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R 在两列中串联每个值_R_Loops_Concat_String Concatenation - Fatal编程技术网

R 在两列中串联每个值

R 在两列中串联每个值,r,loops,concat,string-concatenation,R,Loops,Concat,String Concatenation,我有两列,我需要说明的值。例如,我有颜色栏黑色、蓝色、红色、紫色、黄色和服装、鞋子、T恤、长衬衫。然后我想要黑色连衣裙、黑色鞋子、黑色T恤、黑色长衬衫、蓝色连衣裙、蓝色鞋子、蓝色T恤等等 我尝试过使用for循环和粘贴函数,但它没有按照我希望的方式工作 A <- colour B <- Clothing for(i in count(A)){ paste( print(A[i]) , for (i in count(B)) {print(B[i])}

我有两列,我需要说明的值。例如,我有颜色栏黑色、蓝色、红色、紫色、黄色和服装、鞋子、T恤、长衬衫。然后我想要黑色连衣裙、黑色鞋子、黑色T恤、黑色长衬衫、蓝色连衣裙、蓝色鞋子、蓝色T恤等等

我尝试过使用for循环和粘贴函数,但它没有按照我希望的方式工作

A <- colour
B <- Clothing

 for(i in count(A)){ paste( print(A[i]) ,
             for (i in count(B)) {print(B[i])} 
      )
    }
如果您想要矩阵:

> sapply(A,function(x) paste(x,B))
     Black              Blue              Red              Purple              Yellow             
[1,] "Black Dress"      "Blue Dress"      "Red Dress"      "Purple Dress"      "Yellow Dress"     
[2,] "Black Shoe"       "Blue Shoe"       "Red Shoe"       "Purple Shoe"       "Yellow Shoe"      
[3,] "Black T-shirt"    "Blue T-shirt"    "Red T-shirt"    "Purple T-shirt"    "Yellow T-shirt"   
[4,] "Black Long-shirt" "Blue Long-shirt" "Red Long-shirt" "Purple Long-shirt" "Yellow Long-shirt"
向量:

> as.vector(sapply(A,function(x) paste(x,B)))
 [1] "Black Dress"       "Black Shoe"        "Black T-shirt"     "Black Long-shirt"  "Blue Dress"        "Blue Shoe"        
 [7] "Blue T-shirt"      "Blue Long-shirt"   "Red Dress"         "Red Shoe"          "Red T-shirt"       "Red Long-shirt"   
[13] "Purple Dress"      "Purple Shoe"       "Purple T-shirt"    "Purple Long-shirt" "Yellow Dress"      "Yellow Shoe"      
[19] "Yellow T-shirt"    "Yellow Long-shirt"
我们可以使用外层空间

或者用purrr的地图

数据
只要用cbindA,B来满足你的需求。或者,根据你的需要,选择A,B。
outer(A, B, FUN = paste)
#  [,1]           [,2]          [,3]             [,4]               
#[1,] "Black Dress"  "Black Shoe"  "Black T-shirt"  "Black Long-shirt" 
#[2,] "Blue Dress"   "Blue Shoe"   "Blue T-shirt"   "Blue Long-shirt"  
#[3,] "Red Dress"    "Red Shoe"    "Red T-shirt"    "Red Long-shirt"   
#[4,] "Purple Dress" "Purple Shoe" "Purple T-shirt" "Purple Long-shirt"
#[5,] "Yellow Dress" "Yellow Shoe" "Yellow T-shirt" "Yellow Long-shirt"
library(purrr)
map(A, ~ str_c(.x, B, sep= " ")) %>%
      flatten_chr
#[1] "Black Dress"       "Black Shoe"        "Black T-shirt"     "Black Long-shirt"  "Blue Dress"        "Blue Shoe"        
#[7] "Blue T-shirt"      "Blue Long-shirt"   "Red Dress"         "Red Shoe"          "Red T-shirt"       "Red Long-shirt"   
#[13] "Purple Dress"      "Purple Shoe"       "Purple T-shirt"    "Purple Long-shirt" "Yellow Dress"      "Yellow Shoe"      
#[19] "Yellow T-shirt"    "Yellow Long-shirt"
A <- c("Black", "Blue", "Red", "Purple", "Yellow")
B <- c("Dress", "Shoe", "T-shirt", "Long-shirt")