使用“["方括号作为 R 中 lapply 的函数 [英] Using '[' square bracket as a function for lapply in R

问题描述

我已经看到在 R 中使用 lapply 函数从矩阵列表中存在的矩阵中提取元素.

I've seen the function lapply used in R to extract elements from matrices that exist in a list of matrices.

例如我有一个包含 3 (2x2) 个矩阵的列表，我想从这 3 个矩阵中的每一个中提取元素 [1,2].

E.g. I have a list of 3 (2x2) matrices, and I want to extract element [1,2] from each of those 3 matrices.

代码:list1 = lapply(mylist, '[', 1,2) 工作得很好.它返回一个包含这 3 个元素的列表.

The code: list1 = lapply(mylist, '[', 1,2) works just fine. It returns a list with those 3 elements.

我正在尝试研究这到底是做什么的.谷歌没有提供帮助，在 R 帮助中使用 ?'[' 并不太解释.我不明白 '[' 是 R 中的函数，所以代码不直观.

I am trying to research what this is exactly doing. Google hasn't helped and using ?'[' in the R help isn't too explanatory. I don't see how '[' is a function in R, so the code is not intuitive.

推荐答案

方括号实际上是一个函数，它的第一个参数是被子集化的对象.后续参数是该子集的索引.

The square brackets are in fact a function whose first argument is the object being subsetted. Subsequent arguments are the index to that subset.

# For example, if M is a matrix
M[1, 2]  # extracts the element at row 1, col 2
# is the same as 
`[`(M, 1, 2)
# Try them! 

现在，看看 lapply 的参数:

Now, Have a look at the arguments to lapply:

args(lapply)
# function (X, FUN, ...) 

这些点中表示的所有内容都作为参数传递给函数 FUN.

Everything represented in those dots gets passed on to the function FUN as arguments.

因此，当 FUN="[" 时，"[" 的第一个参数是列表的当前元素(被迭代)，即对象被子集化.而随后的参数是 "["

Thus, when FUN="[", the first argument to "[" is the current element of the list (being iterated over), ie, the object being subsetted. While the subsequent arguments are the indexes to "["

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