Data restructure in R (short lists to binary) - r

Question

I have a data set with this structure:
region1 region2 region3
1 10 5 5
2 8 10 8
3 13 15 12
4 3 17 11
5 17 9
6 12 15
7 4
8 18
9 1
I need:
item region1 region2 region3
1 1 1 0 0
2 3 1 0 0
3 4 1 0 0
4 5 0 1 1
5 8 1 0 1
6 9 0 0 1
7 10 1 1 0
8 11 0 0 1
9 12 1 0 1
10 13 1 0 0
11 15 0 1 1
12 17 1 1 0
13 18 1 0 0
The plan was to get a distinct list of items, left join each of the regions as its own column and replace matches with 1s, missing with 0; but I must be missing a key point of the R merge, dropping the main column of interest. Any advice is greatly appreciated! I'd prefer an R solution, but my next step would be to look into sqldf package.
#read in data
regions <- read.csv("c:/data/regions.csv")
#get unique list of items from all regions
items <- na.omit(unique(stack(regions)[1]))
#merge distinct items with each region, replace matches with 1, missings with 0
merge.test <- merge(items,regions,by.x="values", by.y=c("region1"), all=TRUE)

Answer 1

Helps to provide a reproducible example (i.e. give us an easy copy-paste command to construct your sample data).
You didn't say, so I guess your data is in a list perhaps?
dat <- list(region1=c(10, 8, 3, 17, 12, 4, 18, 1),
region2=c(5,10,15,17),
region3=c(5,8,12,11,9,15))
First find all the items (perhaps no need to sort, I just did it because yours is sorted)
ids <- sort(unique(unlist(dat)))
Then for each region, just see if the list of unique IDs is in that region, coercing the logical TRUE/FALSE to 0 and 1 (you could leave as T/F if that would do for you)
data.frame(ids,
region1=as.integer(ids %in% dat$region1),
region2=as.integer(ids %in% dat$region2),
region3=as.integer(ids %in% dat$region3))
If you have just 3 regions that's OK, if you have more you might want to automate the typing:
cols <- lapply(dat, function (region) as.integer(ids %in% region))
cols$id <- ids
df <- do.call(data.frame, cols)
where do.call calls the data.frame function with the list cols as its (named) arguments, i.e. it just does
data.frame(id=..., region1=..., region2=..., region3=...)
If your original dat was a CSV and each column has NA values you might want to insert na.omit as appropriate.

Answer 2

The existing answers are fine, but they seem to complicated. Just try stack + table instead:
table(stack(dat))
# ind
# values region1 region2 region3
# 1 1 0 0
# 3 1 0 0
# 4 1 0 0
# 5 0 1 1
# 8 1 0 1
# 9 0 0 1
# 10 1 1 0
# 11 0 0 1
# 12 1 0 1
# 15 0 1 1
# 17 1 1 0
# 18 1 0 0
I'm also going to go out on a limb and say that considering your current approach, you actually have a data.frame not a list:
DAT <- dat
Len <- max(sapply(DAT, length))
DAT <- data.frame(lapply(DAT, function(x) { length(x) <- Len; x }))
In that case, the solution is no different:
table(stack(DAT))
# ind
# values region1 region2 region3
# 1 1 0 0
# 3 1 0 0
# 4 1 0 0
# 5 0 1 1
# 8 1 0 1
# 9 0 0 1
# 10 1 1 0
# 11 0 0 1
# 12 1 0 1
# 15 0 1 1
# 17 1 1 0
# 18 1 0 0

Answer 3

Using #mathematical.coffee's example and qdap:
dat <- list(region1=c(10, 8, 3, 17, 12, 4, 18, 1),
region2=c(5,10,15,17),
region3=c(5,8,12,11,9,15))
library(qdap)
matrix2df(t(mtabulate(dat)), "item")
You may need to expand with:
FUN <- function(x) as.numeric(x > 0)
matrix2df(apply(t(mtabulate(dat)), 2, FUN), "item")
If you have more than one item in in a vector.