string 将列表列表转换为字符向量
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Convert a list of lists to a character vector
提问by dan
I have a list of lists of characters. For example:
我有一个字符列表列表。例如：
l <- list(list("A"),list("B"),list("C","D"))
So as you can see some elements are lists of length > 1.
因此，如您所见，某些元素是长度 > 1 的列表。
I want to convert this list of lists to a character vector, but I'd like the lists with length > 1 to appear as a single element in the character vector.
我想将此列表列表转换为字符向量，但我希望长度 > 1 的列表显示为字符向量中的单个元素。
the unlistfunction does not achieve that but rather:
该unlist功能没有实现，而是：
> unlist(l)
[1] "A" "B" "C" "D"
Is there anything faster than:
有什么比：
sapply(l,function(x) paste(unlist(x),collapse=""))
To get my desired result:
为了得到我想要的结果：
"A" "B" "CD"
回答by IRTFM
You can skip the unlist step. You already figured out that paste0needs collapse = TRUEto "bind" sequential elements of a vector together:
您可以跳过取消列出步骤。您已经发现paste0需要collapse = TRUE将向量的顺序元素“绑定”在一起：
> sapply( l, paste0, collapse="")
[1] "A" "B" "CD"
回答by A5C1D2H2I1M1N2O1R2T1
Here's a variation of @thela's suggestion, if you don't mind a multi-line approach:
如果您不介意多行方法，这是@thela 建议的变体：
x <- lengths(l) ## Get the lengths of each list
l[x > 1] <- lapply(l[x > 1], paste0, collapse = "") ## Paste only those together
unlist(l, use.names = FALSE) ## Unlist the result
# [1] "A" "B" "CD"
Alternatively, if you don't mind using a package, look at the "stringi" package, specifically stri_flatten, as suggested by @Jota.
或者，如果您不介意使用包，请查看@Jotastri_flatten建议的“stringi”包，特别是 。
Here's a performance comparison:
下面是性能对比：
l <- list(list("A"), list("B"), list("B"), list("B"), list("B"),
list("C","D"), list("E","F", "G", "H"),
as.list(rep(letters,10)), as.list(rep(letters,2)))
l <- unlist(replicate(1000, l, FALSE), recursive = FALSE)
funop <- function() sapply(l,function(x) paste(unlist(x),collapse=""))
fun42 <- function() sapply(l, paste0, collapse="")
funv <- function() vapply(l, paste0, character(1L), collapse = "")
funam <- function() {
x <- lengths(l)
l[x > 1] <- lapply(l[x > 1], paste0, collapse = "")
unlist(l, use.names = FALSE)
}
funj <- function() sapply(l, stri_flatten)
funamj <- function() {
x <- lengths(l)
l[x > 1] <- lapply(l[x > 1], stri_flatten)
unlist(l, use.names = FALSE)
}
library(microbenchmark)
microbenchmark(funop(), fun42(), funv(), funam(), funj(), times = 20)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# funop() 78.21822 84.79588 85.30055 85.36399 86.90540 90.48321 20 e
# fun42() 56.16938 57.35735 61.60008 58.04969 65.82836 81.46482 20 d
# funv() 54.64101 56.23245 60.07896 57.26049 63.96815 78.58043 20 d
# funam() 45.89760 46.89890 48.99810 47.29617 48.28764 56.92544 20 c
# funj() 28.73405 29.94041 32.00676 30.56711 31.11448 39.93765 20 b
# funamj() 18.64829 19.01328 21.05989 19.12468 19.52516 32.87569 20 a
Note: The relative efficiency of this approach would depend on how many list items are going to have length(x) > 1. If most of them are going to be > 1anyway, then just go with @42-'s approach. stri_flattenonly improves performance if you have long character vectors to paste together as in sample list used for the above benchmark, otherwise, it doesn't help.
注意：这种方法的相对效率取决于将有多少个列表项length(x) > 1。如果他们中的大多数人> 1无论如何都会成为，那么就采用@42- 的方法。 stri_flatten仅当您将长字符向量粘贴在一起时才能提高性能，如用于上述基准测试的示例列表，否则无济于事。
