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 这返回了一个长度为1的R字符向量： 
 
Curve.equation(lmab)
[1] "-26488.6153547735+4928.74310702739*x-217.568666179266*x^2+4.00331698261622*x^3"
 
 这将允许您获取该字符值并将其放入“空白”函数对象中。 不需要eval ，因为parse返回必要的表达式。 它执行测试是否可以从该字符值构建有效的R表达式树： 
 
func <- function(x){}
    body(func) <- parse(text=Curve.equation(lmab))


> func
function (x) 
-26488.6153547735 + 4928.74310702739 * x - 217.568666179266 * 
    x^2 + 4.00331698261622 * x^3
 
 由于func现在是一个带有一个名为“x”的形式参数的函数，因此curve函数很高兴： 
 
png(); curve(func, -10,10); dev.off()
 
  

That returned an R character vector of length 1:
 
Curve.equation(lmab)
[1] "-26488.6153547735+4928.74310702739*x-217.568666179266*x^2+4.00331698261622*x^3"
 
This will allow you to take that character value and place it into a "blank" function object. There is no need for eval, since parse returns the necessary expression. It performs a test whether it's possible to build a valid R expression tree from that character value:
 
func <- function(x){}
    body(func) <- parse(text=Curve.equation(lmab))


> func
function (x) 
-26488.6153547735 + 4928.74310702739 * x - 217.568666179266 * 
    x^2 + 4.00331698261622 * x^3
 
Since func is now a function with a single formal parameter named "x", the curve-function is happy:
 
png(); curve(func, -10,10); dev.off()