I like this very simple version. Note that you don't need as.data.frame().
model.matrix(Freq ~ Hair + Eye, data=haireye,
contrasts.arg=list(Hair=diag(4), Eye=diag(4)))
On Tue, Dec 30, 2014 at 6:05 PM, Michael Friendly <frien...@yorku.ca> wrote:
> In a manuscript, I have the following code to illustrate dummy coding of two
> factors in a contingency table.
>
> It works, but is surely obscured by the method I used, involving outer() to
> find equalities and 0+outer()
> to convert to numeric. Can someone help simplify this code to be more
> comprehensible and give the
> *same* result? I'd prefer a solution that uses base R.
>
> haireye <- margin.table(HairEyeColor, 1:2)
>
> haireye.df <- as.data.frame(haireye)
> dummy.hair <- 0+outer(haireye.df$Hair, levels(haireye.df$Hair), `==`)
> colnames(dummy.hair) <- paste0('h', 1:4)
> dummy.eye <- 0+outer(haireye.df$Eye, levels(haireye.df$Eye), `==`)
> colnames(dummy.eye) <- paste0('e', 1:4)
>
> haireye.df <- data.frame(haireye.df, dummy.hair, dummy.eye)
> haireye.df
>
>> haireye.df
> Hair Eye Freq h1 h2 h3 h4 e1 e2 e3 e4
> 1 Black Brown 68 1 0 0 0 1 0 0 0
> 2 Brown Brown 119 0 1 0 0 1 0 0 0
> 3 Red Brown 26 0 0 1 0 1 0 0 0
> 4 Blond Brown 7 0 0 0 1 1 0 0 0
> 5 Black Blue 20 1 0 0 0 0 1 0 0
> 6 Brown Blue 84 0 1 0 0 0 1 0 0
> 7 Red Blue 17 0 0 1 0 0 1 0 0
> 8 Blond Blue 94 0 0 0 1 0 1 0 0
> 9 Black Hazel 15 1 0 0 0 0 0 1 0
> 10 Brown Hazel 54 0 1 0 0 0 0 1 0
> 11 Red Hazel 14 0 0 1 0 0 0 1 0
> 12 Blond Hazel 10 0 0 0 1 0 0 1 0
> 13 Black Green 5 1 0 0 0 0 0 0 1
> 14 Brown Green 29 0 1 0 0 0 0 0 1
> 15 Red Green 14 0 0 1 0 0 0 0 1
> 16 Blond Green 16 0 0 0 1 0 0 0 1
>>
>
> --
> Michael Friendly Email: friendly AT yorku DOT ca
> Professor, Psychology Dept. & Chair, Quantitative Methods
> York University Voice: 416 736-2100 x66249 Fax: 416 736-5814
> 4700 Keele Street Web:http://www.datavis.ca
> Toronto, ONT M3J 1P3 CANADA
>
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