Using "Apply", Add new column to dataframe in list where value is also stored in list in R

Question

I have two lists of identical length, one that stores some dataframes and one that stores character strings. I am trying to figure out how to add a new column to each dataframe that contains the corresponding value in the list of characters. For example:

#Staring Dataframes
df1 <- data.frame(v1 = c(1,2,3),
                  v2 = c("a","b","c"))
df2 <- data.frame(v1 = c(7,8,9),
                  v2 = c("x","y","z"))

dflist <- list(df1, df2)

#List containing values to add to each dataframe
nameslist <- c('first df value', 'second df value')

#Desired output 
df1$v3 <- nameslist[1]

  v1 v2    v3
 1  a first df value
 2  b first df value
 3  c first df value

df2$v3 <- nameslist[2]

 v1 v2       v3
 7  x second df value
 8  y second df value
 9  z second df value

It seems like this would be easy enought to do with a loop, but how would I accomplish this using one of the "apply" functions?

I have tried the following:

output <- lapply(seq_along(dflist), function(i) dflist[[i]]$test = nameslist[[i]])
#This returns a list of character vectors coresponding to nameslist
output <- mapply(function(df, name) df$test = name, dflist, nameslist)
#Returns a single character vector of the values in nameslist

I feel like one of the above must be reasonably close. If anyone could provide any guidance on this it would be much appreciated. TIA.

Answer 1

Use Map here

Map(cbind, dflist, v3 = nameslist)

-output

[[1]]
  v1 v2             v3
1  1  a first df value
2  2  b first df value
3  3  c first df value

[[2]]
  v1 v2              v3
1  7  x second df value
2  8  y second df value
3  9  z second df value

---

with mapply, SIMPLIFY = TRUE, by default, also, have to return the data if we are doing the assignment <-

mapply(function(df, name) {df$test = name;df}, dflist, nameslist, SIMPLIFY= FALSE)
[[1]]
  v1 v2           test
1  1  a first df value
2  2  b first df value
3  3  c first df value

[[2]]
  v1 v2            test
1  7  x second df value
2  8  y second df value
3  9  z second df value

Answer 2

We could use lapply

lapply(seq_along(dflist), function(i) {
  cbind(dflist[[i]], v3 = nameslist[i])
})

[[1]]
  v1 v2             v3
1  1  a first df value
2  2  b first df value
3  3  c first df value

[[2]]
  v1 v2              v3
1  7  x second df value
2  8  y second df value
3  9  z second df value