有效地将命名列表转换为data.frame [英] Convert named list to a data.frame efficiently

问题描述

我正在寻找一种有效的方法来进行以下转换:

I'm looking for a efficient way for the following transformation:

示例输入:

ob <- list(a = 2, b = 3)

预期输出:

  key value
1   a     2
2   b     3

当前(详细)解决方案:

Current (verbose) solution:

data.frame(key = names(ob), value = unlist(ob, use.names = FALSE))

推荐答案

尝试stack

stack(ob)
#  values ind
#1      2   a
#2      3   b

您将需要更改名称.使用setNames

You would need to change the names though. Use setNames

setNames(stack(ob), nm = c("value", "key"))

---

基准

在处理@Roland的评论时， stack似乎确实更有效.请不要使用stack，而是出于效率原因使用OP的解决方案.

Addressing @Roland's comment, stack seems to be more efficient indeed. please don't use stack but OP's solution for efficiency reasons.

n <- 1e5
lst <- as.list(seq_len(n))
names(lst) <- paste0("a", seq_len(n))

library(microbenchmark)
benchmark <- microbenchmark(
  snoram = snoram(lst),
  markus = markus(lst), times = 50
)
benchmark
#Unit: milliseconds
#   expr      min       lq     mean   median       uq      max neval
# snoram   2.475258   2.594479   2.739639   2.652843   2.715575   5.92216    50
# markus 114.387692 119.028200 134.745626 137.524606 144.045112 162.11510    50

到目前为止使用的功能

snoram <- function(l) {
  data.frame(key = names(l), value = unlist(l, use.names = FALSE),
             stringsAsFactors = FALSE) # this gives a hugh performance gain
                                       # thanks to @Roland
}

markus <- function(l) {
  setNames(stack(l), nm = c("value", "key"))
}

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