[R] quantmod's addTA plotting functions
Hi,
I'm having trouble with quantmod's addTA plotting functions. They seem to
work fine when run from the command line. But when run inside a function,
only the last one run is visible. Here's an example.
test.addTA - function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close - GSPC[,GSPC.Close]
GSPC.EMA.3 - EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10 - EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'), up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10 appears on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser() neither appears
when the browser() interrupt occurs. I can then draw both GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All intended plots are
visible after the function terminates. So it isn't as if one wipes out the
other. This shows that it's possible to get all three lines on the plot, but
I can't figure out how to do it without manual intervention. Any suggestions
are appreciated.
Thanks.
*-- Russ *
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[R] R - problem with my loops which operate on raster data
library(rgdal)
my_asc=dir(~/Pulpit/dods/karol/TVDI
113_121,pattern=.asc,recursive=T,full.names=T)
for (i in 1:length(my_asc))
{
r - readGDAL(my_asc[i])
z - as.matrix(r)
vectordata[i] - mean(z)
vectordatamax[i] - max(z)
vectordatamin[i] - min(z)
vectordev[i] - sd(z,na.rm=True)
hist(z)
png(filename=hist+tostring(i)+.png)
}
liI try to do some modyfication of this loop, but it still doesn't works -
which fragment is incorrect?
liI would like also to use more complicated pattern (to list only files
which contains on the end of it name two numbers), but adding something
like:
pattern=_??.asc seems not works.
liI would like to add one more loop to get list of folders (instead of
manually inserting directories into my_asc variable), but I haven't got Idea
how I can do it?
liI do not know, why my way of creating vectors for mean, max, min and
standard deviation values is not working...
--
View this message in context:
http://r.789695.n4.nabble.com/R-problem-with-my-loops-which-operate-on-raster-data-tp3497533p3497533.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.
[R] memory and bootstrapping
hello, the following questions will without doubt reveal some fundamental ignorance, but hopefully you can still help me out. I'd like to bootstrap a coefficient gained on the basis of the coefficients in a logistic regression model (the mean differences in the predicted probabilities between two groups, where each predict() operation uses as the newdata-argument a dataframe of equal size as the original dataframe).I've got 130,000 rows and 7 columns in my dataframe. The glm-model uses all variables (as well as two 2-way interactions). System: - R-version: 2.12.2 - OS: Windows XP Pro, 32-bit - 3.16Ghz intel dual core processor, 2.9GB RAM I'm using the boot package to arrive at the standard errors for this difference, but even with only 10 replications, this takes quite a long time: 216 seconds (perhaps this is partly also due to my inefficiently programmed function underlying the boot-call, I'm also looking into that). I wanted to try out calculating a bca-bootstrapped confidence interval, which as I understand requires a lot more replications than normal-theory intervals. Drawing on John Fox' Appendix to his An R Companion to Applied Regression, I was thinking of trying out 2000 replications -- but this will take several hours to compute on my system (which isn't in itself a major issue though). My Questions: - let's say I try bootstrapping with 2000 replications. Can I be certain that the memory available to R will be sufficient for this operation? - (this relates to statistics more generally): is it a good idea in your opinion to try bca-bootstrapping, or can it be assumed that a normal theory confidence interval will be a sufficiently good approximation (letting me get away with, say, 500 replications)? Best, Esther __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lapply, if statement and concatenating to a list
Hi R users
I was wondering on how to use lapply co when the applied function has a
conditional statement and the output is a 'growing' object.
See example below:
list1 - list('A','B','C')
list2 - c()
myfun - function(x,list2)
{
one_elem - x
cat('one_elem= ', one_elem, '\n')
random - sample(1:2,1)
show(random)
if(random==2)
{
list2 - c(list2,one_elem)
}else{
list2
}
}
lapply(list1,myfun,list2)
Is there a way to get rid of the 'NULL' elements in the output (when there is
any?), without using a for loop?
Thanks for your help
Lorenzo
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[R] Extraction of columns from two matrices
Hi, I have two matrices with 4885 cols and 36 cols respectively . I would like to extract these 36 columns from the bigger matrix, the column values are different but the column names would be the same.Hence based on the names I would like to perform my operation. So is there any way of extracting this info in an easy way. I have tried names,which,grep,subset none have worked. Could someone help me out here Thanks a ton Nisha [[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] combine lattice plot and standard R plot
Thank you very much for your help and for your quick answers. Lucía Cañás Ferreiro Instituto Español de Oceanografía Centro Oceanográfico de A coruña Paseo Marítimo Alcalde Francisco Vázquez, 10 15001 - A Coruña, Spain Tel: +34 981 218151 Fax: +34 981 229077 lucia.ca...@co.ieo.es http://www.ieo.es [[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] Panels order in lattice graphs
May 04, 2011; 5:50pm Cristina Silva wrote: In lattice graphs, panels are drawn from left to right and bottom to top. The flag as.table=TRUE changes to left to right and top to bottom. Is there any way to change to first top to bottom and then left to right? didn´t find anything neither in Help pages nor Lattice book. Cristina, You have not fully explained your problem. An approach I use for difficult-to-get-right arrangements is the following. Say you have two conditioning variables (13 panels in all) and you want to place the last panel on the top row in the first position on the bottom row, but leave everything else the same, then easiest is the following: ## Note: T.xyplot$index.cond is a __list__, so you need to use [[ T.xyplot - xyplot(Prop ~ OM | interaction(Treatment, Aspect, drop = TRUE), data = myDat) print(T.xyplot) T.xyplot$index.cond [[1]] [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 T.xyplot$index.cond[[1]] - c(13, 1:12) print(T.xyplot) Hope this helps to solve your problem. Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Panels-order-in-lattice-graphs-tp3496022p3497691.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: join tables in R
Thanks a lot to Steve Lianoglou and Peter Savicky for their help! Alfredo -Messaggio originale- Da: Steve Lianoglou [mailto:mailinglist.honey...@gmail.com] I'd to match-merge 2 tables in such a manner that I keep all the rows in table 1, but not the rows that are in both table 1 and 2. master - data.frame(ID=2001:2011) train - data.frame(ID=2004:2006) valid - ??? in this example table valid should have the following str(valid) Year: int 2001 2002 2003 2007 2008 2009 2010 2011 Are you working with only one column at a time? If so: keep - !(master$ID %in% train$ID) valid - master[keep,] -Messaggio originale- Da: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Per conto di Petr Savicky Try the following, which assumes that train is a subset of master. master - data.frame(ID=2001:2011) train - data.frame(ID=2004:2006) valid - master[! (master[, 1] %in% train[ ,1]), , drop=FALSE] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lapply, if statement and concatenating to a list
Hi Lorenzo,
On Thu, May 5, 2011 at 8:38 AM, Lorenzo Cattarino l.cattar...@uq.edu.au wrote:
Hi R users
I was wondering on how to use lapply co when the applied function has a
conditional statement and the output is a 'growing' object.
See example below:
list1 - list('A','B','C')
list2 - c()
myfun - function(x,list2)
{
one_elem - x
cat('one_elem= ', one_elem, '\n')
random - sample(1:2,1)
show(random)
if(random==2)
{
list2 - c(list2,one_elem)
}else{
list2
}
}
lapply(list1,myfun,list2)
Is there a way to get rid of the 'NULL' elements in the output (when there is
any?), without using a for loop?
I don't understand what your example is trying to do and which object
you expect to be growing. list2 ain't growin', and it's not changing
(i.e., it remains NULL) in your code. Perhaps you meant to have a -
there; this would make your list2 growing, if you really want it to,
but in general, that's a bad idea. Lapply goes best with the
functional style where everything your function does is computing and
returning a value but here you're (if I get your intentions correctly)
counting on side effects. If you like side effects, a for (or while)
loop may be more logical choice.
Getting rid of the NULL elements is simple. One way is:
foo - lapply(list1, yourfun)
foo[!sapply(foo, is.null)]
Regards,
Kenn
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Re: [R] memory and bootstrapping
The only reason the boot package will take more memory for 2000 replications than 10 is that it needs to store the results. That is not to say that on a 32-bit OS the fragmentation will not get worse, but that is unlikely to be a significant factor. As for the methodology: 'boot' is support software for a book, so please consult it (and not secondary sources). From your brief description it looks to me as if you should be using studentized CIs. 130,000 cases is a lot, and running the experiment on a 1% sample may well show that asymptotic CIs are good enough. On Thu, 5 May 2011, E Hofstadler wrote: hello, the following questions will without doubt reveal some fundamental ignorance, but hopefully you can still help me out. I'd like to bootstrap a coefficient gained on the basis of the coefficients in a logistic regression model (the mean differences in the predicted probabilities between two groups, where each predict() operation uses as the newdata-argument a dataframe of equal size as the original dataframe).I've got 130,000 rows and 7 columns in my dataframe. The glm-model uses all variables (as well as two 2-way interactions). System: - R-version: 2.12.2 - OS: Windows XP Pro, 32-bit - 3.16Ghz intel dual core processor, 2.9GB RAM I'm using the boot package to arrive at the standard errors for this difference, but even with only 10 replications, this takes quite a long time: 216 seconds (perhaps this is partly also due to my inefficiently programmed function underlying the boot-call, I'm also looking into that). I wanted to try out calculating a bca-bootstrapped confidence interval, which as I understand requires a lot more replications than normal-theory intervals. Drawing on John Fox' Appendix to his An R Companion to Applied Regression, I was thinking of trying out 2000 replications -- but this will take several hours to compute on my system (which isn't in itself a major issue though). My Questions: - let's say I try bootstrapping with 2000 replications. Can I be certain that the memory available to R will be sufficient for this operation? - (this relates to statistics more generally): is it a good idea in your opinion to try bca-bootstrapping, or can it be assumed that a normal theory confidence interval will be a sufficiently good approximation (letting me get away with, say, 500 replications)? Best, Esther __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Str info. Thanks for helping
Hi William! So... many things to say First, your first steps are completely unnecessary. Just do this: data1 - read.csv(file.choose(), header=TRUE) data1$IND - factor(data1$IND) M - manova(as.matrix(data1[,-1])~data1$IND, data=data1) Though I don't understand why this doesn't work: M - manova(.~IND, data=data1) Error in eval(expr, envir, enclos) : object '.' not found Second, this doesn't work either: S-summary(M, test=Pillai) Error in summary.manova(M, test = Pillai) : residuals have rank 11 12 And I don't know what that means since I never do manova, but it's not really where you're stuck, I guess Third, where you're stuck. Is this what you want? SA-summary.aov(M) out - lapply(seq_along(SA), FUN=function(x) unlist(SA[[x]][1,])) mat - do.call(rbind, out) row.names(mat) - names(SA) Fourth, I wouldn't use as.vector(). I'm no R expert, but I have the impression that it is useless in this case. It probably has its use in some situations, but I think as.matrix(), as.data.frame() and so on would be more useful to you. And last, reply to the list as well!! Since attached files are not transferred, copy the output from dput(data1) into the email, like this: data1 - structure(list(IND = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, ... Dep1, Dep2, Dep3, Dep4, Dep5, Dep6, Dep7, Dep8, Dep9, Dep10, Dep11, Dep12), row.names = c(NA, -133L), class = data.frame) ## copy the whole thing, right!! HTH, Ivan Le 5/5/2011 02:41, Reith, William [USA] a écrit : CSV file is attached. I am doing a manova test. Followed by a summary.aov to determine which of the 12 dependent variables are significant and warrant further testing. There is only one independent variable which is categorical with 4 factors. R Code: data1- read.csv(file link, header=TRUE) as.matrix(data1) X-data1[,1] Dep1-data1[,2] Dep2-data1[,3] Dep3-data1[,4] Dep4-data1[,5] Dep5-data1[,6] Dep6-data1[,7] Dep7-data1[,8] Dep8-data1[,9] Dep9-data1[,10] Dep10-data1[,11] Dep11-data1[,12] Dep12-data1[,13] Y-cbind(Dep1, Dep2, Dep3, Dep4, Dep5, Dep6, Dep7, Dep8, Dep9, Dep10, Dep11, Dep12) M- manova(Y ~ as.factor(X), data=data1) S-summary(M, test=Pillai) S1-as.vector(S$stats[1,]) L-mat.or.vec(12,3) SA-summary.aov(M) #Stops working here. I want to save the numbers from each Depi test as a matrix or vector. S1 above works for the manova test, but I don't know how to reference my values for summary.aov SA1-SA[ Response Dep1] Thank you so much for any help you can give, William -Original Message- From: Ivan Calandra [mailto:ivan.calan...@uni-hamburg.de] Sent: Wednesday, May 04, 2011 4:38 PM To: Reith, William [USA] Cc: r-help@r-project.org Subject: Re: Str info. Thanks for helping It looks from str(SA) that Response IPS1 is a data.frame of class anova, which probably cannot be coerced to vector. Maybe you can use unlist() instead of as.vector() Or something like SA[[Response IPS1]][as.factor(WSD),] ## to select the first row only, even maybe with unlist() Without a better REPRODUCIBLE example, I cannot tell more (maybe some others can, that's why I reply to the list) HTH, Ivan Le 4 mai 2011 à 20:17, reith_will...@bah.com a écrit : I am still waiting for this to get posted so I thought I would email it to you. SA gives the output: Response IPS1 : Df Sum Sq Mean Sq F value Pr(F) as.factor(WSD) 3 3.3136 1.10455 23.047 5.19e-12 *** Residuals 129 6.1823 0.04793 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 . . . There are 11 more just like this output. Just increment IPS1 to IPS2, etc. Goal: save 3 3.3136 1.10455 23.047 5.19e-12 as a vector. Str(SA) gives the output: str(SA) str(SA) List of 12 $ Response IPS1 :Classes 'anova' and 'data.frame': 2 obs. of 5 variables: ..$ Df : num [1:2] 3 129 ..$ Sum Sq : num [1:2] 3.31 6.18 ..$ Mean Sq: num [1:2] 1.1045 0.0479 ..$ F value: num [1:2] 23 NA ..$ Pr(F) : num [1:2] 5.19e-12 NA There are several more but they are just repeats of this one only with IPS2, IPS3,... The command: SA1-as.vector(SA$Reponse IPS1) Returns NULL As do several variations I have tried. Any ideas. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Institut und Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and
[R] Options for print()
Hello everyone, I have a few questions about the print() fn: 1) I have the following code that does not center the character string: print(The TITLE,quote=FALSE,justify=center) 2) How can I get R to not print the leading [1], etc. when using print()? (Sorry, I don't know what the leading [1] is called. I tried looking it up in An Introduction, but could not find it). Any help is greatly appreciated! Dan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] combine lattice plot and standard R plot
On May 04, 2011 at 8:26pm Lucia Cañas wrote: I would like to combine lattice plot (xyplot) and standard R plot (plot and plotCI) in an unique figure. Hi Lucia, Combining the two systems can be done. See: Paul Murrell. Integrating grid graphics output with base graphics output. R News, 3(2):7-12, October 2003 http://cran.r-project.org/doc/Rnews/Rnews_2003-2.pdf Hope this helps. Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/combine-lattice-plot-and-standard-R-plot-tp3496409p3497717.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] Options for print()
Hi, Try this, cat(format(The TITLE, width=80, justify=centre)) HTH, baptiste On 5 May 2011 19:28, Dan Abner dan.abne...@gmail.com wrote: Hello everyone, I have a few questions about the print() fn: 1) I have the following code that does not center the character string: print(The TITLE,quote=FALSE,justify=center) 2) How can I get R to not print the leading [1], etc. when using print()? (Sorry, I don't know what the leading [1] is called. I tried looking it up in An Introduction, but could not find it). Any help is greatly appreciated! Dan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Remove all whitespaces
Hi I got a string that looks something like this 1 2 3 4 5 6 7 8 9 ... and I want it to be 123456789... So I want to remove all spaces (or whitespaces) from my string. Anyone know a good way of doing this? //Joel -- View this message in context: http://r.789695.n4.nabble.com/Remove-all-whitespaces-tp3497867p3497867.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] R - problem with my loops which operate on raster data
Stackoverflow users have attempted to help you improve your question -
please don't ignore advice this with an undeclared cross-post here:
http://stackoverflow.com/questions/5892548/problem-with-loops-in-r
For this list
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.
Cheers, Mike.
On Thu, May 5, 2011 at 2:21 PM, mateokkk vkedz...@is.pw.edu.pl wrote:
library(rgdal)
my_asc=dir(~/Pulpit/dods/karol/TVDI
113_121,pattern=.asc,recursive=T,full.names=T)
for (i in 1:length(my_asc))
{
r - readGDAL(my_asc[i])
z - as.matrix(r)
vectordata[i] - mean(z)
vectordatamax[i] - max(z)
vectordatamin[i] - min(z)
vectordev[i] - sd(z,na.rm=True)
hist(z)
png(filename=hist+tostring(i)+.png)
}
liI try to do some modyfication of this loop, but it still doesn't works
-
which fragment is incorrect?
liI would like also to use more complicated pattern (to list only files
which contains on the end of it name two numbers), but adding something
like:
pattern=_??.asc seems not works.
liI would like to add one more loop to get list of folders (instead of
manually inserting directories into my_asc variable), but I haven't got
Idea
how I can do it?
liI do not know, why my way of creating vectors for mean, max, min and
standard deviation values is not working...
--
View this message in context:
http://r.789695.n4.nabble.com/R-problem-with-my-loops-which-operate-on-raster-data-tp3497533p3497533.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Michael Sumner
Institute for Marine and Antarctic Studies, University of Tasmania
Hobart, Australia
e-mail: mdsum...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple General Statistics and R question (with 3 line example) - get z value from pairwise.wilcox.test
On 4 May 2011 15:32, peter dalgaard pda...@gmail.com wrote: On May 4, 2011, at 15:11 , JP wrote: Peter thanks for the fantastically simple and understandable explanation... To sum it up... to find the z values of a number of pairwise wilcox tests do the following: # pairwise tests with bonferroni correction x - pairwise.wilcox.test(a, b, alternative=two.sided, p.adj=bonferroni, exact=F, paired=T) You probably don't want the bonferroni correction there. Rather p.adj=none. You generally correct the p values for multiple testing, not the test statistics. Oh, I see thanks... of course since I have 5 groups (samples) and 10 comparisons I still have to correct when quoting p values... (My sentiment would be to pick apart the stats:::wilcox.test.default function and clone the computation of Z from it, but presumably backtracking from the p value is a useful expedient.) Should this be so onerous for the user [read non-statistician] ? # what is the data structure we got back is.matrix(x$p.value) # p vals x$p.value # z.scores for each z.score - qnorm(x$p.value / 2) Hmm, you're not actually getting a signed z out of this, you might want to try alternative=greater and drop the division by 2 inside qnorm(). (If the signs come out inverted, I meant less not greater...) But I need a two sided test (changing the alternative would change the hypothesis!)... do I still do this? All my z values are negative Is this correct? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Remove all whitespaces
Hi Joel, Try this: x - 1 2 3 4 5 6 7 8 9 gsub( , , x) Ivan Le 5/5/2011 10:50, Joel a écrit : Hi I got a string that looks something like this 1 2 3 4 5 6 7 8 9 ... and I want it to be 123456789... So I want to remove all spaces (or whitespaces) from my string. Anyone know a good way of doing this? //Joel -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Options for print()
Also take a look at sprintf, which offers everything the C-equivalent has. sprintf returns a string which can be sent to the console via cat. Andreas baptiste auguie schrieb: Hi, Try this, cat(format(The TITLE, width=80, justify=centre)) HTH, baptiste On 5 May 2011 19:28, Dan Abner dan.abne...@gmail.com wrote: Hello everyone, I have a few questions about the print() fn: 1) I have the following code that does not center the character string: print(The TITLE,quote=FALSE,justify=center) 2) How can I get R to not print the leading [1], etc. when using print()? (Sorry, I don't know what the leading [1] is called. I tried looking it up in An Introduction, but could not find it). Any help is greatly appreciated! Dan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Andreas Borg Medizinische Informatik UNIVERSITÄTSMEDIZIN der Johannes Gutenberg-Universität Institut für Medizinische Biometrie, Epidemiologie und Informatik Obere Zahlbacher Straße 69, 55131 Mainz www.imbei.uni-mainz.de Telefon +49 (0) 6131 175062 E-Mail: b...@imbei.uni-mainz.de Diese E-Mail enthält vertrauliche und/oder rechtlich geschützte Informationen. Wenn Sie nicht der richtige Adressat sind oder diese E-Mail irrtümlich erhalten haben, informieren Sie bitte sofort den Absender und löschen Sie diese Mail. Das unerlaubte Kopieren sowie die unbefugte Weitergabe dieser Mail und der darin enthaltenen Informationen ist nicht gestattet. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Ruby Koans an amazing platform for teaching programming. Would this work with R?
For those not familiar Ruby Koans is a fantastic platform for teaching many of the basics of programming in the Ruby language. It uses unit tests written for methods each of which describe a component of the Ruby programming language. http://rubykoans.com/ The koans platform has become popular enough that it has been translated to a few other popular languages. Clojure - https://github.com/functional-koans/clojure-koans Python - https://github.com/gregmalcolm/python_koans Javascript - https://github.com/mrdavidlaing/javascript-koans I have been programming with R for around two years now, having come to R as a non-programmer, hoping to use it as a pragmatic solution to handeling and plotting lots of data. My skills have increased but are still certainly limited, and much of the advanced functionality remains a bit of a mystery to me. As I have learned more of other languages, particularly Ruby and Javascript, I am amazed with what can be done and with some concepts I so clearly have not learned while programming R. Take a look at Ruby Koans. Would this work for R? -- Patrick Schmitz Graduate Student Plant Biology 1206 West Gregory Drive RM 1500 [[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] nls problem with R
the dataset's form is changed after my post so I repost it here t 0 0.3403 0.4181 0.4986 0.7451 1.0069 1.5535 1.8049 2.4979 6.4903 13.5049 27.5049 41.5049 V(t) 6.053078443 5.56937391 5.45484486 5.193124598 4.31386722 3.645422269 3.587710965 3.740362689 3.699837726 2.908485019 1.888179494 1.176091259 1.176091259 -- View this message in context: http://r.789695.n4.nabble.com/nls-problem-with-R-tp3494454p3497827.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] nls problem with R
ID1 ID2 t V(t)
1 1 0 6.053078443
2 1 0.3403 5.56937391
3 1 0.4181 5.45484486
4 1 0.4986 5.193124598
5 1 0.7451 4.31386722
6 1 1.0069 3.645422269
7 1 1.5535 3.587710965
8 1 1.8049 3.740362689
9 1 2.4979 3.699837726
10 1 6.4903 2.908485019
11 1 13.5049 1.888179494
12 1 27.5049 1.176091259
13 1 41.5049 1.176091259
The model
(1) V(t)=V0[1-epi+ epi*exp(-c(t-t0))]
(2) V(t)=V0{A*exp[-lambda1(t-t0)]+(1-A)*exp[-lambda2(t-t0)]}
in formula (2) lambda1=0.5*{(c+delta)+[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
lambda2=0.5*{(c+delta)-[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
A=(epi*c-lambda2)/(lambda1-lambda2)
The regression rule :
for formula (1):(t=2,that is) first 8 rows are used for non-linear
regression
epi,c,t0,V0 parameters are obtained
for formula (2):all 13 rows of results are used for non-linear regression
lambda1,lambda2,A (with these parameters, delta can be calculated from them)
Thanks for help
Ster Lesser
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Re: [R] Remove all whitespaces
sed? 1,$s/ //g - Original Message - From: Joel To: r-help@r-project.org Sent: Thursday, May 05, 2011 6:50 PM Subject: [R] Remove all whitespaces Hi I got a string that looks something like this 1 2 3 4 5 6 7 8 9 ... and I want it to be 123456789... So I want to remove all spaces (or whitespaces) from my string. Anyone know a good way of doing this? //Joel -- View this message in context: http://r.789695.n4.nabble.com/Remove-all-whitespaces-tp3497867p3497867.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] issue with strange characters (readHTMLTable)
Thank you. The line of code you give certainly resolves several of the issues. I didn't realize that font support is such a tough matter to realize. Let me express my gratitude to those who provide this for us in R. On 04-05-11, Prof Brian Ripley rip...@stats.ox.ac.uk wrote: Oh, please! This is about the contributed package XML, not R and not Windows. Some of us have worked very hard to provide reasonable font support in R, including on Windows. We are given exceedingly little credit, just the brickbats for things for which we are not responsible. (We even work hard to port XML to Windows for you, again with almost zero credit.) That URL is a page in UTF-8, as its header says. We have provided many ways to work with UTF-8 on Windows, but it seems readHTMLTable() is not making use of them. You need to run iconv() on the strings in your object (which as it has factors, are the levels). When you do so, you will discover that page contains characters not in your native charset (I presume, not having your locale). What you can do, in Rgui only, is for (n in names(Islands)) Encoding(levels(Islands[[n]])) -UTF-8 but likely there are still characters it will not know how to display. On Wed, 4 May 2011, R.T.A.J.Leenders wrote: WinXP-x32, R-21.13.0 Dear list, I have a problem that (I think) relates to the interaction between Windows and R. I am trying to scrape a table with data on the Hawai'ian Islands, This is my code: library(XML) u - [1]http://en.wikipedia.org/wiki/Hawaii; tables - readHTMLTable(u) Islands - tables[[5]] The output is (first set of columns): IslandNickname Islands IslandNickname Location 1 Hawai�ûi[7]The Big Island 19�ð34âÀòN 155�ð30âÀòWïûÿ / ïûÿ19.567 �ðN 155.5�ðWïûÿ / 19.567; -155.5 2 Maui[8]TheValley Isle 20�ð48âÀòN 156�ð20âÀòWïûÿ / ïûÿ20.8�ðN 156.333�ðWïûÿ / 20.8; -156.333 3 Kaho�ûolawe[9] The Target Isle 20�ð33âÀòN 156�ð36âÀòWïûÿ / ïûÿ20.55 �ðN 156.6�ðWïûÿ / 20.55; -156.6 4 L�na�ûi[10] ThePineappleIsle 20�ð50âÀòN 156�ð56âÀòWïûÿ / ïûÿ20.833�ðN 15 6.933�ðWïûÿ / 20.833; -156.933 5 Moloka�ûi[11] TheFriendlyIsle 21�ð08âÀòN 157�ð02âÀòWïûÿ / ïûÿ21.133�ðN 1 57.033�ðWïûÿ / 21.133; -157.033 6O�ûahu[12]TheGatheringPlace 21�ð28âÀòN 157�ð59âÀòWïûÿ / ïûÿ21.467�ðN 1 57.983�ðWïûÿ / 21.467; -157.983 7 Kaua�ûi[13] The Garden Isle 22�ð05âÀòN 159�ð30âÀòWïûÿ / ïûÿ22.083 �ðN 159.5�ðWïûÿ / 22.083; -159.5 8 Ni�ûihau[14]The Forbidden Isle 21�ð54âÀòN 160�ð10âÀòWïûÿ / ïûÿ21.9�ðN 160.167�ðWïûÿ / 21.9; -160.167 As you can see, there are weird characters in there. I have also tried readHTMLTable(u, encoding = UTF-16) and readHTMLTable(u, encoding = UTF-8) but that didn't help. It seems to me that there may be an issue with the interaction of the Windows settings of the character set. sessionInfo() gives sessionInfo() R version 2.13.0 (2011-04-13) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Dutch_Netherlands.1252 LC_CTYPE=Dutch_Netherlands.1252 LC_MONETARY=Dutch_Netherlands.1252 [4] LC_NUMERIC=C LC_TIME=Dutch_Netherlands.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] XML_3.2-0.2 I have also attempted to let R use another setting by entering: Sys.setlocale(LC_ALL, en_US.UTF-8), but this yields the response: Sys.setlocale(LC_ALL, en_US.UTF-8) [1] Warning message: In Sys.setlocale(LC_ALL, en_US.UTF-8) : OS reports request to set locale to en_US.UTF-8 cannot be honored In addition, I have attempted to make the change directly from the windows command prompt, using: chcp 65001 and variations of that, but that didn't change anything. I have searched the list and the web and have found others bringing forth a similar issues, but have not been able to find a solution. I looks like this
Re: [R] Panels order in lattice graphs
On Wed, May 4, 2011 at 9:20 PM, Cristina Silva csi...@ipimar.pt wrote:
Hi all,
In lattice graphs, panels are drawn from left to right and bottom to top.
The flag as.table=TRUE changes to left to right and top to bottom. Is
there any way to change to first top to bottom and then left to right?
didn´t find anything neither in Help pages nor Lattice book.
See ?packet.panel.default. For example,
p - xyplot(mpg ~ disp | factor(carb), mtcars, as.table = TRUE)
print(p, packet.panel = packet.panel.default)
my.packet.panel -
function(layout, condlevels, page, row, column, ...)
{
tlayout - layout[c(2, 1, 3)] # switch row and column
print(packet.panel.default(tlayout, condlevels, page = page,
row = column, column = row, ...))
}
print(p, packet.panel = my.packet.panel)
-Deepayan
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Re: [R] Extraction of columns from two matrices
Hi:
Is this what you have in mind?
m1 - matrix(rpois(100, 10), nrow = 10, dimnames = list(NULL,
paste('V', 1:10, sep = '')))
m2 - matrix(rpois(40, 10), nrow = 10, dimnames = list(NULL,
paste('V', c(2, 5, 7, 10), sep = '')))
colnames(m1)
colnames(m2)
m1[, colnames(m2)]
HTH,
Dennis
On Wed, May 4, 2011 at 11:10 PM, nisha chandran
slamdunkangel...@gmail.com wrote:
Hi,
I have two matrices with 4885 cols and 36 cols respectively . I would like
to extract these 36 columns from the bigger matrix, the column values are
different but the column names would be the same.Hence based on the names I
would like to perform my operation. So is there any way of extracting this
info in an easy way. I have tried names,which,grep,subset none have worked.
Could someone help me out here
Thanks a ton
Nisha
[[alternative HTML version deleted]]
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Re: [R] Remove all whitespaces
If your whitespace does not only contain regular spaces, this might also be useful: x.1 - \t1 2 \n3 4 write(x.1, test.txt) x.2 - readLines(test.txt) x.3 - gsub([[:space:]], , x.2) x - paste(x.3, collapse=) See ?regex for details on regular expressions in R. HTH, Janko On 05.05.2011 11:02, Ivan Calandra wrote: Hi Joel, Try this: x - 1 2 3 4 5 6 7 8 9 gsub( , , x) Ivan Le 5/5/2011 10:50, Joel a écrit : Hi I got a string that looks something like this 1 2 3 4 5 6 7 8 9 ... and I want it to be 123456789... So I want to remove all spaces (or whitespaces) from my string. Anyone know a good way of doing this? //Joel -- *Janko Thyson* janko.thy...@googlemail.com mailto:janko.thy...@googlemail.com Jesuitenstraße 3 D-85049 Ingolstadt Mobile: +49 (0)176 83294257 This e-mail and any attachment is for authorized use by the intended recipient(s) only. It may contain proprietary material, confidential information and/or be subject to legal privilege. It should not be copied, disclosed to, retained or used by any other party. If you are not an intended recipient then please promptly delete this e-mail and any attachment and all copies and inform the sender. [[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] Extraction of columns from two matrices
Yea it did...
Thanks :)
On Thu, May 5, 2011 at 3:07 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
Is this what you have in mind?
m1 - matrix(rpois(100, 10), nrow = 10, dimnames = list(NULL,
paste('V', 1:10, sep = '')))
m2 - matrix(rpois(40, 10), nrow = 10, dimnames = list(NULL,
paste('V', c(2, 5, 7, 10), sep = '')))
colnames(m1)
colnames(m2)
m1[, colnames(m2)]
HTH,
Dennis
On Wed, May 4, 2011 at 11:10 PM, nisha chandran
slamdunkangel...@gmail.com wrote:
Hi,
I have two matrices with 4885 cols and 36 cols respectively . I would
like
to extract these 36 columns from the bigger matrix, the column values are
different but the column names would be the same.Hence based on the names
I
would like to perform my operation. So is there any way of extracting
this
info in an easy way. I have tried names,which,grep,subset none have
worked.
Could someone help me out here
Thanks a ton
Nisha
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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[R] Using functions/loops for repetitive commands
I still need to do some repetitive statistical analysis on some outcomes from a dataset. Take the following as an example; id sex hiv age famsize bmi resprate 1 M Pos 23 2 16 15 2 F Neg 24 5 18 14 3 F Pos 56 14 23 24 4 F Pos 67 3 33 31 5 M Neg 34 2 21 23 I want to know if there are statistically detectable differences in all of the continuous variables in my data set when subdivided by sex or hiv status (ie are age, family size, bmi and resprate different in my male and female patients or in hiv pos/neg patients) Of course I can use wilcoxon or t-tests e.g: wilcox.test( age~sex) wilcox.test(famsize~sex) wilcox.test(bmi~sex) wilcox.test(resprate~sex) wilcox.test( age~hiv) wilcox.test(famsize~hiv) wilcox.test(bmi~hiv) wilcox.test(resprate~hiv) but there must be some easy way of looping/automating this code (i.e. get all the continuous variables analysed one by one by sex, then analysed one by one by hiv status). Obviously my actual dataset is considerably bigger than what is shown here - I have many variables to assess making the longhand instruction to do every test pretty unsatisfactory. I think I can use ‘for’ or some other looping command for this purpose but I can’t work out how. I think I don’t properly understand how loops work yet as I'm still quite new to R. Please could someone help – ideally with an explanation and some quick sample code? Derek -- View this message in context: http://r.789695.n4.nabble.com/Using-functions-loops-for-repetitive-commands-tp3498006p3498006.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] nls problem with R
Apologies, but I don't see a question here ... am I missing something
obvious?
Andrew
On Thu, May 05, 2011 at 01:20:33AM -0700, sterlesser wrote:
ID1 ID2 t V(t)
1 1 0 6.053078443
2 1 0.3403 5.56937391
3 1 0.4181 5.45484486
4 1 0.4986 5.193124598
5 1 0.7451 4.31386722
6 1 1.0069 3.645422269
7 1 1.5535 3.587710965
8 1 1.8049 3.740362689
9 1 2.4979 3.699837726
101 6.4903 2.908485019
111 13.5049 1.888179494
121 27.5049 1.176091259
131 41.5049 1.176091259
The model
(1) V(t)=V0[1-epi+ epi*exp(-c(t-t0))]
(2) V(t)=V0{A*exp[-lambda1(t-t0)]+(1-A)*exp[-lambda2(t-t0)]}
in formula (2) lambda1=0.5*{(c+delta)+[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
lambda2=0.5*{(c+delta)-[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
A=(epi*c-lambda2)/(lambda1-lambda2)
The regression rule :
for formula (1):(t=2,that is) first 8 rows are used for non-linear
regression
epi,c,t0,V0 parameters are obtained
for formula (2):all 13 rows of results are used for non-linear regression
lambda1,lambda2,A (with these parameters, delta can be calculated from them)
Thanks for help
Ster Lesser
--
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--
Andrew Robinson
Program Manager, ACERA
Department of Mathematics and StatisticsTel: +61-3-8344-6410
University of Melbourne, VIC 3010 Australia (prefer email)
http://www.ms.unimelb.edu.au/~andrewpr Fax: +61-3-8344-4599
http://www.acera.unimelb.edu.au/
Forest Analytics with R (Springer, 2011)
http://www.ms.unimelb.edu.au/FAwR/
Introduction to Scientific Programming and Simulation using R (CRC, 2009):
http://www.ms.unimelb.edu.au/spuRs/
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Re: [R] quantmod's addTA plotting functions
On 2011-05-05 0:47, Russ Abbott wrote:
Hi,
I'm having trouble with quantmod's addTA plotting functions. They seem to
work fine when run from the command line. But when run inside a function,
only the last one run is visible. Here's an example.
test.addTA- function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close- GSPC[,GSPC.Close]
GSPC.EMA.3- EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10- EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'), up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10 appears on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser() neither appears
when the browser() interrupt occurs. I can then draw both GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All intended plots are
visible after the function terminates. So it isn't as if one wipes out the
other. This shows that it's possible to get all three lines on the plot, but
I can't figure out how to do it without manual intervention. Any suggestions
are appreciated.
Perhaps you didn't see this NOTE on the ?TA help page:
Calling any of the above methods from within a function
or script will generally require them to be wrapped in a
plot call as they rely on the context of the call to
initiate the actual charting addition.
Peter Ehlers
Thanks.
*-- Russ *
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Re: [R] Using functions/loops for repetitive commands
Hello, Derek,
see below.
On Thu, 5 May 2011, dereksloan wrote:
I still need to do some repetitive statistical analysis on some outcomes
from a dataset.
Take the following as an example;
id sex hiv age famsize bmi resprate
1 M Pos 23 2 16 15
2 F Neg 24 5 18 14
3 F Pos 56 14 23 24
4 F Pos 67 3 33 31
5 M Neg 34 2 21 23
I want to know if there are statistically detectable differences in all
of the continuous variables in my data set when subdivided by sex or hiv
status (ie are age, family size, bmi and resprate different in my male
and female patients or in hiv pos/neg patients) Of course I can use
wilcoxon or t-tests e.g:
wilcox.test( age~sex)
wilcox.test(famsize~sex)
wilcox.test(bmi~sex)
wilcox.test(resprate~sex)
wilcox.test( age~hiv)
wilcox.test(famsize~hiv)
wilcox.test(bmi~hiv)
wilcox.test(resprate~hiv)
[snip]
Define, e. g.,
my.wilcox.tests - function( var.names, groupvar.name, data) {
lapply( var.names,
function( v) {
form - as.formula( paste( v, ~, groupvar.name))
wilcox.test( form, data = data)
} )
}
and call something like
my.wilcox.test( character vector with relevant variable names,
character string with relevant grouping variable,
data = your data set as data frame)
Caveat: untested!
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner
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Re: [R] Panels order in lattice graphs
Thank you very much.
Cristina
On 05/05/2011 10:44, Deepayan Sarkar wrote:
On Wed, May 4, 2011 at 9:20 PM, Cristina Silvacsi...@ipimar.pt wrote:
Hi all,
In lattice graphs, panels are drawn from left to right and bottom to top.
The flag as.table=TRUE changes to left to right and top to bottom. Is
there any way to change to first top to bottom and then left to right?
didn´t find anything neither in Help pages nor Lattice book.
See ?packet.panel.default. For example,
p- xyplot(mpg ~ disp | factor(carb), mtcars, as.table = TRUE)
print(p, packet.panel = packet.panel.default)
my.packet.panel-
function(layout, condlevels, page, row, column, ...)
{
tlayout- layout[c(2, 1, 3)] # switch row and column
print(packet.panel.default(tlayout, condlevels, page = page,
row = column, column = row, ...))
}
print(p, packet.panel = my.packet.panel)
-Deepayan
--
--
Cristina Silva
INRB/L-IPIMAR
Unidade de Recursos Marinhos e Sustentabilidade
Av. de Brasília, 1449-006 Lisboa
Portugal
Tel.: 351 21 3027096
Fax: 351 21 3015948
csi...@ipimar.pt
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[R] Alter a line in a file.
Hi all R users Ive got a file that contains diffrent settings in the manor of: setting1=value1 setting2=value2 setting3=value3 setting4=value4 . . . What I want to do is open the file and change the value of a specific setting like wanna change setting4=value4 - setting4=value5 and then save the file again. setting1=value1 setting2=value2 setting3=value3 setting4=value5 . . . -- View this message in context: http://r.789695.n4.nabble.com/Alter-a-line-in-a-file-tp3498187p3498187.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] Draw a nomogram after glm
Please read the documentation for the rms package, particularly the datadist
function.
Note that in your subject line glm should be lrm.
Frank
Komine wrote:
Hi all R users
I did a logistic regression with my binary variable Y (0/1) and 2
explanatory variables.
Now I try to draw my nomogram with predictive value. I visited the help of
R but I have problem to understand well the example. When I use glm
fonction, I have a problem, thus I use lrm. My code is:
modele-lrm(Y~L+P,data=donnee)
fun- function(x) plogis(x-modele$coef[1]+modele$coef[2])
f - Newlabels(modele,c(L=poids,P=taille))
nomogram(f, fun=list('Prob Y=1'=plogis),
fun.at=c(seq(0,1,by=.1),.95,.99),
lmgp=.1, cex.axis=.6)
fun.at=c(.01,.05,seq(.1,.9,by=.1),.95,.99),
lmgp=.2, cex.axis=.6)
options(Fire=NULL)
Result is bad and I have this following error message:
Erreur dans value.chk(at, i, NA, -nint, Limval, type.range = full) :
variable L does not have limits defined by datadist
Could you help me on the code to draw nomogram.
Nb: my English is low, I apologize.
Thank for your help
Komine
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] Alter a line in a file.
try this:
a - readLines(textConnection('setting1=value1
setting2=value2
setting3=value3
setting4=value4'))
closeAllConnections()
# change values
ac - sub('setting4=value4', 'setting4=value5', a)
writeLines(ac, con='myFile.txt')
On Thu, May 5, 2011 at 8:16 AM, Joel joda2...@student.uu.se wrote:
Hi all R users
Ive got a file that contains diffrent settings in the manor of:
setting1=value1
setting2=value2
setting3=value3
setting4=value4
.
.
.
What I want to do is open the file and change the value of a specific
setting
like wanna change setting4=value4 - setting4=value5 and then save the
file again.
setting1=value1
setting2=value2
setting3=value3
setting4=value5
.
.
.
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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[R] Odp: Alter a line in a file.
Hi r-help-boun...@r-project.org napsal dne 05.05.2011 14:16:04: Joel joda2...@student.uu.se Odeslal: r-help-boun...@r-project.org 05.05.2011 14:16 Hi all R users Ive got a file that contains diffrent settings in the manor of: What file, what is its structure, is it some R object or separate file? What did you try and what went wrong? Regards Petr setting1=value1 setting2=value2 setting3=value3 setting4=value4 . . . What I want to do is open the file and change the value of a specific setting like wanna change setting4=value4 - setting4=value5 and then save the file again. setting1=value1 setting2=value2 setting3=value3 setting4=value5 . . . -- View this message in context: http://r.789695.n4.nabble.com/Alter-a-line- in-a-file-tp3498187p3498187.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] Outlier removal by Principal Component Analysis : error message
Dear Boule,
thank you for your interest in hyperSpec.
In order to look into your *problem* I need some more information.
I suggest that we solve the error off-list. Please note also that
hyperSpec has its own help mailing list:
hyperspec-h...@lists.r-forge.r-project.org
(due to the amount of spam I got to moderate, you need to subscribe
first here:
https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/hyperspec-help)
- Which version of hyperSpec do you use? If it is the version from CRAN,
could you please update to the development version at r-forge with
install.packages(hyperSpec,repos=http://R-Forge.R-project.org;)
?
- Next, if the problem persists with the latest build, could you send me
the raw data file so that I can exactly reproduce your problem?
- Also, for tracking down the exact source of the error, please execute
traceback ()
after you got the error and email me its output.
It is basically impossible to give general recommendations about
*Outlier detection*: a few spectra that are very different from all
other spectra may be outliers or they may be the target of a study...
This is also why the example in the vignette uses a two step procedure:
PCA only identifies suspects, i.e. spectra that have very different
scores than all others for some principal components. The second step is
a manually supervised decision whether the spectrum is really an outlier.
The first step could be replaced by other measures that however depend
on your data. E.g. if you expect/know your data to consist of different
clusters, suspects could be spectra that are too far away from any
cluster. If your data comes from a mixture of a few components, spectra
that cannot be modeled decently by a few PLS components could be
suspicious. Or spectra that require an own component, ...
Some kinds of outliers are actually well-defined in a spectroscopic
sense, e.g. contamination by fluorescent lamp light.
The second step could be replaced by an automatic decision, e.g. with a
distance threshold.
Personally, I rather use the term filtering for such automatic rules.
And there you can think about any number of rules your spectra must
comply with in order to be acceptable: signal to noise ratio, minimal
and maximal intensity, original offset (baseline) less than, ...
Hope that helps,
Claudia
I am currently analysis Raman spectroscopic data with the hyperSpec package.
I consulted the documentation on this package and I found an example
work-flow dedicated to Raman spectroscopy (see the address :
http://hyperspec.r-forge.r-project.org/chondro.pdf)
I am currently trying to remove outliers thanks to PCA just as they did in
the documentation, but I get a message error I can't explain. Here is my
code :
#import the data :
T=read.table('bladder bis concatenation colonne.txt',header=TRUE)
spec=new(hyperSpec,wavelength=T[,1],spc=t(T[,-1]),data=data.frame(sample=colnames(T[,-1])),label=list(.wavelength=Raman
shift (cm-1),spc=Intensity (a.u.)))
#baseline correction of the spectra
spec=spec[,,500~1800]
bl=spc.fit.poly.below(spec)
spec=spec-bl
#normalization of the spectra
spec=sweep(spec,1,apply(spec,1,mean),'/')
#PCA
pca=prcomp(~ spc,data=spec$.,center=TRUE)
scores=decomposition(spec,pca$x,label.wavelength=PC,label.spc=score/a.u.)
loadings=decomposition(spec,t(pca$rotation),scores=FALSE,label.spc=laoding
I/a.u.)
#plot the scores of the first 20 PC against all other to have an idea where
to find the outliers
pairs(scores[[,,1:20]],pch=19,cex=0.5)
#identify the outliers thanks to map.identify
out=map.identify(scores[,,5])
Erreur dans `[.data.frame`(x@data, , j, drop = FALSE) :
undefined columns selected
Does anybody understand where the problem comes from ?
And does anybody know another mean to find spectra outliers ?
Thank you in advance.
Boule
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--
Claudia Beleites
Spectroscopy/Imaging
Institute of Photonic Technology
Albert-Einstein-Str. 9
07745 Jena
Germany
email: claudia.belei...@ipht-jena.de
phone: +49 3641 206-133
fax: +49 2641 206-399
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Re: [R] Alter a line in a file.
Well your question is quite general the solution would involve several steps. Probably the easiest solution would be to read the data in as a dataframe (using read.table()) and using the '=' as the separator of the columns. Then change the desired values in the dataframe and save it back as a *.csv file, again using sep='='. Another option would be to read the data as a text string and use regexpressions to replace certain strings. Hope that gets you started Jannis --- Joel joda2...@student.uu.se schrieb am Do, 5.5.2011: Von: Joel joda2...@student.uu.se Betreff: [R] Alter a line in a file. An: r-help@r-project.org Datum: Donnerstag, 5. Mai, 2011 12:16 Uhr Hi all R users Ive got a file that contains diffrent settings in the manor of: setting1=value1 setting2=value2 setting3=value3 setting4=value4 . . . What I want to do is open the file and change the value of a specific setting like wanna change setting4=value4 - setting4=value5 and then save the file again. setting1=value1 setting2=value2 setting3=value3 setting4=value5 . . . -- View this message in context: http://r.789695.n4.nabble.com/Alter-a-line-in-a-file-tp3498187p3498187.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] Alter a line in a file.
jholtman wrote:
a - readLines(textConnection('setting1=value1
setting2=value2
setting3=value3
setting4=value4'))
closeAllConnections()
# change values
ac - sub('setting4=value4', 'setting4=value5', a)
writeLines(ac, con='myFile.txt')
Problem is that I dont know the value on all the settings that I wanna
change otherwise that looks like something to continue on.
Petr Pikal wrote:
What file, what is its structure, is it some R object or separate file?
What did you try and what went wrong?
Regards
Petr
Just a normal textfile nothing fancy
Ive tried diffrent kind of ways of useing my OS witch is linux by the system
command to do it for me but Im not good enought on regexp to get it to work
properly.
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Re: [R] Alter a line in a file.
Hi
r-help-boun...@r-project.org napsal dne 05.05.2011 15:13:34:
Joel joda2...@student.uu.se
Odeslal: r-help-boun...@r-project.org
05.05.2011 15:13
jholtman wrote:
a - readLines(textConnection('setting1=value1
setting2=value2
setting3=value3
setting4=value4'))
closeAllConnections()
# change values
ac - sub('setting4=value4', 'setting4=value5', a)
writeLines(ac, con='myFile.txt')
Problem is that I dont know the value on all the settings that I wanna
change otherwise that looks like something to continue on.
But in that case how would you like to select the setting?
Petr Pikal wrote:
What file, what is its structure, is it some R object or separate
file?
What did you try and what went wrong?
Regards
Petr
Just a normal textfile nothing fancy
Ive tried diffrent kind of ways of useing my OS witch is linux by the
system
command to do it for me but Im not good enought on regexp to get it to
work
properly.
I read the simple text file by read.table
data
V1
1 setting1=value1
2 setting2=value2
3 setting3=value3
4 setting4=value4
grep(4, data$V1)
[1] 4
Regards
Petr
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Re: [R] Simple General Statistics and R question (with 3 line example) - get z value from pairwise.wilcox.test
On May 5, 2011, at 10:58 , JP wrote: On 4 May 2011 15:32, peter dalgaard pda...@gmail.com wrote: On May 4, 2011, at 15:11 , JP wrote: Peter thanks for the fantastically simple and understandable explanation... To sum it up... to find the z values of a number of pairwise wilcox tests do the following: # pairwise tests with bonferroni correction x - pairwise.wilcox.test(a, b, alternative=two.sided, p.adj=bonferroni, exact=F, paired=T) You probably don't want the bonferroni correction there. Rather p.adj=none. You generally correct the p values for multiple testing, not the test statistics. Oh, I see thanks... of course since I have 5 groups (samples) and 10 comparisons I still have to correct when quoting p values... (My sentiment would be to pick apart the stats:::wilcox.test.default function and clone the computation of Z from it, but presumably backtracking from the p value is a useful expedient.) Should this be so onerous for the user [read non-statistician] ? My main reservation is that if the p value gets very small or close to 1 (for 1-sided tests), then it might not be possible to reconstruct the underlying Z (qnorm(pnorm(Z)) breaks down around Z = -37 and Z=+8). If you get sensible values, there's probably not a problem. # what is the data structure we got back is.matrix(x$p.value) # p vals x$p.value # z.scores for each z.score - qnorm(x$p.value / 2) Hmm, you're not actually getting a signed z out of this, you might want to try alternative=greater and drop the division by 2 inside qnorm(). (If the signs come out inverted, I meant less not greater...) But I need a two sided test (changing the alternative would change the hypothesis!)... do I still do this? For the z's, yes. Hmm, the asymmetry of qnorm(pnorm(...)) indicates that you might want to be a bit more careful. All my z values are negative Is this correct? No, that's the problem. The sign of Z should depend on whether the signed ranks are predominantly positive or negative. If you do a two-sided p-value and divide by two, then by definition you get something less than .5 and qnorm of that will be negative. So if the 2-sided p is 0.04, the one-sided p will be either 0.02 or 0.98, depending on the alternative and on whether the V statistic is above or below its expectation. Notice that this corresponds to Z statistics of opposite sign: qnorm(.02) [1] -2.053749 qnorm(.98) [1] 2.053749 -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using functions/loops for repetitive commands
Your code may be untested but it works - also helping me slowly to start
understanding how to write functions. Thank you.
However I still have difficulty. I also have some categorical variables to
analyse by age hiv status - i.e. my dataset expands to (for example);
id sex hiv age famsize bmi resprate smoker alcohol
1 M Pos 23 2 16 15 Y
Y
2 F Neg 24 5 18 14 Y
Y
3 F Pos 56 14 23 24 Y
N
4 F Pos 67 3 33 31 N
N
5 M Neg 34 2 21 23 N
N
Using the template for the code you sent me I thought I could analyse the
categorical variables by sex hiv status using a chiq-squared test;
Long-hand this would be;
chisq.test(smoker,sex)
chisq.test(alcohol,sex)
chisq.test(smoker,hiv)
chisq.test(alcohol,hiv)
Again I wanted to use a function to loop automate it and thought I could
write;
categ-c(smoker,alcohol)
group.name-c(sex,hiv)
bl.chisq-function(categ,group.name,dataframe name){
lapply(categ,
function(y){
form2-as.formula(paste(y,group.name))
chisq.test(form2,dataframe name)
})
}
bl.chisq(categ,group.name,data frame name)
but I get an error message:
Error in parse(text = x) : unexpected symbol in smoker sex
What is wrong with the code? Is is because the wilcox.test is a formula
(with a ~ symbol for modelling) whilst the chisq.test simply requires me to
list raw data? If so how can I change my code to automate the chisq.test in
the same way I did for the wilcox.test?
Many thanks for any help!
Derek
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Re: [R] Simple General Statistics and R question (with 3 line example) - get z value from pairwise.wilcox.test
Thanks once again Peter, I understand your points -- I fiddled and googled and read some more and found an eas(ier) route: library(coin) t - wilcoxsign_test(a ~ b, alternative = two.sided, distribution = exact()) # This is equivalent to paired wilcox.test pval - pvalue(t) sweet.zscore - statistic(t) # signed and everything # correct for multiple testing if you are doing a number of the above... I would never have got here without your guidance - so kudos to you. JP On 5 May 2011 14:52, peter dalgaard pda...@gmail.com wrote: On May 5, 2011, at 10:58 , JP wrote: On 4 May 2011 15:32, peter dalgaard pda...@gmail.com wrote: On May 4, 2011, at 15:11 , JP wrote: Peter thanks for the fantastically simple and understandable explanation... To sum it up... to find the z values of a number of pairwise wilcox tests do the following: # pairwise tests with bonferroni correction x - pairwise.wilcox.test(a, b, alternative=two.sided, p.adj=bonferroni, exact=F, paired=T) You probably don't want the bonferroni correction there. Rather p.adj=none. You generally correct the p values for multiple testing, not the test statistics. Oh, I see thanks... of course since I have 5 groups (samples) and 10 comparisons I still have to correct when quoting p values... (My sentiment would be to pick apart the stats:::wilcox.test.default function and clone the computation of Z from it, but presumably backtracking from the p value is a useful expedient.) Should this be so onerous for the user [read non-statistician] ? My main reservation is that if the p value gets very small or close to 1 (for 1-sided tests), then it might not be possible to reconstruct the underlying Z (qnorm(pnorm(Z)) breaks down around Z = -37 and Z=+8). If you get sensible values, there's probably not a problem. # what is the data structure we got back is.matrix(x$p.value) # p vals x$p.value # z.scores for each z.score - qnorm(x$p.value / 2) Hmm, you're not actually getting a signed z out of this, you might want to try alternative=greater and drop the division by 2 inside qnorm(). (If the signs come out inverted, I meant less not greater...) But I need a two sided test (changing the alternative would change the hypothesis!)... do I still do this? For the z's, yes. Hmm, the asymmetry of qnorm(pnorm(...)) indicates that you might want to be a bit more careful. All my z values are negative Is this correct? No, that's the problem. The sign of Z should depend on whether the signed ranks are predominantly positive or negative. If you do a two-sided p-value and divide by two, then by definition you get something less than .5 and qnorm of that will be negative. So if the 2-sided p is 0.04, the one-sided p will be either 0.02 or 0.98, depending on the alternative and on whether the V statistic is above or below its expectation. Notice that this corresponds to Z statistics of opposite sign: qnorm(.02) [1] -2.053749 qnorm(.98) [1] 2.053749 -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about error of non-numeric argument to binary operator
On May 5, 2011, at 1:31 AM, Maximo Polanco wrote:
I have been trying to do a nls model and gives me the error of a
nonnumeric
argument
This is my data set
TT
FFT
V
C
table(file=c:/tt2.txt,header=T)
fit.model - nls(TT~60*(1+alpha*(v/c)^beta),data=tt2,
start=list(alpha=1,
beta=3, v=1000))
Error in v/c : non-numeric argument to binary operator
You have (perhaps) defined a column name of C and then attempted to
reference it with c. Since `c` is a (rather fundamental) function in
R, the interpreter has no problem accessing `c` but trying to take
the ratio of a scalar to a function fails.
You can try changing c to C above and it may work, but in the
future, please heed the issues raised at the bottom of this trimmed
message!
(It would be much better to not use either C or c for object names
since they are both function names.)
?C
C {stats}
R Documentation
Sets Contrasts for a FactorDescription
Sets the contrasts attribute for the factor.
Usage
C(object, contr, how.many, ...)
is.numeric(tt2)
[1] FALSE
is.character(tt2)
[1] FALSE
as.numeric(tt2)
Error: (list) object cannot be coerced to type 'double'
This is my data set
TT
FFT
V
C
1
70.475
60
snipped long ... incorrectly formatted data due to failing to post in
plain text.
[[alternative HTML version deleted]]
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###
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West Hartford, CT
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Re: [R] nls problem with R
Date: Thu, 5 May 2011 01:20:33 -0700
From: sterles...@hotmail.com
To: r-help@r-project.org
Subject: Re: [R] nls problem with R
ID1 ID2 t V(t)
1 1 0 6.053078443
2 1 0.3403 5.56937391
3 1 0.4181 5.45484486
4 1 0.4986 5.193124598
5 1 0.7451 4.31386722
6 1 1.0069 3.645422269
7 1 1.5535 3.587710965
8 1 1.8049 3.740362689
9 1 2.4979 3.699837726
10 1 6.4903 2.908485019
11 1 13.5049 1.888179494
12 1 27.5049 1.176091259
13 1 41.5049 1.176091259
The model
(1) V(t)=V0[1-epi+ epi*exp(-c(t-t0))]
A=Vo, B-Vo*epi, C=exp(-c*t0)
V(t)=A-B+B*C*exp(-ct)
or further, D=A-B, F=B*C,
V(t)=D+F*exp(-ct)
this model only really has 3 attriubtes: initial value, final value,
and decay constant yet you ask for 4 parameters. There is no
way to get a unique answer. For some reason this same form comes up
a lot here, I think this is about third time I've sene this in last few weeks.
I guess when fishing or shopping for forms to fit, it is tempting to
throw a bunch of parameteres into your model but this can create intractable
ambiguities.
Indeed, if I just remove t0 and use your first 8 points I get this
( random starting values, but convewrged easily you still need to plot etc)
[1] 1 v= 8.77181162126362 epi= 0.672516376478598 cl= 1.90973175223917 t0= 0
.643481321167201
summary(nls2)
Formula: V2 ~ v0 * (1 - epi + epi * exp(-cl * (T2)))
Parameters:
Estimate Std. Error t value Pr(|t|)
v0 6.2901 0.3384 18.585 8.3e-06 ***
epi 0.5430 0.1373 3.955 0.0108 *
cl 0.9684 0.5491 1.763 0.1381
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.3579 on 5 degrees of freedom
Number of iterations to convergence: 11
Achieved convergence tolerance: 4.057e-06
(2) V(t)=V0{A*exp[-lambda1(t-t0)]+(1-A)*exp[-lambda2(t-t0)]}
in formula (2) lambda1=0.5*{(c+delta)+[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
lambda2=0.5*{(c+delta)-[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
A=(epi*c-lambda2)/(lambda1-lambda2)
The regression rule :
for formula (1):(t=2,that is) first 8 rows are used for non-linear
regression
epi,c,t0,V0 parameters are obtained
for formula (2):all 13 rows of results are used for non-linear regression
lambda1,lambda2,A (with these parameters, delta can be calculated from them)
Thanks for help
Ster Lesser
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[R] Draw a nomogram after glm
Hi all R users
I did a logistic regression with my binary variable Y (0/1) and 2
explanatory variables.
Now I try to draw my nomogram with predictive value. I visited the help of R
but I have problem to understand well the example. When I use glm fonction,
I have a problem, thus I use lrm. My code is:
modele-lrm(Y~L+P,data=donnee)
fun- function(x) plogis(x-modele$coef[1]+modele$coef[2])
f - Newlabels(modele,c(L=poids,P=taille))
nomogram(f, fun=list('Prob Y=1'=plogis),
fun.at=c(seq(0,1,by=.1),.95,.99),
lmgp=.1, cex.axis=.6)
fun.at=c(.01,.05,seq(.1,.9,by=.1),.95,.99),
lmgp=.2, cex.axis=.6)
options(Fire=NULL)
Result is bad and I have this following error message:
Erreur dans value.chk(at, i, NA, -nint, Limval, type.range = full) :
variable L does not have limits defined by datadist
Could you help me on the code to draw nomogram.
Nb: my English is low, I apologize.
Thank for your help
Komine
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Re: [R] Convert a presence/ absence matrix to a list of presence only
Thank you, it perfectly worked, except I had to modify some codes on the rep(, maxLen - length(sp)) because it gave me some error, thanks -- View this message in context: http://r.789695.n4.nabble.com/Convert-a-presence-absence-matrix-to-a-list-of-presence-only-tp3468479p3498116.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 error in histograms
Hello! I am trying to produce a histogram of measurement data (orientation of archaeological structures) that are a subject to measurement error. The normal histogram just computes frequencies, but does not take into account that a particular value is spread over a range of values (in my case the spread is different for reach measurement and is larger than the bin size). The closest approach is kernel density estimation (in the image is a comparison of histogram and KDE): http://en.wikipedia.org/wiki/Kernel_density_estimation http://en.wikipedia.org/wiki/File:Comparison_of_1D_histogram_and_KDE.png However in my case the kernel size is different for each value. I wrote a program in IDL that performs the plotting, but am just wondering if such a function is available in R. Basically it is a problem of summing (and later plotting) several data distributions. I would appreciate also any hint to a book that might be dealing with the problem. I am not an expert in statistics and I might not be using use the correct terminology in web searches. Regards, Kristof __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Insert values to histogram
I'm trying to add the exact value on top of each column of an Histogram, i have been trying with the text function but it doesn't work. The problem is that the program it self decides the exact value to give to each column, and ther is not like in a bar-plot that I know exactly which values are been plotting. If anyone have any new idea on how to do this Thanks Matias -- View this message in context: http://r.789695.n4.nabble.com/Insert-values-to-histogram-tp3498140p3498140.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] lm weight adj r-sq
Dear all, First of all apologies, I'm pretty newbie, and secondly I know this is not a question for the forum but I'd be very grateful if you'd help me. I had problems with normality assumptions, and I tried with weights in the linear regression this way: modelw -lm(log(v1)~v2+log(v2)+log(v3)+v4, data=dat, weight=1/group.var^2) I'd like to know how interpret/compute adj R-sq/mse from modelw, in order to compare it to the adj R-sq/mse from model (model - lm(log(v1)~v2+log(v2)+log(v3)+v4, data=dat) Thanks in advance, u...@host.com pd I set the weights following this notes: (I was reading here: statistics.unl.edu/faculty/bilder/stat870/schedule/Chapter%2011.doc http://statistics.unl.edu/faculty/bilder/) -- View this message in context: http://r.789695.n4.nabble.com/lm-weight-adj-r-sq-tp3498020p3498020.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] heatmap.3
I am intesrested by your heatmap function (allowing matrix in ColSideColors option). Can you give your complete code of your function? thanks -- View this message in context: http://r.789695.n4.nabble.com/heatmap-3-tp1566584p3498156.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] distance matrix
Hello all, I am wondering if there is anyway to create distance matrix for replicated data for example, I have a data like sample pop id var1var2var3var4 1.1 1 a 1 1 0 1 1.2 1 a 0 0 1 0 1.3 1 a 1 1 0 1 2.1 2 b 0 0 1 0 2.2 2 b 1 1 1 1 2.3 2 b 0 1 0 0 2.4 2 b 1 0 1 1 3.1 3 a 0 1 0 1 3.2 3 a 1 1 1 0 3.3 3 a 0 0 0 0 dist(data[,c(4:7)] gives the distance of samples, but I also need the distances of pop ie (1,2,3) and also id (a,b) how can I achieve this?? Thanks -- View this message in context: http://r.789695.n4.nabble.com/distance-matrix-tp3498033p3498033.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] Null
This is probably a very simple question but I am completely stumped!I am trying to do shapiro.wilk(x) test on a relatively small dataset(75) and each time my variable and keeps coming out as 'NULL', and shapiro.test(fcv) Error in complete.cases(x) : no input has determined the number of cases my text file looks like this: case 1.600972896 1.534026106 1.633468456 1.69019608 1.686636269 1.713490543 1.460897843 1.604226053 1.547774705 1.575187845 1.50242712 1.489958479 1.555094449 1.56937391 1.46686762 1.583198774 1.59439255 1.627365857 1.596597096 1.598790507 1.596597096 1.613841822 1.607455023 1.586587305 1.72427587 1.668385917 1.743509765 1.5774918 1.709269961 1.507855872 1.650307523 1.670245853 1.721810615 1.613841822 1.586587305 1.658011397 1.595496222 1.662757832 1.521138084 1.564666064 1.515873844 1.596597096 1.617000341 1.621176282 1.598790507 1.73479983 1.498310554 1.571708832 1.426511261 1.698970004 1.534026106 1.5774918 1.682145076 1.689308859 1.654176542 1.526339277 1.545307116 1.658964843 1.638489257 1.557507202 1.604226053 1.627365857 1.651278014 1.627365857 1.559906625 1.720159303 1.64738297 1.62324929 1.698970004 1.704150517 1.57863921 1.558708571 1.681241237 1.539076099 1.5132176 Any ideas? -- View this message in context: http://r.789695.n4.nabble.com/Null-tp3498261p3498261.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fit a random data into Beta distribution?
Hi, @Steven: Since Beta distribution is a generic distribution by which i mean that by varying the parameter of alpha and beta we can fit any distribution. So to check this i generated a random data from Normal distribution like x.norm-rnorm(n=100,mean=10,sd=10); Now i want to estimate the paramters alpha and beta of the beta distribution which will fit the above generated random data. That's what i want to do. @Ali: When you said you drafted your own procedure, do you mean that you are calculate the parameters using MLE or bayesian..???Can you please give me some more ideas into this? Thanks and Regards, Som Shekhar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Remove all whitespaces
A more elegant way would be: myString-1 2 3 4 5 myString-paste(unlist(strsplit(myString, )),collapse=) The output will be 12345 Regards, Som Shekhar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] functions pandit and treebase in the package apTreeshape
Hello. I'm trying to use the functions pandit and treebase. They are in the package apTreeshape. Once I've loaded the package, R responses: - no function pandit/treebase. Somebody knows why or what is the reason? Thanks, Arnau. Arnau Mir Torres Edifici A. Turmeda Campus UIB Ctra. Valldemossa, km. 7,5 07122 Palma de Mca. tel: (+34) 971172987 fax: (+34) 971173003 email: arnau@uib.es URL: http://dmi.uib.es/~arnau [[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] Using functions/loops for repetitive commands
Hi Derek,
You can accomplish your loop jobs by following means:
(a) use for loop
(b) use while loop
(c) use lapply, tapply, or sapply. (i feel lapply is the elegant
way )
---For Loop-
for loops are pretty simple to use and is almost similar to any
other scripting languages you know.( I am referring to Matlab)
(Example 1) lets say you know that you have to run 10 iterations then
you can run it as
for(i in 1:10) print(i)
//it will print the number from 1 to 10
(Example 2) You don't know how many iterations you need to run. Only
thing you have is some vector and you want to do some operation on
that vector. You can do something like this:
myVector-c(20,45,23,45,89)
for(i in seq_along(myVector)) print(myVector[i]
-Using lapply-
In lapply you need to provide mainly two things:
(1)First parameter: vectors or some sequence of numbers
(2)Second parameter: A function which could be user defined function
or some other inbuilt function.
lapply will call the function for every number given in the First
parameter of the function)
For example:
x-c(10,20,20)
lapply(seq_along(x),function(i) {//your logic})
if you see the first parameter i have sent seq_along(x). The outcome
of seq_along(x) will be 1, 2,3.
Now lapply will take each of these numbers and call the function. That
means lapply is calling the function thrice for the current data set
something like this
function(1) { //your logic}
function(2) { }
function(3) { //)
That means your logic inside the function will be executed for each
and every value specified in the first parameter of the lapply
function.
I hope it helps you in some way.
For your problem, i am making a guess that you are using data frame or
matrix to store the data and then you want to automate the data right?
You can try using lapply, i think that would be efficient..Let me
also try ..
Regards,
Som Shekhar
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[R] Confidence interval for difference in Harrell's c statistics (or equivalently Somers' D statistics)
Dear All, I am trying to calculate a 95% confidence interval for the difference in two c statistics (or equivalently D statistics). In Stata I gather that this can be done using the lincom command. Is there anything similar in R? As you can see below I have two datasets (that are actually two independent subsets of the same data) and the respective c statistics for the variables in both cases. What I would now like to do is to prove that there is no statistically significant difference between the statistics (between the dev and val datasets.) Any help would be much appreciated. rdev - rcorrcens(Surv(stimes1,eind1)~gendat1+neurodat1) rdev Somers' Rank Correlation for Censored DataResponse variable:Surv(stimes1, eind1) CDxy aDxySDZ Pn gendat1 0.534 0.069 0.069 0.017 3.98 0.0001 1500 neurodat1 0.482 -0.036 0.036 0.011 3.18 0.0015 1500 rval - rcorrcens(Surv(stimes2,eind2)~gendat2+neurodat2) rval Somers' Rank Correlation for Censored DataResponse variable:Surv(stimes2, eind2) CDxy aDxySDZ Pn gendat2 0.543 0.085 0.085 0.017 4.94 0e+00 1500 neurodat2 0.481 -0.038 0.038 0.011 3.44 6e-04 1500 Many thanks, Laura P.S. I'm using Windows XP, R 2.9.2 [[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] RV: R question
which is the maximum large of digits that R has?, because SQL work with 50 digits I think. and I need a software that work with a lot of digits. Thanks. Pamela Santelices Elgueta Estadístico Instituto Nacional de Estadísticas Fono: (56-2) 7962491 Paseo Bulnes 209 oficina 82 Santiago [[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] two-way group mean prediction in survreg with three factors
Oops, I hope not too. Don't know why I had the brackets around B+C. My model is actually A*B+C. And I'm not sure how to obtain the two-way prediction of AB with C marginalized. Thanks. Pang -Original Message- From: Andrew Robinson [mailto:a.robin...@ms.unimelb.edu.au] Sent: Wednesday, May 04, 2011 10:13 PM To: Pang Du Cc: r-help@r-project.org Subject: Re: [R] two-way group mean prediction in survreg with three factors I hope not! Facetiousness aside, the model that you have fit contains C, and, indeed, an interaction between A and C. So, the effect of A upon the response variable depends on the level of C. The summary you want must marginalize C somehow, probably by a weighted or unweighted average across its levels. What does that summary really mean? Can you meaningfully average across the levels of a predictor that is included in the model as a main and an interaction term? Best wishes Andrew On Wed, May 04, 2011 at 12:24:50PM -0400, Pang Du wrote: I'm fitting a regression model for censored data with three categorical predictors, say A, B, C. My final model based on the survreg function is Surv(..) ~ A*(B+C). I know the three-way group mean estimates can be computed using the predict function. But is there any way to obtain two-way group mean estimates, say estimated group mean for (A1, B1)-group? The sample group means don't incorporate censoring and thus may not be appropriate here. Pang Du Virginia Tech [[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. -- Andrew Robinson Program Manager, ACERA Department of Mathematics and StatisticsTel: +61-3-8344-6410 University of Melbourne, VIC 3010 Australia (prefer email) http://www.ms.unimelb.edu.au/~andrewpr Fax: +61-3-8344-4599 http://www.acera.unimelb.edu.au/ Forest Analytics with R (Springer, 2011) http://www.ms.unimelb.edu.au/FAwR/ Introduction to Scientific Programming and Simulation using R (CRC, 2009): http://www.ms.unimelb.edu.au/spuRs/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fit a random data into Beta distribution?
Beta is not as general as you think. Its support is limited to [0,1], but you are trying to fit data that lies outside of its support. Please read about the beta distribution from a basic stats/prob book. Ravi. From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Shekhar [shekhar2...@gmail.com] Sent: Thursday, May 05, 2011 6:54 AM To: r-help@r-project.org Subject: Re: [R] How to fit a random data into Beta distribution? Hi, @Steven: Since Beta distribution is a generic distribution by which i mean that by varying the parameter of alpha and beta we can fit any distribution. So to check this i generated a random data from Normal distribution like x.norm-rnorm(n=100,mean=10,sd=10); Now i want to estimate the paramters alpha and beta of the beta distribution which will fit the above generated random data. That's what i want to do. @Ali: When you said you drafted your own procedure, do you mean that you are calculate the parameters using MLE or bayesian..???Can you please give me some more ideas into this? Thanks and Regards, Som Shekhar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Conditional distribution plot using Model-based Recursive Partitioning
Hello, I am using the party module to estimate the relationship between the probability of being a student and number of siblings (alive). However, I need to include a number of relevant covariates. My code is below: fm3 - mob(Student ~ age + alive + sex2 + cwa + cha + cym | Religion+Servant + Literacy, control = ctrl, data = samp2, model = glinearModel, family =binomial()) plot(fm3, tp_args = list(which = alive),tnex = 2, type = simple ) Which works fine. However, the plots generated do not show the conditional relationship between the dependent and independent variable at each node. Basically, there are spinogram leafs at nodes where the unconditional relationship is the opposite to the conditional relationship -- as given in the summary tables. I have tried playing around with the plot function arguements, yet to no avail. Can anybody suggest to me a way which I would be able to plot the tree and have either a conditional spinogram in the leaf or even just text with a number showing a marginal effect or coefficient please? Thanks, Alan -- Alan Fernihough IRCHSS Scholar UCD School of Economics [[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] Using functions/loops for repetitive commands
On May 5, 2011, at 10:01 AM, dereksloan wrote:
Your code may be untested but it works - also helping me slowly to
start
understanding how to write functions. Thank you.
However I still have difficulty. I also have some categorical
variables to
analyse by age hiv status - i.e. my dataset expands to (for
example);
id sex hiv age famsize bmi resprate smoker alcohol
1 M Pos 23 2 16 15 Y
Y
2 F Neg 24 5 18 14 Y
Y
3 F Pos 56 14 23 24 Y
N
4 F Pos 67 3 33 31 N
N
5 M Neg 34 2 21 23 N
N
Using the template for the code you sent me I thought I could
analyse the
categorical variables by sex hiv status using a chiq-squared test;
Long-hand this would be;
chisq.test(smoker,sex)
chisq.test(alcohol,sex)
chisq.test(smoker,hiv)
chisq.test(alcohol,hiv)
Again I wanted to use a function to loop automate it and thought I
could
write;
categ-c(smoker,alcohol)
group.name-c(sex,hiv)
bl.chisq-function(categ,group.name,dataframe name){
lapply(categ,
function(y){
form2-as.formula(paste(y,group.name))
I haven't tested it but I suspect you failed to note that Eichner used
sep=~ in his paste argument to as.formula().
chisq.test(form2,dataframe name)
})
}
bl.chisq(categ,group.name,data frame name)
but I get an error message:
Error in parse(text = x) : unexpected symbol in smoker sex
What is wrong with the code? Is is because the wilcox.test is a
formula
(with a ~ symbol for modelling) whilst the chisq.test simply
requires me to
list raw data? If so how can I change my code to automate the
chisq.test in
the same way I did for the wilcox.test?
Many thanks for any help!
Derek
David Winsemius, MD
West Hartford, CT
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Re: [R] Confidence interval for difference in Harrell's c statistics (or equivalently Somers' D statistics)
On May 5, 2011, at 8:20 AM, Laura Bonnett wrote:
Dear All,
I am trying to calculate a 95% confidence interval for the
difference in two
c statistics (or equivalently D statistics). In Stata I gather that
this
can be done using the lincom command. Is there anything similar in R?
Have you looked at rcorrp.cens {Hmisc}?
snipped
[[alternative HTML version deleted]]
Please post in plain text.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] distance matrix
Em 5/5/2011 07:34, antu escreveu: Hello all, I am wondering if there is anyway to create distance matrix for replicated data for example, I have a data like sample pop id var1var2var3var4 1.1 1 a 1 1 0 1 1.2 1 a 0 0 1 0 1.3 1 a 1 1 0 1 2.1 2 b 0 0 1 0 2.2 2 b 1 1 1 1 2.3 2 b 0 1 0 0 2.4 2 b 1 0 1 1 3.1 3 a 0 1 0 1 3.2 3 a 1 1 1 0 3.3 3 a 0 0 0 0 dist(data[,c(4:7)] gives the distance of samples, but I also need the distances of pop ie (1,2,3) and also id (a,b) how can I achieve this?? Just a doubt: does the idea of comparing (ordering) the variable pop and id make sense? Or expressed in more direct way: what would a distance between b and a mean (same for the pop labels)? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Tone in mailing lists (was issue with strange characters (readHTMLTable))
I did read How To Ask Questions The Smart Way http://www.catb.org/%7Eesr/faqs/smart-questions.html and I don't have a problem with calling me stupid and/or a winer for posting this... What's the purpose of an R Mailing list (especially r-help), after all? IMHO it should be about R users being able to ask questions and getting help from those who would like to provide some. Unless a post has been written in an clearly offensive tone and/or shows a major amount of ignorance on the part of the person asking, why would you want to not just provide an answer to the question posed but deliver on top either sarcasm, personal criticism, sharp remarks or the like? I also believe that it's in the nature of things that R users will most likely rather post about problems/bugs they encountered than to simply drop a line of acknowledgment for the people who continuously develop R and its packages. It's nice if they do explicitly express their gratitude in their posts (and this *does* happen a lot also), but it is not *demanded* and if they don't, it doesn't automatically mean that they are not grateful... Please understand me right: all the people working on pushing R forward do fantastic work and I greatly appreciate the inputs from the entire community, I sometimes just wonder about the tone in the mailing lists. But that's probably just me , maybe it's just the way the cookie crumbles in technical mailing lists ... Regards, Janko On 05.05.2011 11:33, R.T.A.J.Leenders wrote: Thank you. The line of code you give certainly resolves several of the issues. I didn't realize that font support is such a tough matter to realize. Let me express my gratitude to those who provide this for us in R. On 04-05-11, Prof Brian Ripleyrip...@stats.ox.ac.uk wrote: Oh, please! This is about the contributed package XML, not R and not Windows. Some of us have worked very hard to provide reasonable font support in R, including on Windows. We are given exceedingly little credit, just the brickbats for things for which we are not responsible. (We even work hard to port XML to Windows for you, again with almost zero credit.) That URL is a page in UTF-8, as its header says. We have provided many ways to work with UTF-8 on Windows, but it seems readHTMLTable() is not making use of them. You need to run iconv() on the strings in your object (which as it has factors, are the levels). When you do so, you will discover that page contains characters not in your native charset (I presume, not having your locale). What you can do, in Rgui only, is for (n in names(Islands)) Encoding(levels(Islands[[n]]))-UTF-8 but likely there are still characters it will not know how to display. On Wed, 4 May 2011, R.T.A.J.Leenders wrote: WinXP-x32, R-21.13.0 Dear list, I have a problem that (I think) relates to the interaction between Windows and R. I am trying to scrape a table with data on the Hawai'ian Islands, This is my code: library(XML) u- [1]http://en.wikipedia.org/wiki/Hawaii; tables- readHTMLTable(u) Islands- tables[[5]] The output is (first set of columns): IslandNickname Islands IslandNickname Location 1 Hawai�?�»i[7]The Big Island 19�?�°34�¢�¤�²N 155�?�°30�¢�¤�²W�¯�»�¿ / �¯�»�¿19.567 �?�°N 155.5�?�°W�¯�»�¿ / 19.567; -155.5 2 Maui[8]TheValley Isle 20�?�°48�¢�¤�²N 156�?�°20�¢�¤�²W�¯�»�¿ / �¯�»�¿20.8�?�°N 156.333�?�°W�¯�»�¿ / 20.8; -156.333 3 Kaho�?�»olawe[9] The Target Isle 20�?�°33�¢�¤�²N 156�?�°36�¢�¤�²W�¯�»�¿ / �¯�»�¿20.55 �?�°N 156.6�?�°W�¯�»�¿ / 20.55; -156.6 4 L�?na�?�»i[10] ThePineappleIsle 20�?�°50�¢�¤�²N 156�?�°56�¢�¤�²W�¯�»�¿ / �¯�»�¿20.833�?�°N 15 6.933�?�°W�¯�»�¿ / 20.833; -156.933 5 Moloka�?�»i[11] TheFriendlyIsle 21�?�°08�¢�¤�²N 157�?�°02�¢�¤�²W�¯�»�¿ / �¯�»�¿21.133�?�°N 1 57.033�?�°W�¯�»�¿ / 21.133; -157.033 6O�?�»ahu[12]TheGatheringPlace 21�?�°28�¢�¤�²N 157�?�°59�¢�¤�²W�¯�»�¿ / �¯�»�¿21.467�?�°N 1 57.983�?�°W�¯�»�¿ / 21.467; -157.983 7 Kaua�?�»i[13] The Garden Isle 22�?�°05�¢�¤�²N 159�?�°30�¢�¤�²W�¯�»�¿ / �¯�»�¿22.083 �?�°N 159.5�?�°W�¯�»�¿ / 22.083; -159.5 8 Ni�?�»ihau[14]The Forbidden
[R] Boxplot in order
Hi,
I need construct box plot graph, but I want keep Groups
order
karla = data.frame(
Groups =
factor(rep(c('CPre','SPre','C7','S7','C14','S14','C21','S21'),
11)),
Time = rep(c(0,7,14,21), 11),
Resp = valor
)
boxplot(Resp~Groups, order=T)
doesn't work.
How do this?
--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
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[R] cross-correlation table with subscript or superscript to indicate significant differences
Hi, I wonder whether the following is possible with R, and whether anyone has done that and can share his/her code with me. I have a correlation matrix, and I want to create a correlation table that I can copy to Microsoft Word with a superscript above each correlation, indicating significant differences in the same row. That is, when correlations in the same row do not share superscript, it means that they are significantly different from each other. thanks,yoav [[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] Null
There were no problems when I repeated the test with data provided. I simply created a vector -c(1.600972896,1.534026106,1.633468456,1.69019608,1.686636269,1.713490543,1.460897843,1.604226053,1.547774705,1.575187845,1.50242712,1.489958479,1.555094449,1.56937391,1.46686762,1.583198774,1.59439255,1.627365857,1.596597096,1.598790507,1.596597096,1.613841822,1.607455023,1.586587305,1.72427587,1.668385917,1.743509765,1.5774918,1.709269961,1.507855872,1.650307523,1.670245853,1.721810615,1.613841822,1.586587305,1.658011397,1.595496222,1.662757832,1.521138084,1.564666064,1.515873844,1.596597096,1.617000341,1.621176282,1.598790507,1.73479983,1.498310554,1.571708832,1.426511261,1.698970004,1.534026106,1.5774918,1.682145076,1.689308859,1.654176542,1.526339277,1.545307116,1.658964843,1.638489257,1.557507202,1.604226053,1.627365857,1.651278014,1.627365857,1.559906625,1.720159303,1.64738297,1.62324929,1.698970004,1.704150517,1.57863921,1.558708571,1.681241237,1.539076099,1.5132176) and performed shapiro.test() with following results: W = 0.9876, p-value = 0.677 As far as I understand R, complete.cases has nothing to do with normality check of single vector. pcc wrote: This is probably a very simple question but I am completely stumped!I am trying to do shapiro.wilk(x) test on a relatively small dataset(75) and each time my variable and keeps coming out as 'NULL' -- View this message in context: http://r.789695.n4.nabble.com/Null-tp3498261p3498607.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] Vermunt's LEM in R
Hello- Does anyone know of packages that could emulate what J. Vermunt's LEM does ? What is the closest relative in R ? I use both R and LEM but have trouble transforming my multiway tables in R into a .dat file compatible with LEM. Thanks, David Joubert [[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] Compiling a FORTRAN program under Windows 7
Hi, I am trying to compile a FORTRAN program to call from R under Windows 7 but I am having problem in the compiling step. To demonstrate this is the program testit.f: -- subroutine TESTIT(x,n,m) dimension x(n) do 10 i=1,n 10 x(i)=x(i)**m end When I compile it with gfortran I get the following error: -- c:\MinGW\programsgfortran testit.f -o testit.o c:/mingw/bin/../lib/gcc/mingw32/4.5.2/../../../libmingw32.a(main.o):main.c:(.tex t+0xd2): undefined reference to `WinMain@16' collect2: ld returned 1 exit status. I should add that a program like the following hello.f compiles with no problem. -- READ (*, *) YOURNAME WRITE (*, 200) YOURNAME 200 FORMAT(//,' Hello ',A/) STOP END -- I realize that this is not directly a question about R but I guess there are some people here who have compiled FORTRAN programs under Windows 7 to call from R. I appreciate any help to fix the problem. /Mikael [[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] Confidence interval for difference in Harrell's c statistics (or equivalently Somers' D statistics)
Dear David,
Thank you for your reply. I have come across rcorrp.cens before. However,
I'm not sure it does quite what I want it to. It seems to compare whether
one predictor is more concordant than another within the same survival
function. I want to see whether one predictor is more concordant than
another over two survival functions hence I fitted two rcorrcens functions.
E.g. if I have a development data set with a variable for age and a
validation data set for age then I want to know if the concordance is the
same over the development and validation data sets.
Thank you,
Laura
On 5 May 2011 16:09, David Winsemius dwinsem...@comcast.net wrote:
On May 5, 2011, at 8:20 AM, Laura Bonnett wrote:
Dear All,
I am trying to calculate a 95% confidence interval for the difference in
two
c statistics (or equivalently D statistics). In Stata I gather that this
can be done using the lincom command. Is there anything similar in R?
Have you looked at rcorrp.cens {Hmisc}?
snipped
[[alternative HTML version deleted]]
Please post in plain text.
--
David Winsemius, MD
West Hartford, CT
[[alternative HTML version deleted]]
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[R] perspective plot
Hello, I`m tryint to plot some variable over x and y coordinate, but can`t figure out hot to do it ( I could do a simple scatterplot3d, but there I can`t change the viewing angle). My dataset looks something like this (dataframe): X Y Q95 21 2628711 1104437 0.7723994 22 2628721 1104437 0.5961789 23 2628731 1104437 1.2013182 24 2628741 1104437 1.3468632 25 2628751 1104437 1.1035517 26 2628761 1104437 1.0528809 Is there a way to plot Q95 over x and y? Something like persp(x,y,z) would be nice, but can`t figure out the right input format for this function. Thanks a lot, Mauro -- View this message in context: http://r.789695.n4.nabble.com/perspective-plot-tp3498754p3498754.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] cross-correlation table with subscript or superscript to indicate significant differences
On May 5, 2011, at 10:48 AM, yoav baranan wrote: Hi, I wonder whether the following is possible with R, and whether anyone has done that and can share his/her code with me. I have a correlation matrix, and I want to create a correlation table that I can copy to Microsoft Word with a superscript above each correlation, indicating significant differences in the same row. That is, when correlations in the same row do not share superscript, it means that they are significantly different from each other. thanks,yoav [[alternative HTML version deleted]] An example with data and the desired result might help focus the discussion. This shows how to set up an example showing how extract the row numbers from a correlation matrix with absolute values above 0.5 but less than 1 (to exclude the trivial cases). set.seed(123) X - matrix(rnorm(100), 10) apply(cor(X), 2, function(x) which(abs(x) 0.5 x 1) ) [[1]] [1] 2 4 8 [[2]] [1] 1 3 [[3]] [1] 2 6 9 [[4]] [1] 1 7 [[5]] integer(0) [[6]] [1] 3 10 [[7]] [1] 4 [[8]] [1] 1 [[9]] [1] 3 [[10]] [1] 6 This would extract the rownames if they are letters[1:10] lapply( apply(cor(X), 2, function(x) which(abs(x) 0.5 x 1) ), function(x) rownames(X)[x]) [[1]] [1] b d h [[2]] [1] a c [[3]] [1] b f i [[4]] [1] a g [[5]] character(0) [[6]] [1] c j [[7]] [1] d [[8]] [1] a [[9]] [1] c [[10]] [1] f Exactly how we are supposed to pass this to MS Word does not seem to be a proper question for this mailing list. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] simulate AR(1) process
Dear R users,
May any of you tell me how to simulate data on:
y_t = a+b*y_{t-1} + u_t
where u_t~N(0,sigma^2), b1, and for some constant a.
Many thanks
Tsegaye tseg...@exeter.ac.uk
[[alternative HTML version deleted]]
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Re: [R] Compiling a FORTRAN program under Windows 7
You are compiling a subroutine not a program and you compile line should read: gfortran testit.f -c testit.o You then reference that object code testit.o in your final loading stage after compiling other routiens and the main program. -- Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Thu, 5 May 2011, Mikael Anderson wrote: Hi, I am trying to compile a FORTRAN program to call from R under Windows 7 but I am having problem in the compiling step. To demonstrate this is the program testit.f: -- subroutine TESTIT(x,n,m) dimension x(n) do 10 i=1,n 10 x(i)=x(i)**m end When I compile it with gfortran I get the following error: -- c:\MinGW\programsgfortran testit.f -o testit.o c:/mingw/bin/../lib/gcc/mingw32/4.5.2/../../../libmingw32.a(main.o):main.c:(.tex t+0xd2): undefined reference to `WinMain@16' collect2: ld returned 1 exit status. I should add that a program like the following hello.f compiles with no problem. -- READ (*, *) YOURNAME WRITE (*, 200) YOURNAME 200 FORMAT(//,' Hello ',A/) STOP END -- I realize that this is not directly a question about R but I guess there are some people here who have compiled FORTRAN programs under Windows 7 to call from R. I appreciate any help to fix the problem. /Mikael [[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] simulate AR(1) process
?arima.sim
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Thu, 5 May 2011, Alemtsehai Abate wrote:
Dear R users,
May any of you tell me how to simulate data on:
y_t = a+b*y_{t-1} + u_t
where u_t~N(0,sigma^2), b1, and for some constant a.
Many thanks
Tsegaye tseg...@exeter.ac.uk
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Null
Maybe you were doing something like
fcv - read.csv('fcv.csv')
instead of
fcv - read.csv('fcv.csv')[1]
(I haven't tested this.)
Tom
On Thu, May 5, 2011 at 8:48 AM, pcc polly...@hotmail.com wrote:
This is probably a very simple question but I am completely stumped!I am
trying to do shapiro.wilk(x) test on a relatively small dataset(75) and each
time my variable and keeps coming out as 'NULL', and
shapiro.test(fcv)
Error in complete.cases(x) : no input has determined the number of cases
my text file looks like this:
case
1.600972896
1.534026106
1.633468456
1.69019608
1.686636269
1.713490543
1.460897843
1.604226053
1.547774705
1.575187845
1.50242712
1.489958479
1.555094449
1.56937391
1.46686762
1.583198774
1.59439255
1.627365857
1.596597096
1.598790507
1.596597096
1.613841822
1.607455023
1.586587305
1.72427587
1.668385917
1.743509765
1.5774918
1.709269961
1.507855872
1.650307523
1.670245853
1.721810615
1.613841822
1.586587305
1.658011397
1.595496222
1.662757832
1.521138084
1.564666064
1.515873844
1.596597096
1.617000341
1.621176282
1.598790507
1.73479983
1.498310554
1.571708832
1.426511261
1.698970004
1.534026106
1.5774918
1.682145076
1.689308859
1.654176542
1.526339277
1.545307116
1.658964843
1.638489257
1.557507202
1.604226053
1.627365857
1.651278014
1.627365857
1.559906625
1.720159303
1.64738297
1.62324929
1.698970004
1.704150517
1.57863921
1.558708571
1.681241237
1.539076099
1.5132176
Any ideas?
--
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Re: [R] Using error in histograms
The logspline package does density estimation in a different way than KDE, but it does allow for interval censored data (I know this value is between a and b, but not where in that range) using the oldlogspline function. This may be what you are looking for. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Kristof Ostir Sent: Thursday, May 05, 2011 4:16 AM To: r-help@r-project.org Subject: [R] Using error in histograms Hello! I am trying to produce a histogram of measurement data (orientation of archaeological structures) that are a subject to measurement error. The normal histogram just computes frequencies, but does not take into account that a particular value is spread over a range of values (in my case the spread is different for reach measurement and is larger than the bin size). The closest approach is kernel density estimation (in the image is a comparison of histogram and KDE): http://en.wikipedia.org/wiki/Kernel_density_estimation http://en.wikipedia.org/wiki/File:Comparison_of_1D_histogram_and_KDE.pn g However in my case the kernel size is different for each value. I wrote a program in IDL that performs the plotting, but am just wondering if such a function is available in R. Basically it is a problem of summing (and later plotting) several data distributions. I would appreciate also any hint to a book that might be dealing with the problem. I am not an expert in statistics and I might not be using use the correct terminology in web searches. Regards, Kristof __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] quantmod's addTA plotting functions
Thanks. You're right. I didn't see that. I read the ?addTA help page, which
(annoyingly) didn't mention that feature, but I didn't read the ?TA page.
(That page was mentioned as a see also, but not as a must see.)
I don't know what it means to wrap these calls in a plot call. I tried to
put the addTA calls into a function and call that function from the higher
level function, but that didn't work either. Would you tell me what it means
to wrap these calls in a plot call.
Thanks
*-- Russ *
P.S. Pardon my irritation, but I continually find that many of the help
files assume one already knows the information one is looking for. If you
don't know it, the help files are not very helpful. This is a good example.
In fact, it's two good examples. I didn't know that I had to look at
another page, and I (still) don't know what it means to wrap plot calls in
another plot call.
On Thu, May 5, 2011 at 3:39 AM, P Ehlers ehl...@ucalgary.ca wrote:
On 2011-05-05 0:47, Russ Abbott wrote:
Hi,
I'm having trouble with quantmod's addTA plotting functions. They seem to
work fine when run from the command line. But when run inside a function,
only the last one run is visible. Here's an example.
test.addTA- function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close- GSPC[,GSPC.Close]
GSPC.EMA.3- EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10- EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'), up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10 appears
on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser() neither
appears
when the browser() interrupt occurs. I can then draw both GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All intended plots
are
visible after the function terminates. So it isn't as if one wipes out the
other. This shows that it's possible to get all three lines on the plot,
but
I can't figure out how to do it without manual intervention. Any
suggestions
are appreciated.
Perhaps you didn't see this NOTE on the ?TA help page:
Calling any of the above methods from within a function
or script will generally require them to be wrapped in a
plot call as they rely on the context of the call to
initiate the actual charting addition.
Peter Ehlers
Thanks.
*-- Russ *
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compiling a FORTRAN program under Windows 7
Mikael Anderson wrote: Hi, I am trying to compile a FORTRAN program to call from R under Windows 7 but I am having problem in the compiling step. To demonstrate this is the program testit.f: -- subroutine TESTIT(x,n,m) dimension x(n) do 10 i=1,n 10 x(i)=x(i)**m end In addition to the previous remarks, you would do yourself a favour by utilizing the implicit-none option of gfortran. That will force you to declare every variable (and its type) which will avoid many nasty bugs. In your case: x is a REAL single precision but with R you will preferably need double precision. So I would advise you to use gfortran -fimplicit-none .. Berend -- View this message in context: http://r.789695.n4.nabble.com/Compiling-a-FORTRAN-program-under-Windows-7-tp3498663p3498839.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] Null
with(fcv,shapiro.test(case))
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600
USPS: PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274
On Thu, 5 May 2011, Thomas Levine wrote:
Maybe you were doing something like
fcv - read.csv('fcv.csv')
instead of
fcv - read.csv('fcv.csv')[1]
(I haven't tested this.)
Tom
On Thu, May 5, 2011 at 8:48 AM, pcc polly...@hotmail.com wrote:
This is probably a very simple question but I am completely stumped!I am
trying to do shapiro.wilk(x) test on a relatively small dataset(75) and each
time my variable and keeps coming out as 'NULL', and
shapiro.test(fcv)
Error in complete.cases(x) : no input has determined the number of cases
my text file looks like this:
case
1.600972896
1.534026106
1.633468456
1.69019608
1.686636269
1.713490543
1.460897843
1.604226053
1.547774705
1.575187845
1.50242712
1.489958479
1.555094449
1.56937391
1.46686762
1.583198774
1.59439255
1.627365857
1.596597096
1.598790507
1.596597096
1.613841822
1.607455023
1.586587305
1.72427587
1.668385917
1.743509765
1.5774918
1.709269961
1.507855872
1.650307523
1.670245853
1.721810615
1.613841822
1.586587305
1.658011397
1.595496222
1.662757832
1.521138084
1.564666064
1.515873844
1.596597096
1.617000341
1.621176282
1.598790507
1.73479983
1.498310554
1.571708832
1.426511261
1.698970004
1.534026106
1.5774918
1.682145076
1.689308859
1.654176542
1.526339277
1.545307116
1.658964843
1.638489257
1.557507202
1.604226053
1.627365857
1.651278014
1.627365857
1.559906625
1.720159303
1.64738297
1.62324929
1.698970004
1.704150517
1.57863921
1.558708571
1.681241237
1.539076099
1.5132176
Any ideas?
--
View this message in context:
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Insert values to histogram
Are you really sure that you want to do that? Read the discussion starting with this post: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22858.html for reasons why you probably don't (yes, the question is about bar plots not histograms, but much of it will still apply). Near the end of the discussion there are examples of alternatives and some ways to add that may apply to your question if you still feel the need. Part of the answer depends on how you are creating your histogram in the first place, 3 different functions pop to my mind that create histograms (and I am sure there are plenty more if I looked), how to add numbers in each case would be very different, so any help we offered (beyond generalities mentioned above) could be more misleading than helpful without that information. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of matibie Sent: Thursday, May 05, 2011 5:50 AM To: r-help@r-project.org Subject: [R] Insert values to histogram I'm trying to add the exact value on top of each column of an Histogram, i have been trying with the text function but it doesn't work. The problem is that the program it self decides the exact value to give to each column, and ther is not like in a bar-plot that I know exactly which values are been plotting. If anyone have any new idea on how to do this Thanks Matias -- View this message in context: http://r.789695.n4.nabble.com/Insert- values-to-histogram-tp3498140p3498140.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] quantmod's addTA plotting functions
There is a struggle in documentation that revolves around being too brief to
be useful and too verbose which then is often ignored. In general, R
proper is far less verbose than quantmod docs - so if you have trouble
with quantmod...
from ?addTA
Value:
addTA will invisibly return an S4 object of class chobTA. If
this function is called interactively, the chobTA object will be
evaluated and added to the current chart.
newTA will return a function object that can either be assigned
or evaluated. Evaluating this function will follow the logic of
any standard addTA-style call, returning invisibly a chobTA
object, or adding to the chart.
In addition there is a see also that references ?TA
One can also use the archives to see this problem discussed over and over.
It isn't for lack of documentation typically - it is for lack of looking.
That said, could the docs be more clear/verbose? Certainly. Does that take
time? Certainly. Does answering the same question in a public forum take
time? Certainly. What the latter does though is provide for an excellent
reference for those who misunderstand or don't understand the 'Value' aspect
of the docs (which here explains quite clearly that nothing is printed by
the call itself).
A suitably phrased patch for the docs is one way to contribute in a
productive manner - though it may not always be applied by those you submit
it to (to me or any of the other 3000 package author/maintainers/etc).
HTH
Jeff
On Thu, May 5, 2011 at 11:42 AM, Russ Abbott russ.abb...@gmail.com wrote:
Thanks. You're right. I didn't see that. I read the ?addTA help page,
which (annoyingly) didn't mention that feature, but I didn't read the ?TA
page. (That page was mentioned as a see also, but not as a must see.)
I don't know what it means to wrap these calls in a plot call. I tried to
put the addTA calls into a function and call that function from the higher
level function, but that didn't work either. Would you tell me what it means
to wrap these calls in a plot call.
Thanks
*-- Russ *
P.S. Pardon my irritation, but I continually find that many of the help
files assume one already knows the information one is looking for. If you
don't know it, the help files are not very helpful. This is a good example.
In fact, it's two good examples. I didn't know that I had to look at
another page, and I (still) don't know what it means to wrap plot calls in
another plot call.
On Thu, May 5, 2011 at 3:39 AM, P Ehlers ehl...@ucalgary.ca wrote:
On 2011-05-05 0:47, Russ Abbott wrote:
Hi,
I'm having trouble with quantmod's addTA plotting functions. They seem
to
work fine when run from the command line. But when run inside a function,
only the last one run is visible. Here's an example.
test.addTA- function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close- GSPC[,GSPC.Close]
GSPC.EMA.3- EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10- EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'), up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10 appears
on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser() neither
appears
when the browser() interrupt occurs. I can then draw both GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All intended plots
are
visible after the function terminates. So it isn't as if one wipes out
the
other. This shows that it's possible to get all three lines on the plot,
but
I can't figure out how to do it without manual intervention. Any
suggestions
are appreciated.
Perhaps you didn't see this NOTE on the ?TA help page:
Calling any of the above methods from within a function
or script will generally require them to be wrapped in a
plot call as they rely on the context of the call to
initiate the actual charting addition.
Peter Ehlers
Thanks.
*-- Russ *
[[alternative HTML version deleted]]
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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.
--
Jeffrey Ryan
jeffrey.r...@lemnica.com
www.lemnica.com
[[alternative HTML version deleted]]
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Re: [R] Boxplot in order
Hi:
Try this:
karla = data.frame(
Groups = factor(rep(c('CPre','SPre','C7','S7','C14','S14','C21','S21'), 11),
levels =
c('CPre','SPre','C7','S7','C14','S14','C21','S21')),
Time = rep(c(0,7,14,21), 11),
Resp = rnorm(88)
)
boxplot(Resp~Groups, data = karla)
Since you didn't have a variable valor defined, I substituted in
random normal deviates.
HTH,
Dennis
On Thu, May 5, 2011 at 7:41 AM, Silvano silv...@uel.br wrote:
Hi,
I need construct box plot graph, but I want keep Groups order
karla = data.frame(
Groups = factor(rep(c('CPre','SPre','C7','S7','C14','S14','C21','S21'),
11)),
Time = rep(c(0,7,14,21), 11),
Resp = valor
)
boxplot(Resp~Groups, order=T)
doesn't work.
How do this?
--
Silvano Cesar da Costa
Departamento de Estatística
Universidade Estadual de Londrina
Fone: 3371-4346
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Re: [R] Using functions/loops for repetitive commands
Thanks David, I did notice that and I got his code to work using wilcox.test for the continuous variables. The problem is that when I tried to alter the code to do chisq.test on my categorical variables there is something wrong with the syntax and I don't know what. Derek -- View this message in context: http://r.789695.n4.nabble.com/Using-functions-loops-for-repetitive-commands-tp3498006p3498896.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] perspective plot
The persp function expects z to be a matrix, so you could reshape your data so that z is a matrix (the reshape function or package may help). Or the wireframe function in the lattice package expects data more like what you show, that may be the easiest solution. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Mauro Sent: Thursday, May 05, 2011 10:09 AM To: r-help@r-project.org Subject: [R] perspective plot Hello, I`m tryint to plot some variable over x and y coordinate, but can`t figure out hot to do it ( I could do a simple scatterplot3d, but there I can`t change the viewing angle). My dataset looks something like this (dataframe): X Y Q95 21 2628711 1104437 0.7723994 22 2628721 1104437 0.5961789 23 2628731 1104437 1.2013182 24 2628741 1104437 1.3468632 25 2628751 1104437 1.1035517 26 2628761 1104437 1.0528809 Is there a way to plot Q95 over x and y? Something like persp(x,y,z) would be nice, but can`t figure out the right input format for this function. Thanks a lot, Mauro -- View this message in context: http://r.789695.n4.nabble.com/perspective-plot-tp3498754p3498754.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] quantmod's addTA plotting functions
Russ,
All you have to do is replace
addTA(GSPC.EMA.3, on = 1, col = #ff)
with
plot(addTA(GSPC.EMA.3, on = 1, col = #ff))
etc.
I can sympathize with the documentation frustration, but I think
that much of the documentation in R and in many R packages is
actually very good. I get much more frustrated with the attempts
at 'non-technical' explanations I find in other software. It
does take a bit of getting used to always looking at the Value
section and, if in doubt, checking some of the See Alsos, but
it's worth it. I don't know quantmod very well, but even a
cursory look at the pdf file shows that the docs are quite
good.
As Jeff points out, good documentation is not easy. More good
examples are always better, but that's mighty time-consuming.
Peter Ehlers
On 2011-05-05 10:42, Russ Abbott wrote:
Thanks. You're right. I didn't see that. I read the ?addTA help page,
which (annoyingly) didn't mention that feature, but I didn't read the
?TA page. (That page was mentioned as a see also, but not as a must see.)
I don't know what it means to wrap these calls in a plot call. I tried
to put the addTA calls into a function and call that function from the
higher level function, but that didn't work either. Would you tell me
what it means to wrap these calls in a plot call.
Thanks
/-- Russ /
P.S. Pardon my irritation, but I continually find that many of the help
files assume one already knows the information one is looking for. If
you don't know it, the help files are not very helpful. This is a good
example. In fact, it's two good examples. I didn't know that I had to
look at another page, and I (still) don't know what it means to wrap
plot calls in another plot call.
On Thu, May 5, 2011 at 3:39 AM, P Ehlers ehl...@ucalgary.ca
mailto:ehl...@ucalgary.ca wrote:
On 2011-05-05 0:47, Russ Abbott wrote:
Hi,
I'm having trouble with quantmod's addTA plotting functions.
They seem to
work fine when run from the command line. But when run inside a
function,
only the last one run is visible. Here's an example.
test.addTA- function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close- GSPC[,GSPC.Close]
GSPC.EMA.3- EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10- EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'),
up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10
appears on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser()
neither appears
when the browser() interrupt occurs. I can then draw both
GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All
intended plots are
visible after the function terminates. So it isn't as if one
wipes out the
other. This shows that it's possible to get all three lines on
the plot, but
I can't figure out how to do it without manual intervention. Any
suggestions
are appreciated.
Perhaps you didn't see this NOTE on the ?TA help page:
Calling any of the above methods from within a function
or script will generally require them to be wrapped in a
plot call as they rely on the context of the call to
initiate the actual charting addition.
Peter Ehlers
Thanks.
*-- Russ *
[[alternative HTML version deleted]]
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Re: [R] quantmod's addTA plotting functions
Sorry, I forgot to give you a quick example of 'wrapping in plot'
plot(addTA(...))
The objects returned by most of the charting functions in quantmod results
from the desire for the functions to be syntactically identical whether
called from inside of chartSeries (e.g. TA=addMACD() ) as they are from
outside:
R addMACD()
The code the simplicity of this (and the lack of need for documenting 2
different functions, and learning all the details of 2 different functions
(2x the work)) you need to be a bit clever in how R works. Describing that
is 1) not simple and 2) ironically, not very well documented in base R. For
me to write it I just had to get around number 2.
HTH
Jeff
On Thu, May 5, 2011 at 11:55 AM, Jeff Ryan jeff.a.r...@gmail.com wrote:
There is a struggle in documentation that revolves around being too brief
to be useful and too verbose which then is often ignored. In general, R
proper is far less verbose than quantmod docs - so if you have trouble
with quantmod...
from ?addTA
Value:
addTA will invisibly return an S4 object of class chobTA. If
this function is called interactively, the chobTA object will be
evaluated and added to the current chart.
newTA will return a function object that can either be assigned
or evaluated. Evaluating this function will follow the logic of
any standard addTA-style call, returning invisibly a chobTA
object, or adding to the chart.
In addition there is a see also that references ?TA
One can also use the archives to see this problem discussed over and
over. It isn't for lack of documentation typically - it is for lack of
looking.
That said, could the docs be more clear/verbose? Certainly. Does that
take time? Certainly. Does answering the same question in a public forum
take time? Certainly. What the latter does though is provide for an
excellent reference for those who misunderstand or don't understand the
'Value' aspect of the docs (which here explains quite clearly that nothing
is printed by the call itself).
A suitably phrased patch for the docs is one way to contribute in a
productive manner - though it may not always be applied by those you submit
it to (to me or any of the other 3000 package author/maintainers/etc).
HTH
Jeff
On Thu, May 5, 2011 at 11:42 AM, Russ Abbott russ.abb...@gmail.comwrote:
Thanks. You're right. I didn't see that. I read the ?addTA help page,
which (annoyingly) didn't mention that feature, but I didn't read the ?TA
page. (That page was mentioned as a see also, but not as a must see.)
I don't know what it means to wrap these calls in a plot call. I tried to
put the addTA calls into a function and call that function from the higher
level function, but that didn't work either. Would you tell me what it means
to wrap these calls in a plot call.
Thanks
*-- Russ *
P.S. Pardon my irritation, but I continually find that many of the help
files assume one already knows the information one is looking for. If you
don't know it, the help files are not very helpful. This is a good example.
In fact, it's two good examples. I didn't know that I had to look at
another page, and I (still) don't know what it means to wrap plot calls in
another plot call.
On Thu, May 5, 2011 at 3:39 AM, P Ehlers ehl...@ucalgary.ca wrote:
On 2011-05-05 0:47, Russ Abbott wrote:
Hi,
I'm having trouble with quantmod's addTA plotting functions. They seem
to
work fine when run from the command line. But when run inside a
function,
only the last one run is visible. Here's an example.
test.addTA- function(from = 2010-06-01) {
getSymbols(^GSPC, from = from)
GSPC.close- GSPC[,GSPC.Close]
GSPC.EMA.3- EMA(GSPC.close, n=3, ratio=NULL)
GSPC.EMA.10- EMA(GSPC.close, n=10, ratio=NULL)
chartSeries(GSPC.close, theme=chartTheme('white'), up.col=black,
dn.col=black)
addTA(GSPC.EMA.3, on = 1, col = #ff)
addTA(GSPC.EMA.10, on = 1, col = #ff)
# browser()
}
When I run this, GSPC.close always appears. But only GSPC.EMA10 appears
on
the plot along with it. If I switch the order of the addTA calls,
only GSPC.EMA3 appears. If I uncomment the call to browser() neither
appears
when the browser() interrupt occurs. I can then draw both GSPC.EMA.3 and
GSPC.EMA10 manually, and let the function terminate. All intended plots
are
visible after the function terminates. So it isn't as if one wipes out
the
other. This shows that it's possible to get all three lines on the plot,
but
I can't figure out how to do it without manual intervention. Any
suggestions
are appreciated.
Perhaps you didn't see this NOTE on the ?TA help page:
Calling any of the above methods from within a function
or script will generally require them to be wrapped in a
plot call as they rely on the context of the call to
initiate the actual charting addition.
Peter Ehlers
Thanks.
*-- Russ *
[[alternative HTML
Re: [R] Using functions/loops for repetitive commands
On May 5, 2011, at 1:08 PM, dereksloan wrote: Thanks David, I did notice that and I got his code to work using wilcox.test for the continuous variables. The problem is that when I tried to alter the code to do chisq.test on my categorical variables there is something wrong with the syntax and I don't know what. Right ?chisq.test # No mention of a formula argument seen ?chisq.test.formula No documentation for 'chisq.test.formula' in specified packages and libraries: you could try '??chisq.test.formula' `chisq.test` doesn't have a formula method, so sending it a formula will fail. Why aren't you sending it the arguments instead of turning them into strings? Derek -- View this message in context: http://r.789695.n4.nabble.com/Using-functions-loops-for-repetitive-commands-tp3498006p3498896.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 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] Using functions/loops for repetitive commands
Thanks a lot, I understand what you say but I'm having problems - maybe with the syntax or the specific command. You are right - I have a dataframe to store the data and want to automate the analysis. i.e. I want do a chisq.test with to know if alcohol intake (Y/N) differs between sexes, then if smoking (Y/N) differs between sexes, then if alcohol intake or smoking differ by hiv status. The command within my data frame for each individual comparison is e.g. chisq.test(alcohol,sex)... then repeat it for all combination of variables. but using lapply I'm still unsure how to design the loop. I'll keep trying - let me know if you have more ideas. Derek -- View this message in context: http://r.789695.n4.nabble.com/Using-functions-loops-for-repetitive-commands-tp3498006p3499001.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] distance matrix
I don't know whether I understood your question, but 1.1, 1.2 , 1.3 all are subsample of 1 , so, rather than comparing 1000 subsample, comparison of 20 pop level makes more sense in my case. thanks for query - Ananta Acharya Graduate Student -- View this message in context: http://r.789695.n4.nabble.com/distance-matrix-tp3498033p3498890.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] cross-correlation table with subscript or superscript to indicate significant differences
On May 5, 2011, at 12:40 PM, yoav baranan wrote:
Here is an example for my earlier question.
Say you have a 3x3 correlation matrix:
corrs - matrix(c(0.25,0.32,0.66,0.14,0.24,0.34,0.44,0.34,0.11),
nrow=3, ncol=3, dimnames = list(c('varA','varB', 'varC'),
c('varA','varB', 'varC')))
And another matrix for the sample size of each correlation:
sizes - matrix(c(44,68,313,142,144,207,201,100,99), nrow=3, ncol=3,
dimnames = list(c('varA','varB', 'varC'), c('varA','varB', 'varC')))
corrs looks like this:
varA varB varC
varA 0.05 0.14 0.44
varB 0.32 0.24 0.34
varC 0.66 0.57 0.50
sizes:
varA varB varC
varA 44 142 201
varB 68 144 100
varC 313 207 99
i.e., the correlation between variables A and C was 0.66 with sample
size of 313. (I got these tables from rcorr).
Why not offer the result of dput() on the result from rcorrs cone on 3
variables. Then we should have the necessary building blocks for your
original request. This way we do not have the matrix of p-values.
And as the Posting Guide clearly says ...Please post in plain text. In
your case it is particularly annoying because I do not get the
filtered version but rather you html version and the text is almost
unreadable at a font size of 10 in whatever font it is specifying!
What I want to do is to compare the correlations in each row
(probably using r.test), and then create a correlation table with
subscripts or superscripts indicating the significance
group (again: correlations with different superscripts, in the
same row, are significantly different from each other).
Something like this:
varA varB varC
varA 0.05b 0.14b 0.44a
varB 0.32a 0.24a 0.34a
varC 0.66a 0.57ab 0.50b
At the moment we have no way of determining what values should be post-
pended with which letters.
Of course, I don't have a 3x3 table. I have about 20 tables of at
least 7x7 each, so this is why I'm looking for methods to automate
the process.
Thanks,
Yoav
CC: r-help@r-project.org
From: dwinsem...@comcast.net
To: ybara...@hotmail.com
Subject: Re: [R] cross-correlation table with subscript or
superscript to indicate significant differences
Date: Thu, 5 May 2011 12:17:25 -0400
On May 5, 2011, at 10:48 AM, yoav baranan wrote:
Hi, I wonder whether the following is possible with R, and whether
anyone has done that and can share his/her code with me. I have a
correlation matrix, and I want to create a correlation table
that I
can copy to Microsoft Word with a superscript above each
correlation, indicating significant differences in the same row.
That is, when correlations in the same row do not share
superscript,
it means that they are significantly different from each other.
thanks,yoav
[[alternative HTML version deleted]]
An example with data and the desired result might help focus the
discussion.
This shows how to set up an example showing how extract the row
numbers from a correlation matrix with absolute values above 0.5 but
less than 1 (to exclude the trivial cases).
snipped
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Using $ accessor in GAM formula
This is not mission critical, but it's bothering me. I'm getting inconsistent results when I use the $ accessor in the gam formula *In window #1:* library(mgcv) dat=data.frame(x=1:100,y=sin(1:100/50)+rnorm(100,0,.05)) str(dat) gam(dat$y~s(dat$x)) Error in eval(expr, envir, enclos) : object 'x' not found *In window #2:* gm = gam(dat$cf~s(dat$s)) gm Family: gaussian Link function: identity Formula: dat$cf ~ s(dat$s) Estimated degrees of freedom: 8.7757 total = 9.77568980091 GCV score: 302.551417213 Has anyone else seen the same thing? In both cases I'm using Windows 7 and R 2.13.0 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cross-correlation table with subscript or superscript to indicate significant differences
Here is an example for my earlier question. Say you have a 3x3 correlation
matrix:corrs - matrix(c(0.25,0.32,0.66,0.14,0.24,0.34,0.44,0.34,0.11), nrow=3,
ncol=3, dimnames = list(c('varA','varB', 'varC'), c('varA','varB', 'varC')))And
another matrix for the sample size of each correlation:sizes -
matrix(c(44,68,313,142,144,207,201,100,99), nrow=3, ncol=3, dimnames =
list(c('varA','varB', 'varC'), c('varA','varB', 'varC')))corrs looks like this:
varA varB varCvarA 0.05 0.14 0.44varB 0.32 0.24 0.34varC 0.66 0.57
0.50sizes: varA varB varCvarA 44 142 201varB 68 144 100varC 313
207 99i.e., the correlation between variables A and C was 0.66 with sample
size of 313. (I got these tables from rcorr).What I want to do is to compare
the correlations in each row (probably using r.test), and then create a
correlation table with subscripts or superscripts indicating the significance
group (again: correlations with different superscripts, in the same row, are
signif!
icantly different from each other). Something like this: varA varB
varCvarA 0.05b 0.14b 0.44avarB 0.32a 0.24a 0.34avarC 0.66a 0.57ab 0.50bOf
course, I don't have a 3x3 table. I have about 20 tables of at least 7x7 each,
so this is why I'm looking for methods to automate the process. Thanks,Yoav
CC: r-help@r-project.org
From: dwinsem...@comcast.net
To: ybara...@hotmail.com
Subject: Re: [R] cross-correlation table with subscript or superscript to
indicate significant differences
Date: Thu, 5 May 2011 12:17:25 -0400
On May 5, 2011, at 10:48 AM, yoav baranan wrote:
Hi, I wonder whether the following is possible with R, and whether
anyone has done that and can share his/her code with me. I have a
correlation matrix, and I want to create a correlation table that I
can copy to Microsoft Word with a superscript above each
correlation, indicating significant differences in the same row.
That is, when correlations in the same row do not share superscript,
it means that they are significantly different from each other.
thanks,yoav
[[alternative HTML version deleted]]
An example with data and the desired result might help focus the
discussion.
This shows how to set up an example showing how extract the row
numbers from a correlation matrix with absolute values above 0.5 but
less than 1 (to exclude the trivial cases).
set.seed(123)
X - matrix(rnorm(100), 10)
apply(cor(X), 2, function(x) which(abs(x) 0.5 x 1) )
[[1]]
[1] 2 4 8
[[2]]
[1] 1 3
[[3]]
[1] 2 6 9
[[4]]
[1] 1 7
[[5]]
integer(0)
[[6]]
[1] 3 10
[[7]]
[1] 4
[[8]]
[1] 1
[[9]]
[1] 3
[[10]]
[1] 6
This would extract the rownames if they are letters[1:10]
lapply( apply(cor(X), 2, function(x) which(abs(x) 0.5 x 1) ),
function(x) rownames(X)[x])
[[1]]
[1] b d h
[[2]]
[1] a c
[[3]]
[1] b f i
[[4]]
[1] a g
[[5]]
character(0)
[[6]]
[1] c j
[[7]]
[1] d
[[8]]
[1] a
[[9]]
[1] c
[[10]]
[1] f
Exactly how we are supposed to pass this to MS Word does not seem to
be a proper question for this mailing list.
--
David Winsemius, MD
West Hartford, CT
[[alternative HTML version deleted]]
__
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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] Using functions/loops for repetitive commands
On May 5, 2011, at 1:45 PM, dereksloan wrote: Thanks a lot, I understand what you say but I'm having problems - maybe with the syntax or the specific command. You are right - I have a dataframe to store the data and want to automate the analysis. i.e. I want do a chisq.test with to know if alcohol intake (Y/N) differs between sexes, then if smoking (Y/N) differs between sexes, then if alcohol intake or smoking differ by hiv status. The command within my data frame for each individual comparison is e.g. chisq.test(alcohol,sex)... then repeat it for all combination of variables. I don't generally answer questions that support shotgun approaches to manufacturing p-values for fear of encouraging unprincipled data- ming ... unless it is clear that the questioner understands what he are doing from a statistical point of view. So my apologies. I probably shouldn't have even posted in this case. I misunderstood the question and thought it was just a quick syntactic fix. I now understand it to be more involved and really demands more care and respect than I was giving it. but using lapply I'm still unsure how to design the loop. I'll keep trying - let me know if you have more ideas. Derek -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using $ accessor in GAM formula
On May 5, 2011, at 1:08 PM, Gene Leynes wrote: This is not mission critical, but it's bothering me. I'm getting inconsistent results when I use the $ accessor in the gam formula *In window #1:* library(mgcv) dat=data.frame(x=1:100,y=sin(1:100/50)+rnorm(100,0,.05)) str(dat) gam(dat$y~s(dat$x)) Error in eval(expr, envir, enclos) : object 'x' not found I get the same error, but using the standard R approach to passing data and formulae to regression functions I have not difficulty: gam(y~s(x), data=dat) Family: gaussian Link function: identity Formula: y ~ s(x) Estimated degrees of freedom: 4.1552 total = 5.155229 GCV score: 0.002242771 *In window #2:* gm = gam(dat$cf~s(dat$s)) gm Family: gaussian Link function: identity Formula: dat$cf ~ s(dat$s) Estimated degrees of freedom: 8.7757 total = 9.77568980091 GCV score: 302.551417213 Has anyone else seen the same thing? In both cases I'm using Windows 7 and R 2.13.0 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R CMD check warning
Dear All,
I am trying to build a package for a set of functions. I am
able to build the package and its working fine. When I check it with
R CMD check
I get a following warning : no visible global function
definition for biocLite
I have used biocLite to load a user defined library from
within a function if that library is not pre-installed
if(is.element(annotpkg, installed.packages()[,1]) == FALSE){
source(http://www.bioconductor.org/biocLite.R;)
biocLite(annotpkg)
library(annotpkg,character.only=TRUE)
}
Should I ignore this error or there is a workaround for the
warning. My package is working fine though still I guess the warning has to
have significance. Please help in clarifying this
warning.
[[alternative HTML version deleted]]
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[R] ANOVA
Hello R-Help How can i exctact and store the within group mean squared difference from an anova summary table into a varible. [[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.
