10

I have a data set that looks something like this:

     id1  id2   size
1   5400 5505      7
2   5033 5458      1
3   5452 2873     24
4   5452 5213      2
5   5452 4242     26
6   4823 4823      4
7   5505 5400     11

Where id1 and id2 are unique nodes in a graph, and size is a value assigned to the directed edge connecting them from id1 to id2. This data set is fairly large (a little over 2 million rows). What I would like to do is sum the size column, grouped by unordered node pairs of id1 and id2. For example, in the first row, we have id1=5400 and id2=5505. There exists another row in the data frame where id1=5505 and id2=5400. In the grouped data, the sum of the size columns for these two rows would be added to a single row. So in other words I want to summarize the data where I'm grouping on an (unordered) set of (id1,id2). I've found a way to do this using apply with a custom function that checks for the reversed column pair in the full data set, but this works excruciatingly slow. Does anyone know of a way to do this another way, perhaps with plyr or with something in the base packages that would be more efficient?

4

2 回答 2

12

pmax一种方法是使用and pminofid1和创建额外的列，id2如下所示。我将data.table在这里使用解决方案。

require(data.table)
DT <- data.table(DF)
# Following mnel's suggestion, g1, g2 could be used directly in by
# and it could be even shortened by using `id1` and id2` as their names
DT.OUT <- DT[, list(size=sum(size)), 
        by=list(id1 = pmin(id1, id2), id2 = pmax(id1, id2))]
#     id1  id2 size
# 1: 5400 5505   18
# 2: 5033 5458    1
# 3: 5452 2873   24
# 4: 5452 5213    2
# 5: 5452 4242   26
# 6: 4823 4823    4
于 2013-03-18T21:37:04.417 回答
7

另一种方法：

R> library(igraph)
R> DF
   id1  id2 size
1 5400 5505    7
2 5033 5458    1
3 5452 2873   24
4 5452 5213    2
5 5452 4242   26
6 4823 4823    4
7 5505 5400   11
R> g  <- graph.data.frame(DF, directed=F)
R> g  <- simplify(g, edge.attr.comb="sum", remove.loops=FALSE)
R> DF <- get.data.frame(g)
R> DF
   id1  id2 size
1 5400 5505   18
2 5033 5458    1
3 5452 2873   24
4 5452 5213    2
5 5452 4242   26
6 4823 4823    4
于 2013-04-20T21:10:31.373 回答