Re: [R] define subset argument for function lm as variable?
Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit : Hi Rainer, You could try: subs - expression(dead==FALSE recTreat==FALSE) lme(formula, subset = eval(subs)) Not tested, but something along those lines should work. Out of curiosity, why isn't subset (and weights, which is very similar in that regard) evaluated in the data environment, just like the formula? Is this for historical reasons, or are there drawbacks to such a feature? It seems very common to pass a data frame via the data argument, and use variables from it for subsetting and/or weighting. Regards Josh On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr wrote: Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit : Hi Rainer, You could try: subs - expression(dead==FALSE recTreat==FALSE) lme(formula, subset = eval(subs)) Not tested, but something along those lines should work. Out of curiosity, why isn't subset (and weights, which is very similar in that regard) evaluated in the data environment, just like the formula? Is this for historical reasons, or are there drawbacks to such a feature? I am not sure about weights offhand, but subset is evaluated in the data environmentthat is why that solution works. The original question was how to setup the expression as an object that was passed to subset. The trick is to avoid having the logical expression evaluated when the object is created, which I avoided by using expression, and then in lme() forcing the evaluation of the object. It seems very common to pass a data frame via the data argument, and use variables from it for subsetting and/or weighting. Regards Josh On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] define subset argument for function lm as variable?
Le mercredi 29 août 2012 à 04:01 -0700, Joshua Wiley a écrit : On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr wrote: Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit : Hi Rainer, You could try: subs - expression(dead==FALSE recTreat==FALSE) lme(formula, subset = eval(subs)) Not tested, but something along those lines should work. Out of curiosity, why isn't subset (and weights, which is very similar in that regard) evaluated in the data environment, just like the formula? Is this for historical reasons, or are there drawbacks to such a feature? I am not sure about weights offhand, but subset is evaluated in the data environmentthat is why that solution works. The original question was how to setup the expression as an object that was passed to subset. The trick is to avoid having the logical expression evaluated when the object is created, which I avoided by using expression, and then in lme() forcing the evaluation of the object. OK, my phrasing was not really correct. What I meant (and what triggered the OP question) was : why doesn't the subset argument behave the same in lm() and in subset.data.frame()? Is there any advantage to evaluating the argument at the object creation? AFAICS, subset.data.frame() merely uses this trick: e - substitute(subset) r - eval(e, x, parent.frame()) I'm probably missing something... ;-) It seems very common to pass a data frame via the data argument, and use variables from it for subsetting and/or weighting. Regards Josh On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
Le mercredi 29 août 2012 à 14:26 +0200, Milan Bouchet-Valat a écrit : Le mercredi 29 août 2012 à 04:01 -0700, Joshua Wiley a écrit : On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr wrote: Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit : Hi Rainer, You could try: subs - expression(dead==FALSE recTreat==FALSE) lme(formula, subset = eval(subs)) Not tested, but something along those lines should work. Out of curiosity, why isn't subset (and weights, which is very similar in that regard) evaluated in the data environment, just like the formula? Is this for historical reasons, or are there drawbacks to such a feature? I am not sure about weights offhand, but subset is evaluated in the data environmentthat is why that solution works. The original question was how to setup the expression as an object that was passed to subset. The trick is to avoid having the logical expression evaluated when the object is created, which I avoided by using expression, and then in lme() forcing the evaluation of the object. OK, my phrasing was not really correct. What I meant (and what triggered the OP question) was : why doesn't the subset argument behave the same in lm() and in subset.data.frame()? Is there any advantage to evaluating the argument at the object creation? Nevermind, forget this silly question. This works exactly as I describe it, it's just that I did not get the OP's problem right, and for an unexplained reason in my testing this did not work. But now I realize, as you said, the problem is just that the OP wanted to store the subset in an object first. Sorry for the noise - at least I learned I can specify weights the easy way... ;-) AFAICS, subset.data.frame() merely uses this trick: e - substitute(subset) r - eval(e, x, parent.frame()) I'm probably missing something... ;-) It seems very common to pass a data frame via the data argument, and use variables from it for subsetting and/or weighting. Regards Josh On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
[R] define subset argument for function lm as variable?
Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
Hi Rainer, You could try: subs - expression(dead==FALSE recTreat==FALSE) lme(formula, subset = eval(subs)) Not tested, but something along those lines should work. Cheers, Josh On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] define subset argument for function lm as variable?
?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
What is wrong with what I suggested initially? subst - expression(trt 0) lm(weight ~ group, data=dat, subset=eval(subst)) ?? On Tue, Aug 21, 2012 at 8:41 AM, Rainer M Krug r.m.k...@gmail.com wrote: On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] define subset argument for function lm as variable?
On 21/08/12 17:54, Joshua Wiley wrote: What is wrong with what I suggested initially? subst - expression(trt 0) lm(weight ~ group, data=dat, subset=eval(subst)) That it does not work? ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) rm(ctl) rm(trt) subst - expression(trt 0) lm(weight ~ group, data=dat, subset=eval(subst)) # output: Error in eval(expr, envir, enclos) : object 'trt' not found and lm(weight ~ group, data=dat, subset=subst) # output: Error in xj[i] : invalid subscript type 'expression' also does not work. Rainer ?? On Tue, Aug 21, 2012 at 8:41 AM, Rainer M Krug r.m.k...@gmail.com wrote: On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list
Re: [R] define subset argument for function lm as variable?
Josh's solution should be fine. just one more note, trt0 may not work as intended since trt is a factor in your example. Just check subset(dat,trt0) wich is just 'dat'. Am 21.08.2012 17:41, schrieb Rainer M Krug: On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] define subset argument for function lm as variable?
Sorry - it is working as suggested by Joshua. Thanks a lot and sorry for the horrible confusion and examples, Rainer On 21/08/12 18:12, Rainer M Krug wrote: On 21/08/12 17:54, Joshua Wiley wrote: What is wrong with what I suggested initially? subst - expression(trt 0) lm(weight ~ group, data=dat, subset=eval(subst)) That it does not work? ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) rm(ctl) rm(trt) subst - expression(trt 0) lm(weight ~ group, data=dat, subset=eval(subst)) # output: Error in eval(expr, envir, enclos) : object 'trt' not found and lm(weight ~ group, data=dat, subset=subst) # output: Error in xj[i] : invalid subscript type 'expression' also does not work. Rainer ?? On Tue, Aug 21, 2012 at 8:41 AM, Rainer M Krug r.m.k...@gmail.com wrote: On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21
