title: Dominance-directed (tournament) graph metrics
body: <p>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.</p>
  
  <pre><code>library(igraph)
  </code></pre>
  
  <p>create a data frame of edges</p>
  
  <pre><code>the.froms &lt;- c(1,1,1,2,2,3)
  
  the.tos &lt;- c(2,3,4,3,4,4)
  
  the.set &lt;- data.frame(the.froms, the.tos)
  
  set.graph &lt;- graph.data.frame(the.set)
  
  plot(set.graph)
  </code></pre>
  
  <p>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. </p>
  
  <p>However, I see no easy way to actually calculate a dominance hierarchy as in the page: <a href="https://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/GraphTheory/GraphTheory_9_17/node11.html" rel="nofollow">https://www.math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/GraphTheory/GraphTheory_9_17/node11.html</a> . <B> 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? </B></p>
  
  <p>Moreover, in my real case, I actually have a sparse matrix that is missing some interactions, e.g.</p>
  
  <pre><code>incomplete.set &lt;- the.set[-2, ]
  
  incomplete.graph &lt;- graph.data.frame(incomplete.set)
  
  plot(incomplete.graph)
  </code></pre>
  
  <p>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.</p>
  
  <p>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.</p>
  
  <p>Thanks in advance!</p>