Question

我想转置一个嵌套列表。假设给出了以下嵌套列表x：

a <- list(c("a","b","c","d"))
b <- list(c("d","c","b","a"))
c <- list(c("4","3","2","1"))
d <- list(c("1","2","3","4"))

x <- list(a,b,c,d)

结果应该是一个嵌套列表，其中原始列表x的第一列是第一个嵌套列表元素，即“a”，“d”，“4”，“1”，第二列是第二列嵌套列表元素，即“b”，“c”，“3”，“2”等。最后，结构是原始结构的转置。怎么能在R？

中完成

Answer 1

我们也可以不用lapply（使用矩阵）：

relist(matrix(unlist(x), ncol = 4, byrow = T), skeleton = x)

<强>基准

library(microbenchmark)
a <- list(c("a","b","c","d"))
b <- list(c("d","c","b","a"))
c <- list(c("4","3","2","1"))
d <- list(c("1","2","3","4"))
x <- list(a,b,c,d)

f_akrun <- function(x) {m1 <- do.call(rbind, lapply(x, function(y) do.call(rbind, y)));relist(m1, skeleton = x);}
f_m0h3n <- function(x) {relist(matrix(unlist(x), ncol = length(x[[1]][[1]]), byrow = T), skeleton = x)}

setequal(f_akrun(x), f_m0h3n(x))
# [1] TRUE
microbenchmark(f_akrun(x), f_m0h3n(x))

# Unit: microseconds
   # expr     min      lq     mean  median      uq     max neval
 # f_akrun(x) 135.591 137.301 144.3545 138.585 148.422 334.484   100
 # f_m0h3n(x) 110.782 111.638 116.5477 112.493 117.412 212.153   100

Answer 2

我们可以尝试

m1 <- do.call(rbind, lapply(x, function(y) do.call(rbind, y)))
relist(m1, skeleton = x)

转置嵌套列表

2 个答案: