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copied!<p>Here's a sort of answer; there is no way that I know of to formulate this model directly by the formula ...</p> <p>Construct data as above:</p> <pre><code>d <- data.frame(A = rep(c("a1", "a2"), each = 50), B = c("b1", "b2"), value = rnorm(100)) options(contrasts=c('contr.sum','contr.poly')) </code></pre> <p>Confirm original finding that just subtracting the factor from the formula doesn't work:</p> <pre><code>m1 <- lm(value ~ A * B, data = d) coef(m1) ## (Intercept) A1 B1 A1:B1 ## -0.23766309 0.04651298 -0.13019317 -0.06421580 m2 <- update(m1, .~. - A) coef(m2) ## (Intercept) B1 Bb1:A1 Bb2:A1 ## -0.23766309 -0.13019317 -0.01770282 0.11072877 </code></pre> <p>Formulate the new model matrix:</p> <pre><code>X0 <- model.matrix(m1) ## drop Intercept column *and* A from model matrix X1 <- X0[,!colnames(X0) %in% "A1"] </code></pre> <p><code>lm.fit</code> allows direct specification of the model matrix:</p> <pre><code>m3 <- lm.fit(x=X1,y=d$value) coef(m3) ## (Intercept) B1 A1:B1 ## -0.2376631 -0.1301932 -0.0642158 </code></pre> <p>This method only works for a few special cases that allow the model matrix to be specified explicitly (e.g. <code>lm.fit</code>, <code>glm.fit</code>).</p> <p>More generally:</p> <pre><code>## need to drop intercept column (or use -1 in the formula) X1 <- X1[,!colnames(X1) %in% "(Intercept)"] ## : will confuse things -- substitute something inert colnames(X1) <- gsub(":","_int_",colnames(X1)) newf <- reformulate(colnames(X1),response="value") m4 <- lm(newf,data=data.frame(value=d$value,X1)) coef(m4) ## (Intercept) B1 A1_int_B1 ## -0.2376631 -0.1301932 -0.0642158 </code></pre> <p>This approach has the disadvantage that it won't recognize multiple input variables as stemming from the same predictor (i.e., multiple factor levels from a more-than-2-level factor).</p>

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