Question

我似乎无法使应用函数访问/修改在外部声明的变量...什么给出了？

    x = data.frame(age=c(11,12,13), weight=c(100,105,110))
    x

    testme <- function(df) {
        i <- 0
        apply(df, 1, function(x) {
            age <- x[1]
            weight <- x[2]
            cat(sprintf("age=%d, weight=%d\n", age, weight))
            i <- i+1   #this could not access the i variable in outer scope
            z <- z+1   #this could not access the global variable
        })
        cat(sprintf("i=%d\n", i))
        i
    }

    z <- 0
    y <- testme(x)
    cat(sprintf("y=%d, z=%d\n", y, z))

结果：

    age=11, weight=100
    age=12, weight=105
    age=13, weight=110
    i=0
    y=0, z=0

Answer 1

使用<<-运算符可以写入外部作用域中的变量：

x = data.frame(age=c(11,12,13), weight=c(100,105,110))
x

testme <- function(df) {
    i <- 0
    apply(df, 1, function(x) {
        age <- x[1]
        weight <- x[2]
        cat(sprintf("age=%d, weight=%d\n", age, weight))
        i <<- i+1   #this could not access the i variable in outer scope
        z <<- z+1   #this could not access the global variable
    })
    cat(sprintf("i=%d\n", i))
    i
}

z <- 0
y <- testme(x)
cat(sprintf("y=%d, z=%d\n", y, z))

结果如下：

age=11, weight=100
age=12, weight=105
age=13, weight=110
i=3
y=3, z=3

请注意，<<-的使用是危险的，因为您打破了范围。只有在真的有必要时才这样做，如果你这样做，请清楚地记录这种行为（至少在更大的脚本中）

Answer 2

在申请中尝试以下内容。试验n的值。我相信对于i，它应该比z少一个。

         assign("i", i+1, envir=parent.frame(n=2))
         assign("z", z+1, envir=parent.frame(n=3))

<小时/>

testme <- function(df) {
    i <- 0
    apply(df, 1, function(x) {
        age <- x[1]
        weight <- x[2]
        cat(sprintf("age=%d, weight=%d\n", age, weight))

        ## ADDED THESE LINES
         assign("i", i+1, envir=parent.frame(2))
         assign("z", z+1, envir=parent.frame(3))

    })
    cat(sprintf("i=%d\n", i))
    i
}

输出

> z <- 0
> y <- testme(x)
age=11, weight=100
age=12, weight=105
age=13, weight=110
i=3
> cat(sprintf("y=%d, z=%d\n", y, z))
y=3, z=3

如何从R apply函数访问全局/外部范围变量？

2 个答案:

输出