线性回归[R]：如何根据分类变量的出现为同一预测变量计算多个系数

Question

I have a forecasting with linear regression problem. In this problem the days of the week matter. At the moment I use:

lm.mod <- lm(y ~ x + monday + tuesday + thursday + friday + saturday + sunday, data=train)

Where y and x are continuous variables and the days of the week are dummy variables (they can be either 0 or 1). In this way the week dependence is given by different intercepts (the coefficients in front of the dummies). However, I would like to calculate a different coefficient for x for each day of the week.

I can do this operation when I use GAM (library: mgcv) inside the spline function, where "day" is a categorical variable containing the name of the day of the week

gam.mod <- mgcv::gam(y ~ s(x, bs='cs', by=day) + monday + tuesday + thursday + friday + saturday + sunday, data = train, method="REML", select=TRUE)

I show a few lines of the data frame train

Date        | y          | x          | day       | Monday | Tuesday | Wednesday |
---------------------------------------------------------------------------------
2013-01-01  | 0.87604858 | 0.07339450 | Tuesday   | 0      | 1       | 0         |
2013-01-02  | 0.90190414 | 0.16513761 | Wednesday | 0      | 0       | 1         |

With mgcv I obtain a different spline for each day of the week (each value of the factor variable "day"), with a linear model I would like to obtain many coefficients for x as the number of values of factor variables. Is it possible? Any workaround?

Answer 1

Maybe I'm missing something, but it appears to me you are asking for the interaction between x and the week days?

Ie simplified a bit, something like this:

# Toy data
n <- 100
train <- data.frame(replicate(5, rnorm(n)))
names(train) <- c("x", "y", "mon", "tue", "wed")

lm.mod <- lm(y ~ x*(mon + tue + wed), data=train)

Answer 2

You want to avoid creating the binary terms yourself. In fact, the way the mgcv notation implies a spline by day , you ant to include day as a factor in the model, not all those separate terms.

So, the gam model would be:

gam(y ~ s(x, bs='cs', by=day) + day, data = train, method="REML", select=TRUE)

where day is a factor with levels c('Monday','Tuesday', ....) .

Then the linear model becomes:

lm(y ~ x * day, data = train)

You have to work a little harder to get the estimated means for each day; use predict() for the gam() model with newdata and one row per day and type = 'terms' and you can add the intercept to the day contribution (effect). For the lm() model you can most easily do this using the multcomp package.

You could also just drop the intercept (add + 0 to the model formula). There are other ways to potentially parameterise the model to model easily give you the estimates you may want.

That your models are even fitting is because R internally is dropping some effects; you can't fit an intercept and all those day terms because one of the separate day variables is linearly dependent on the intercept and thus cannot be uniquely identified.