorm(y,mu6[i-1],sqrt(var[i-1])))
m=as.numeric(na.omit(ms))
B = m/sum(m)
B1 = B[1:size]
B2 = B[size+1:size*2]
B3 = B[size*2+1:size*3]
B4 = B[size*3+1:size*4]
B5 = B[size*4+1:size*5]
B6 = B[size*5+1:size*6]
mu1[i]=sum(B1*y)/sum(B1)
mu2[i]=sum(B2*y)/sum(B2)
mu3[i]=sum(B3*y)/sum(B3)
mu4[i]=sum(B4*y)/sum(B4)
mu5[i]=sum(B5*y)/sum(B5)
mu6[i]=sum(B6*y)/sum(B6)
a1[i]=sum(B1)/size
a2[i]=sum(B2)/size
a3[i]=sum(B3)/size
a4[i]=sum(B4)/size
a5[i]=sum(B5)/size
a6[i]=sum(B6)/size
var[i]=sum((B1*(y-mu1[i])^2+ B2*(y-mu2[i])^2 + B3*(y-mu3[i])^2+
B4*(y-mu4[i])^2 + B5*(y-mu5[i])^2 + B6*(y-mu6[i])^2),na.rm=TRUE)/size
if(abs(mylog(y,mu1[i],mu2[i],mu3[i],mu4[i],mu5[i],mu6[i],var[i],
a1[i],a2[i],a3[i],a4[i],a5[i],a6[i])-
mylog(y,mu1[i-1],mu2[i-1],mu3[i-1],mu4[i-1],mu5[i-1],mu6[i-1],var[i-1],
a1[i-1],a2[i-1],a3[i-1],a4[i-1],a5[i-1],a6[i-1]))
=tol)
break()
}
f[j]=mylog(y,mu1s[j],mu2s[j],mu3s[j],mu4s[j],mu5s[j],mu6s[j],vs[j],
as1[j],as2[j],as3[j],as4[j],as5[j],as6[j])
AIC[v]=f[j]-v
BIC[v]=f[j]-(v/2)*log(size)
CAIC[v]=f[j]-(v/2)*log(size)-(v/2)
if (AIC[v] AIC[v+1]) # i need to record the value of v with minimum AIC
if (BIC[v] BIC[v+1]) # i need to record the value of v with minimum AIC
if (CAIC[v] CAIC[v+1]) #i need to record the value of v with minimum AIC
}
Aicno[j]=v#storage for the value of v when minimum AIC was obtained
Bicno[j]=v#storage for the value of v when minimum AIC was obtained
Caicno[j]=v #storage for the value of v when minimum AIC was obtained
as1[j]=a1[i]
as2[j]=a2[i]
as3[j]=a3[i]
as4[j]=a4[i]
as5[j]=a5[i]
as6[j]=a6[i]
mu1s[j]=mu1[i]
mu2s[j]=mu2[i]
mu3s[j]=mu3[i]
mu4s[j]=mu4[i]
mu5s[j]=mu5[i]
mu6s[j]=mu6[i]
vs[j]=var[i]
}
Aicno# so for this, I wish to obtain 300 values of k, where k can
be
equal to 1,2,3,4,5,6, corresponding to the component with lowest AIC
Bicno# so for this, I wish to obtain 300 values of k, where k can
be
equal to 1,2,3,4,5,6, corresponding to the
[R] calculus using R
can anyone suggest how can i do calculus (e.g.limit,differentiation,integrate,mean value theorem,definite integral,convergence,maxima minima of functions,checking continuity) using R?? -- View this message in context: http://n4.nabble.com/calculus-using-R-tp1676727p1676727.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] Applying SVM model to a new data
Hi I'm using Libsvm, I made a function to save models, confusion matrix of different models but when I want to use saved model, I can't let suppose output of function is : models=list(model1,model2,) but how can I use these models? using predict(models[[1]],y) causes error message: Error in UseMethod(predict) : no applicable method for 'predict' applied to an object of class list thanks Regards -- View this message in context: http://n4.nabble.com/Applying-SVM-model-to-a-new-data-tp1676864p1676864.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] sapply, lattice functions
Dear R-gurus..
How do I implement the following:
a) Overlay frequency(instead of density) with line of density plot, vertical
lines of confidence intervals and reference levels?
b) Control the breaks (using nint?), order of the panel, and the layout,
place units for each conditioning variable?
Is there more efficient way than lines provided below? Please feel free to
suggest other ways of displaying the above information.
library(reshape)
# The data frame below is an example of a subset of results from bootstrap
runs. From the results of bootstrap, I would like to # display
frequency histogram with overlay of density,confidence intervals, and point
estimates.
aa - data.frame(id=seq(20),a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
aa1 - rbind(aa,c( -99, -0.02,1.09, 0.23)) # record of reference values for
each distribution, set id = -99
ab - melt(aa1,measure=names(aa1)[-match(id,names(aa1)])
uns - unlist(list(id=,a1=mL/h,b1=L,c1=L))
uns1 - data.frame(var=names(uns),uni=(uns))
ab1 - merge(ab,uns1,by=variable,all.x=T)
ab2 - ab1[order(ab1$variable,ab1$id),]
histogram(~value|paste(as.character(variable),
(,uni,),sep=),breaks=NULL,nint=10,ab2,layout=c(2,2),as.table=T,type=density,scales=list(relation='free'),
panel=function(x,lqp=c(0.025,0.975),...) { # lqp indicate
confidence intervals
x1 - x[-1]
vs - x[1]
panel.histogram(x1,col='light blue',...)
panel.densityplot(x1,col.line='blue',lwd=1.75,...)
panel.abline(v=c(quantile(as.vector(x1),prob=lqp,na.rm = T),vs),
col=c(blue,blue,dark green),lwd=2,lty=c(2,2,1))},
strip=strip.custom( strip.names=F, strip.levels=T)
)
Thanks so much for your help!
Santosh
On Sat, Mar 20, 2010 at 10:56 AM, Sundar Dorai-Raj sdorai...@gmail.comwrote:
You're right. It's necessary for xyplot though to prevent grouping.
On Mar 20, 2010 10:43 AM, Dieter Menne dieter.me...@menne-biomed.de
wrote:
Sundar Dorai-Raj-2 wrote:
Or perhaps more clearly,
histogram(~a1 + b1 + c1, data = aa, o...
Why outer=TRUE? Looks same for me without:
Dieter
library(lattice)
aa - data.frame(a1=rnorm(20),b1=rnorm(20,0.8),c1=rnorm(20,0.5))
histogram(~a1 + b1 + c1, data = aa)
--
View this message in context:
http://n4.nabble.com/sapply-lattice-functions-tp1618134p1676043.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] calculus using R
On Mar 21, 2010, at 11:26 AM, ATANU wrote: can anyone suggest how can i do calculus (e.g.limit,differentiation,integrate,mean value theorem,definite integral,convergence,maxima minima of functions,checking continuity) using R?? ?D ?integrate ?pmax ?pmin package:RYacas Continuity? Limits? R is of necessity being run on discrete machines and R is not a symbolic algebra machine. -- David. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using a condition given as string in subset function - how?
On Sun, 21 Mar 2010, Albert-Jan Roskam wrote: Hi, Is eval always used in conjunction with parse? Based on other languages, I'd expect the expression already to work without the use of parse(), but indeed it doesn't, or at least not as intended. Just a newbie question.. parse() turns a character string into R code, and eval() evaluates R code. parse() usually requires eval(), because there's rarely much point parsing a string if you aren't going to evaluate it, but eval() usually doesn't require parse(). That is, most uses of eval() are not on strings but on expressions. In your case it is not clear whether someone else hands you a condition in string form, in which care parse() is necessary, or whether you build it up your self, in which case parse() is not necessary, and eval() may not be necessary. -thomas Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ --- On Sun, 3/21/10, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] using a condition given as string in subset function - how? To: Mark Heckmann mark.heckm...@gmx.de Cc: r-help@r-project.org Date: Sunday, March 21, 2010, 2:33 AM I know that if you have to resort to 'parse(text=...)', you should look for another way (it is a 'fortune'), but it is getting late, and at least it works: eval(parse(text=subset(df, A==1 B==1))) ?? A B 1 1 1 On Sat, Mar 20, 2010 at 9:09 PM, Mark Heckmann mark.heckm...@gmx.de wrote: df - data.frame(A=c(1,2), B=c(1,1)) I have a string containing a condition for a subset function, like: conditionAsString - paste(names(df), df[1,], sep===, collapse= ) ?? conditionAsString ?? A==1 B==1 Now I want to use this string in the subset call, like subset(df, conditionAsString) I do not exactly now how to combine substitute, expression, parse and so on to get what I want, which would be: subset(df, A==1 B==1) but using the string conditionAsString. Thanks, Mark ? Mark Heckmann Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.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.htmlhttp://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? ?? [[alternative HTML version deleted]] -Inline Attachment Follows- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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]] 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] calculus using R
Try this: library(Ryacas) # http://ryacas.googlecode.com Loading required package: XML x - Sym(x) h - Sym(h) # limit Limit(1/(1-x), x, Infinity) [1] Starting Yacas! expression(0) # differentiation deriv(x^3, x) expression(3 * x^2) # integration: indefinite and definite Integrate(x^3, x) expression(x^4/4) Integrate(x^3, x, 0, 1) expression(1/4) # mean value Limit((cos(x+h)-cos(x))/h, h, 0) expression(-sin(x)) # min/max Solve(deriv(x^2-x, x) == 0, x) expression(list(x == 1/2)) # continuity Limit(sin(x+h^2) - sin(x-h^2), h, 0) expression(0) On Sun, Mar 21, 2010 at 11:26 AM, ATANU ata.s...@gmail.com wrote: can anyone suggest how can i do calculus (e.g.limit,differentiation,integrate,mean value theorem,definite integral,convergence,maxima minima of functions,checking continuity) using R?? -- View this message in context: http://n4.nabble.com/calculus-using-R-tp1676727p1676727.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multiple logical comparisons
Hi, I would like to compare a column of data with a vector. I have this data.frame for example; x - data.frame(A = c(1:5), B = c(1,1,2,2,2)) Now I have a search vector: search - c(1,3,5) when I now try to get all the data-rows which have a 1, a 3, or a 5 in column A and a 2 in column B, I tried this: x[x$B == 2 x$A == search,] I hoped to get 3 2 5 2 as output. But I get an empty set. Can someone help me on that? _ search - c(1,3,5) x - data.frame(A = c(1:5), B = c(1,1,2,2,2)) x[x$B == 2 x$A == search,] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem specifying Gamma distribution in lme4/glmer
Dieter Menne dieter.menne at menne-biomed.de writes: Ben Bolker wrote: 3. zero-inflated data may not be particularly well-represented by a Gamma distribution: if you actually have a significant number of exactly-zero values, you may want to analyze your data in two stages, first as a presence-absence problem and then as a conditional density (i.e., what is the distribution of the non-zero values)? [...] Do you know of a example where this was done (independent of lmer)? [...] Nothing springs to mind, but it seems sensible. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using a condition given as string in subset function - how?
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Albert-Jan Roskam Sent: Sunday, March 21, 2010 2:25 AM To: Mark Heckmann; jim holtman Cc: r-help@r-project.org Subject: Re: [R] using a condition given as string in subset function - how? Hi, Is eval always used in conjunction with parse? Based on other languages, I'd expect the expression already to work without the use of parse(), but indeed it doesn't, or at least not as intended. Just a newbie question.. When eval is given a parse tree (e.g., the output of parse() or call()) it recursively ascends the tree, effectively calling eval() on the subtrees. E.g., parse(text='paste(No, 1)')[[1]] returns a call object of length 3, the elements being the name object `paste`, the character object No, and the numeric object 1. Evaluating a name object returns its value and evaluating a character or numeric object returns the object (unchanged). I.e., in this case eval(`paste`) - function(..., sep= , collapse=NULL) eval(No) - No eval(1) - 1 The root of the parse tree is the call object and evaluating that means applying the function given by the first argument to the rest of the argumeents. If eval(No) meant the same as eval(parse(text=No)) then this scheme would break down, since you wouldn't be able to distinguish between character objects and name objects, so it would not be possible to say you wanted the string No or the value of the object called No. You might say that eval(String) should do one thing when called directly, whatever that means, and another when called by a recursive call to eval(), but I think functions should act the same no matter who calls them. There are also times when you want the raw parse tree, as when processing model formulae like log(response)~groupId/predictor or trellis/lattice formulae like log(y)~predictor|groupId. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ --- On Sun, 3/21/10, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] using a condition given as string in subset function - how? To: Mark Heckmann mark.heckm...@gmx.de Cc: r-help@r-project.org Date: Sunday, March 21, 2010, 2:33 AM I know that if you have to resort to 'parse(text=...)', you should look for another way (it is a 'fortune'), but it is getting late, and at least it works: eval(parse(text=subset(df, A==1 B==1)))  A B 1 1 1 On Sat, Mar 20, 2010 at 9:09 PM, Mark Heckmann mark.heckm...@gmx.de wrote: df - data.frame(A=c(1,2), B=c(1,1)) I have a string containing a condition for a subset function, like: conditionAsString - paste(names(df), df[1,], sep===, collapse= )  conditionAsString  A==1 B==1 Now I want to use this string in the subset call, like subset(df, conditionAsString) I do not exactly now how to combine substitute, expression, parse and so on to get what I want, which would be: subset(df, A==1 B==1) but using the string conditionAsString. Thanks, Mark ––––––––––––––––––––†“–––––––––––––––––– Mark Heckmann Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.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.htmlhttp://www.r-proje ct.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?    [[alternative HTML version deleted]] -Inline Attachment Follows- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] multiple logical comparisons
Martin Batholdy wrote: Hi, I would like to compare a column of data with a vector. I have this data.frame for example; x - data.frame(A = c(1:5), B = c(1,1,2,2,2)) Now I have a search vector: search - c(1,3,5) when I now try to get all the data-rows which have a 1, a 3, or a 5 in column A and a 2 in column B, I tried this: x[x$B == 2 x$A == search,] I hoped to get 3 2 5 2 as output. See ?%in% x[x$B == 2 x$A %in% search, ] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem specifying Gamma distribution in lme4/glmer
On Mar 21, 2010, at 5:16 PM, Ben Bolker wrote: Dieter Menne dieter.menne at menne-biomed.de writes: Ben Bolker wrote: 3. zero-inflated data may not be particularly well-represented by a Gamma distribution: if you actually have a significant number of exactly-zero values, you may want to analyze your data in two stages, first as a presence-absence problem and then as a conditional density (i.e., what is the distribution of the non-zero values)? [...] Do you know of a example where this was done (independent of lmer)? [...] Nothing springs to mind, but it seems sensible. I thought this was what hurdle and ZIF models were supposed to handle gracefully? -- David. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] different forms of nls recommendations
On Sat, 2010-03-20 at 14:55 -0800, emorway wrote:
Hello,
Using this data:
http://n4.nabble.com/file/n1676330/US_Final_Values.txt US_Final_Values.txt
and the following code i got the image at the end of this message:
US.final.values-read.table(c:/tmp/US_Final_Values.txt,header=T,sep= )
US.nls.1-nls(US.final.values$ECe~a*US.final.values$WTD^b+c,data=US.final.values,start=list(a=2.75,b=-0.95,c=0.731),trace=TRUE)
f.US1-function(x){coef(US.nls.1)[a]*x^coef(US.nls.1)[b]+coef(US.nls.1)[c]}
xvals.US1-seq(min(US.final.values$WTD),max(US.final.values$WTD),length.out=75)
yvals.US1-f.US1(xvals.US1)
Rsq.nls.1-sum((predict(US.nls.1)-mean(US.final.values$ECe))^2/sum((US.final.values$ECe-mean(US.final.values$ECe))^2))
plot(US.final.values$WTD,US.final.values$ECe,col=red,pch=19,cex=.75)
lines(xvals.US1,yvals.US1,col=blue)
but the r^2 wasn't so hot.
Rsq.nls.1
[1] 0.2377306
So I wanted to try a different equation of the general form a/(b+c*x^d)
US.nls.2-nls(US.final.values$ECe~(a/(b+c*US.final.values$WTD^d)),data=US.final.values,start=list(a=100.81,b=73.7299,c=0.0565,d=-6.043),trace=TRUE,algorithm=port)
but that ended with Convergence failure: false convergence (8). I tried
Hi emorway,
Do you have 657 obs and 4 parameters to fit.
In my opinion you have few obs...
I think do you fit in steps:
US.nls.2-nls(ECe~(a/(b + c *
WTD^d)),data=US.final.values,start=list(a=100.81,b=73.7299,c=0.0565,d=-6.043),trace=TRUE,algorithm=port)
temp_nls1 - nls(ECe~(100/(73 + .05 *
WTD^d)),data=US.final.values,start=list(d=-6.043),trace=TRUE,algorithm=port)
temp_nls2 - nls(ECe~(100/(73 + .05 *
WTD^d)),data=US.final.values,start=list(d=-1.01613),trace=TRUE,algorithm=port)
temp_nls3 - nls(ECe~(100/(73 + c *
WTD^(-1.01613))),data=US.final.values,start=list(c=0.05),trace=TRUE,algorithm=port)
temp_nls4 - nls(ECe~(100/(73 + c *
WTD^(-1.01613))),data=US.final.values,start=list(c=-14.7127),trace=TRUE,algorithm=port)
temp_nls5 - nls(ECe~(100/(b-14.7127 *
WTD^(-1.01613))),data=US.final.values,start=list(b=73),trace=TRUE,algorithm=port)
temp_nls6 - nls(ECe~(100/(b-14.7127 *
WTD^(-1.01613))),data=US.final.values,start=list(b=70.4936),trace=TRUE,algorithm=port)
temp_nls7 - nls(ECe~(a/(70.4936-14.7127 *
WTD^(-1.01613))),data=US.final.values,start=list(a=100),trace=TRUE,algorithm=port)
0: 2243.9898: 100.000
1: 2122.8218: 106.219
2: 1359.8819: 187.530
3: 1359.8819: 187.530
--
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil
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Re: [R] multiple logical comparisons
thanks! Now I have one more question; How can I do the reverse? when %in% is == (for two vectors of different lengths); what is the equivalent to != ? On 21.03.2010, at 22:33, Erik Iverson wrote: Martin Batholdy wrote: Hi, I would like to compare a column of data with a vector. I have this data.frame for example; x - data.frame(A = c(1:5), B = c(1,1,2,2,2)) Now I have a search vector: search - c(1,3,5) when I now try to get all the data-rows which have a 1, a 3, or a 5 in column A and a 2 in column B, I tried this: x[x$B == 2 x$A == search,] I hoped to get 3 2 5 2 as output. See ?%in% x[x$B == 2 x$A %in% search, ] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multiple logical comparisons
Martin Batholdy wrote: thanks! Now I have one more question; How can I do the reverse? when %in% is == (for two vectors of different lengths); what is the equivalent to != ? a %in% b returns a logical vector, so !a %in% b returns its negation. See order of precedence in ?Syntax. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice grob
What's wrong with using grid.grabExpr?
p1 - xyplot(1:10 ~ 1:10)
g1 - grid.grabExpr(print(p1))
I can imagine there would be potential problems to do with the
plot-time aspect and layout calculations...
On 19 March 2010 21:51, baptiste auguie baptiste.aug...@googlemail.com wrote:
Dear list,
I'm trying to arrange various grid objects on a page using a
frameGrob. It works fine with basic grobs (textGrob, gTree, etc.), and
also with ggplot2 objects using the ggplotGrob() function. I am
however stuck with lattice. As far as I understand, lattice produces a
list of class trellis, which is eventually displayed using the
plot.trellis method. I am not sure if/how one can convert this list
into a high-level grob. I tried the following,
latticeGrob - function(p, ...){
grob(p=p, ..., cl=lattice)
}
drawDetails.lattice - function(x, recording=FALSE){
lattice:::plot.trellis(x$p)
}
p1 - xyplot(1:10 ~ 1:10)
g1 - latticeGrob(p1)
grid.draw(g1) # works fine
but,
fg - frameGrob(layout = grid.layout(1,1))
fg - placeGrob(fg, g1, row = 1, col = 1)
grid.draw(fg)
Error in UseMethod(depth) :
no applicable method for 'depth' applied to an object of class NULL
Ideas are most welcome,
Best regards,
baptiste
sessionInfo()
R version 2.10.1 RC (2009-12-06 r50690)
i386-apple-darwin9.8.0
locale:
[1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] grid tools stats graphics grDevices utils
datasets methods base
other attached packages:
[1] ggplot2_0.8.7 digest_0.4.1 reshape_0.8.3 plyr_0.1.9
proto_0.3-8 gridExtra_0.5 lattice_0.17-26 gtools_2.6.1
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--
Felix Andrews / 安福立
Postdoctoral Fellow
Integrated Catchment Assessment and Management (iCAM) Centre
Fenner School of Environment and Society [Bldg 48a]
The Australian National University
Canberra ACT 0200 Australia
M: +61 410 400 963
T: + 61 2 6125 4670
E: felix.andr...@anu.edu.au
CRICOS Provider No. 00120C
--
http://www.neurofractal.org/felix/
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[R] using lmer weights argument to represent heteroskedasticity
Hi-
I want to fit a model with crossed random effects and heteroskedastic
level-1 errors where inferences about fixed effects are of primary
interest. The dimension of the random effects is making the model
computationally prohibitive using lme() where I could model the
heteroskedasticity with the weights argument. I am aware that the weights
argument to lmer() cannot be used to estimate a heteroskedastic variance
function with unknown parameters. However my data situation is such that I
am able to assign each unit uniquely to one of four groups and I am willing
to treat the relative error variances of the groups as known. I.e. I would
be able to specify the model I want to fit in lme() using a weights
statement of the form weights = varIdent(form = ~1 | group, fixed =
knownSDratios). I am trying to figure out how I can concoct a weights
argument for lmer() that will fit this same model. In particular, since the
reported inferences about fixed effects parameters are not invariant to how
the weights passed to lmer() are scaled, I am trying to understand how to
construct a weights argument that not only contains the information about
the relative precisions of the errors in the different groups, but also
provides valid inferences about the fixed effects. In a simple example
below with balanced, nested data, scaling the weights to sum to the number
of cases in the data makes lmer() and lme() agree on the inferences about
the fixed effects. However, at the end of the example, it is also clear
that this correspondence does not hold when the data are not balanced. My
question boils down to this: is there a general way to scale lmer()'s
weights argument that will cause it to always correspond to what lme()
would report about fixed effects inferences when passed the same fixed
relative precision information?
Thanks for any advice.
J.R.
###
## EXAMPLE: generate balanced nested data with heteroskedastic errors of
## variance 0.5, 1, 2, or 4
###
ngroup- 50
npergroup - 20
n - ngroup*npergroup
set.seed(5046)
d - data.frame(unitid = 1:n, groupid = rep(1:ngroup,
each=npergroup), verror = sample(c(0.5, 1, 2, 4), size=n, replace=T),
x = 0.1*rnorm(n))
groupeffx - data.frame(groupid = 1:ngroup, theta = rnorm(ngroup, sd = 0.25))
d - merge(d, groupeffx)
d$etrue - rnorm(n, sd = sqrt(d$verror))
d$y - 5 + d$x + d$theta + d$etrue
d$verrorf - factor(paste(v,d$verror,sep=))
print(tapply(d$etrue, d$verrorf, var))
## function to collect pieces from lme() output
sumLME - function(o){
tab - summary(o)$tTable
r - c(tab[1,1:2], tab[2,1:2], as.numeric(VarCorr(o))[3:4])
names(r) - c(b0,b0sd,b1,b1sd,sdgroup,sdresid)
return(r)
}
## function to collect pieces from lmer() output
sumLMER - function(o){
tab- summary(o)@coefs
r - c(tab[1,1:2], tab[2,1:2],
as.numeric(attributes(VarCorr(o)$groupid)$stddev),
as.numeric(attr(VarCorr(o), sc)))
names(r) - c(b0,b0sd,b1,b1sd,sdgroup,sdresid)
return(r)
}
res - vector(0, mode=list)
library(nlme)
## lme, ignoring heteroskedasticity
res[[lme.nohet]] - sumLME( lme(fixed = y ~ x, data = d, random = ~1
| groupid) )
## lme, heteroskedastic model with variance weights estimated
res[[lme.esthet]] - sumLME( lme(fixed = y ~ x, data = d, random =
~1 | groupid, weights = varIdent(form = ~1|verrorf)) )
## lme, heteroskedastic model with true fixed weights
v - c(v0.5 = 0.5, v1 = 1, v2 = 2, v4 = 4)
sdrats - sqrt(v / v[1])[-1]
res[[lme.fixedhet]] - sumLME( lme(fixed = y ~ x, data = d, random =
~1 | groupid, weights = varIdent(form = ~1|verrorf, fixed = sdrats)) )
detach(package:nlme)
library(lme4)
## lmer, ignoring heteroskedasticity
res[[lmer.nohet]] - sumLMER( lmer(y ~ x + (1|groupid), data = d) )
## essentially matches res[[lme.nohet]], makes sense
## lmer, with fixed weights equal to known precisions
d$w1 - 1 / d$verror
res[[lmer.w1]] - sumLMER( lmer(y ~ x + (1|groupid), data = d, weights = w1) )
## only b0 and b1 matches res[[lme.fixedhet]]
## lmer, with fixed weights proportional to known precisions but weights sum to
## number of records as they would with the unweighted case.
d$w2 - nrow(d) * d$w1 / sum(d$w1)
res[[lmer.w2]] - sumLMER( lmer(y ~ x + (1|groupid), data = d, weights = w2) )
## this is extremely close to res[[lme.fixedhet]] except for sdresid
do.call(rbind, res)
## is the case that matched only because of balance?
dsub - d[sample(1:nrow(d), size= nrow(d)/2, replace=F),]
dsub$w2 - nrow(dsub) * dsub$w1 / sum(dsub$w1)
sumLMER( lmer(y ~ x + (1|groupid), data = dsub, weights = w2) )
detach(package:lme4)
library(nlme)
sumLME( lme(fixed = y ~ x, data = dsub, random = ~1 | groupid, weights
= varIdent(form = ~1|verrorf, fixed = sdrats)) )
## these no longer match
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[R] add euro sign to a plot
hi, Is it possible to add special characters like the euro sign to a plot? thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add euro sign to a plot
Hi Martin, here as example code that may help you to get what you want (once you didn't specified were in the plot you want the symbol). plot(1:10) title(main= \u20ac, font = 5) Ps.: This coding may differ between operational system, this one worked in a Windows Vista 32 bits. This works for all (I guess) the special character present in the windows Character Map, just copy the code and use as \SymbolCode. Hope it helps. Rodrigo. 2010/3/21 Martin Batholdy batho...@googlemail.com hi, Is it possible to add special characters like the euro sign to a plot? thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] calling external .EXE file in R macOSX
Hi All,
I am currently working on an analysis which requires a call to an
external FORTRAN routine contained within a file called MCDS.EXE. This
file is usually called from within a WINDOWS program called DISTANCE. I
have some R script from the developers of the original software which
apparently makes a call to this .EXE file which i need to re-engineer
slightly, but i am having difficulty interpreting how the script makes
this call and passes to the EXE file the relevant input command file and
data file for processing. I am also unsure if making such a call will
work within the OSX environment for a (windows?) executable file. Below
is an example from the script which is supposed to make the call. Any
suggestions would be much appreciated.
Cheers
A
cds - function (key, adj, L, w, A=NA, xi, zi,
file.base,ext.files=F,bootstrap=F) {
#Purpose: Driver function to run the mcds.exe engine from R
#Updated by Tiago on 19/12/2004 so that it can take an external data and
command input files
#Usefull if you want to do a non-standard bootstrap analysis of some
data analised in distance
#and for wich distance is not capable of producing variance estimates
#This way, you can call this function inside a bootstrap procedure, as
long as you update the
#data file at each resample
#Inputs:
# key- vector string containing key functions
# adj- vector string containing adjustment terms
# L - total survey effort
# A - survey area
# xi - vector of perpendicular distances
# zi - vector of cluster sizes
# file.base - if specified and engine is cds, then the cds input and
# output files are written into the current directory, with the
# cds.file.base as a prefix and '.txt' as a suffix. E.g., setting
# cds.file.base to 'cds' produces 'cds.cmd.txt', 'cds.data.txt',
# 'cds.log.txt','cds.stat.txt','cds.plot.txt' and 'cds.boot.txt'
# If not specified, these files are created in a temp location and
# are deleted at the end.
# ext.files - If true, then the funtion uses the external files
'file.base'+'.cms.'+txt' as command file
# and 'file.base'+'.data.'+.txt' as data file. Typicaly these
files would be the result of
# runing Distance in debug mode, and should be placed in the
working directory for R.
# By default ext.files is false, so the function looks for the
files produced by functions
# 'create.data.file' and 'create.command.file'
# You need to change the input comand file in order for the
mcds engine to produce the files
# that the function 'read.stats.file' expects, and that means
that inside the command file you should define
# file names with prefix = file.base
# bootstrap - If true, procedure is being called inside a bootstrap
routine, and intermediate files
# are deleted at each loop step, except the command file
#Outputs: list, containing
# Nhat.grp, Nhat.ind, mu, nL, Es, LnL, AIC, status
# Note - status is an integer:
# 1=OK, 2=warnings, 3=errors, 4=file errors, 5=some other problem
(e.g., program crash)
run.cds-function(cmd.file.name) {
#Purpose: runs the MCDS.exe engine and waits for it to finish
# *Note* that mcds.exe needs to be in the working directory, or in the
# PATH windows environment
# variable for this to work, as it makes no attempt to find the
# location of the file
#Inputs:
# cmd.file.name - name of the command file to run
#Returns:
# A status integer - 1=OK, 2=warnings, 3=errors, 4=file errors, 5=some
other problem (e.g., program crash)
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[R] fixed effects regression
Hi All: I am trying to move a model from Stata to R. It is a linear regression model with about 90,000 indicator variables. What is the best approach to follow in R? - Roy [[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] Find a rectangle of maximal area
On 03/21/2010 10:12 PM, Hans W Borchers wrote: For an application in image processing -- using R for statistical purposes -- I need to solve the following task: Given n (e.g. n = 100 or 200) points in the unit square, more or less randomly distributed. Find a rectangle of maximal area within the square that does not contain any of these points in its interior. If a, b are height and width of the rectangel, other constraints may have to be imposed such as a, b= 0.5 and/or 0.5= a/b= 2.0 . The rectangle is allowed to touch the border of the square. For each new image the points will be identified by the application, like all stars of a certain brightness on an astronomical picture. So the task will have to be performed several times. I assume this problem is computationally hard. I would like to find a solution that is reasonably fast for n = 100..200 points. Exhaustive search along the x, y coordinates of the points will not be fast enough. I know this request is not about R syntax and does not contain 'repro-code'. But perhaps a somehow similar problem has been solved before. Hi Hans, The emptyspace function in the plotrix package is a fairly crude implementation of this, but the code might give you some ideas. 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] calling external .EXE file in R macOSX
Alex Anderson wrote:
Hi All,
I am currently working on an analysis which requires a call to an
external FORTRAN routine contained within a file called MCDS.EXE. This
file is usually called from within a WINDOWS program called DISTANCE. I
have some R script from the developers of the original software which
apparently makes a call to this .EXE file which i need to re-engineer
slightly, but i am having difficulty interpreting how the script makes
this call and passes to the EXE file the relevant input command file and
data file for processing. I am also unsure if making such a call will
work within the OSX environment for a (windows?) executable file. Below
is an example from the script which is supposed to make the call. Any
suggestions would be much appreciated.
Cheers
A
{snip}
Hrm, I really can't tell what that function you posted does-- it seems to be
mostly comments. But I would imagine MCDS.EXE is being run using something
like a call to the system() function.
The problem with running the program on OS X is that MCDS.EXE is a Windows
binary. You can't just run a binary compiled for Winows on OS X or Linux
more than you could run a binary compiled for OS X or Linux on Windows.
Your options include:
1. Get the Fortran source code, recompile MCDS on OS X , and run the
program using something like system('mcds')
2. Get the Fortran source code, refactor and recompile MCDS as a shared
library, load directly into R, and run the routines using .Fortran()
3. Try using an emulator, such as Wine (www.winehq.ort), and run the
program using something like system('wine MCDS.EXE')
Number three would require the least amount of hassle, but is not guaranteed
to work. Other people may have better suggestions and you might try asking
on the mac-specific R mailing list:
https://stat.ethz.ch/mailman/listinfo/r-sig-mac
The people there may be able to give you better answers concerning R and OS
X
Good Luck!
-Charlie
-
Charlie Sharpsteen
Undergraduate-- Environmental Resources Engineering
Humboldt State University
--
View this message in context:
http://n4.nabble.com/calling-external-EXE-file-in-R-macOSX-tp1677148p1677162.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] multiple logical comparisons
Just read the help page: ?%in% %w/o% - function(x,y) x[!x %in% y] #-- x without y (1:10) %w/o% c(3,7,12) -- David. On Mar 21, 2010, at 5:57 PM, Martin Batholdy wrote: thanks! Now I have one more question; How can I do the reverse? when %in% is == (for two vectors of different lengths); what is the equivalent to != ? On 21.03.2010, at 22:33, Erik Iverson wrote: Martin Batholdy wrote: Hi, I would like to compare a column of data with a vector. I have this data.frame for example; x - data.frame(A = c(1:5), B = c(1,1,2,2,2)) Now I have a search vector: search - c(1,3,5) when I now try to get all the data-rows which have a 1, a 3, or a 5 in column A and a 2 in column B, I tried this: x[x$B == 2 x$A == search,] I hoped to get 3 2 5 2 as output. See ?%in% x[x$B == 2 x$A %in% search, ] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Problem using ESS. Error message: cannot change working directory
Greetings R wizards. I have the following problem using Emacs Speaks Statistics: When trying to set a working directory, or when trying to read a file that is in a directory that has the character á (note that it has a spanish accent symbol: ´ ) R being ran from emacs tells me that the directory doesn't exist, or that it can't be found, or: setwd(c:/Documents and Settings/Nicolás) Error in setwd(c:/Documents and Settings/Nicolás) : cannot change working directory I do NOT have the same problem if I try to do this directly with R GUI. In my opinion, emacs is telling R Nicolás when it should tell it Nicolás. Any thoughts?? Thanks for your help. Thanks for R, and greetings to the hole community of users, developers, etc., Nicolás Leveroni __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem specifying Gamma distribution in lme4/glmer
On Mar 21, 2010, at 5:16 PM, Ben Bolker wrote: Dieter Menne dieter.menne at menne-biomed.de writes: Ben Bolker wrote: 3. zero-inflated data may not be particularly well-represented by a Gamma distribution: if you actually have a significant number of exactly-zero values, you may want to analyze your data in two stages, first as a presence-absence problem and then as a conditional density (i.e., what is the distribution of the non-zero values)? [...] Do you know of a example where this was done (independent of lmer)? [...] Nothing springs to mind, but it seems sensible. I thought this was what hurdle and ZIF models were supposed to handle gracefully? -- David. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice grob
Hi
baptiste auguie wrote:
Dear list,
I'm trying to arrange various grid objects on a page using a
frameGrob. It works fine with basic grobs (textGrob, gTree, etc.), and
also with ggplot2 objects using the ggplotGrob() function. I am
however stuck with lattice. As far as I understand, lattice produces a
list of class trellis, which is eventually displayed using the
plot.trellis method. I am not sure if/how one can convert this list
into a high-level grob. I tried the following,
latticeGrob - function(p, ...){
grob(p=p, ..., cl=lattice)
}
drawDetails.lattice - function(x, recording=FALSE){
lattice:::plot.trellis(x$p)
Try ...
lattice:::plot.trellis(x$p, newpage=FALSE)
Paul
}
p1 - xyplot(1:10 ~ 1:10)
g1 - latticeGrob(p1)
grid.draw(g1) # works fine
but,
fg - frameGrob(layout = grid.layout(1,1))
fg - placeGrob(fg, g1, row = 1, col = 1)
grid.draw(fg)
Error in UseMethod(depth) :
no applicable method for 'depth' applied to an object of class NULL
Ideas are most welcome,
Best regards,
baptiste
sessionInfo()
R version 2.10.1 RC (2009-12-06 r50690)
i386-apple-darwin9.8.0
locale:
[1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] grid tools stats graphics grDevices utils
datasets methods base
other attached packages:
[1] ggplot2_0.8.7 digest_0.4.1reshape_0.8.3 plyr_0.1.9
proto_0.3-8 gridExtra_0.5 lattice_0.17-26 gtools_2.6.1
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--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
p...@stat.auckland.ac.nz
http://www.stat.auckland.ac.nz/~paul/
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Re: [R] Find a rectangle of maximal area
Hans W Borchers hwborchers at googlemail.com writes: For an application in image processing -- using R for statistical purposes -- I need to solve the following task: Given n (e.g. n = 100 or 200) points in the unit square, more or less randomly distributed. Find a rectangle of maximal area within the square that does not contain any of these points in its interior. If a, b are height and width of the rectangel, other constraints may have to be imposed such as a, b = 0.5 and/or 0.5 = a/b = 2.0 . The rectangle is allowed to touch the border of the square. For each new image the points will be identified by the application, like all stars of a certain brightness on an astronomical picture. So the task will have to be performed several times. I assume this problem is computationally hard. I would like to find a solution that is reasonably fast for n = 100..200 points. Exhaustive search along the x, y coordinates of the points will not be fast enough. I know this request is not about R syntax and does not contain 'repro-code'. But perhaps a somehow similar problem has been solved before. Thanks in advance for any suggestions, Hans Werner I solved this with a simple minded MINLP formulation using BARON (a global solver). This seems to produce solutions relatively quickly (somewhat slower for n=200). Actually this solved easier than I expected. See: http://yetanothermathprogrammingconsultant.blogspot.com/2010/03/looks-difficult- to-me-2.html Erwin Kalvelagen Amsterdam Optimization Modeling Group er...@amsterdamoptimization.com http://amsterdamoptimization.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.
