在列表的每个数据框中创建新列，并根据位置 (R) 从字符向量中填充字符串

Question

I have a list of data frames and a character vector with strings. The number of dataframes and the number of strings in the chr are the same.

I'd like to populate a specific column in each dataframe in the list with the string at the corresponding position in the character vector

dfs<-list(mtcars[,1:4], iris[,1:4])
dfs <- lapply(dfs, function(x) transform(x, mycol=""))

z <- c("red", "blue")

As the final output I'd like

dfs[[1]]$mycol to be populated with red and

dfs[[2]]$mycol to be populated with blue

Conceptually, I think I need to do something like this:

dfs <- lapply(dfs, function(n) dfs[[n]]$mycol <- z[n]) , but I get the error

Error in z[n] : invalid subscript type 'list'

The real data is a list of 97 elements

Answer 1

You can also try creating directly mycol with mapply() :

#Data
dfs<-list(mtcars[,1:4], iris[,1:4])
z <- c("red", "blue")
#Code
L <- mapply(function(x,y) {x$mycol<-y;return(x)},x=dfs,y=z,SIMPLIFY = F)

Answer 2

This is what you're after

lapply(1:n, function(x) transform(dfs[x], mycol = z[x]))

When you want to perform an apply by passing an index on several objects, simply apply on 1:n then pass this as an argument to the different objects within the anonymous function.

---

EDIT transform only works with all objects contained in the same environment. So it throws the error object not found with the previous code because dfs exists the .GlobalEnv while x exists only in the function environments.

The below code works

lapply(c(1:n), function(x) {
  toreplace <<- z[x] # forcing to parent envir using <<-
  base::transform(dfs[x], mycol = toreplace)
})

Answer 3

We could use Map with transform in base R

Map(transform, dfs, mycol = z)

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Or map2 from purrr

library(purrr)
library(dplyr)
map2(dfs, z, ~ .x %>%
                   mutate(mycol = .y))