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 开发者_Python百科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.
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
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 =
.
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
.
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
```
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
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