通过引用在 R 中的向量上进行子分配 [英] Sub-assign by reference on vector in R

问题描述

我能否以某种方式在原子向量上使用通过引用进行子分配?
当然不用将它包装在 1 列 data.table 中以使用 :=.

Can I use sub-assign by reference on atomic vectors somehow?
Of course without wrapping it in 1 column data.table to use :=.

library(data.table)
N <- 5e7
x <- sample(letters, N, TRUE)
X <- data.table(x = x)
upd_i <- sample(N, 1L, FALSE)
system.time(x[upd_i] <- NA_character_)
#    user  system elapsed 
#    0.11    0.06    0.17 
system.time(X[upd_i, x := NA_character_])
#    user  system elapsed 
#    0.00    0.00    0.03 

如果 R6 可以提供帮助，我愿意接受 R6 解决方案，因为它已经是我的部门之一.
我已经检查了 <- 里面的 R6 对象仍然复制:要点.

If R6 can help on that I'm open for R6 solution as it is one of my dep already.
I've already checked that <- inside R6 object still makes copy: gist.

推荐答案

在最近的 R 版本(3.1-3.1.2+ 左右)中，对向量的赋值不会复制.但是，您不会通过运行 OP 的代码看到这一点，原因如下.因为您重用 x 并将其分配给其他对象，所以不会通知 R 在此时复制 x ，并且必须假设它不会被复制(在上面的特殊情况，我认为最好在 data.table::data.table 中更改它并通知 R 已制作副本，但这是一个单独的问题 - data.frame 遇到同样的问题)，因此它在第一次使用时复制 x .如果你稍微改变一下命令的顺序，你会发现没有区别:

In most recent R versions (3.1-3.1.2+ or so), assignment to a vector does not copy. You will not see that by running OP's code though, and the reason for that is the following. Because you reuse x and assign it to some other object, R is not notified that x is copied at that point, and has to assume that it won't be (in the particular case above, I think it'll be good to change it in data.table::data.table and notify R that a copy has been made, but that's a separate issue - data.frame suffers from same issue), and because of that it copies x on first use. If you change the order of the commands a bit, you'd see no difference:

N <- 5e7
x <- sample(letters, N, TRUE)
upd_i <- sample(N, 1L, FALSE)
# no copy here:
system.time(x[upd_i] <- NA_character_)
#   user  system elapsed 
#      0       0       0 
X <- data.table(x = x)
system.time(X[upd_i, x := NA_character_])
#   user  system elapsed 
#      0       0       0 

# but now R will copy:
system.time(x[upd_i] <- NA_character_)
#   user  system elapsed 
#   0.28    0.08    0.36 

<小时>

(旧答案，主要留作好奇)

---

(old answer, mostly left as a curiosity)

您实际上可以使用 data.table := 运算符来修改您的向量(我认为您需要 R 版本 3.1+ 以避免在 列表):

You actually can use the data.table := operator to modify your vector in place (I think you need R version 3.1+ to avoid the copy in list):

modify.vector = function (v, idx, value) setDT(list(v))[idx, V1 := value]

v = 1:5
address(v)
#[1] "000000002CC7AC48"

modify.vector(v, 4, 10)
v
#[1]  1  2  3 10  5

address(v)
#[1] "000000002CC7AC48"

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