I have a list with dataframes:
df1 <- data.frame(id = seq(1:10), name = LETTERS[1:10])
df2 <- data.frame(id = seq(11:20), name = LETTERS[11:20])
mylist <- list(df1, df2)
I want to remove rows from each dataframe in the list based on a condition (in this case, the value stored in column id
). I create an empty vector where I will store the ids:
ids_to_remove <- c()
Then I apply my function:
sapply(mylist, function(df) {
rows_above_th <- df[(df$id > 8),] # select the rows from each df above a threshold
a <- rows_above_th$id # obtain the ids of the rows above the threshold
ids_to_remove <- append(ids_to_remove, a) # append each id to the vector
},
simplify = T
)
However, with or without simplify = T
, this returns a matrix, while my desired output (ids_to_remove
) would be a vector containing the ids
, like this:
ids_to_remove <- c(9,10,9,10)
Because lastly I would use it in this way on single dataframes:
for(i in 1:length(ids_to_remove)){
mylist[[1]] <- mylist[[1]] %>%
filter(!id == ids_to_remove[i])
}
And like this on the whole list (which is not working and I don´t get why):
i = 1
lapply(mylist,
function(df) {
for(i in 1:length(ids_to_remove)){
df <- df %>%
filter(!id == ids_to_remove[i])
i = i 1
}
} )
I get the errors may be in the append
part of the sapply
and maybe in the indexing of the lapply
. I played around a bit but couldn´t still find the errors (or a better way to do this).
CodePudding user response:
If you are using sapply/lapply
you want to avoid trying to change the values of global variables. Instead, you should return the values you want. For example generate a vector if IDs to remove for each item in the list as a list
ids_to_remove <- lapply(mylist, function(df) {
rows_above_th <- df[(df$id > 8),] # select the rows from each df above a threshold
rows_above_th$id # obtain the ids of the rows above the threshold
})
And then you can use that list with your data list and mapply
to iterate the two lists together
mapply(function(data, ids) {
data %>% dplyr::filter(!id %in% ids)
}, mylist, ids_to_remove, SIMPLIFY=FALSE)
CodePudding user response:
Using base R
Map(\(x, y) subset(x, !id %in% y), mylist, ids_to_remove)