如何在str_split之后连接字符串

Question

Given this data frame

column_1     column_2
A            w,x
B            z 
C            q,r,s

My desired output would be

"Aw", "Ax", "Bz", "Cq", "Cr", "Cs"

I've tried

paste0(df$column_1, strsplit(df$column_2, ","))

But the output is

"Ac(\"w\", \"x\")"  "Bz"  "Cc(\"q\", \"r\", \"s\")"

Answer 1

We can split column_2 on "," and paste them with column_1 using mapply

unlist(mapply(paste0, df$column_1,strsplit(df$column_2, ",")))
#[1] "Aw" "Ax" "Bz" "Cq" "Cr" "Cs"

Answer 2

We can rep licate the 'column_1' by the lengths of list output from strsplit and then do the paste

 lst1 <- strsplit(df$column_2, ",")
 paste0(rep(df$column_1, lengths(lst1)), unlist(lst1))
 #[1] "Aw" "Ax" "Bz" "Cq" "Cr" "Cs"

NOTE: The above is a vectorized approach as we are not looping through the list

---

Or use stack to create a two column data.frame from list and then paste

 do.call(paste0, stack(setNames(lst1, df$column_1))[2:1])
 #[1] "Aw" "Ax" "Bz" "Cq" "Cr" "Cs"

stack ing to a two column data.frame approach may be a bit less efficient compared to the first approach

---

Or with tidyverse , split the 'column_2' to long format with separate_rows , then unite the two columns and pull it to vector

library(tidyverse)
df %>% 
    separate_rows(column_2) %>%
    unite(newcol, column_1, column_2, sep="") %>%
    pull(newcol)
#[1] "Aw" "Ax" "Bz" "Cq" "Cr" "Cs"

---

The issue in the OP's approach is based on the fact that the strsplit output is a list of vector s. We need a function to loop over the list ( lapply/sapply/vapply ) or unlist the list into a vector while replicating the 'column_1' (to make the length during paste ing)

data

df <- structure(list(column_1 = c("A", "B", "C"), column_2 = c("w,x", 
 "z", "q,r,s")), class = "data.frame", row.names = c(NA, -3L))

Answer 3

This can also be achieved using below code. Although not very idiomatic

df <- data.frame(column_1 = c("A", "B", "C"), column_2 = c("w,x", "z", "q,r,s"))
l_vals <- strsplit(as.character(df$column_2), split = ",", perl =TRUE)
l_append = list()
for(i in seq_along(l_vals)){
  l_append <- c(l_append,paste0(df$column_1[i], l_vals[[i]]))
}

unlist(l_append)