在 R/Rcpp 中转置列表的最快方法 [英] Fastest way to transpose a list in R / Rcpp

问题描述

我有一个清单:

ls <- list(c("a", "b", "c"), c("1", "2", "3"), c("foo", "bar", "baz"))
ls

#> [[1]]
#> [1] "a" "b" "c"

#> [[2]]
#> [1] "1" "2" "3"

#> [[3]]
#> [1] "foo" "bar" "baz"

我希望转置"给:

resulting_ls

#> [[1]]
#> [1] "a"   "1"   "foo"

#> [[2]]
#> [1] "b"   "2"   "bar"

#> [[3]]
#> [1] "c"   "3"   "baz"

我可以通过以下方式实现:

I can achieve this with:

mat <- matrix(unlist(ls), ncol = 3, byrow = TRUE)
resulting_ls <- lapply(1:ncol(mat), function(i) mat[, i])

但是对于我的真实数据，它非常慢......(我需要为许多列表执行此操作，每个列表都比上面的示例大得多)

But with my real data it's very slow...(and I need to do this for many lists each of which are much larger than example above)

对于大列表length(ls) 和/或length(ls[[i]])，最快的方法是什么?

What is the fastest way to do this for a large list length(ls) and/or length(ls[[i]])?

1. 在 R 中(如果不是这样的话)
2. 使用Rcpp

1. in R (if this is not the case already)
2. with Rcpp

推荐答案

在 data.table 包中，有一个 transpose() 函数就是这样做的.为了速度，它是用 C 实现的.

In the data.table package, there's a transpose() function which does exactly this. It is implemented in C for speed.

require(data.table) # v1.9.6+
transpose(ls)
# [[1]]
# [1] "a"   "1"   "foo"

# [[2]]
# [1] "b"   "2"   "bar"

# [[3]]
# [1] "c"   "3"   "baz"

如果列表元素的长度不相等，它还会自动填充 NA，并自动强制到最高的 SEXPTYPE.如有必要，您可以为 fill 参数提供不同的值.检查 ?transpose.

It also fills automatically with NA in case the list elements are not of equal lengths, and also coerces automatically to the highest SEXPTYPE. You can provide a different value to the fill argument if necessary. Check ?transpose.

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