我希望在对象parLapply()
内的窗口上使用R6
并注意到(至少在某些情况下)我不需要将 R6 函数或数据导出到节点。
这是一个示例,我可以在其中访问私有方法parLapply()
:
require(R6);require(parallel)
square <-
R6Class("square",
public = list(
numbers = NA,
squares = NA,
initialize = function(numbers,integer) {
self$numbers <- numbers
squares <- private$square.numbers()
}
),
private = list(
square = function(x) {
return(x^2)
},
square.numbers = function() {
cl <- makeCluster(detectCores())
self$squares <- parLapply(cl,
self$numbers,
function (x) private$square(x)
)
stopCluster(cl)
}
))
##Test
test <- square$new(list(1,2,3))
print(test$squares)
# [[1]]
# [1] 1
#
# [[2]]
# [1] 4
#
# [[3]]
# [1] 9
第二个例子,我也可以访问公共成员:
square2 <-
R6Class("square2",
public = list(
numbers = NA,
squares = NA,
integer = NA,
initialize = function(numbers,integer) {
self$numbers <- numbers
self$integer <- integer
squares <- private$square.numbers()
}
),
private = list(
square = function(x) {
return(x^2)
},
square.numbers = function() {
cl <- makeCluster(detectCores())
self$squares <- parLapply(cl,
self$numbers,
function (x) private$square(x)+self$integer
)
stopCluster(cl)
}
))
##Test
test2 <- square2$new(list(1,2,3),2)
print(test2$squares)
#[[1]]
#[1] 3
#
#[[2]]
#[1] 6
#
#[[3]]
#[1] 11
我的问题是双重的:(1)R6 是什么让这成为可能,这样我就不需要导出数据对象和函数;(2)我可以依赖这种行为还是这些具体例子的产物?
更新:
在实例化对象后,这种行为似乎也适用于公共方法和成员:
square3 <- R6Class(
classname = "square3",
public = list(
numbers = NA,
squares = NA,
integer = NA,
square = function(x) {
return(x^2)
},
square.numbers = function() {
cl <- makeCluster(detectCores())
self$squares <- parLapply(cl,
self$numbers,
function (x) self$square(x)+self$integer
)
stopCluster(cl)
},
initialize = function(numbers,integer) {
self$numbers <- numbers
self$integer <- integer
}
)
)
test3.obj <- square3$new(list(1,2,3),2)
test3.obj$square.numbers()
test3.obj$squares
# [[1]]
# [1] 3
#
# [[2]]
# [1] 6
#
# [[3]]
# [1] 11