Hi Sam,
Have a look at the function prop.part: it is the working function of
consensus and other similar functions.
EP
Sam Brown s_d_j_br...@hotmail.com a écrit :
Hello one and all
I'm wondering if there is any way of obtaining the consensus
frequencies of nodes in majority-rule consensus trees where p=0.5.
i.e. given:
t1 - ((t4,t8),((t10,((t2,t5),t3)),((t6,t9),(t1,t7;
t2 - ((t4,t8),((t10,((t2,t5),t3)),((t6,t9),(t1,t7;
t3 - ((t4,t8),((t10,((t2,t5),t3)),((t6,t9),(t1,t7;
t4 - ((t2,t10),((t5,((t4,t3),t8)),((t6,t9),(t1,t7;
t5 - ((t10,t5),((t8,((t2,t4),t3)),((t6,t9),(t1,t7;
t6 - ((t4,t10),((t8,((t3,t5),t2)),((t6,t9),(t1,t7;
t7 - ((t4,t3),((t5,((t2,t10),t8)),((t6,t9),(t1,t7;
t8 - ((t4,t2),((t10,((t8,t5),t3)),((t6,t9),(t1,t7;
t9 - ((t4,t3),((t8,((t10,t5),t2)),((t6,t9),(t1,t7;
t10 - ((t10,t8),((t4,((t2,t3),t5)),((t6,t9),(t1,t7;
tr-read.tree(text=c(t1,t2,t3,t4,t5,t6,t7,t8,t9,t10))
conTR-consensus(tr, p=0.5)
plot(conTR)
The node (t3,t2,t5) will have a consensus frequency of 0.5 (50%) and
the three nodes ((t6,t9),(t1,t7)) will each have a frequency of 1
(100%).
Are there any quick ways of getting this information?
Thanks for any help/pointers
Sam
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