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I am interested in deriving dominance metrics (as in a dominance hierarchy) for nodes in a dominance directed graph, aka a tournament graph. I can use R and the package igraph to easily construct such graphs, e.g.

library(igraph)

create a data frame of edges

the.froms <- c(1,1,1,2,2,3)

the.tos <- c(2,3,4,3,4,4)

the.set <- data.frame(the.froms, the.tos)

set.graph <- graph.data.frame(the.set)

plot(set.graph)

This plotted graph shows that node 1 influences nodes 2, 3, and 4 (is dominant to them), that 2 is dominant to 3 and 4, and that 3 is dominant to 4.

However, I see no easy way to actually calculate a dominance hierarchy as in the page: https://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/GraphTheory/GraphTheory_9_17/node11.html . So, my first and main question is does anyone know how to derive a dominance hierarchy/node-based dominance metric for a graph like this using some hopefully already coded solution in R?

Moreover, in my real case, I actually have a sparse matrix that is missing some interactions, e.g.

incomplete.set <- the.set[-2, ]

incomplete.graph <- graph.data.frame(incomplete.set)

plot(incomplete.graph)

In this plotted graph, there is no connection between species 1 and 3, however making some assumptions about transitivity, the dominance hierarchy is the same as above.

This is a much more complicated problem, but if anyone has any input about how I might go about deriving node-based metrics of dominance for sparse matrices like this, please let me know. I am hoping for an already coded solution in R, but I'm certainly MORE than willing to code it myself.

Thanks in advance!

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不确定这是否完美或我完全理解这一点，但它似乎可以通过一些试验和错误来正常工作：

library(relations)
result <- relation_consensus(endorelation(graph=the.set),method="Borda")
relation_class_ids(result)
#1 2 3 4 
#1 2 3 4

method=处理关系等有很多潜在的选择- 请参阅?relation_consensus更多信息。使用method="SD/L"which is a linear order 可能最适合您的数据，尽管由于更复杂示例中的冲突，它可以建议多种可能的解决方案。但是，对于当前的简单数据，情况并非如此-尝试：

result <- relation_consensus(endorelation(graph=the.set),method="SD/L",
                             control=list(n="all"))
result
#An ensemble of 1 relation of size 4 x 4.

lapply(result,relation_class_ids)
#[[1]]
#1 2 3 4 
#1 2 3 4

中的示例再次提供了处理此问题的方法?relation_consensus。

于 2013-09-24T01:04:15.237 回答