Reputation: 11
I have a dataframe with two colmns:
C1 <- c("abcd > de > efg", "hij > kl > iiii", "aa", "a > bbb")
C2 <- c("1980","1982","1989","1989")
df <- data.frame(C1, C2, stringsAsFactors = FALSE)
My goal is concatenate the arguments of the 2 of them like this:
result <- c("1980abcd > 1980de > 1980efg", "1982hij > 1982kl > 1982iiii", "1989aa", "1989a > 1989bbb")
How can i do that? Thanks.
Upvotes: 0
Views: 69
Reputation: 70256
Here's an approach that doesn't require splitting each string and pasting back together:
mapply(function(x,y) gsub("(^|\\s)(?=[a-z]+)", paste0("\\1", y), x, perl = TRUE),
df$C1, df$C2, USE.NAMES = FALSE)
#[1] "1980abcd > 1980de > 1980efg" "1982hij > 1982kl > 1982iiii"
#[3] "1989aa" "1989a > 1989bbb"
The regular expression pattern (^|\\s)(?=[a-z]+)
matches either the beginning of the string or a space followed by a lower case character and then replaces it with the corresponding C2-entry.
Here's a purrr alternative:
library(purrr)
strsplit(df$C1, " > ") %>% map2_chr(df$C2, ~paste(.y, .x, sep = "", collapse=" > "))
#[1] "1980abcd > 1980de > 1980efg" "1982hij > 1982kl > 1982iiii"
#[3] "1989aa" "1989a > 1989bbb"
Upvotes: 1
Reputation: 51582
One way via base R is to use split the C1
vector and use mapply
to paste with C2
, i.e.
v1 <- mapply(function(x, y) paste(paste0(x, y), collapse = ' > '), C2, strsplit(C1, ' > '))
unname(v1)
#[1] "1980abcd > 1980de > 1980efg" "1982hij > 1982kl > 1982iiii" "1989aa" "1989a > 1989bbb"
NOTE: The result of mapply
(i.e. v1
) is a named vector. Hence I used unname
to get to your desired structure. However, note that a named vector is still a vector and will behave as such.
Upvotes: 1
Reputation: 1998
Using strsplit, apply and paste :
library(dplyr)
df <- tibble(C1=strsplit(C1," > "),C2)
res <- unlist(apply(df,1,function(y){paste(paste(x$C2,x$C1,sep=""),collapse=" > ")}))
# [1] "1980abcd > 1980de > 1980efg" "1982hij > 1982kl > 1982iiii" "1989aa"
# [4] "1989a > 1989bbb"
Upvotes: 0