对R中的函数的操作符重载奇怪的行为

本文介绍了对R中的函数的操作符重载奇怪的行为的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 不幸的是像（f + g）（3）其中f和g都是一元函数在R中不工作。因此我试图重载+运算符用于一元函数：Unfortunately things like (f+g)(3) where f and g are both unary functions do not work in R. Hence I tried to overload the "+" operator for unary functions in the following way:"+.function" = function(e1, e2){ return(function(x) e1(x) + e2(x))}但是如果我尝试使用这个，这什么也没有。代码But if I try to use this, this does nothing. The code a = function(x) 2*x (a+a)(2)产生与 +。function 不同的错误甚至定义。produces the same error as if +.function is not even defined.通过一段时间我发现，事实上有一个可能性，以这种方式添加函数：如果函数是一个引用类的成员函数，这工作！即，以下代码（以及上面的+定义）By some time playing around I found out that there is in fact a possibility to add functions in this way: If the functions are member functions of an reference class, this works! I.e., the following code (together with the "+" definition from above)clsA = setRefClass("clsA", methods = list( b = function(x) 2*x ))inst_a = clsA$new()(inst_a$b + inst_a$b)(2)返回8因此，我已经有一些解决我的问题的解决方法。现在我的问题是：returns "8" (as expected). Hence I already have some kind of a workaround for my problem. Now my questions are:这种奇怪行为的原因是什么？为什么不 +。function 关心通常函数，但类成员函数？有没有人知道如何扩展操作符到通常的功能？What is the reason for this strange behavior? Why doesn´t +.function care about "usual" function but class member functions? Has anyone an idea how "expand" the operator to usual functions?推荐答案如果你重新定义 a （例如 class（a）< - ownfunction < -c（ownfunction，function），并使你的+。function as + .ownfunction，然后（a + a）（2）工作。If you redifine the class of a, for example like class(a)<-"ownfunction" (or better yet class(a)<-c("ownfunction","function"), and make your "+.function" as "+.ownfunction", then (a+a)(2) works. 看起来函数类以某种特殊方式处理：如果运行 debug（+。function）; + a）（2）你会看到+。function甚至不被调用。It seems that the function class is treated in some special way: If you run debug("+.function");(a+a)(2) you see that "+.function" is not even called.编辑：见注释。 这篇关于对R中的函数的操作符重载奇怪的行为的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！