CODEWITHSUNDEEP
rlistlm
Thomas Neitmann
Thomas Neitmann

Reputation: 2722

lm on element of a list

Let say I have a list of three data frames

set.seed(55)
df1 <- data.frame(a=rnorm(6), b=rnorm(6), c=rnorm(6))
df2 <- data.frame(a=rnorm(6), b=rnorm(6), c=rnorm(6))
df3 <- data.frame(a=rnorm(6), b=rnorm(6), c=rnorm(6))
l <- list(df1, df2, df3)

I now like to do a linear model with a and c of every data frame. I tried the following

for(i in l) {
  x <- (lm(l[[i]]$a~l[[i]]$c))
}

However I get the following error

Error in l[[i]] : invalid subscript type 'list'

I'd like to have a list with every element an lm of a and c. Any help would be appreciated. Thank you very much!

Upvotes: 0

Views: 1936

Answers (3)

IRTFM
IRTFM

Reputation: 263481

The accepted answer is to be preferred but it might be useful for people coming from a for-loop mentality to be told that the lapply solution ( at least the one where it's result is assigned to 'x') is equivalent to this:

x<-list();
for(i in seq_along(l) ) {
      x[[i]] <- lm(a~c, data=l[[i]])
    }

Upvotes: 0

Jake
Jake

Reputation: 520

If you want to stay with the format you proposed (the for loop) you need to use something like 1:length(l) in your for loop statement.

Also, if you keep assigning things to x you will keep overwriting it. In what I have below I am assigning the linear model to the 24th, 25th, and 26th letters of the alphabet, so you have a lm called x, y, and z. Other suggested solutions are much cleaner.

for(i in 1:length(l)) {
 assign(letters[24:26][i], (lm(l[[i]]$a~l[[i]]$c)))
}

Upvotes: 0

A5C1D2H2I1M1N2O1R2T1
A5C1D2H2I1M1N2O1R2T1

Reputation: 193687

You can do this with a simple lapply, like this:

lapply(l, lm, formula = a ~ c)

Upvotes: 6

Related Questions