如何编写分配给调用环境对象的R函数?
我有一个包含多个矩阵的特定类的对象,我想构建一个函数来访问并可能修改这些矩阵的子集。例如:如何编写分配给调用环境对象的R函数?,r,function,global-variables,subset,assignment-operator,R,Function,Global Variables,Subset,Assignment Operator,我有一个包含多个矩阵的特定类的对象,我想构建一个函数来访问并可能修改这些矩阵的子集。例如: foo<-list(x=diag(1:4),y=matrix(1:8,2,4)) class(foo)<-"bar" attr(foo,"m")<-4 attr(foo,"p")<-2 rownames(foo$x)<-colnames(foo$x)<-colnames(foo$y)<-c("a.1","b.1","b.2","c.1") attr(foo,"ty
foo<-list(x=diag(1:4),y=matrix(1:8,2,4))
class(foo)<-"bar"
attr(foo,"m")<-4
attr(foo,"p")<-2
rownames(foo$x)<-colnames(foo$x)<-colnames(foo$y)<-c("a.1","b.1","b.2","c.1")
attr(foo,"types")<-c("a","b","b","c")
foo有一点需要注意,这通常是不赞成的,您可以使用以下方法:
在目标环境中,为环境设置一个变量,然后将该变量作为参数传递给您的函数,您可以在assign
、get
等中使用该函数
en <- environment()
myfunc <- function(..., en=en) {
. etc .
assign("varname", envir=en)
}
好的,我自己找到了一个解决方案
foo如果您只是在更改属性,那么内置的attr将非常感谢,但问题是原始对象的名称,即“varname”可以是任意的。@Hemmo,这是个什么问题?在assign
中使用一个不带引号的变量如果我在调用环境中有一个名为myvar的对象,并且我运行函数myfunc(var,en),那么函数myfunc不知道我需要为名为myvar的变量赋值。但这并不重要,我找到了另一个解决方案,我将很快发布。
modify<-function(object,element,types){
# check that the object is proper class,
# and the element and the types are found in the object
# this returns to the local copy of the corresponding subset:
object[[element]][attr(object,"types")%in%types,attr(object,"types")%in%types]
}
modify(foo,"x",c("c","b"))<-matrix(5,3,3)
Error in modify(foo, "x", c("c", "b")) <- matrix(5, 3, 3) :
could not find function "modify<-
en <- environment()
myfunc <- function(..., en=en) {
. etc .
assign("varname", envir=en)
}
setattr(x,name,value)
foo<-list(x=diag(1:4),y=matrix(1:8,2,4))
class(foo)<-"bar"
attr(foo,"m")<-4
attr(foo,"p")<-2
rownames(foo$x)<-colnames(foo$x)<-colnames(foo$y)<-c("a.1","b.1","b.2","c.1")
attr(foo,"types")<-c("a","b","b","c")
`modify<-` <- function(x, element, subset,value) UseMethod("modify<-")
`modify<-.bar` <- function(x, element, subset,value) {
x[[element]][,attr(foo,"types")%in%subset] <- value
x }
modify(foo,"x",c("c","b"))<-matrix(5,3,3)
foo$x
a.1 b.1 b.2 c.1
a.1 1 0 0 0
b.1 0 5 5 5
b.2 0 5 5 5
c.1 0 5 5 5