Counting the number of same elements in each class - r

Question

According to the following table, I have many different teachers(10,11,12,...) with different ideas(1,2,... for example 1:Very Good, 2:Good,... ) of each class (1,2,3,...).
Some teachers don't have any idea about some classes.
class Teacher-code Opinion
1 12 1
1 13 1
1 14 1
2 11 3
2 13 1
3 10 1
3 11 2
3 12 1
3 13 1
This is a sample of my table but I have many records. I want to have a symmetric matrix of teachers with counts of their same ideas about classes. for example, teacher 12 and 13 have the same idea in the class of 1 and 3, then the intersection elements of them are 2. Or teacher codes of 14 and 13 have just one same idea about the first class. I want to get the following matrix:
[10] [11] [12] [13] [14]
[10] 0 0 1 1 0
[11] 0 0 0 0 0
[12] 1 0 0 2 1
[13] 1 0 2 0 1
[14] 0 0 1 1 0

Answer 1

This is a base R solution based on a general approach taken to find common rows between data frames. Maybe this could be helpful.
Create a function that will find overlap in your dataframe between teachers that share other common values in certain columns (in this case, class and Opinion). With merge you can identify overlap, and nrow to count the overlapping rows.
Using outer you can generate a matrix of all teachers. The function passed to the product needs to be vectorized.
the_teachers <- sort(unique(df$Teacher_code))
get_num_classes <- function(x, y) {
nrow(
merge(
df[df$Teacher_code == x, c("class", "Opinion")],
df[df$Teacher_code == y, c("class", "Opinion")]
)
)
}
mat <- outer(the_teachers, the_teachers, Vectorize(get_num_classes))
diag(mat) <- 0
dimnames(mat) <- list(the_teachers, the_teachers)
mat
Output
10 11 12 13 14
10 0 0 1 1 0
11 0 0 0 0 0
12 1 0 0 2 1
13 1 0 2 0 1
14 0 0 1 1 0
Edit: Based on comment, there is interest in identifying the fraction of (teacher pairs sharing same opinions in same class) / (teacher pairs sharing same class). Building off the same logic, you could modify the function as below. A separate merge will determine the number of teachers sharing the same class. If this number is not zero, it will determine the number of opinions shared between the teacher pair. If there are no classes shared, the function will just return zero. Depending on size of data and concordance between teachers this may be optimized further.
get_num_classes <- function(x, y) {
same_class <- nrow(
merge(
df[df$Teacher_code == x, "class", drop = F],
df[df$Teacher_code == y, "class", drop = F]
)
)
if (same_class != 0) {
same_opinion <- nrow(
merge(
df[df$Teacher_code == x, c("class", "Opinion")],
df[df$Teacher_code == y, c("class", "Opinion")]
)
)
return(same_opinion / same_class)
} else {
return(0)
}
}

Answer 2

Here is a base R option by defining a user function f, where aggregate + pmin + vecsets::vintersect are applied:
library(vecsets)
f <- function(df) {
u <- aggregate(. ~ Teacher_code, df, I)
res <- do.call(
pmin,
lapply(
u[c("class", "Opinion")],
function(x) outer(x, x, FUN = function(...) lengths(Vectorize(vintersect)(...)))
)
)
`dimnames<-`(`diag<-`(res, 0), rep(list(u[["Teacher_code"]]), 2))
}
and you will see
> f(df)
10 11 12 13 14
10 0 0 1 1 0
11 0 0 0 0 0
12 1 0 0 2 1
13 1 0 2 0 1
14 0 0 1 1 0

Answer 3

I mean its some pretty hideous code (I'm sure someone can do something better) but I think it gets you the result you need (this isn't homework is it -.- ?). Also in the future it makes life easier if you provide data using dput()
library(dplyr)
library(tidyr)
dat <- tibble(
class = c( 1,1,1,2,2,3,3,3,3),
Teacher_code = c(12,13,14,11,13,10,11,12,13),
Opinion = c(1,1,1,3,1,1,2,1,1)
)
dat2 <- complete(dat, class, Teacher_code)
classes <- unique(dat2$class)
teachers <- unique(dat2$Teacher_code)
len_teachers <- length(teachers)
mat <- matrix(nrow = len_teachers, ncol = len_teachers)
for(i in seq_along(teachers)){
for( j in seq_along(teachers)){
same_opinion <- 0
for(k in classes){
opinion_i <- dat2 %>% filter(Teacher_code == teachers[[i]] , class == k) %>% pull(Opinion)
opinion_j <- dat2 %>% filter(Teacher_code == teachers[[j]] , class == k) %>% pull(Opinion)
same_opinion <- same_opinion + (opinion_i == opinion_j & !(is.na(opinion_i) | is.na(opinion_j)))
}
mat[i,j] <- same_opinion
}
}