Subsetting a dataframe by the amount of repetition [duplicate]

Question

If I have a dataframe like this:

neu <- data.frame(test1 = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14), 
                  test2 = c("a","b","a","b","c","c","a","c","c","d","d","f","f","f"))
neu
   test1 test2
1      1     a
2      2     b
3      3     a
4      4     b
5      5     c
6      6     c
7      7     a
8      8     c
9      9     c
10    10     d
11    11     d
12    12     f
13    13     f
14    14     f

and I would like to select only those values where the level of the factor test2 appears more than let's say three times, what would be the fastest way?

Thanks very much, didn't really find the right answer in the previous questions.

Answer 1

Find the rows using:

z <- table(neu$test2)[table(neu$test2) >= 3] # repeats greater than or equal to 3 times

Or:

z <- names(which(table(neu$test2)>=3))

Then subset with:

subset(neu, test2 %in% names(z))

Or:

neu[neu$test2 %in% names(z),]

Answer 2

Here's another way:

 with(neu, neu[ave(seq(test2), test2, FUN=length) > 3, ])

#   test1 test2
# 5     5     c
# 6     6     c
# 8     8     c
# 9     9     c

Answer 3

I'd use count from the plyr package to perform the counting:

library(plyr)
count_result = count(neu, "test2")
matching = with(count_result, test2[freq > 3])
with(neu, test1[test2 %in% matching])
[1] 5 6 8 9

Answer 4

The (better scaling) data.table way:

library(data.table)
dt = data.table(neu)

dt[dt[, .I[.N >= 3], by = test2]$V1]

Note: hopefully, in the future, the following simpler syntax will be the fast way of doing this:

dt[, .SD[.N >= 3], by = test2]

(c.f. Subset by group with data.table)