r - 计算 value 和 value - x 之间的值的数量，按变量

Question

我有一些数据:

library(data.table)
set.seed(1)
df1 <- data.frame(let=sample(sample(letters,2),5, replace=TRUE),
                  num=sample(1:10,5))
setDT(df1)
   let num
1:   j   7
2:   j   6
3:   g   1
4:   j   2
5:   j  10

我想计算 num的数量小于或等于 num AND 大于或等于 num - 4, by let .使用 data.table 包会更可取，但任何使用 dplyr 或 base r 的解决方案也可以。
输出将如下所示:

   let num countNumByLet
1:   j   7             2
2:   j   6             2
3:   g   1             1
4:   j   2             1
5:   j  10             3

Answer 1

这也可以使用 non-equi 解决加入:

dt <- data.table(let = sample(sample(letters, n_let), n_let * n_per_grp, replace = T),
                 num = sample (20, n_let * n_per_grp, replace = T))

dt[, .(let, high = num + 4L, num)
   ][dt,
     on = .(let,
            num <= num,
            high >= num),
     .(countNumByLet = .N),
     by = .EACHI
     ][, high:= NULL][]

   let num countNumByLet
1:   j   7             2
2:   j   6             2
3:   g   1             1
4:   j   2             1
5:   j  10             3

对于 5 的数据集，方法无关紧要。但是在扩大规模时，非对等联接确实有帮助:

n_let <- 26
n_per_grp <- 1E1

dt <- data.table(let = sample(sample(letters, n_let), n_let * n_per_grp, replace = T),
                 num = sample (20, n_let * n_per_grp, replace = T))

# 260 observations; 26 groups
# A tibble: 2 x 13
  expression     min median `itr/sec` mem_alloc
  <bch:expr>  <bch:> <bch:>     <dbl> <bch:byt>
1 dt_sapply   2.41ms 2.67ms      364.    53.9KB
2 dt_non_equi 5.08ms 5.66ms      170.   223.7KB

#2,600 observations; 26 groups
# A tibble: 2 x 13
  expression      min  median `itr/sec` mem_alloc
  <bch:expr>  <bch:t> <bch:t>     <dbl> <bch:byt>
1 dt_sapply   11.49ms 12.15ms      80.3    4.67MB
2 dt_non_equi  6.39ms  7.25ms     117.    398.8KB

#26,000 observations; 26 groups
# A tibble: 2 x 13
  expression      min median `itr/sec` mem_alloc
  <bch:expr>  <bch:t> <bch:>     <dbl> <bch:byt>
1 dt_sapply   404.1ms  404ms      2.47  403.46MB
2 dt_non_equi  24.2ms   25ms     39.8     2.09MB

#260,000 observations; 26 groups
# A tibble: 2 x 13
  expression      min  median `itr/sec` mem_alloc
  <bch:expr>  <bch:t> <bch:t>     <dbl> <bch:byt>
1 dt_sapply     38.6s   38.6s    0.0259    38.8GB
2 dt_non_equi 524.2ms 524.2ms    1.91      19.1MB