How to find intersection of values that meet multiple conditions in R?

Since the years are distinct and in the desired set we only have to count how many there are for each company. (If that is not true, in general, then apply the solutions below to df2 <- unique(merge(df, data.frame(year = 2010:2012))) in place of df. Also if we did not know the value 3 and we wanted it to equal the number of unique years in the data then we could compute it using length(unique(df$year)).

Now, using that idea here are several alternatives. We can use table to get their frequencies and then keep those with frequency 3 or in the last case we can use dplyr's count and then filter to get those with a count of 3.

tab <- table(df$company)
names(tab)[tab == 3]
## [1] "a" "c"

names(Filter(function(x) x == 3, table(df$company)))
## [1] "a" "c"

library(dplyr)   
df %>%
  count(company) %>%
  filter(n == 3) %>%
  select(company)
##   company
## 1       a
## 2       c

To use the intersect idea of the question split company by year and then use Reduce to apply intersect repeatedly:

 with(df, Reduce(intersect, split(company, year)))
 ## [1] "a" "c"

We can visualize this using table like this. Any column having no zero values corresponds to a company having all years.

table(df)
##       company
## year   a b c
##   2010 1 1 1
##   2011 1 1 1
##   2012 1 0 1

or a heatmap where green is present and red is absent. Any column that is entirely green has all years.

heatmap(table(df), Rowv = NA, Colv = NA, col = 2:3, scale = "none")

Here is a base R solution with ave and unique.

n <- with(df, ave(year, company, FUN = length))
unique(df$company[n == 3])
#[1] "a" "c"

Since you have received some excellent answers, I thought a very different approach with a little bit of imagination might also be fine.

I visualized your problem set up as a bipartite graph where we have 2 distinct set of nodes where one is company names and the other is the years, while there is no connection between companies and also between year.

library(igraph)

# Creating a graph object but first I alternate the columns of your data set
df[, c(2, 1)] |>
  graph_from_data_frame() -> g

# Then we create a type object to distinguish between 2 sets of nodes, Type FALSE
# refers to company name and type TRUE refers to years
V(g)$type <- bipartite.mapping(g)$type

# Then we extract those nodes whose degree are equal to 3 while they are of type FALSE
V(g)[degree(g, V(g)) == length(unique(df$year)) & V(g)$type == FALSE]
+ 2/6 vertices, named, from c172916:
[1] a c

In case you would like to see how the graph looks like:

plot(g,
     vertex.color = ifelse(V(g)$type, "lightblue", "salmon"),
     vertex.shape = ifelse(V(g)$type, "circle", "square"),
     vertex.size = 25,
     edge.color = "grey",
     layout = layout.bipartite)

This won't work in general to compute arbitrary intersections, but (¿ I think ?) does what you specified above:

(df 
   %>% group_by(company)
   %>% filter(all(2010:2012 %in% year))
   %>% select(company)
   %>% distinct()
)

Just nest and reduce:

df <- data.frame(year=c(2010,2011,2012,2010,2011,2010,2011,2012), company = c("a","a","a","b","b","c","c","c"))
df %>% 
    tidyr::nest(data = -year) %>% 
    magrittr::use_series(data) %>% 
    purrr::reduce(dplyr::intersect)
# A tibble: 2 x 1
  company
  <chr>  
1 a      
2 c 

Or split-map-reduce:

split.data.frame(df, df$year) %>% 
    purrr::map(magrittr::use_series, company) %>% 
    purrr::reduce(dplyr::intersect)
[1] "a" "c"

Expounding on the answers given, this here would work in general:

df %>%
  mutate(rn = list(seq(min(year), max(year))))%>%
  group_by(company) %>%
  summarise(rn = all(unlist(rn) %in% year)) %>%
  filter(rn) %>%
  select(company)

# A tibble: 2 x 1
  company
  <chr>  
1 a      
2 c  

Another igraph solution

> g <- graph_from_data_frame(df)

> names(which(degree(g, V(g)[names(V(g)) %in% df$company]) == length(unique(df$year))))
[1] "a" "c"