yep That worked Jorge. Thanks!

summary(pb)$"Pr(>F)1"
NULL
> summ<-summary(pb)
> class(unlist(summ))

[1] "numeric"
> unlist(summ)["Pr(>F)1"]

   Pr(>F)1
0.02533637
> names(summ)
NULL 



Jorge Ivan Velez wrote:
>
> Hi Paul,
>
> Perhaps
>
> # Data set
> set.seed(123)
> x=rnorm(100)
> y=x+2*rnorm(100)
>
> # ANOVA
> AOV=anova(lm (y ~ x))
> p1=AOV$"Pr(>F)"[1]
> p1
>  
> or
>
> # ANOVA 2
> AOV1=aov(y ~ x)
> p2=unlist(summary(AOV1))["Pr(>F)1"]
> p2
>
>
> HTH,
>
> Jorge
>
>
>
> On Wed, May 7, 2008 at 8:47 PM, Paul Benton <hpbenton@scripps.edu 
> <mailto:hpbenton@scripps.edu>> wrote:
>
>     hello all,
>
>     Quick question, how do I get the p value out of the anova?
>
>     Thanks,
>
>     Paul
>
>     > pb<-aov(as.numeric(diff[5,16:33]) ~ grF)
>     > summary(pb)
>                Df     Sum Sq    Mean Sq F value  Pr(>F)
>     grF          3 2.7860e+10 9.2867e+09  4.2236 0.02534 *
>     Residuals   14 3.0783e+10 2.1988e+09
>     ---
>     Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>     > str(summary(pb))
>     List of 1
>      $ :Classes 'anova' and 'data.frame':   2 obs. of  5 variables:
>      ..$ Df     : num [1:2] 3 14
>      ..$ Sum Sq : num [1:2] 2.79e+10 3.08e+10
>      ..$ Mean Sq: num [1:2] 9.29e+09 2.20e+09
>      ..$ F value: num [1:2] 4.22   NA
>      ..$ Pr(>F) : num [1:2] 0.0253     NA
>      - attr(*, "class")= chr [1:2] "summary.aov" "listof"
>     > attr(summary(pb), "Pr(>F)")
>     NULL
>
>     ______________________________________________
>     R-help@r-project.org <mailto: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.
>

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