将list1转换为list2以及将list2转换为list1的最优雅方式是什么？
list1<- c('a','b','c','d','e','f','g','h','i')
list2<- c('abc','def','ghi')
即：以三个为一组连接元素。
谢谢：D
答案 0 :(得分：3)
设list1 <- letters[1:10]（显示当矢量的长度不是3的倍数时它是如何工作的）。然后，试试这个：
# 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版本的运行时间与sapply+split相似（基准测试未显示）。
# method 2
pattern <- "(?<=[[:alnum:]]{3})(?=[[:alnum:]])"
strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]]
# [1] "abc" "def" "ghi" "j"
如果你想将最后一部分连接到最后一部分（这里是j到ghi）那么，请执行：
pattern <- "(?<=[[:alnum:]]{3})(?=[[:alnum:]]{3})"
strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]]
# [1] "abc" "def" "ghij"
unlist(strsplit(list2, ""), use.names=FALSE)
# [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j"
以下是method1，method2和eddi的基准测试：
list1 <- sample(letters, 1e5, replace=TRUE)
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)))
}
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
答案 1 :(得分：2)
这是另一个版本，它比@ Arun的两种方法都快（imo以牺牲可读性为代价，与他的方法1相比，遗憾的是它比他的方法2或者慢得多）[编辑：经过一些基准测试后看起来像Arun的第一种方法虽然在中小尺寸上表现不佳，实际上可以更好地扩展很多，赢得更大的尺寸] [[另一个编辑： Grothendieck解决方案是另一个在小尺寸上表现不佳的解决方案，但是比Arun的第一种方法更好地扩展]]：
substring(paste(list1, collapse = ""),
seq(1, length(list1), 3),
pmin(seq(3, length(list1)+2, 3), length(list1)))
<强>基准
list1 = sample(letters, 10000, replace = T)
microbenchmark(eddi=substring(paste(list1, collapse = ""),seq(1, length(list1), 3),pmin(seq(3, length(list1)+2, 3), length(list1))),
Arun1=sapply(split(list1, (seq_along(list1)-1) %/% 3), paste, collapse = ""),
Arun2=strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]],
Grothendieck=apply(matrix(c(list1, rep("", (3 - length(list1) %% 3) %% 3)), 3), 2, paste, collapse = ""),
times = 100)
#Unit: milliseconds
# expr min lq median uq max neval
# eddi 8.804764 10.17807 11.33133 11.58993 12.69495 100
# Arun1 51.287326 61.74937 65.51151 67.15510 73.98805 100
# Arun2 12.305300 13.52000 14.65123 15.00816 17.20151 100
# Grothendieck 25.043657 29.15488 29.87843 31.02118 45.85889 100
基准继续 这有点令人感兴趣，在1e5，Arun1实际上略微突破了另外两个：
list1 = sample(letters, 1e5, replace = T)
microbenchmark(eddi=substring(paste(list1, collapse = ""),seq(1, length(list1), 3),pmin(seq(3, length(list1)+2, 3), length(list1))),
Arun1=sapply(split(list1, (seq_along(list1)-1) %/% 3), paste, collapse = ""),
Arun2=strsplit(paste(list1, collapse=""), pattern, perl=TRUE)[[1]],
Grothendieck=apply(matrix(c(list1, rep("", (3 - length(list1) %% 3) %% 3)), 3), 2, paste, collapse = ""),
times = 30)
#Unit: milliseconds
# expr min lq median uq max neval
# eddi 417.5631 452.6823 480.4397 528.6187 681.0612 30
# Arun1 363.0641 401.6795 420.8844 475.2225 587.3645 30
# Arun2 426.9462 466.5132 506.1106 552.9374 778.7303 30
# Grothendieck 178.2272 206.0161 216.2643 246.3848 280.7988 30
大N长凳
list1 = sample(letters, 1e6, replace = T)
microbenchmark(Arun1=sapply(split(list1, (seq_along(list1)-1) %/% 3), paste, collapse = ""),
+ Grothendieck=apply(matrix(c(list1, rep("", (3 - length(list1) %% 3) %% 3)), 3), 2, paste, collapse = ""), times = 10)
#Unit: seconds
# expr min lq median uq max neval
# Arun1 5.829132 7.654288 8.582664 8.779793 9.168519 10
# Grothendieck 3.196645 3.416421 3.533622 3.725822 3.951419 10
答案 2 :(得分：2)
1）试试这个：
apply(matrix(list1, 3), 2, paste, collapse = "")
2）以及即使list1的长度不是3的倍数也有效的变体。3 * ceiling(n/3)的长度是m的长度，我们减去n从中获得仍需填补的职位数量：
n <- length(list1)
k <- 3 * ceiling(n / 3) - n
m <- matrix(c(list1, rep("", k)), 3)
apply(m, 2, paste, collapse = "")
3）这里有一个不同的解决方案，如果n不是3的倍数，这里的第二个解决方案也有效：
n <- length(list1)
tapply(list1, c(gl(n, 3, n)), paste, collapse = "")
更新：添加了处理长度不是3的倍数的变体以及不同的解决方案。