Reputation: 476
I have a data.frame with several variables. I would like to get a list where each item is a variable of the data.frame filtered with a condition.
For example, let's say I have something like this:
df <- tribble(
~ var1, ~ var2, ~ var3,
4, 0, 0,
2, 3, 1,
0, 4, 0
)
# var1 var2 var3
# <dbl> <dbl> <dbl>
# 1 4 0 0
# 2 2 3 1
# 3 0 4 0
And I want to get the list of the variable filtered >0
# $var1
# [1] 4 2
#
# $var2
# [1] 3 4
#
# $var3
# [1] 1
I tried several things but the closest I can get for now is something like
df %>% map(~filter(df, .>0))
and I would like to include a dplyr::select
to get only the variable filtered. But I can't figure out how to do that.
Thanks for help and sorry for bad English, I hope it's still understandable.
Upvotes: 1
Views: 3718
Reputation: 886938
We can loop through the names
. Note that filter
expects a data.frame/tbl_df
. With map
, we are looping through the columns and it is a vector
. So, in order to make filter
work, map
through the names
, subset the column, apply the filter
and unlist
map(names(df), ~ df %>%
select(.x) %>%
filter(. >0) %>%
unlist(., use.names = FALSE))
Or with split
split.default(df, names(df)) %>%
map(~ .x %>%
filter(. > 0) %>%
pull(1))
NOTE: The OP's question is How to use dplyr::filter inside purrr::map
Other ways without using dplyr::filter
are
map(df, ~ keep(.x, .x != 0))
Or
map(df, setdiff, 0)
Or
map(df, ~ discard(.x, .x == 0))
Or using base R
lapply(df, setdiff, 0)
#$var1
#[1] 4 2
#$var2
#[1] 3 4
#$var3
#[1] 1
Upvotes: 3
Reputation: 388807
Using purrr::map
we can do
purrr::map(df, ~.[.!= 0])
#$var1
#[1] 4 2
#$var2
#[1] 3 4
#$var3
#[1] 1
Base R approach with lapply
could be
lapply(df, function(x) x[x!= 0])
Upvotes: 2