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从R中的向量累计选择元素

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我正在使用以下内容进行累积（即，首先选择前2个元素，每次添加1个；请参见下面的输出）从下面的向量<img 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R中是否有更快或更简单的方法？

mods
xty2004 回答：从R中的向量累计选择元素

您可以使用Reduce：

Reduce(c,n,accumulate=TRUE)

或者，从前两个元素开始：

Reduce(c,tail(n,-2),n[1:2],accumulate=TRUE)

IMHO比较简单，但速度不快：

microbenchmark(
    lapply=lapply(seq_along(n)[-1],function(i)n[seq_len(i)]),Reduce=Reduce(c,accumulate=TRUE)
)

Unit: microseconds
   expr  min   lq   mean median   uq    max neval
 lapply 10.2 10.6 22.056   10.8 11.0 1119.8   100
 Reduce 10.1 10.4 10.865   10.6 10.8   17.5   100
,

这里是rep和split的选项

split(n[sequence(seq_along(n))],rep(seq_along(n),seq_along(n)))
,

我建议使用runner软件包和window_run函数，它完全可以满足您的要求。您还可以在每个窗口中指定要返回的元素数（k。您还可以指定lag并根据日期创建窗口。这是示例：

library(runner)
window_run(names(mtcars))

# [[1]]
# [1] "mpg"
# 
# [[2]]
# [1] "mpg" "cyl"
# 
# [[3]]
# [1] "mpg"  "cyl"  "disp"
# 
# [[4]]
# [1] "mpg"  "cyl"  "disp" "hp"

由于cpp的实现，它在基准测试中与Reduce解决方案非常相似：

microbenchmark::microbenchmark(
  window_run = window_run(names(mtcars)),Reduce = Reduce(c,accumulate = TRUE),lapply = lapply(seq_along(n)[-1],function(i) n[seq_len(i)]),times = 1000000L
)
# Unit: microseconds
#        expr    min     lq     mean median     uq       max neval
#  window_run 10.296 12.782 15.45241 13.543 14.467  98018.87 1e+06
#      Reduce 10.979 13.048 15.80921 13.882 14.847  98881.51 1e+06
#      lapply 11.797 14.540 17.42722 15.518 16.729 116346.98 1e+06

有关更多详细信息，请查看documentation

dataframefunctionlapplyloopsr
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