chunjin - 2 months ago 5x

R Question

`x1 <- data.frame(id=1:3,item=c("A","B","A","B","C","D"))`

x1[order(x1$item),]

id item

1 1 A

3 3 A

2 2 B

4 1 B

5 2 C

6 3 D

`id1=c(1,2,1,3,2,3)`

id2 = c(2,1,3,1,3,2)

A=c(0,0,1,1,0,0)

B=c(1,1,0,0,0,0)

C = 0

D=0

datawanted <- data.frame(id1,id2,A,B,C,D)

id1 id2 A B C D

1 1 2 0 1 0 0

2 2 1 0 1 0 0

3 1 3 1 0 0 0

4 3 1 1 0 0 0

5 2 3 0 0 0 0

6 3 2 0 0 0 0

if person1 and person2 both have B,then in the

Can someone give me some suggestions or functions in R,to deal with this problem?

Answer

Cool question. You have a bipartite graph, so following Gabor's tutorial...

```
library(igraph)
g = graph_from_edgelist(as.matrix(x1))
V(g)$type = grepl("[A-Z]", V(g)$name)
```

For OP's desired output, first we can extract the incidence matrix:

```
gi = get.incidence(g)
# A B C D
# 1 1 1 0 0
# 2 0 1 1 0
# 3 1 0 0 1
```

Note (thanks @thelatemail), that if you don't want to use igraph, you can get to `gi`

as `table(x1)`

.

Then, we look at the combinations of ids:

```
res = t(combn(nrow(gi), 2, function(x) c(
as.integer(rownames(gi)[x]),
pmin( gi[x[1], ], gi[x[2], ] )
)))
dimnames(res) <- list( NULL, c("id1", "id2", colnames(gi)))
# id1 id2 A B C D
# [1,] 1 2 0 1 0 0
# [2,] 1 3 1 0 0 0
# [3,] 2 3 0 0 0 0
```

This essentially is the OP's desired output. They had included redundant rows (e.g., 1,2 and 2,1).

Fun reason to use a graph (ht Chris):

```
V(g)$color <- ifelse(V(g)$type, "red", "light blue")
V(g)$x <- (1:2)[ V(g)$type + 1 ]
V(g)$y <- ave(seq_along(V(g)), V(g)$type, FUN = seq_along)
plot(g)
```

Or, apparently this can be done more or less like

```
plot(g, layout = layout.bipartite(g)[,2:1])
```

Source (Stackoverflow)

Comments