How can I use attr<- with lapply?

问题

Or put it more general: How can I add multiple attributes to the elements of list?

I am stuck trying to set an attribute to elements of a list all of which are data.frames. In the end I would like to add names(myList) as a varying attribute to every data.frame inside. But I even cannot get a static attribute for all list elements to go.

lapply(myList,attr,which="myname") <- "myStaticName"

This does not work because lapply does not work with lapply<-. If I had at least an idea how to do this, maybe I could figure out how to do it with varying attributes like the name of the list.

回答1:

I don't recommend it, but you could do: lapply(myList, 'attr<-', which='myname', value='myStaticName'). An old fashioned for loop is probably the clearest way to perform this task---or do this assignment upstream when the objects are created.

for (i in seq_along(myList)) attr(myList[[i]], 'myname') <- 'myStaticName'

EDIT:

As @mnel points out in the comments, setattr in the data.table package is also an efficient option, since it assigns by reference.

Edit: @mnel -- don't use setattr with lapply. This is one case where the for loop is much faster.

library(microbenchmark)
library(data.table)
myList <- as.list(1:10000)

`lapply.attr<-` <- 
    function() 
    lapply(myList, 'attr<-', which='myname', value='myStaticName')

`for.attr<-` <- 
    function() 
    for (i in seq_along(myList)) 
        attr(myList[[i]], 'myname') <- 'myStaticName'

lapply.setattr <- 
  function() 
  lapply(myList, setattr, name='myname', value='myStaticName')

for.setattr <- function() 
  for (i in seq_along(myList)) 
  setattr(myList[[i]], name = 'myname', value = 'myStaticName')

result <- microbenchmark(`lapply.attr<-`(), `for.attr<-`(), lapply.setattr(), for.setattr())
plot(result)

回答2:

Based on this answer by Thierry I found a solution on my own. Actually I have been close with several tries but did not return the WHOLE list which is key.

myList <- lapply(names(myList),function(X){
attr(myList[[X]],"myname") <- X
myList[[X]]
})

My mistake was not to return the whole list but only the second line of the function, i.e. the attribute. Thus I was not able to replace the initial list.

@Matthew Plourde: what's strange: your benchmark looks somewhat different on my machine: RStudio, OS X, 2.5 Ghz Intel Core i7, 16GB RAM.

来源：https://stackoverflow.com/questions/13614399/how-can-i-use-attr-with-lapply