# How to interpret these custom contrasts?

I am doing a one way ANOVA (per species) with custom contrasts.

     [,1] [,2] [,3] [,4]
0.5    -1    0    0    0
5       1   -1    0    0
12.5    0    1   -1    0
25      0    0    1   -1
50      0    0    0    1


where I compare intensity 0.5 against 5, 5 against 12.5 and so on. These is the data I'm working on

with the following results

Generalized least squares fit by REML
Model: dark ~ intensity
Data: skofijski.diurnal[skofijski.diurnal$species == "niphargus", ] AIC BIC logLik 63.41333 67.66163 -25.70667 Coefficients: Value Std.Error t-value p-value (Intercept) 16.95 0.2140872 79.17334 0.0000 intensity1 2.20 0.4281744 5.13809 0.0001 intensity2 1.40 0.5244044 2.66970 0.0175 intensity3 2.10 0.5244044 4.00454 0.0011 intensity4 1.80 0.4281744 4.20389 0.0008 Correlation: (Intr) intns1 intns2 intns3 intensity1 0.000 intensity2 0.000 0.612 intensity3 0.000 0.408 0.667 intensity4 0.000 0.250 0.408 0.612 Standardized residuals: Min Q1 Med Q3 Max -2.3500484 -0.7833495 0.2611165 0.7833495 1.3055824 Residual standard error: 0.9574271 Degrees of freedom: 20 total; 15 residual  16.95 is the global mean for "niphargus". In intensity1, I'm comparing means for intensity 0.5 against 5. If I understood this right, the coefficient for intensity1 of 2.2 should be half the difference between means of intensity levels 0.5 and 5. However, my hand calculations don't match those of the summary. Can anyone chip in what am I doing wrong? ce1 <- skofijski.diurnal$intensity
levels(ce1) <- c("0.5", "5", "0", "0", "0")
ce1 <- as.factor(as.character(ce1))
tapply(skofijski.diurnal$dark, ce1, mean) 0 0.5 5 14.500 11.875 13.000 diff(tapply(skofijski.diurnal$dark, ce1, mean))/2
0.5       5
-1.3125  0.5625

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Could you provide the lm() function from R that you used to estimate. How exactly did you use the contrasts function? –  Philippe Sep 11 '12 at 0:50

The matrix you specified for the contrasts is correct in principle. To convert it into an appropriate contrast matrix, you need to calculate the generalized inverse of your original matrix.

If M is your matrix:

M

#     [,1] [,2] [,3] [,4]
#0.5    -1    0    0    0
#5       1   -1    0    0
#12.5    0    1   -1    0
#25      0    0    1   -1
#50      0    0    0    1


Now, calculate the generalized inverse using ginv and transpose the result using t:

library(MASS)
t(ginv(M))

#     [,1] [,2] [,3] [,4]
#[1,] -0.8 -0.6 -0.4 -0.2
#[2,]  0.2 -0.6 -0.4 -0.2
#[3,]  0.2  0.4 -0.4 -0.2
#[4,]  0.2  0.4  0.6 -0.2
#[5,]  0.2  0.4  0.6  0.8


The result is identical to the one of @Greg Snow. Use this matrix for your analysis.

This is a much easier way than doing it manually.

There is an even easier way to generate a matrix of sliding differences (a.k.a. repeated contrasts). This can be done with the function contr.sdif and the number of factor levels as a parameter. If you have five factor levels, like in your example:

library(MASS)
contr.sdif(5)

#   2-1  3-2  4-3  5-4
#1 -0.8 -0.6 -0.4 -0.2
#2  0.2 -0.6 -0.4 -0.2
#3  0.2  0.4 -0.4 -0.2
#4  0.2  0.4  0.6 -0.2
#5  0.2  0.4  0.6  0.8

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If the matrix at the top is how you are encoding the dummy variables (what you are passing to the C or contrast function in R) then they the first one is comparing the 1st level to the others (actually 0.8 times the 1st subtracted from 0.2 times the sum of the others).

The second term compares the 1st 2 levels to the last 3. The 3rd compares the 1st 3 levels to the last2 and the 4th compares the 1st 4 levels to the last one.

If you want to do the comparisons that you describe (compare each pair) then the dummy variable encoding that you want is:

      [,1] [,2] [,3] [,4]
[1,] -0.8 -0.6 -0.4 -0.2
[2,]  0.2 -0.6 -0.4 -0.2
[3,]  0.2  0.4 -0.4 -0.2
[4,]  0.2  0.4  0.6 -0.2
[5,]  0.2  0.4  0.6  0.8

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Is the use of this generalized inverese matrix also necessary when using aov() instead of lm()? I'm asking, because I have read several tutorials, in which contrast matrices for aov() are constructed just as the one given by Roman. E.g. see Chapter 5 in cran.r-project.org/doc/contrib/Vikneswaran-ED_companion.pdf –  crash Oct 18 '12 at 16:14
The aov function calls the lm function to do the main computations, so things like contrast matrices will have the same effect in both. –  Greg Snow Oct 18 '12 at 16:26
That's what I thought. Thanks for your quick answer. –  crash Oct 18 '12 at 20:57