我们可以删除第一个和最后一个元素，并通过使用Map循环相应的元素来连接

Map(c, x[-length(x)], x[-1])
$a
[1] "a" "b"

$b
[1] "b" "c"

$c
[1] "c" "b"

或者将cbind转换为matrix，并使用asplit按行拆分

asplit(cbind(x[-length(x)], x[-1]), 1)
[[1]]
[1] "a" "b"

[[2]]
[1] "b" "c"

[[3]]
[1] "c" "b"

如果n的值可以大于2，我们也可以用shift来实现

library(data.table)
Filter(\(x) all(complete.cases(x)), 
  data.table::transpose(shift(x, seq_len(n)-1, type = 'lead')))
[[1]]
[1] "a" "b"

[[2]]
[1] "b" "c"

[[3]]
[1] "c" "b"

Vectorized BaseR解决方案（我不使用embed或asplit;它们内部具有for-循环）。
第一个
你能简化这个函数吗？删除format。
直截了当的一行字：

split(x[sequence(rep(length(x) - n + 1, n), 1:n)], seq_len(length(x) - n + 1))

这里有些好玩的。

fun1 <- function (x, n) asplit(embed(x, n)[, n:1], 1)

fun2 <- function (x, n) split(x[sequence(rep(length(x) - n + 1, n), 1:n)], seq_len(length(x) - n + 1))

fun3 <- function (x, n) lapply(1:(length(x) - n + 1), \(i) x[i:(i + n - 1)])

library(microbenchmark)

x <- 1:10000
microbenchmark("for" = fun1(x, 2), "split" = fun2(x, 2), "lapply" = fun3(x, 2))
#Unit: milliseconds
#   expr       min        lq      mean    median        uq       max neval cld
#    for 30.536090 39.196876 49.400427 48.541195 55.481533 107.46441   100   c
#  split  6.453484  7.049844  7.765709  7.647299  7.904683  13.63022   100 a  
# lapply 16.070532 21.959815 26.988959 28.482102 31.133325  45.47318   100  b 

microbenchmark("for" = fun1(x, 10), "split" = fun2(x, 10), "lapply" = fun3(x, 10))
#Unit: milliseconds
#   expr       min        lq      mean    median        uq       max neval cld
#    for 34.115408 34.826142 39.136366 35.631689 37.200893 200.63875   100   c
#  split  8.566762  8.780026  9.255456  9.057524  9.641736  12.67383   100 a  
# lapply 17.343556 17.845281 19.289687 18.301174 18.833777  28.19920   100  b 

microbenchmark("for" = fun1(x, 20), "split" = fun2(x, 20), "lapply" = fun3(x, 20))
#Unit: milliseconds
#   expr      min       lq     mean   median       uq      max neval cld
#    for 38.33747 38.90368 40.61395 39.72388 40.64009 51.51035   100   c
#  split 11.29013 11.39768 12.07148 11.48208 12.13088 17.46919   100 a  
# lapply 18.77825 18.94005 20.88440 19.33751 19.93676 42.35469   100  b

lapply(1:(length(x) - n + 1), \(i) x[i:(i + n - 1)])

# [[1]]
# [1] "a" "b"

# [[2]]
# [1] "b" "c"

# [[3]]
# [1] "c" "b"