Delete a column in a data frame within a list

Question

I made a list out of my dataframe, based on the factor levels in column A. In the list I would like to remove that column. My head is saying lapply, but not anything else :P

Answer 1

We can efficiently use the bracket function "[" here.

Example

L <- replicate(3, iris[1:3, 1:4], simplify=FALSE)  # example list

Delete columns by numbers

lapply(L, "[", -c(2, 3))

Delete columns by names

lapply(L, "[", -grep(c("Sepal.Width|Petal.Length"), names(L[[1]])))

Result

# [[1]]
#   Sepal.Length Petal.Width
# 1          5.1         0.2
# 2          4.9         0.2
# 3          4.7         0.2
# 
# [[2]]
#   Sepal.Length Petal.Width
# 1          5.1         0.2
# 2          4.9         0.2
# 3          4.7         0.2

Answer 2

If you had a data frame that didn't contain the ID column, you could use map_if to remove it only where it exists.

myList <- list(A = data.frame(ID = c("A", "A"), 
                          Test = c(1, 1), 
                          Value = 1:2), 
           B = data.frame(ID = c("B", "B", "B"), 
                          Test = c(1, 3, 5), 
                          Value = 1:3),
           C = data.frame(Test = c(1, 3, 5), 
                          Value = 1:3))
map_if(myList, ~ "ID" %in% names(.x), ~ .x %>% select(-ID), .depth = 2)

Answer 3

Assuming your list is called myList, something like this should work:

lapply(myList, function(x) { x["ID"] <- NULL; x })

Update

For a more general solution, you can also use something like this:

# Sample data
myList <- list(A = data.frame(ID = c("A", "A"), 
                              Test = c(1, 1), 
                              Value = 1:2), 
               B = data.frame(ID = c("B", "B", "B"), 
                              Test = c(1, 3, 5), 
                              Value = 1:3))
# Keep just the "ID" and "Value" columns
lapply(myList, function(x) x[(names(x) %in% c("ID", "Value"))])
# Drop the "ID" and "Value" columns
lapply(myList, function(x) x[!(names(x) %in% c("ID", "Value"))])

Answer 4

If you are tidyverse user there is an alternative solution, which utilizes map function from purrr package.

# Create same sample data as above
myList <- list(A = data.frame(ID = c("A", "A"), 
                              Test = c(1, 1), 
                              Value = 1:2), 
               B = data.frame(ID = c("B", "B", "B"), 
                              Test = c(1, 3, 5), 
                              Value = 1:3))
# Remove column by name in each element of the list
map(myList, ~ (.x %>% select(-ID)))