从两个数据集中创建唯一列名称的组合，而无需循环

Question

我有两个向量：

a <- c(1,2,3)
b <- c(11,12,13)

我想创建列名称（3 * 3 = 9）的组合，以便它们使用来自两个名称的值：

paper1grid11 
paper1grid12
paper1grid13
paper2grid11
paper2grid12
paper2grid13
paper3grid11
paper3grid12
paper3grid13

我试过使用paste0命令，但没有组合。 而是只创建3个列名。

paste0("paper", a,"grid", b)

我不想使用for循环

Answer 1

您可以使用expand.grid进行此expand.grid 。

a <- c(1,2,3)
b <- c(11,12,13)
do.call(paste0, expand.grid("paper", a, "grid", b))

Answer 2

您可以使用outer ，即

outer(a, b, function(x, y) paste0('paper', x, 'grid', y))
#     [,1]           [,2]           [,3]          
#[1,] "paper1grid11" "paper1grid12" "paper1grid13"
#[2,] "paper2grid11" "paper2grid12" "paper2grid13"
#[3,] "paper3grid11" "paper3grid12" "paper3grid13"

或将其环绕在c以获得答案，例如

c(outer(a, b, function(x, y) paste0('paper', x, 'grid', y)))
#[1] "paper1grid11" "paper2grid11" "paper3grid11" "paper1grid12" "paper2grid12" "paper3grid12" "paper1grid13" "paper2grid13" "paper3grid13"

Answer 3

这篇文章用data.table标记，因此这是一个data.table解决方案：

CJ(a, b)[, paste0("paper", a, "grid", b)]

# [1] "paper1grid11" "paper2grid11" "paper3grid11" "paper1grid12" "paper2grid12"
# [6] "paper3grid12" "paper1grid13" "paper2grid13" "paper3grid13

Answer 4

apply(expand.grid("paper", a, "grid", b), 1, paste, collapse="")

Answer 5

您可以使用crossing（）

library(tidyverse)

crossing("paper",a,"grid",b)%>%unite(`"paper"`,a,`"grid"`,b, col = "col1",sep="")%>%pull(col1)

[1] "paper1grid11" "paper1grid12" "paper1grid13" "paper2grid11" "paper2grid12" "paper2grid13"
[7] "paper3grid11" "paper3grid12" "paper3grid13"