Re: [igraph] degree according to vertex attributes

[Dominik Santner (CRIE)]

Hello,

thank you very much for your suggestion. Unfortunately there are two 
problems:

first:
when using your example graph created by

> g <- grg.game(100, 0.2)
> V(g)$city <- LETTERS[(0:99 %/% 10) + 1]

I recieve a warning when doing

> degree(g)-degree(internal.graph)

telling me that the two objects are not of the same length

> length(degree(internal.graph))
[1] 96
> length(degree(g))
[1] 100


the second problem occures when applying this procedure to my data:

in your example you use V(g)$city when doing
 
> internal.edges <- which(V(g)$city[el[,1]] == V(g)$city[el[,2]])

whith the result:
> internal.edges
  [1]   1   5   6   7   8 ...
 
when checking for
> V(g)$city
  [1] "A" "A" "A" "A" "A" ...

> V(g)$city[el[,2]]
  [1] "A" "B" "B" "B" "A" "A" ...

everything is fine.


But with my dataset this happens:

> internal.edges <- which(V(xxx)$st_AMR[el[,1]] == V(xxx)$st_AMR[el[,2]])
> internal.edges
integer(0)

> V(xxx)$st_AMR
  [1] "München"                "Darmstadt"              "Regensburg"             
"Halle" ...

> V(xxx)$st_AMR[el[,2]]
  [1] NA NA NA NA NA ...

Maybe it has something to do with my vertex labels and/or my edge list, but I 
have no idea what it might be:

The edge list of your example:

> head(el)
     [,1] [,2]
[1,]    1    7
[2,]    1   11
[3,]    1   12
[4,]    1   13
[5,]    2    3
[6,]    2    4

The edge list of my dataset:

> head(el)
     [,1]
[1,] "Company A (6411000)"
[2,] "Company A (6411000)"
[3,] "Company A (6411000)"
[4,] "Company A (6411000)"
[5,] "Company A (6411000)"
[6,] "Company A (6411000)"
     [,2]
[1,] "University A (8416041)"
[2,] "University B (8416041)"
[3,] "Company B (5562024)"
[4,] "Company C (5358024)"
[5,] "Company D (5358024)"
[6,] "Company E (8215101)"

Best regards
Dominik


Am 28.09.2012 23:44, schrieb Tamás Nepusz:
> Hello,
>
> Suppose that the city is encoded in the "city" vertex attribute of the graph; 
> for instance:
>
> > g <- grg.game(100, 0.2)
> > V(g)$city <- LETTERS[(0:99 %/% 10) + 1]
> First, we get the edge list of the graph:
>
> > el <- get.edgelist(g)
> Next, we select those edges (i.e. rows) from the edge list where the endpoints 
> are in the same city:
>
> > internal.edges <- which(V(g)$city[el[,1]] == V(g)$city[el[,2]])
> We then create a subgraph which consists of these edges only:
>
> > internal.graph <- subgraph.edges(g, internal.edges)
> degree(internal.graph) will then give you how many friends each person has in 
> the same city, and degree(g)-degree(internal.graph) will give you how many 
> friends each person has in different cities.
>
> Best,