[R] How to combine the results of different survey?
I have results of different surveys (years) for different species that I called NAMA. I want to combining the results of those different surveys to produce one overall result by using meta-analysis I tried to implement my data using the R package metaplus. I used as Vector of observed effect sizes corresponding to each study (yi) the years (colomns). I used as Vector of character strings corresponding to each study (slab) the species (lines). note: I don't have Vector of observed standard errors corresponding to each study (sei) NAMA.meta <- metaplus(yi= 2011 ~ 2019, slab = 1 ~ 113, data=NAMA) The expected results should be the parameter estimates & forest plot. However, I receive an error message: Error in terms.formula(object, data = data) : terme incorrect dans une formule de mod�le In addition: Warning message: In metaplus(yi = 2011 ~ 2019, slab = 1 ~ 113, data = NAMA) : Very few studies. Solution may be unstable. Thanks for your help, Damien [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] flexmix - concomitant model and significance of variables
Dear All, I am new to this forum and to flexmix. I am using flexmix to make a cluster analysis (Model Based). The data for the clustering are all continuous (although all between 0 and 1). I also want to see the correlation between found clusters and some socio economic variables, so I am using a concomitant model The formulation is: Conc<- FLXmultinom(~factor(Area)+factor(income)+Gender+factor(Education)) f2c <- flexmix(cbind(ECO, SOC, ENV, CULT)~1, k=5, model=FLXMCmvnorm(), concomitant=Conc, data=data) To recover the influence of socio-economic variables from the output I am using: f2c@concomitant@coef However, how do I know which variables are significant? Any help is welcomed! Best Damien -- View this message in context: http://r.789695.n4.nabble.com/flexmix-concomitant-model-and-significance-of-variables-tp4712809.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] incrementation within ifelse
Dear R-helper,
I am working on a very large data frame and I am trying to add a new column
and write in it with certain conditions. I have try to use this code with
the data frame p :
ID = 0
p[,newColumn]-
ifelse (p$flagFoehn3_durr == 1,
ifelse(p$Guetsch == 0,
ID - ID ++
,
ID
)
,
0
)
What I am trying to do is to increment the ID when p$Guetsch == 0 and to
put this result in the column. The problem is that ID does not increment
itself.
An other way is to use a loop for like this example :
ID = 0
for (s in 1:(nrow(z))){
z[s,newColumn]-
if (z$flagFoehn3_durr[s] == 1){
if(z$flagFoehn3_durr[s-1] == 0){
ID -ID+1
}else{
ID
}
}else{
0
}
}
This work perfectly, but the problem is that it will take me more than a
month to run it.
Is there a way to increment with the first code I used or a way of running
the second code faster (I have more than 1 million rows)
Thanks!
Cheers,
Damien
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[R] customize packages' help index ( 00index.html file )
Hi all, I'm writing my packages helps files and I'm not really satisfied by the visual results. I'm would like to make subsections in a package function help index file. I would like for example to put all S4 object documentation link together, then all the getters function.. and so on.. Does someone know if it's possible to do it? Is it possible to define by myself the html/00index.html file that will be use in my package? If it's not possible, how could I add the alphabetic subsections that exist in most of packages index help files? Best, Damien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] random between two values
Hi, I would obtain a random value between two (for example between 40.15 and 56.58 I would have only one value). I'm looking for a package/a function which could do this. Could anybody help me please? Cordialement Damien Landais __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] strange dlply behavior
I'm running R 2.9.1 on winXP, using the library plyr. Can anyone explain to me what is going wrong in this code? (in particular see lines marked with **) Trying to modify objects in a list created using dlply seems to corrupt the objects in the list. library(plyr) d=as.data.frame(cbind(c(1,1,1,2,2,2),c(1,2,3,4,5,6))) d V1 V2 1 1 1 2 1 2 3 1 3 4 2 4 5 2 5 6 2 6 c=dlply(d,.(V1)) c [[1]] V1 V2 1 1 1 2 1 2 3 1 3 [[2]] V1 V2 4 2 4 5 2 5 6 2 6 ## display an element from the second data frame c[[2]][2,2] [1] 5 ## change element in the second data from c[[2]][2,2]=10 c [[1]] V1 V2 21 2** 2.1 1 2 ** What happened to V2? 2.2 1 2 ** [[2]] V1 V2 4 2 4 NA NA NA ** 6 2 6 ##Try again with first data frame c=dlply(d,.(V1)) c[[1]][2,2]=10 ** c [[1]] NULL * YIKES! ##Try again but copy c into a new list k c=dlply(d,.(V1)) k=list(c[[1]],c[[2]]) k[[1]] V1 V2 1 1 1 2 1 2 3 1 3 k[[2]][2,2]=10 k [[1]] V1 V2 1 1 1 2 1 2 3 1 3 [[2]] V1 V2 4 2 4 5 2 10 *** 6 2 6 k[[1]][2,2]=10 k [[1]] V1 V2 1 1 1 2 1 10 *** 3 1 3 [[2]] V1 V2 4 2 4 5 2 10 6 2 6 [[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] plm Issues
Duh! Thanks and good advice. I was using 2.7.2 because it was, until recently, the latest version working with RPy (http://rpy.sourceforge.net/). Also didn't realize plm was still actively developed. Interesting that since plm now correctly handles diff and lag operations, it actually breaks with the behavior of lm: a=ts(c(1,2,4)) lm(a~diff(a)) Error in model.frame.default(formula = a ~ diff(a), drop.unused.levels = TRUE) : variable lengths differ (found for 'diff(a)') To regress a on its difference, one needs the more laborious: a=ts(c(1,2,4)) adata=as.data.frame(cbind(a,diff(a))) colnames(adata)=c(a,diffa) lm(a~diffa,data=adata) Call: lm(formula = a ~ diffa, data = adata) Coefficients: (Intercept)diffa 02 From the R help Fitting Linear ModelsUsing time series Considerable care is needed when using lm with time series. Unless na.action = NULL, the time series attributes are stripped from the variables before the regression is done. (This is necessary as omitting NAs would invalidate the time series attributes, and if NAs are omitted in the middle of the series the result would no longer be a regular time series.) Even if the time series attributes are retained, they are not used to line up series, so that the time shift of a lagged or differenced regressor would be ignored. It is good practice to prepare a data argument by ts.intersectts.union.html(..., dframe = TRUE), then apply a suitable na.action to that data frame and call lm with na.action = NULL so that residuals and fitted values are time series. On Sat, Jul 11, 2009 at 10:53 PM, milton ruser milton.ru...@gmail.comwrote: The first think one need to do when has a so old version, is update it :-) After, if the problem remain, try get help with the colleagues. best milton On Thu, Jul 9, 2009 at 10:58 AM, Damien Moore damienlmo...@gmail.comwrote: Hi List I'm having difficulty understanding how plm should work with dynamic formulas. See the commands and output below on a standard data set. Notice that the first summary(plm(...)) call returns the same result as the second (it shouldn't if it actually uses the lagged variable requested). The third call results in error (trying to use diff'ed variable in regression) Other info: I'm running R 2.7.2 on WinXP cheers *data(Gasoline,package=Ecdat) Gasoline_plm-plm.data(Gasoline,c(country,year)) pdim(Gasoline_plm) **Balanced Panel: n=18, T=19, N=342 * *summary(plm(lgaspcar~lincomep,data=Gasoline_plm**)) **Oneway (individual) effect Within Model Call: plm(formula = lgaspcar ~ lincomep, data = Gasoline_plm) Balanced Panel: n=18, T=19, N=342 Residuals : Min. 1st Qu. Median 3rd Qu. Max. -0.40100 -0.08410 -0.00858 0.08770 0.73400 Coefficients : Estimate Std. Error t-value Pr(|t|) lincomep -0.761830.03535 -21.551 2.2e-16 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Total Sum of Squares: 17.061 Residual Sum of Squares: 6.9981 Multiple R-Squared: 0.58981 F-statistic: 464.442 on 323 and 1 DF, p-value: 0.036981 ** summary(plm(lgaspcar~lag(lincomep),data=Gasoline_plm)) **Oneway (individual) effect Within Model Call: plm(formula = lgaspcar ~ lag(lincomep), data = Gasoline_plm) Balanced Panel: n=18, T=19, N=342 Residuals : Min. 1st Qu. Median 3rd Qu. Max. -0.40100 -0.08410 -0.00858 0.08770 0.73400 Coefficients : Estimate Std. Error t-value Pr(|t|) lag(lincomep) -0.761830.03535 -21.551 2.2e-16 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Total Sum of Squares: 17.061 Residual Sum of Squares: 6.9981 Multiple R-Squared: 0.58981 F-statistic: 464.442 on 323 and 1 DF, p-value: 0.036981 * *summary(plm(lgaspcar~diff(lincomep),data=Gasoline_plm))* *Error in model.frame.default(formula = lgaspcar ~ diff(lincomep), data = mydata, : variable lengths differ (found for 'diff(lincomep)') * [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] plm Issues
Hi List I'm having difficulty understanding how plm should work with dynamic formulas. See the commands and output below on a standard data set. Notice that the first summary(plm(...)) call returns the same result as the second (it shouldn't if it actually uses the lagged variable requested). The third call results in error (trying to use diff'ed variable in regression) Other info: I'm running R 2.7.2 on WinXP cheers *data(Gasoline,package=Ecdat) Gasoline_plm-plm.data(Gasoline,c(country,year)) pdim(Gasoline_plm) **Balanced Panel: n=18, T=19, N=342 * *summary(plm(lgaspcar~lincomep,data=Gasoline_plm**)) **Oneway (individual) effect Within Model Call: plm(formula = lgaspcar ~ lincomep, data = Gasoline_plm) Balanced Panel: n=18, T=19, N=342 Residuals : Min. 1st Qu. Median 3rd Qu. Max. -0.40100 -0.08410 -0.00858 0.08770 0.73400 Coefficients : Estimate Std. Error t-value Pr(|t|) lincomep -0.761830.03535 -21.551 2.2e-16 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Total Sum of Squares: 17.061 Residual Sum of Squares: 6.9981 Multiple R-Squared: 0.58981 F-statistic: 464.442 on 323 and 1 DF, p-value: 0.036981 ** summary(plm(lgaspcar~lag(lincomep),data=Gasoline_plm)) **Oneway (individual) effect Within Model Call: plm(formula = lgaspcar ~ lag(lincomep), data = Gasoline_plm) Balanced Panel: n=18, T=19, N=342 Residuals : Min. 1st Qu. Median 3rd Qu. Max. -0.40100 -0.08410 -0.00858 0.08770 0.73400 Coefficients : Estimate Std. Error t-value Pr(|t|) lag(lincomep) -0.761830.03535 -21.551 2.2e-16 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Total Sum of Squares: 17.061 Residual Sum of Squares: 6.9981 Multiple R-Squared: 0.58981 F-statistic: 464.442 on 323 and 1 DF, p-value: 0.036981 * *summary(plm(lgaspcar~diff(lincomep),data=Gasoline_plm))* *Error in model.frame.default(formula = lgaspcar ~ diff(lincomep), data = mydata, : variable lengths differ (found for 'diff(lincomep)') * [[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] to raise in a loop more than 1
I would raise x,y and z in a loop but I won't raise of 1. I tried this but it
doesn't work
mydata=matrix(nrow=1500,ncol=3)
i=1
for(x in 0:10){
for(y in 0:20){
for(z in 0:10){
mydata[i,]=c(x,y,z)
i=i+1
z=z+2}
y=y+4}
x=x+2}
And I would have something like that
x y z
0 0 0
0 0 2
0 0 4
0 0 6
0 0 8
0 0 10
0 4 0
0 4 2
...
Could anybody help me?
I don't work on a special package to do it...
Thanks
Cordialement
Damien Landais
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[R] Variogram fitted by Cauchy
Hi I'm using R(2.9.0) and gstat package under Windows to plot sample variograms. When I want to fit them, I don't have the possibility to choose Cauchy functions. Under Mathematica for example, we have it and it seems to be the best one to fit the variograms I have I tested the others functions, they can fit but I would,if possible, use the Cauchy one. If anybody knows something about it, I would know, even with another package To show what I want to do, here is my code mydata=read.table(97MHz_15cm_V3.txt) colnames(mydata)=c(x,y,z,E,taille) g=gstat(id=zinc,formula=mydata[,4]~1,locations=~x+y+z,data=mydata) g.var=variogram(g,cutoff=2500,width=100) modele=vgm(model=Exp,range=2000) vario=fit.variogram(g.var,modele) Thanks! Cordialement Damien Landais __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] plot two variograms on a same graph
Hi, I would know how to plot two variograms on a same graph. I can plot one by one but I would draw both on the same one. Is it possible? Do i need any special package? Thanks! Cordialement Damien Landais __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read from url requiring authentication?
René Sachse wrote: Damien schrieb: I'm looking into opening an url on a server which requires authentication. Under a Windows Operating System you could try to start R with the --internet2 option. This worked in my case. Thanks René it did the trick for me too! Best Regards, Damien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Read from url requiring authentication?
On 8 Sep, 20:15, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Mon, 8 Sep 2008, Damien wrote: Hi all, I'm looking into opening an url on a server which requires authentication. After failing to find some kind of connection structure to fill in I turned to explicitly stating the credentials in the url itself (e.g. http://username:[EMAIL PROTECTED]). Sadly this didn't do the trick either and both source() and url() failed trying to resolve the username () Is there anything I missed in the documentation/internet/groups? If not could I maybe add to the existing R functions as it doesn't seem too far of a stretch to allow the username and password in the url string fed to the web server? Look at the RCurl package: it is more like download.file than url, though, and you could perhaps wse the wget method of download.file. Thank you for the quick reply, it seems that the argument --internet2 did solve my immediate problem but I'll have a look at RCurl too. Best Regards, Damien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Read from url requiring authentication?
Hi all, I'm looking into opening an url on a server which requires authentication. After failing to find some kind of connection structure to fill in I turned to explicitly stating the credentials in the url itself (e.g. http://username:[EMAIL PROTECTED]). Sadly this didn't do the trick either and both source() and url() failed trying to resolve the username () Is there anything I missed in the documentation/internet/groups? If not could I maybe add to the existing R functions as it doesn't seem too far of a stretch to allow the username and password in the url string fed to the web server? Thanks, Damien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
