将值分配给R中另一个环境中的对象的一部分

Question

假设我在环境 e 中定义了两个相似的大对象 x、y（数据表）。 我想使用函数 f 以类似的方式更改 x 或 y 的很大一部分，而无需在 f 的执行环境中创建 x 或 y 的副本。示例：

e <- new.env()
e$x <- c(1,2,3) # imagine this to be BIG (ie. dataframe with 200k vars each 500k rows)
e$y <- c(4,5,6) # same here
e$v <- 2        # minor variables 

f <- function(var_str, env, input){

    # do some computation on parts of var_str which is either "x" or "y"
    # and store these right back into e$x or e$y, respectively.
    # ie

    str <- paste0(var_str,"[2:3] <- (",var_str,"[2:3])^2 + rep(v,2) + ", deparse(input1),"^3/c(100,101)")
    eval(parse(text=str), envir= e)

    # this does work but I can image there is an easier/more elegant way 
    # of doing this.
}

我想在全局环境中定义函数并将这个函数应用到 e$x 和 e$y 输入不同的变量。 IE。执行

f("x", e, c(1,2))
f("y", e, c(3,4))

有没有人对此有一个优雅的解决方案。

Answer 1

出于好奇，我尝试使用 the simple example in the R6 tutorial 并想出了这个（复制为控制台记录）。老实说，我不知道它是否符合请求的非复制要求，但它似乎确实就地修改了对象。

# Assumes one has created an R6 constructor named `RC` ...
# slightly extended from the example in section
RC <- setRefClass("RC",
  fields = list(x = 'ANY'),
  methods = list(
    getx = function() x,
    setx = function(value) x <<- value,
    setsub = function(i,j,val) x[i,j] <<- val
  )
)

#--- execution --- 
rc <- RC$new()
rc$setx(matrix(1:20, 4,5))
rc
# --- result ---    
Reference class object of class "RC"

Field "x":
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    5    9   13   17
[2,]    2    6   10   14   18
[3,]    3    7   11   15   19
[4,]    4    8   12   16   20

测试setsub类特定函数：

rc$setsub(4,5,0)   #Test of setting a single element within an object to a new value
rc
#-------    
Reference class object of class "RC"
Field "x":
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    5    9   13   17
[2,]    2    6   10   14   18
[3,]    3    7   11   15   19
[4,]    4    8   12   16    0

rc$setsub(1:4,1:4,0)    #Test of setting a range of elements within an object to a new value
rc
# ---------
Reference class object of class "RC"
Field "x":
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    0   17
[2,]    0    0    0    0   18
[3,]    0    0    0    0   19
[4,]    0    0    0    0    0

因此，通过setsub 实现的[<--函数的这个（非常简单的测试）确实成功了。使用self 引用将其扩展到示例的一次尝试确实失败了。

Answer 2

eval(parse()) 是要避免的。您可以像这样引用环境：

e <- new.env()
e$x <- c(1,2,3) # imagine this to be BIG (ie. dataframe with 200k vars each 500k rows)
e$y <- c(4,5,6) # same here
e$v <- 2        # minor variables 

f <- function(var_str, env, input){

  # do some computation on parts of var_str which is either "x" or "y"
  # and store these right back into e$x or e$y, respectively.
  # ie
  env[[var_str]][2:3] <- (env[[var_str]][2:3])^2 + rep(env$v,2) + input^3/input

}


f("x", e, 1:2)
e$x
#[1]  1  7 15