Suppose I have a response variable and a data containing three covariates (as a toy example):
y = c(1,4,6)
d = data.frame(x1 = c(4,-1,3), x2 = c(3,9,8), x3 = c(4,-4,-2))
I want to fit a linear regression to the data:
fit = lm(y ~ d$x1 + d$x2 + d$y2)
Is there a way to write the formula, so that I don't have to write out each individual covariate? For example, something like
fit = lm(y ~ d)
(I want each variable in the data frame to be a covariate.) I'm asking because I actually have 50 variables in my data frame, so I want to avoid writing out x1 + x2 + x3 + etc.
There is a special identifier that one can use in a formula to mean all the variables, it is the . identifier.
y <- c(1,4,6)
d <- data.frame(y = y, x1 = c(4,-1,3), x2 = c(3,9,8), x3 = c(4,-4,-2))
mod <- lm(y ~ ., data = d)
You can also do things like this, to use all variables but one (in this case x3 is excluded):
mod <- lm(y ~ . - x3, data = d)
Technically, . means all variables not already mentioned in the formula. For example
lm(y ~ x1 * x2 + ., data = d)
where . would only reference x3 as x1 and x2 are already in the formula.
You can check the package leaps and in particular the function regsubsets()
functions for model selection. As stated in the documentation:
Model selection by exhaustive search, forward or backward stepwise, or sequential replacement
An extension of juba's method is to use reformulate, a function which is explicitly designed for such a task.
## Create a formula for a model with a large number of variables:
xnam <- paste("x", 1:25, sep="")
reformulate(xnam, "y")
y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 +
x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 +
x22 + x23 + x24 + x25
For the example in the OP, the easiest solution here would be
# add y variable to data.frame d
d <- cbind(y, d)
reformulate(names(d)[-1], names(d[1]))
y ~ x1 + x2 + x3
or
mod <- lm(reformulate(names(d)[-1], names(d[1])), data=d)
Note that adding the dependent variable to the data.frame in d <- cbind(y, d) is preferred not only because it allows for the use of reformulate, but also because it allows for future use of the lm object in functions like predict.
Yes of course, just add the response y as first column in the dataframe and call lm() on it:
d2<-data.frame(y,d)
> d2
y x1 x2 x3
1 1 4 3 4
2 4 -1 9 -4
3 6 3 8 -2
> lm(d2)
Call:
lm(formula = d2)
Coefficients:
(Intercept) x1 x2 x3
-5.6316 0.7895 1.1579 NA
Also, my information about R points out that assignment with <- is recommended over =.
A slightly different approach is to create your formula from a string. In the formula help page you will find the following example :
## Create a formula for a model with a large number of variables:
xnam <- paste("x", 1:25, sep="")
fmla <- as.formula(paste("y ~ ", paste(xnam, collapse= "+")))
Then if you look at the generated formula, you will get :
R> fmla
y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 +
x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20 + x21 +
x22 + x23 + x24 + x25
I build this solution, reformulate does not take care if variable names have white spaces.
add_backticks = function(x) {
paste0("`", x, "`")
}
x_lm_formula = function(x) {
paste(add_backticks(x), collapse = " + ")
}
build_lm_formula = function(x, y){
if (length(y)>1){
stop("y needs to be just one variable")
}
as.formula(
paste0("`",y,"`", " ~ ", x_lm_formula(x))
)
}
# Example
df <- data.frame(
y = c(1,4,6),
x1 = c(4,-1,3),
x2 = c(3,9,8),
x3 = c(4,-4,-2)
)
# Model Specification
columns = colnames(df)
y_cols = columns[1]
x_cols = columns[2:length(columns)]
formula = build_lm_formula(x_cols, y_cols)
formula
# output
# "`y` ~ `x1` + `x2` + `x3`"
# Run Model
lm(formula = formula, data = df)
# output
Call:
lm(formula = formula, data = df)
Coefficients:
(Intercept) x1 x2 x3
-5.6316 0.7895 1.1579 NA
```
Source: Stackoverflow.com