I have a data set with individuals ( ID ) that can be part of more than one group.
Example:
library(data.table)
DT <- data.table(
ID = rep(1:5, c(3:1, 2:3)),
Group = c("A", "B", "C", "B",
"C", "A", "A", "C",
"A", "B", "C")
)
DT
# ID Group
# 1: 1 A
# 2: 1 B
# 3: 1 C
# 4: 2 B
# 5: 2 C
# 6: 3 A
# 7: 4 A
# 8: 4 C
# 9: 5 A
# 10: 5 B
# 11: 5 C
I want to know the sum of identical individuals for 2 groups.
The result should look like this:
Group.1 Group.2 Sum
A B 2
A C 3
B C 3
Where Sum indicates the number of individuals the two groups have in common.
Here's my version:
# size-1 IDs can't contribute; skip
DT[ , if (.N > 1)
# simplify = FALSE returns a list;
# transpose turns the 3-length list of 2-length vectors
# into a length-2 list of 3-length vectors (efficiently)
transpose(combn(Group, 2L, simplify = FALSE)), by = ID
][ , .(Sum = .N), keyby = .(Group.1 = V1, Group.2 = V2)]
With output:
# Group.1 Group.2 Sum
# 1: A B 2
# 2: A C 3
# 3: B C 3
As of version 1.9.8 (on CRAN 25 Nov 2016), data.table
has gained the ability to do non-equi joins . So, a self non-equi join can be used:
library(data.table) # v1.9.8+
setDT(DT)[, Group:= factor(Group)]
DT[DT, on = .(ID, Group < Group), nomatch = 0L, .(ID, x.Group, i.Group)][
, .N, by = .(x.Group, i.Group)]
x.Group i.Group N 1: AB 2 2: AC 3 3: BC 3
The non-equi join on ID, Group < Group
is a data.table
version of combn()
(but applied group-wise):
DT[DT, on = .(ID, Group < Group), nomatch = 0L, .(ID, x.Group, i.Group)]
ID x.Group i.Group 1: 1 AB 2: 1 AC 3: 1 BC 4: 2 BC 5: 4 AC 6: 5 AB 7: 5 AC 8: 5 BC
We self-join with the same dataset on
'ID', subset the rows where the 'Group' columns are different, get the nrows ( .N
), grouped by the 'Group' columns, sort the 'Group.1' and 'Group.2' columns by row using pmin/pmax
and get the unique
value of 'N'.
library(data.table)#v1.9.6+
DT[DT, on='ID', allow.cartesian=TRUE][Group!=i.Group, .N ,.(Group, i.Group)][,
list(Sum=unique(N)) ,.(Group.1=pmin(Group, i.Group), Group.2=pmax(Group, i.Group))]
# Group.1 Group.2 Sum
#1: A B 2
#2: A C 3
#3: B C 3
Or as mentioned in the comments by @MichaelChirico and @Frank, we can convert 'Group' to factor
class, subset the rows based on as.integer(Group) < as.integer(i.Group)
, group by 'Group', 'i.Group' and get the nrow ( .N
)
DT[, Group:= factor(Group)]
DT[DT, on='ID', allow.cartesian=TRUE][as.integer(Group) < as.integer(i.Group), .N,
by = .(Group.1= Group, Group.2= i.Group)]
yet another solution (base R):
tmp <- split(DT, DT[, 'Group'])
ans <- apply(combn(LETTERS[1 : 3], 2), 2, FUN = function(ind){
out <- length(intersect(tmp[[ind[1]]][, 1], tmp[[ind[2]]][, 1]))
c(group1 = ind[1], group2 = ind[2], sum_ = out)
}
)
data.frame(t(ans))
# group1 group2 sum_
#1 A B 2
#2 A C 3
#3 B C 3
first split data into list of groups, then for each unique pairwise combinations of two groups see how many subjects in common they have, using length(intersect(...
.
Great answers above. Just an alternative using dplyr
in case you, or someone else, is interested.
library(dplyr)
cmb = combn(unique(dt$Group),2)
data.frame(g1 = cmb[1,],
g2 = cmb[2,]) %>%
group_by(g1,g2) %>%
summarise(l=length(intersect(DT[DT$Group==g1,]$ID,
DT[DT$Group==g2,]$ID)))
# g1 g2 l
# (fctr) (fctr) (int)
# 1 A B 2
# 2 A C 3
# 3 B C 3
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