Hi all:
Can someone advice me on how to hold the residuals
Mean sq value on a string
so it can be used in other calculations.
I was trying something like this:
Msquare<-dfr$Mean sq but fails..Thanks
dfr <- read.table(textConnection("percentQ
Efficiency
1.565 0.0125
1.94 0.0213
0.876 0.003736
1.027 0.006
1.536 0.0148
1.536 0.0162
2.607 0.02
1.456 0.0157
2.16 0.0103
1.698 0.0196
1.64 0.0098684
1.814 0.0183
2.394 0.0107
2.469 0.0221
3.611 0.0197
3.466 0.0155
1.877 0.0283
2.893 0.0189
1.851 0.009772
2.834 0.0285
1.923 0.022
2.581 0.0159
2.361 0.0053591
2.43 0.0185
1.66 0.0151
2.285 0.0084034
2.285 0.0124
2.37 0.0122
2.392 0.0146
2.244 0.0175"), header=TRUE)
anova(lm(Efficiency~percentQ,data=dfr))
Analysis of Variance Table
Response: Efficiency
Df Sum Sq Mean Sq F value Pr(>F)
percentQ 1 0.00014432 0.00014432 4.3774 0.04561 *
Residuals 28 0.00092312 0.00003297
Felipe D. Carrillo
Fishery Biologist
US Fish & Wildlife Service
California, USA
____________________________________________________________________________________
Looking for last minute shopping deals?
Assign your anova object a name, like fm1 <- anova(lm(Efficiency~percentQ,data=dfr)) Then do names(fm1) and try fm1$"Mean Sq" Notice you were trying to find "Mean Sq" in dfr, your original data.frame. -Erik Iverson Felipe Carrillo wrote:> Hi all: > Can someone advice me on how to hold the residuals > Mean sq value on a string > so it can be used in other calculations. > I was trying something like this: > Msquare<-dfr$Mean sq but fails..Thanks > > dfr <- read.table(textConnection("percentQ > Efficiency > 1.565 0.0125 > 1.94 0.0213 > 0.876 0.003736 > 1.027 0.006 > 1.536 0.0148 > 1.536 0.0162 > 2.607 0.02 > 1.456 0.0157 > 2.16 0.0103 > 1.698 0.0196 > 1.64 0.0098684 > 1.814 0.0183 > 2.394 0.0107 > 2.469 0.0221 > 3.611 0.0197 > 3.466 0.0155 > 1.877 0.0283 > 2.893 0.0189 > 1.851 0.009772 > 2.834 0.0285 > 1.923 0.022 > 2.581 0.0159 > 2.361 0.0053591 > 2.43 0.0185 > 1.66 0.0151 > 2.285 0.0084034 > 2.285 0.0124 > 2.37 0.0122 > 2.392 0.0146 > 2.244 0.0175"), header=TRUE) > > anova(lm(Efficiency~percentQ,data=dfr)) > Analysis of Variance Table > > Response: Efficiency > Df Sum Sq Mean Sq F value Pr(>F) > percentQ 1 0.00014432 0.00014432 4.3774 0.04561 * > Residuals 28 0.00092312 0.00003297 > > > > Felipe D. Carrillo > Fishery Biologist > US Fish & Wildlife Service > California, USA > > > > ____________________________________________________________________________________ > Looking for last minute shopping deals? > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
Hi Felipe,
Try this
dfr <- read.table(
textConnection("percentQ Efficiency
1.565 0.0125
1.94 0.0213
0.876 0.003736
1.027 0.006
1.536 0.0148
1.536 0.0162
2.607 0.02
1.456 0.0157
2.16 0.0103
1.698 0.0196
1.64 0.0098684
1.814 0.0183
2.394 0.0107
2.469 0.0221
3.611 0.0197
3.466 0.0155
1.877 0.0283
2.893 0.0189
1.851 0.009772
2.834 0.0285
1.923 0.022
2.581 0.0159
2.361 0.0053591
2.43 0.0185
1.66 0.0151
2.285 0.0084034
2.285 0.0124
2.37 0.0122
2.392 0.0146
2.244 0.0175"), header=TRUE)
temp=anova(lm(Efficiency~percentQ,data=dfr))
names(temp)
temp$"Mean Sq" # First entry is MS percentQ and second one is MS
residuals
HTH
Jorge
On Thu, Mar 6, 2008 at 5:49 PM, Felipe Carrillo <mazatlanmexico@yahoo.com>
wrote:
> Hi all:
> Can someone advice me on how to hold the residuals
> Mean sq value on a string
> so it can be used in other calculations.
> I was trying something like this:
> Msquare<-dfr$Mean sq but fails..Thanks
>
> dfr <- read.table(textConnection("percentQ
> Efficiency
> 1.565 0.0125
> 1.94 0.0213
> 0.876 0.003736
> 1.027 0.006
> 1.536 0.0148
> 1.536 0.0162
> 2.607 0.02
> 1.456 0.0157
> 2.16 0.0103
> 1.698 0.0196
> 1.64 0.0098684
> 1.814 0.0183
> 2.394 0.0107
> 2.469 0.0221
> 3.611 0.0197
> 3.466 0.0155
> 1.877 0.0283
> 2.893 0.0189
> 1.851 0.009772
> 2.834 0.0285
> 1.923 0.022
> 2.581 0.0159
> 2.361 0.0053591
> 2.43 0.0185
> 1.66 0.0151
> 2.285 0.0084034
> 2.285 0.0124
> 2.37 0.0122
> 2.392 0.0146
> 2.244 0.0175"), header=TRUE)
>
> anova(lm(Efficiency~percentQ,data=dfr))
> Analysis of Variance Table
>
> Response: Efficiency
> Df Sum Sq Mean Sq F value Pr(>F)
> percentQ 1 0.00014432 0.00014432 4.3774 0.04561 *
> Residuals 28 0.00092312 0.00003297
>
>
>
> Felipe D. Carrillo
> Fishery Biologist
> US Fish & Wildlife Service
> California, USA
>
>
>
>
>
____________________________________________________________________________________
> Looking for last minute shopping deals?
>
> ______________________________________________
> 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.
>
--
Jorge Velez
Industrial Engineer
Master in Sciences (Candidate)
School of Statistics
National University of Colombia at Medellin
e-mail: jivelez@unal.edu.co
Visiting Fellow
Medical Genetics Branch, MGB
National Human Genome Research Institute, NHGRI
National Institutes of Health, NIH
Department of Health and Human Services, DHHS
35 Convent Drive, MSC 3717
Building 35, Room 1B-209
Bethesda, MD, USA 20892-3717
Office: +1 (301) 451-74-06
Mobile: +1 (301) 300-37-32
Fax: +1 (301) 496-71-84
e-mail: velezjo@mail.nih.gov
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