使用“ $”功能

Question

I was going through some examples in hadley's guide to functionals, and came across an unexpected problem.

Suppose I have a list of model objects,

x=1:3;y=3:1; bah <- list(lm(x~y),lm(y~x))

and want to extract something from each (as suggested in hadley's question about a list called "trials"). I was expecting one of these to work:

lapply(bah,`$`,i='call') # or...
lapply(bah,`$`,call)

However, these return nulls. It seems like I'm not misusing the $ function, as these things work:

`$`(bah[[1]],i='call')
`$`(bah[[1]],call)

Anyway, I'm just doing this as an exercise and am curious where my mistake is. I know I could use an anonymous function, but think there must be a way to use syntax similar to my initial non-solution. I've looked through the places $ is mentioned in ?Extract , but didn't see any obvious explanation.

I just realized that this works:

lapply(bah,`[[`,i='call')

and this

lapply(bah,function(x)`$`(x,call))

Maybe this just comes down to some lapply voodoo that demands anonymous functions where none should be needed? I feel like I've heard that somewhere on SO before.

Answer 1

This is documented in ?lapply , in the "Note" section (emphasis mine):

For historical reasons, the calls created by lapply are unevaluated, and code has been written (eg bquote ) that relies on this. This means that the recorded call is always of the form FUN(X[[0L]], ...) , with 0L replaced by the current integer index. This is not normally a problem, but it can be if FUN uses sys.call or match.call or if it is a primitive function that makes use of the call. This means that it is often safer to call primitive functions with a wrapper, so that eg lapply(ll, function(x) is.numeric(x)) is required in R 2.7.1 to ensure that method dispatch for is.numeric occurs correctly.