R:如何为函数中的一系列数据创建循环? [英] R: How to create a loop for, for a range of data in a function?

问题描述

我有此参数:

L_inf <- seq(17,20,by=0.1)

和此功能:

fun <- function(x){
  L_inf*(1-exp(-B*(x-0)))}

我将在L_inf值范围内应用此功能. 我尝试使用循环for，如下所示:

I would to apply this function for a range of value of L_inf. I tried with loop for, like this:

A <- matrix() # maybe 10 col and 31 row or vice versa
for (i in L_inf){
A[i] <- fun(1:10)
}

Bur R回应:longer object length is not a multiple of shorter object length. 我的预期输出是一个矩阵(或数据帧或列表)，其中每个向量L_inf的值(长度= 31)有10个结果(fun(1:10)). 该怎么办?

Bur R respond: longer object length is not a multiple of shorter object length. My expected output is a matrix (or data frame, or list maybe) with 10 result (fun(1:10)) for each value of the vector L_inf (lenght=31). How can to do it?

推荐答案

您正在尝试将10个元素的向量放入一个矩阵单元中.您想将其分配给矩阵行(您可以使用A[i,]访问第i行).

You are trying to put a vector of 10 elements into one of the matrix cell. You want to assign it to the matrix row instead (you can access the ith row with A[i,]).

但是在这种情况下使用for循环效率低下，并且使用"apply"功能之一非常简单. Apply函数通常返回一个列表(这是用途最广泛的容器，因为基本上没有约束).

But using a for loop in this case is inefficient and it is quite straightforward to use one of the "apply" function. Apply functions typically return a list (which is the most versatile container since there is basically no constraint).

此处sapply是Apply函数，它试图 S 将其结果放大为方便的数据结构.在这种情况下，由于所有结果的长度都相同(10)，因此sapply会将结果简化为矩阵.

Here sapply is an apply function which tries to Simplify its result to a convenient data structure. In this case, since all results have the same length (10), sapply will simplify the result to a matrix.

请注意，我修改了您的函数以使其显式依赖于L_inf.否则，它将无法执行您认为应该做的事情(如果您需要更多信息，请参见关键字"closures").

Note that I modified your function to make it explicitly depend on L_inf. Otherwise it will not do what you think it should do (see keyword "closures" if you want more info).

L_inf_range <- seq(17,20,by=0.1)
B <- 1

fun <- function(x, L_inf) {
    L_inf*(1-exp(-B*(x-0)))
}

sapply(L_inf_range, function(L) fun(1:10, L_inf=L))

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