分配给具有重复索引的矩阵的子集 [英] Assignment to subset of a matrix with repeated indices

问题描述

不确定这是否符合 R-Inferno 中的条目，但有人可以评论以下替换工作方式背后的逻辑吗?

Not sure this qualifies for an entry in the R-Inferno, but can someone comment on the logic behind the way the following replacement works?

foo<-matrix(1:6,2)
bar<-foo[2,c(1,3,1)]
bar
# [1] 2 6 2
foo[2,c(1,3,1)]<-foo[2,c(1,3,1)]+5
foo
#      [,1] [,2] [,3]
# [1,]    1    3    5
# [2,]    7    4   11

我的问题是:在生成bar时，重复的坐标导致输出中有重复的元素，但是修改foo时，重复的坐标没有 导致重复的加法运算.(相比之下， for(j in c(1,3,1) ) foo[2,j]<-foo[2,j]+5 确实如此).为什么&[<- 究竟是如何忽略重复索引的?

My question is: when generating bar, the repeated coordinate results in a repeated element in the output, but when modifying foo, the repeated coordinate does not result in a repeated addition operation. (By comparison, for(j in c(1,3,1) ) foo[2,j]<-foo[2,j]+5 does). Why & how exactly does [<- essentially ignore the repeated index?

推荐答案

From help("[<-"):

子赋值是按顺序完成的，所以如果指定一个索引更多将导致索引的最新分配值不止一次.

Subassignment is done sequentially, so if an index is specified more than once the latest assigned value for an index will result.

foo<-matrix(1:6,2)

foo[1,rep(1,2)] <- c(1,42)

#     [,1] [,2] [,3]
#[1,]   42    3    5
#[2,]    2    4    6

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