r - 串联成组的矢量元素

Question

将 list1 转换为 list2，以及将 list2 转换为 list1 的最优雅方法是什么？

list1<- c('a','b','c','d','e','f','g','h','i')
list2<- c('abc','def','ghi')

即:接触三个一组的元素。

谢谢:D

Answer 1

让list1 <- letters[1:10] (展示当向量的长度不是 3 的倍数时它是如何工作的)。然后，试试这个:

list1 到 list2

# method 1 (seems to be the fastest so far, 
# my suspicions about loop being slower were wrong)
list2 <- sapply(split(list1, (seq_along(list1)-1) %/% 3), paste, collapse = "")
# alternatively as @flodel mentions
list2 <- tapply(list1, (seq_along(list1)-1) %/% 3, paste, collapse = "")

tapply版本运行时间与 sapply+split 相似(基准测试未显示)。

更进一步，使用@JoshOBrien 在 this post 中的想法

# method 2
pattern <- "(?<=[[:alnum:]]{3})(?=[[:alnum:]])"
strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]]
# [1] "abc" "def" "ghi" "j"  

如果你想把最后一部分连接到最后一个(这里是 j 到 ghi )，那么，做:

pattern <- "(?<=[[:alnum:]]{3})(?=[[:alnum:]]{3})"
strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]]
# [1] "abc"  "def"  "ghij"

list2 到 list1

unlist(strsplit(list2, ""), use.names=FALSE)
#  [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"

---

这是 method1 的基准测试, method2和埃迪的:

数据:

list1 <- sample(letters, 1e5, replace=TRUE)

功能:

arun <- function() {
    pattern <- "(?<=[[:alnum:]]{3})(?=[[:alnum:]])"
    strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]]
}

arun2 <- function() {
    unname(sapply(split(list1, (seq_along(list1)-1) %/% 3), paste, collapse = ""))
}

eddi <- function() {
    substring(paste(list1, collapse = ""),
          seq(1, length(list1), 3),
          pmin(seq(3, length(list1)+2, 3), length(list1)))    
}

基准测试:

require(microbenchmark)
microbenchmark(t1 <- arun(), t2 <- eddi(), t3 <- arun2(), times=10)
identical(t1, t2) # TRUE
identical(t1, t3) # TRUE

# Unit: milliseconds
#           expr       min        lq    median        uq       max neval
#   t1 <- arun() 3352.9867 3400.8627 3512.7037 3585.6499 3635.2182    10
#   t2 <- eddi() 3302.0925 3318.4184 3356.2109 3409.9728 3487.7220    10
#  t3 <- arun2()  474.9235  494.7407  539.4406  641.2605  907.9072    10