我似乎无法使应用功能访问/修改声明以外的变量...怎么办?
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
使用<<-
运营商,你可以写在外部范围变量:
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))
这里的结果:
age=11, weight=100
age=12, weight=105
age=13, weight=110
i=3
y=3, z=3
注意,使用<<-
是危险的,因为你打破了作用域。 做到这一点只有在真正必要的,如果你这样做,清楚地记录该行为(至少在更大的脚本)
里面试试你的应用如下。 与n的值实验。 我认为, i
应该比少一个z
。
assign("i", i+1, envir=parent.frame(n=2))
assign("z", z+1, envir=parent.frame(n=3))
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
}
OUTPUT
> 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