Patrick Meyer <patrick.meyer <at> internet.lu> writes:
:
: It does not exactly do what it is meant to... Instead of breaking down,
: my code is looping forever now... I think I will have a deeper look at
: it tomorrow...
:
: In fact, I have two lists: let's say L and N.
:
: An elements (or set) of N must be added to L, if and only if it is not a
: subset of an element of L.
:
: I think that the code you wrote does that, for each element of separately.
If both arguments are lists, not a list and vector as you previously
described, then just lapply over the previous formula (here encapsulated
in the function) using a second outer lapply:
L <- c(L, N[unlist(lapply(N, function(x)
!any(unlist(lapply(L, setequal, x)))))])
:
: Couldn't this be a solution: add all the elements of N to L, and
: afterwards eliminate the sets which are present more than once? How
: could one eliminate these sets?
:
Yes but it would involve unnecessary comparisons between pairs in
L as well as unnecessary comparisons between pairs in N.
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