Count numbers that are lower by constant from current number

Question

Imagine that I have a list of numbers (i.e. numbers column in data.table/data.frame).

1
5
5
10
11
12

for each number in a list a want to count how many unique numbers are there which are lower than that particular number + 5.

The explanation for upper case, first number = 1, search range is 1+5 = 6, so three numbers are in range, less than or equal to: c(1,5,5), and then count unique is 2. This is all by assuming we've got the additional condition, that the number must not only be lower than current_number + 5, but also its index in the list must be >= that of current_number.

The result in this case would be:

2
2
2
3
2
1

Note: Is there a fast solution for huge dataset, in data.frame or data.table? My dataset is rather huge, 10+M rows.

Answer 1

The fastest way I can think of in base R (works if x is sorted):

findInterval(x + 5, unique(x)) - cumsum(!duplicated(x)) + 1L
#[1] 2 2 2 3 2 1

---

edit: no problem with the sorting because with data.table, sorting integers is trivial:

nr <- 1e7
nn <- nr/2
set.seed(0L)
DT <- data.table(X=sample(nn, nr, TRUE))
#DT <- data.table(X=c(1,5,5,10,11,12))

system.time(
    DT[order(X), 
        COUNT := findInterval(X + 5L, unique(X)) - cumsum(!duplicated(X)) + 1L
    ]
)
#   user  system elapsed 
#   1.73    0.17    1.53 

2s for 10million rows.

Answer 2

Try this:

x <- c(1,5,5,10,11,12)

sapply(seq_along(x), function(i)
  sum(unique(x[i:length(x)]) <= (x[i] + 5)))
# [1] 2 2 2 3 2 1

Answer 3

One option is to use a sql self-join

library(sqldf)


df$r <- seq(nrow(df))

sqldf('
select    a.V1
          , count(distinct b.V1) as n
from      df a
          left join df b
            on  b.V1 <= a.V1 + 5
                and b.r >= a.r
group by  a.r
')

#   V1 n
# 1  1 2
# 2  5 2
# 3  5 2
# 4 10 3
# 5 11 2
# 6 12 1

Data used:

df <- structure(list(V1 = c(1L, 5L, 5L, 10L, 11L, 12L)), row.names = c(NA, 
-6L), class = "data.frame")