You really really need to go through one of the many R (web) tutorials
where this is discussed in detail. Or see the Intro to R tutorial that
ships with R. This is not the proper venue for learning how R handles
linear models and how its formula interface works. A terse treatment
can be found in ?lm, ?aov, and ?formula and links therein, but a
suitable tutorial would probably be a better alternative.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Jun 25, 2017 at 10:05 PM, Jinsong Zhao <jsz...@yeah.net> wrote:
> Hi there,
>
> I have a experimental design related question. I have done a experiment with
> 3 factors. The design matrix is similar to:
>
>> data.frame(Factor.1 = rep(rep(0:2, each = 3),3), Factor.2 =
>
> rep(c("U","S","N"), 9), Factor.3 = rep(0:2, each = 9))
> Factor.1 Factor.2 Factor.3
> 1 0 U 0
> 2 0 S 0
> 3 0 N 0
> 4 1 U 0
> 5 1 S 0
> 6 1 N 0
> 7 2 U 0
> 8 2 S 0
> 9 2 N 0
> 10 0 U 1
> 11 0 S 1
> 12 0 N 1
> 13 1 U 1
> 14 1 S 1
> 15 1 N 1
> 16 2 U 1
> 17 2 S 1
> 18 2 N 1
> 19 0 U 2
> 20 0 S 2
> 21 0 N 2
> 22 1 U 2
> 23 1 S 2
> 24 1 N 2
> 25 2 U 2
> 26 2 S 2
> 27 2 N 2
>
> The Factor.2 indicates the type of a fertilizer, such as urea, and the
> Factor.3 indicates the weight of the fertilizer. So the treatments with
> Factor.3 == 0 at each levels of Factor.1 are same. That means No.1, No.2 and
> No.3 are same, No.4, No.5 and No.6 are same, and No.7, No.8 and No.9 are
> same. Thus, in the real experiment, those treatments were not performed as
> stated in the design matrix. For instance, only No.1, No.5 and No.9 are
> performed.
>
> In the case, how to code the factor for ANOVA? I really appreciate any
> comments and suggestions. Thanks in advance.
>
> Best regards,
> Jinsong
>
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______________________________________________
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