英文:
Best way to programatically extend and re-order array in R?
问题
使用paste函数可以得到cols3，而不需要使用循环。以下是代码示例：
cols3 <- paste(cols1, cols2)cols3
这将生成与所需cols3相同的输出。
英文:
Very simple R question:
What simplest operation should I use with cols1 and cols2 to obtain cols3 below?
(using loop is not the answer)
> cols1 <- paste0(letters[1:3]);cols1[1] "a" "b" "c"> cols2 <- paste0(cols1, ".new"); cols2[1] "a.new" "b.new" "c.new"> cols3 = c("a", "a.new" , "b" ,"b.new" , "c", "c.new"); cols3[1] "a" "a.new" "b" "b.new" "c" "c.new"</details># 答案1**得分**: 3使用`c` + `rbind`：```rc(rbind(cols1, cols2))#[1] "a" "a.new" "b" "b.new" "c" "c.new"
FYI，这有时被称为“交错”或“交织”。还有内置的vctrs::vec_interleave：
vctrs::vec_interleave(cols1, cols2)#[1] "a" "a.new" "b" "b.new" "c" "c.new"
英文:
With c + rbind
c(rbind(cols1, cols2))#[1] "a" "a.new" "b" "b.new" "c" "c.new"
FYI, this is sometimes called "interleaving", or "interlacing".
There is also the built-in vctrs::vec_interleave:
vctrs::vec_interleave(cols1, cols2)#[1] "a" "a.new" "b" "b.new" "c" "c.new"
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论