Which pattern occurs the most in a matrix - R (UPDATE) - c++

Question

UPDATE 2
*I've added some code (and explanation) I wrote myself at the end of this question, this is however a suboptimal solution (both in coding efficiency as resulting output) but kind of manages to make a selection of items that adhere to the constraints. If you have any ideas on how to improve it (again both in efficiency as resulting output) please let me know.
1. Updated Post
Please look below for the initial question and sample code. Thx to alexis_laz his answer the problem was solved for a small number of items. However when the number of items becomes to large the combn function in R cannot calculate it anymore because of the invalid 'ncol' value (too large or NA) error. Since my dataset has indeed a lot of items, I was wondering whether replacing some of his code (shown after this) with C++ provides a solution to this, and if this is the case what code I should use for this? Tnx!
This is the code as provided by alexis_laz;
ff = function(x, No_items, No_persons)
{
do.call(rbind,
lapply(No_items:ncol(x),
function(n) {
col_combs = combn(seq_len(ncol(x)), n, simplify = F)
persons = lapply(col_combs, function(j) rownames(x)[rowSums(x[, j, drop = F]) == n])
keep = unlist(lapply(persons, function(z) length(z) >= No_persons))
data.frame(persons = unlist(lapply(persons[keep], paste, collapse = ", ")),
items = unlist(lapply(col_combs[keep], function(z) paste(colnames(x)[z], collapse = ", "))))
}))
}
2. Initial Post
Currently I'm working on a set of data coming from adaptive measurement, which means that not all persons have made all of the same items. For my analysis however I need a dataset that contains only items that have been made by all persons (or a subset of these persons).
I have a matrix object in R with rows = persons (100000), and columns = items(220), and a 1 in a cell if the person has made the item and a 0 if the person has not made the item.
How can I use R to determine which combination of at least 15 items, is made by the highest amount of persons?
Hopefully the question is clear (if not please ask me for more details and I will gladly provide those).
Tnx in advance.
Joost
Edit:
Below is a sample matrix with the items (A:E) as columns and persons (1:5) as rows.
mat <- matrix(c(1,1,1,0,0,1,1,0,1,1,1,1,1,0,1,0,1,1,0,0,1,1,1,1,0),5,5,byrow=T)
colnames(mat) <- c("A","B","C","D","E")
rownames(mat) <- 1:5
> mat
A B C D E
"1" 1 1 1 0 0
"2" 1 1 0 1 1
"3" 1 1 1 0 1
"4" 0 1 1 0 0
"5" 1 1 1 1 0
mat[1,1] = 1 means that person 1 has given a response to item 1.
Now (in this example) I'm interested in finding out which set of at least 3 items is made by at least 3 people. So here I can just go through all possible combinations of 3, 4 and 5 items to check how many people have a 1 in the matrix for each item in a combination.
This will result in me choosing the item combination A, B and C, since it is the only combination of items that has been made by 3 people (namely persons 1, 3 and 5).
Now for my real dataset I want to do this but then for a combination of at least 10 items that a group of at least 75 people all responded to. And since I have a lot of data preferably not by hand as in the example data.
I'm thus looking for a function/code in R, that will let me select the minimal amount of items, and questions, and than gives me all combinations of items and persons that adhere to these constraints or have a greater number of items/persons than the constrained.
Thus for the example matrix it would be something like;
f <- function(data,no.items,no.persons){
#code
}
> f(mat,3,3)
no.item no.pers items persons
1 3 3 A, B, C 1, 3, 5
Or in case of at least 2 items that are made by at least 3 persons;
> f(mat,2,3)
no.item no.pers items persons
1 2 4 A, B 1, 2, 3, 5
2 2 3 A, C 1, 3, 5
3 2 4 B, C 1, 3, 4, 5
4 3 3 A, B, C 1, 3, 5
Hopefully this clears up what my question actually is about. Tnx for the quick replies that I already received!
3. Written Code
Below is the code I've written today. It takes each item once as a starting point and then looks to the item that has been answered most by people who also responded to the start item. It the takes these two items and looks to a third item, and repeats this until the number of people that responded to all selected questions drops below the given limit. One drawback of the code is that it takes some time to run, (it goes up somewhat exponentially when the number of items grows). The second drawback is that this still does not evaluate all possible combinations of items, in the sense that the start item, and the subsequently chosen item may have a lot of persons that answered to these items in common, however if the chosen item has almost no similarities with the other (not yet chosen) items, the sample might shrink very fast. While if an item was chosen with somewhat less persons in common with the start item, and this item has a lot of connections to other items, the final collection of selected items might be much bigger than the one based on the code used below. So again suggestions and improvements in both directions are welcome!
set.seed(512)
mat <- matrix(rbinom(1000000, 1, .6), 10000, 100)
colnames(mat) <- 1:100
fff <- function(data,persons,items){
xx <- list()
for(j in 1:ncol(data)){
d <- matrix(c(j,length(which(data[,j]==1))),1,2)
colnames(d) <- c("item","n")
t = persons+1
a <- j
while(t >= persons){
b <- numeric(0)
for(i in 1:ncol(data)){
z <- c(a,i)
if(i %in% a){
b[i] = 0
} else {
b[i] <- length(which(rowSums(data[,z])==length(z)))
}
}
c <- c(which.max(b),max(b))
d <- rbind(d,c)
a <- c(a,c[1])
t <- max(b)
}
print(j)
xx[[j]] = d
}
x <- y <- z <- numeric(0)
zz <- matrix(c(0,0,rep(NA,ncol(data))),length(xx),ncol(data)+2,byrow=T)
colnames(zz) <- c("n.pers", "n.item", rep("I",ncol(data)))
for(i in 1:length(xx)){
zz[i,1] <- xx[[i]][nrow(xx[[i]])-1,2]
zz[i,2] <- length(unname(xx[[i]][1:nrow(xx[[i]])-1,1]))
zz[i,3:(zz[i,2]+2)] <- unname(xx[[i]][1:nrow(xx[[i]])-1,1])
}
zz <- zz[,colSums(is.na(zz))<nrow(zz)]
zz <- zz[which((rowSums(zz,na.rm=T)/rowMeans(zz,na.rm=T))-2>=items),]
zz <- as.data.frame(zz)
return(zz)
}
fff(mat,110,8)
> head(zz)
n.pers n.item I I I I I I I I I I
1 156 9 1 41 13 80 58 15 91 12 39 NA
2 160 9 2 27 59 13 81 16 15 6 92 NA
3 158 9 3 59 83 32 25 80 14 41 16 NA
4 160 9 4 24 27 71 32 10 63 42 51 NA
5 114 10 5 59 66 27 47 13 44 63 30 52
6 158 9 6 13 56 61 12 59 8 45 81 NA
#col 1 = number of persons in sample
#col 2 = number of items in sample
#col 3:12 = which items create this sample (NA if n.item is less than 10)

Answer 1

to follow up on my comment, something like:
set.seed(1618)
mat <- matrix(rbinom(1000, 1, .6), 100, 10)
colnames(mat) <- sample(LETTERS, 10)
rownames(mat) <- sprintf('person%s', 1:100)
mat1 <- mat[rowSums(mat) > 5, ]
head(mat1)
# A S X D R E Z K P C
# person1 1 1 1 0 1 1 1 1 1 1
# person3 1 0 1 1 0 1 0 0 1 1
# person4 1 0 1 1 1 1 1 0 1 1
# person5 1 1 1 1 1 0 1 1 0 0
# person6 1 1 1 1 0 1 0 1 1 0
# person7 0 1 1 1 1 1 1 1 0 0
table(rowSums(mat1))
# 6 7 8 9
# 24 23 21 5
tab <- table(sapply(1:nrow(mat1), function(x)
paste(names(mat1[x, ][mat1[x, ] == 1]), collapse = ',')))
data.frame(tab[tab > 1])
# tab.tab...1.
# A,S,X,D,R,E,P,C 2
# A,S,X,D,R,E,Z,P,C 2
# A,S,X,R,E,Z,K,C 3
# A,S,X,R,E,Z,P,C 2
# A,S,X,Z,K,P,C 2

Answer 2

Here is another idea that matches your output:
ff = function(x, No_items, No_persons)
{
do.call(rbind,
lapply(No_items:ncol(x),
function(n) {
col_combs = combn(seq_len(ncol(x)), n, simplify = F)
persons = lapply(col_combs, function(j) rownames(x)[rowSums(x[, j, drop = F]) == n])
keep = unlist(lapply(persons, function(z) length(z) >= No_persons))
data.frame(persons = unlist(lapply(persons[keep], paste, collapse = ", ")),
items = unlist(lapply(col_combs[keep], function(z) paste(colnames(x)[z], collapse = ", "))))
}))
}
ff(mat, 3, 3)
# persons items
#1 1, 3, 5 A, B, C
ff(mat, 2, 3)
# persons items
#1 1, 2, 3, 5 A, B
#2 1, 3, 5 A, C
#3 1, 3, 4, 5 B, C
#4 1, 3, 5 A, B, C