how to update a value in a list with lapply

On May 10, 2015, at 6:11 AM, ce wrote:

yes indeed :

foo <- lapply(foo, function(x) if(x[1] == 1 ) {x[2] <- 0; x
}else{x} )

would work. But if the list is too long, would it be time consuming
rather than just updating elements that meet the if condition?

Any change to an object will require copying the entire object. That
is the computing model that R uses. If you had presented a
modification strategy that used logical or numeric indexing to effect
only targeted nodes of a list, it still would have ended up copying
the whole object.

This used to be true, but is no longer the case (from R-3.1.0 if my
memory serves me right).  Consider the following example.

lst <- list(c(1L, 2L), 3.0)
lst

[[1]]
[1] 1 2

[[2]]
[1] 3

.Internal(inspect(lst))

@1fecce8 19 VECSXP g1c2 [MARK,NAM(1)] (len=2, tl=0)
   @26996a8 13 INTSXP g1c1 [MARK] (len=2, tl=0) 1,2
   @2699648 14 REALSXP g1c1 [MARK,NAM(2)] (len=1, tl=0) 3

Note the memory addresses (the ones starting with @).  Now, let's modify
the first list element:

lst[[1]][1] <- 4L
.Internal(inspect(lst))

@1fecce8 19 VECSXP g1c2 [MARK,NAM(1)] (len=2, tl=0)
   @26996a8 13 INTSXP g1c1 [MARK] (len=2, tl=0) 4,2
   @2699648 14 REALSXP g1c1 [MARK,NAM(2)] (len=1, tl=0) 3

As you can see, 'lst' changed but the memory address is still the same. 
  However, if we assign a double instead of an integer in the same 
position, copying must take place since the internal representation is 
changed:

lst[[1]][1] <- 4.0
.Internal(inspect(lst))

@1fecce8 19 VECSXP g1c2 [MARK,NAM(1)] (len=2, tl=0)
   @12685a0 14 REALSXP g0c2 [] (len=2, tl=0) 4,2
   @2699648 14 REALSXP g1c1 [MARK,NAM(2)] (len=1, tl=0) 3

But note that *only* the address of the first list element changed. 
This list itself and the second list element remain at the same 
addresses as before.


Henric Winell

The data.table package was invented in large part to get around that
design concern.