Dear R-group,
I am trying desperately to get this Tukey test working. Its my first time
here so I hope my question is not too stupid, but I couldn't find anything
helpful in the help or in the forum.
I am analysing a dataset of treated grasses. I want to find out, if the
grasses from different Provenances react differently.
In the aov test I found a significance for the combination Treatment and
provenance:
summary(PAMaov<-aov(PAMval~Treatmentf*Pretreatmentf*Provenancef+Error(Datef/Code)))
Treatmentf:Provenancef p-value: 0.008023 **
In the Linear fixed effects model lme, I can see that there is a
significance for two provenances (HU and ES)
summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf,
random~1|Datef/Code,na.action=na.omit))
Value Std.Error DF t-value
p-value
(Intercept) 0.6890317 0.06117401 994 11.263473
0.0000
TreatmentfF -0.2897619 0.05484590 467 -5.283201
0.0000
ProvenancefDE 0.0105873 0.05484590 467 0.193037
0.8470
TreatmentfF:ProvenancefES 0.1647302 0.08226884 467 2.002340
0.0458
TreatmentfF:ProvenancefHU 0.1569524 0.07756381 467 2.023526
0.0436
No the big mystery is the Tukey test. I just can't find the mistake, it
keeps telling me, that there are " less than two groups"
summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey")))
Fehler in contrMat(table(mf[[nm]]), type = types[pm]) :
less than two groups
I guess its important to know that I made factors out of some of the data.
Here is the code:
PAMdata$provenance[PAMdata$provenance == "5"] = "ES"
PAMdata$provenance[PAMdata$provenance == "6"] = "HU"
# etc.
Treatmentf <- factor(PAMdata$treatment,
levels=c("C","F"))
Datef <- factor(PAMdata$Date, levels=c( "25.05.10
14:00","26.05.10
19:00","27.05.2010 7:30","27.05.10
14:00","01.06.10 14:00","02.06.10
19:00","23.06.10 12:30"),ordered=TRUE)
Pretreatmentf <- as.factor(PAMdata$pretreatment)
Provenancef <- as.factor(PAMdata$provenance)
Greenhousef <- as.factor(PAMdata$greenhouse)
Individualf <- as.factor(PAMdata$individual)
PAMval <- (PAMdata$DataPAM)
Code<-(PAMdata$code)
Thank you for any hint! That Tukey test seems so easy, I just can't find the
mistake....
Thank you very much fpr your help and greetings from Tanzania,
Lilith
--
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Comments in-line below> -----Original Message----- > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Lilith > Sent: December-02-10 9:39 AM > To: r-help at r-project.org > Subject: [R] Tukey Test, lme, error: less than two groups > > > Dear R-group, > > I am trying desperately to get this Tukey test working. Its my first time > here so I hope my question is not too stupid, but I couldn't find anything > helpful in the help or in the forum. > > I am analysing a dataset of treated grasses. I want to find out, if the > grasses from different Provenances react differently. > In the aov test I found a significance for the combination Treatment and > provenance: > > summary(PAMaov<-aov(PAMval~Treatmentf*Pretreatmentf*Provenancef+Error(Datef/Code))) > > Treatmentf:Provenancef p-value: 0.008023 ** > > In the Linear fixed effects model lme, I can see that there is a > significance for two provenances (HU and ES) > > summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf, random> ~1|Datef/Code,na.action=na.omit)) > > Value Std.Error DF t-value > p-value > (Intercept) 0.6890317 0.06117401 994 11.263473 > 0.0000 > TreatmentfF -0.2897619 0.05484590 467 -5.283201 > 0.0000 > ProvenancefDE 0.0105873 0.05484590 467 0.193037 > 0.8470 > > TreatmentfF:ProvenancefES 0.1647302 0.08226884 467 2.002340 > 0.0458 > TreatmentfF:ProvenancefHU 0.1569524 0.07756381 467 2.023526 > 0.0436 > > No the big mystery is the Tukey test. I just can't find the mistake, it > keeps telling me, that there are " less than two groups" > > summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey"))) > > Fehler in contrMat(table(mf[[nm]]), type = types[pm]) : > less than two groups > > I guess its important to know that I made factors out of some of the data. > Here is the code: > > > PAMdata$provenance[PAMdata$provenance == "5"] = "ES" > PAMdata$provenance[PAMdata$provenance == "6"] = "HU" > # etc. > > Treatmentf <- factor(PAMdata$treatment, levels=c("C","F")) > Datef <- factor(PAMdata$Date, levels=c( "25.05.10 14:00","26.05.10 > 19:00","27.05.2010 7:30","27.05.10 14:00","01.06.10 14:00","02.06.10 > 19:00","23.06.10 12:30"),ordered=TRUE) > > > Pretreatmentf <- as.factor(PAMdata$pretreatment) > Provenancef <- as.factor(PAMdata$provenance) > Greenhousef <- as.factor(PAMdata$greenhouse) > Individualf <- as.factor(PAMdata$individual) > > PAMval <- (PAMdata$DataPAM) > Code<-(PAMdata$code)I suspect the problem is the creation of all these individual variables. Try instead PAMdata$Treatmentf <- factor(PAMdata$treatment, levels=c("C","F")) PAMdata$Datef <- factor(PAMdata$Date, levels=c( "25.05.10 14:00","26.05.10 19:00","27.05.2010 7:30","27.05.10 14:00","01.06.10 14:00","02.06.10 19:00","23.06.10 12:30"),ordered=TRUE) ... PAMdata$PAMval <- (PAMdata$DataPAM) PAMdata$Code<-(PAMdata$code) etc. so that all of your required variables are variables in the dataframe PAMdata. When you pass off fitted model objects to additional functions, the additional functions often require access to the dataframe used in the initial modeling. Then call lme with summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf, random ~1|Datef/Code, data = PAMdata, na.action=na.omit)) then try summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey"))) again. (If you still get an error, run the traceback() command and provide that information.) I'm also wondering why no term for Pretreatmentf shows in your model output. After setting up the factor variables in the PAMdata dataframe, what does the command with(PAMdata, table(Pretreatmentf, Provenancef, Treatmentf)) show? Is Pretreatmentf even needed in the model? The output of the command sessionInfo() is also useful to help people figure out such issues. Also, if you can share the data, or a mock-up of it, others will be able to run code examples, and not just guess. HTH Steve McKinney> > Thank you for any hint! That Tukey test seems so easy, I just can't find the > mistake.... > Thank you very much fpr your help and greetings from Tanzania, > Lilith > > -- > View this message in context: http://r.789695.n4.nabble.com/Tukey-Test-lme-error-less-than-two-groups- > tp3069789p3069789.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at 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.
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Data:
Just to close this thread, Lilith provided the data which was in a .csv text file and had multiple lines of blank data at the end species;code;treatment;pretreatment;provenance;greenhouse;individual;leaf;Date;DataPAM Ae;c-ae-1-1-3;C;C;1;1;3;1;25.05.10 14:00; 0.665 . . . Ae;w-ae-6-3-4;C;W;6;3;4;3;23.06.10 12:30; 0.622 ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; ;;;;;;;;; Removing the blank (NA) rows of data and keeping all variables in the dataframe resolves the issue. (Free floating variables yield this error> summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey")))Error in `[.data.frame`(mf, nhypo[checknm]) : undefined columns selected) Steven McKinney ________________________________________ From: r-help-bounces at r-project.org [r-help-bounces at r-project.org] On Behalf Of Steven McKinney [smckinney at bccrc.ca] Sent: December 2, 2010 2:03 PM To: 'Lilith'; r-help at r-project.org Subject: Re: [R] Tukey Test, lme, error: less than two groups Comments in-line below> -----Original Message----- > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Lilith > Sent: December-02-10 9:39 AM > To: r-help at r-project.org > Subject: [R] Tukey Test, lme, error: less than two groups > > > Dear R-group, > > I am trying desperately to get this Tukey test working. Its my first time > here so I hope my question is not too stupid, but I couldn't find anything > helpful in the help or in the forum. > > I am analysing a dataset of treated grasses. I want to find out, if the > grasses from different Provenances react differently. > In the aov test I found a significance for the combination Treatment and > provenance: > > summary(PAMaov<-aov(PAMval~Treatmentf*Pretreatmentf*Provenancef+Error(Datef/Code))) > > Treatmentf:Provenancef p-value: 0.008023 ** > > In the Linear fixed effects model lme, I can see that there is a > significance for two provenances (HU and ES) > > summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf, random> ~1|Datef/Code,na.action=na.omit)) > > Value Std.Error DF t-value > p-value > (Intercept) 0.6890317 0.06117401 994 11.263473 > 0.0000 > TreatmentfF -0.2897619 0.05484590 467 -5.283201 > 0.0000 > ProvenancefDE 0.0105873 0.05484590 467 0.193037 > 0.8470 > > TreatmentfF:ProvenancefES 0.1647302 0.08226884 467 2.002340 > 0.0458 > TreatmentfF:ProvenancefHU 0.1569524 0.07756381 467 2.023526 > 0.0436 > > No the big mystery is the Tukey test. I just can't find the mistake, it > keeps telling me, that there are " less than two groups" > > summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey"))) > > Fehler in contrMat(table(mf[[nm]]), type = types[pm]) : > less than two groups > > I guess its important to know that I made factors out of some of the data. > Here is the code: > > > PAMdata$provenance[PAMdata$provenance == "5"] = "ES" > PAMdata$provenance[PAMdata$provenance == "6"] = "HU" > # etc. > > Treatmentf <- factor(PAMdata$treatment, levels=c("C","F")) > Datef <- factor(PAMdata$Date, levels=c( "25.05.10 14:00","26.05.10 > 19:00","27.05.2010 7:30","27.05.10 14:00","01.06.10 14:00","02.06.10 > 19:00","23.06.10 12:30"),ordered=TRUE) > > > Pretreatmentf <- as.factor(PAMdata$pretreatment) > Provenancef <- as.factor(PAMdata$provenance) > Greenhousef <- as.factor(PAMdata$greenhouse) > Individualf <- as.factor(PAMdata$individual) > > PAMval <- (PAMdata$DataPAM) > Code<-(PAMdata$code)I suspect the problem is the creation of all these individual variables. Try instead PAMdata$Treatmentf <- factor(PAMdata$treatment, levels=c("C","F")) PAMdata$Datef <- factor(PAMdata$Date, levels=c( "25.05.10 14:00","26.05.10 19:00","27.05.2010 7:30","27.05.10 14:00","01.06.10 14:00","02.06.10 19:00","23.06.10 12:30"),ordered=TRUE) ... PAMdata$PAMval <- (PAMdata$DataPAM) PAMdata$Code<-(PAMdata$code) etc. so that all of your required variables are variables in the dataframe PAMdata. When you pass off fitted model objects to additional functions, the additional functions often require access to the dataframe used in the initial modeling. Then call lme with summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf, random ~1|Datef/Code, data = PAMdata, na.action=na.omit)) then try summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey"))) again. (If you still get an error, run the traceback() command and provide that information.) I'm also wondering why no term for Pretreatmentf shows in your model output. After setting up the factor variables in the PAMdata dataframe, what does the command with(PAMdata, table(Pretreatmentf, Provenancef, Treatmentf)) show? Is Pretreatmentf even needed in the model? The output of the command sessionInfo() is also useful to help people figure out such issues. Also, if you can share the data, or a mock-up of it, others will be able to run code examples, and not just guess. HTH Steve McKinney> > Thank you for any hint! That Tukey test seems so easy, I just can't find the > mistake.... > Thank you very much fpr your help and greetings from Tanzania, > Lilith > > -- > View this message in context: http://r.789695.n4.nabble.com/Tukey-Test-lme-error-less-than-two-groups- > tp3069789p3069789.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at 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 at 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.