In R, when I try to assign a function via ifelse, I get the following error:
在R中,当我尝试通过ifelse分配函数时,我收到以下错误:
> my.func <- ifelse(cond, sqrt, identity)
Error in rep(yes, length.out = length(ans)) :
attempt to replicate an object of type 'builtin'
If cond is FALSE, the error looks equivalent, R complains about an
如果cond为FALSE,则错误看起来相同,R抱怨一个
attempt to replicate an object of type 'closure'
What can I do to assign one of two functions to a variable and what is going on here?
如何将两个函数中的一个赋值给变量以及此处发生了什么?
2 个解决方案
#1
5
Because ifelse is vectorized and does not provide special cases for non-vectorized conditions, the arguments are replicated with rep(...). rep(...) fails for closures such as in the example though.
因为ifelse是矢量化的,并且不为非矢量化条件提供特殊情况,所以使用rep(...)复制参数。 rep(...)无法用于闭包,例如在示例中。
A workaround would be to temprarily wrap the functions:
解决方法是临时包装函数:
my.func <- ifelse(cond, c(sqrt), c(identity))[[1]]
#2
1
@Joshua Ulrich's comment is a proper answer. The correct way to accomplish conditional function assignment is a classic if...else rather than the vectorized ifelse method:
@Joshua Ulrich的评论是一个正确的答案。完成条件函数赋值的正确方法是经典的if ... else而不是vectorized ifelse方法:
my.func <- if (cond) sqrt else identity
#1
5
Because ifelse is vectorized and does not provide special cases for non-vectorized conditions, the arguments are replicated with rep(...). rep(...) fails for closures such as in the example though.
因为ifelse是矢量化的,并且不为非矢量化条件提供特殊情况,所以使用rep(...)复制参数。 rep(...)无法用于闭包,例如在示例中。
A workaround would be to temprarily wrap the functions:
解决方法是临时包装函数:
my.func <- ifelse(cond, c(sqrt), c(identity))[[1]]
#2
1
@Joshua Ulrich's comment is a proper answer. The correct way to accomplish conditional function assignment is a classic if...else rather than the vectorized ifelse method:
@Joshua Ulrich的评论是一个正确的答案。完成条件函数赋值的正确方法是经典的if ... else而不是vectorized ifelse方法:
my.func <- if (cond) sqrt else identity