Annette
I understand your problem.
I think you may find 'blockparts(rep(5,5),5)' helpful.
I'm working on permutations of multisets right now
and expect to have functionality in partitions package
as soon as I finish the Other Ten Thousand Things
On My Things To Do List.arts
Perhaps we could talk offline.
best wishes
Robin
Annette Heisswolf wrote:> Hi all,
>
> I try to generate sets of strategies that contain probability
> distributions for a defined number of elements, e.g. imagine an
> animal that can produce 5 different types of offspring and I want to
> figure out which percentage of each type it should produce in order to
> maximize its fitness. In order to do so, I need to calculate the fitness
> for all potential strategies. As an example, if I assume the probability
> levels to be 0, 0.2, 0.4, 0.6, 0.8, 1, I can generate all possible
> probability distributions for the 5 types of offspring by using the
> following R-code:
>
> library(partitions)
> library(crank)
>
> n=5
> P=parts(n)
> L=length(P[1,])
> for (i in 1:L){
> ma=unique(permute(P[,i]))
> if (i==1) m=ma else m=rbind(m,ma)
> }
> m=m/n
>
>> m
> [,1] [,2] [,3] [,4] [,5]
> [1,] 1.0 0.0 0.0 0.0 0.0
> [2,] 0.0 1.0 0.0 0.0 0.0
> [3,] 0.0 0.0 1.0 0.0 0.0
> [4,] 0.0 0.0 0.0 1.0 0.0
> [5,] 0.0 0.0 0.0 0.0 1.0
> [6,] 0.8 0.2 0.0 0.0 0.0
> [7,] 0.8 0.0 0.2 0.0 0.0
> [8,] 0.8 0.0 0.0 0.2 0.0
> [9,] 0.8 0.0 0.0 0.0 0.2
> [10,] 0.2 0.8 0.0 0.0 0.0
> [11,] 0.2 0.0 0.8 0.0 0.0
> [12,] 0.2 0.0 0.0 0.8 0.0
> [13,] 0.2 0.0 0.0 0.0 0.8
> [14,] 0.0 0.8 0.2 0.0 0.0
> [15,] 0.0 0.8 0.0 0.2 0.0
> [16,] 0.0 0.8 0.0 0.0 0.2
> [17,] 0.0 0.2 0.8 0.0 0.0
> [18,] 0.0 0.2 0.0 0.8 0.0
> [19,] 0.0 0.2 0.0 0.0 0.8
> [20,] 0.0 0.0 0.8 0.2 0.0
> [21,] 0.0 0.0 0.8 0.0 0.2
> [22,] 0.0 0.0 0.2 0.8 0.0
> [23,] 0.0 0.0 0.2 0.0 0.8
> [24,] 0.0 0.0 0.0 0.8 0.2
> [25,] 0.0 0.0 0.0 0.2 0.8
> [26,] 0.6 0.4 0.0 0.0 0.0
> [27,] 0.6 0.0 0.4 0.0 0.0
> .
> .
> .
> [106,] 0.4 0.2 0.2 0.2 0.0
> [107,] 0.4 0.2 0.2 0.0 0.2
> [108,] 0.4 0.2 0.0 0.2 0.2
> [109,] 0.4 0.0 0.2 0.2 0.2
> [110,] 0.2 0.4 0.2 0.2 0.0
> [111,] 0.2 0.4 0.2 0.0 0.2
> [112,] 0.2 0.4 0.0 0.2 0.2
> [113,] 0.2 0.2 0.4 0.2 0.0
> [114,] 0.2 0.2 0.4 0.0 0.2
> [115,] 0.2 0.2 0.2 0.4 0.0
> [116,] 0.2 0.2 0.2 0.0 0.4
> [117,] 0.2 0.2 0.0 0.4 0.2
> [118,] 0.2 0.2 0.0 0.2 0.4
> [119,] 0.2 0.0 0.4 0.2 0.2
> [120,] 0.2 0.0 0.2 0.4 0.2
> [121,] 0.2 0.0 0.2 0.2 0.4
> [122,] 0.0 0.4 0.2 0.2 0.2
> [123,] 0.0 0.2 0.4 0.2 0.2
> [124,] 0.0 0.2 0.2 0.4 0.2
> [125,] 0.0 0.2 0.2 0.2 0.4
> [126,] 0.2 0.2 0.2 0.2 0.2
>
>
> Using the "unique()" function is necessary, as the
"permute()" function
> doesn't check whether some elements within the vector to be permuted
are
> identical, e.g. permute(c(1,0,0)) gives:
>
> [,1] [,2] [,3]
> [1,] 1 0 0
> [2,] 1 0 0
> [3,] 0 1 0
> [4,] 0 0 1
> [5,] 0 1 0
> [6,] 0 0 1
>
> In my above example, this means that permuting the first column of P,
>
>> P[,1]
> [1] 5 0 0 0 0
>
> gives 5! = 120 permutations, although there are in fact only 5
> unique ones, namely:
>
> 5 0 0 0 0
> 0 5 0 0 0
> 0 0 5 0 0
> 0 0 0 5 0
> 0 0 0 0 5
>
> On my computer (Windows XP SP3, Pentium(R) 4 CPU 2.40GHz, 504 MB RAM -
> but I've also tried it on another computer with 1 GB RAM) this leads to
> the problem that as soon as the vector to be permuted has more than 9
> elements, the resulting matrix get's too big and R won't execute
the
> permutation.
>
> Thus, even when I aim to reduce the total number of unique permutations,
> e.g. by using less probability levels than there are elements, the above
> code doesn't work. For instance, I've used 10 elements but only
the
> probability levels 0, 0.2, 0.4, 0.6, 0.8, 1 as above, thus, P looks like
> this:
>
>> P
> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
> [1,] 5 4 3 3 2 2 1
> [2,] 0 1 2 1 2 1 1
> [3,] 0 0 0 1 1 1 1
> [4,] 0 0 0 0 0 1 1
> [5,] 0 0 0 0 0 0 1
> [6,] 0 0 0 0 0 0 0
> [7,] 0 0 0 0 0 0 0
> [8,] 0 0 0 0 0 0 0
> [9,] 0 0 0 0 0 0 0
> [10,] 0 0 0 0 0 0 0
>
> In this case, permute(P[,1]) would give 10! = 3 628 800 permutations,
> although there are only 10 unique ones. There are more then 10 unique
> permutations for the other columns, but still their numbers should be
> small enough.
>
> Thus: Has anyone been working on a similar problem and found a better
> way to generate such probability distributions? I'd appreciate any kind
> of hint how to solve my problem. Thanks!
>
> Annette
>
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877