在列表中追加两个连续项目[重复]

Question

如何使用另一个列表长度在 R 列表中附加两个连续的项目？

a <- list( "a", "b", "c")
audit <- list()
j <- 0
for (i in 1:length(a)){
 pre <- paste(a[i], "_present")
 nonpre <- paste(a[i], "_non_present")
 i <- j+1
 audit[[i]]<- pre
 j <- i+1
 audit[[j]] <- nonpre
}

Result:audit["a_present, "a_non_present", "b_present", "b_non_present", "c_present", "c_non_present"]

我的解决方案很脏。任何更聪明的方式都值得赞赏。追加到列表的其他方法是什么？

Answer 1

您可以使用outer：

outer(a, c('_present', '_non_present'), paste0)

#           [,1]        [,2]           
#[1,] "a_present" "a_non_present"
#[2,] "b_present" "b_non_present"
#[3,] "c_present" "c_non_present"

如果你想以列表形式输出：

as.list(t(outer(a, c('_present', '_non_present'), paste0)))


#[[1]]
#[1] "a_present"

#[[2]]
#[1] "a_non_present"

#[[3]]
#[1] "b_present"

#[[4]]
#[1] "b_non_present"

#[[5]]
#[1] "c_present"

#[[6]]
#[1] "c_non_present"

Answer 2

您可以使用 purrr::map 从 a 创建一个元素的元组，并与 _present' & '_not_present', then just unlist` 连接得到一个向量的结果:)

library(purrr)

a <- list("a", "b", "c")
unlist(purrr::map(a, function(i){paste0(i, c('_present','_not_present'))}))

这会在向量中为您提供所需的输出：

[1] "a_present"     "a_not_present" "b_present"     "b_not_present" "c_present"     "c_not_present"

Answer 3

您可以使用sapply。这里我使用了paste0 而不是paste。

as.list(sapply(a, function(i) paste0(i, c('_present','_non_present'))))

[[1]]
[1] "a_present"

[[2]]
[1] "a_non_present"

[[3]]
[1] "b_present"

[[4]]
[1] "b_non_present"

[[5]]
[1] "c_present"

[[6]]
[1] "c_non_present"

Answer 4

你可以使用sapply和Reduce来获得一个简单的向量：

Reduce(`c`, sapply(a, function(x) paste0(x, c('_present','_not_present'))))

输出

[1] "a_present"     "a_not_present" "b_present"     "b_not_present" "c_present"     "c_not_present

更新

在意识到我的解决方案可能非常慢之后，我又添加了两个函数并进行了基准测试：

fun1 <- function(y) Reduce(`c`, sapply(y, function(x) paste0(x, c('_prfun1 <- function(y) Reduce(`c`, sapply(y, function(x) paste0(x, c('_present','_not_present'))))
fun2 <- function(y) unlist(purrr::map(y, function(i){paste0(i, c('_present','_not_present'))}))
fun3 <- function(y) as.list(sapply(y, function(i) paste0(i, c('_present','_non_present'))))
fun4 <- function(y) array(sapply(y, function(x) paste0(x, c('_present','_not_present'))))
fun5 <- function(y) as.vector(sapply(y, function(x) paste0(x, c('_present','_not_present'))))

fun6 <- function(y){
audit <- list()
j <- 0
for (i in 1:length(y)){
  pre <- paste(y[i], "_present")
  nonpre <- paste(y[i], "_non_present")
  i <- j+1
  audit[[i]]<- pre
  j <- i+1
  audit[[j]] <- nonpre
}
audit
}


x <- list(sample(letters, 1000, replace = TRUE))
microbenchmark::microbenchmark(fun1(x), fun2(x), fun3(x), fun4(x), fun5(x), fun6(x), times = 10L)

结果

    Unit: microseconds
    expr      min       lq      mean    median       uq      max neval
 fun1(x) 2886.633 3107.855 3673.7590 3191.4295 3965.469 6969.991    10
 fun2(x)  225.562  235.774  553.8192  277.9055  407.876 2839.410    10
 fun3(x)  237.601  245.103  643.7252  261.1835  407.830 3840.739    10
 fun4(x)  222.445  230.426  533.4771  249.6535  261.322 3075.166    10
 fun5(x)  219.845  232.682  508.5585  253.7870  298.801 2775.606    10
 fun6(x) 1531.715 1597.888 2229.4907 1616.1170 1913.850 7245.782    10

fun2 - fun5 的速度大致相同，fun4 和 fun5 稍快一些。最初的解决方案（出乎意料）很慢。