Dummy variable as slope shifter without intercept

Question

This is my first time to ask here.

I have trouble generating the slope dummy variables only(without intercept dummy). However, if I multiply dummy variable by independent variable as shown below, both slope dummy and intercept dummy results are represented.

I want to incorporate slope dummy only and exclude intercept dummy.

I will appreciate your help. Bests, yjkim

reg <- lm(year ~ as.factor(age)*log(v1269)) 
Call: 
lm(formula = year ~ as.factor(age) * log(v1269)) 

Residuals: 
   Min     1Q Median     3Q    Max 
-6.083 -1.177  1.268  1.546  3.768 

Coefficients: 
                            Estimate Std. Error t value Pr(>|t|)   
(Intercept)                 5.18076    2.16089   2.398   0.0167 * 
as.factor(age)2             1.93989    2.75892   0.703   0.4821   
as.factor(age)3             2.46861    2.39393   1.031   0.3027   
as.factor(age)4            -0.56274    2.30123  -0.245   0.8069   
log(v1269)                 -0.06788    0.23606  -0.288   0.7737   
as.factor(age)2:log(v1269) -0.15628    0.29621  -0.528   0.5979   
as.factor(age)3:log(v1269) -0.14961    0.25809  -0.580   0.5622   
as.factor(age)4:log(v1269)  0.16534    0.24884   0.664   0.5065

Do you want to get rid of the `(Intercept)` term or the three `as.factor(age)2`, `as.factor(age)3`, and `as.factor(age)4` terms? — Gavin Simpson, Apr 14 '16 at 19:10

score 0 · Answer 1 · answered Apr 14 '16 at 19:06

0

Just need a -1 within the formaula

reg <- lm(year ~ as.factor(age)*log(v1269) -1)

answered Apr 14 '16 at 19:06

Adam Birenbaum

940
9
23

score 0 · Answer 2 · answered Apr 15 '16 at 16:29

If you want to estimate a different slope in each level of age, the you can use the %in% operator in the formula

set.seed(1)
df <- data.frame(age = factor(sample(1:4, 100, replace = TRUE)),
                 v1269 = rlnorm(100),
                 year = rnorm(100))

m <- lm(year ~ log(v1269) %in% age, data = df)
summary(m)

This gives (for this entirely random , dummy, silly data set)

> summary(m)

Call:
lm(formula = year ~ log(v1269) %in% age, data = df)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.93108 -0.66402 -0.05921  0.68040  2.25244 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)
(Intercept)      0.02692    0.10705   0.251    0.802
log(v1269):age1  0.20127    0.21178   0.950    0.344
log(v1269):age2 -0.01431    0.24116  -0.059    0.953
log(v1269):age3 -0.02588    0.24435  -0.106    0.916
log(v1269):age4  0.06019    0.21979   0.274    0.785

Residual standard error: 1.065 on 95 degrees of freedom
Multiple R-squared:  0.01037,   Adjusted R-squared:  -0.0313 
F-statistic: 0.2489 on 4 and 95 DF,  p-value: 0.9097

Note that this fits a single constant term plus 4 different effects of log(v1269), one per level of age. Visually, this is sort of what the model is doing

pred <- with(df,
             expand.grid(age = factor(1:4),
                         v1269 = seq(min(v1269), max(v1269), length = 100)))
pred <- transform(pred, fitted = predict(m, newdata = pred))

library("ggplot2")
ggplot(df, aes(x = log(v1269), y = year, colour = age)) + 
  geom_point() +
  geom_line(data = pred, mapping = aes(y = fitted)) +
  theme_bw() + theme(legend.position = "top")

Clearly, this would only be suitable if there was no significant difference in the mean values of year (the response) in the different age categories.

Note that a different parameterisation of the same model can be achieved via the / operator:

m2 <- lm(year ~ log(v1269)/age, data = df)

> m2

Call:
lm(formula = year ~ log(v1269)/age, data = df)

Coefficients:
    (Intercept)       log(v1269)  log(v1269):age2  log(v1269):age3  
        0.02692          0.20127         -0.21559         -0.22715  
log(v1269):age4  
       -0.14108

Note that now, the first log(v1269) term is for the slope for age == 1, whilst the other terms are the adjustments required to be applied to the the log(v1269) term to get the slope for the indicated group:

coef(m)[-1]
coef(m2)[2] + c(0, coef(m2)[-(1:2)])

> coef(m)[-1]
log(v1269):age1 log(v1269):age2 log(v1269):age3 log(v1269):age4 
     0.20127109     -0.01431491     -0.02588106      0.06018802 
> coef(m2)[2] + c(0, coef(m2)[-(1:2)])
                log(v1269):age2 log(v1269):age3 log(v1269):age4 
     0.20127109     -0.01431491     -0.02588106      0.06018802

But they work out to the same estimated slopes.

Dummy variable as slope shifter without intercept

2 Answers2