library(tidyverse)
df <- tibble(Date = c(rep(as.Date("2020-01-01"), 3), NA),
col1 = 1:4,
thisCol = c(NA, 8, NA, 3),
thatCol = 25:28,
col999 = rep(99, 4))
#> # A tibble: 4 x 5
#> Date col1 thisCol thatCol col999
#> <date> <int> <dbl> <int> <dbl>
#> 1 2020-01-01 1 NA 25 99
#> 2 2020-01-01 2 8 26 99
#> 3 2020-01-01 3 NA 27 99
#> 4 NA 4 3 28 99
My actual R data frame has hundreds of columns that aren't neatly named, but can be approximated by the df data frame above.
I want to replace all values of NA with 0, with the exception of several columns (in my example I want to leave out the Date column and the thatCol column. I'd want to do it in this sort of fashion:
df %>% replace(is.na(.), 0)
#> Error: Assigned data `values` must be compatible with existing data.
#> i Error occurred for column `Date`.
#> x Can't convert <double> to <date>.
#> Run `rlang::last_error()` to see where the error occurred.
And my unsuccessful ideas for accomplishing the "everything except" replace NA are shown below.
df %>% replace(is.na(c(., -c(Date, thatCol)), 0))
df %>% replace_na(list([, c(2:3, 5)] = 0))
df %>% replace_na(list(everything(-c(Date, thatCol)) = 0))
Is there a way to select everything BUT in the way I need to? There's hundred of columns, named inconsistently, so typing them one by one is not a practical option.
You can use mutate_at :
library(dplyr)
Remove them by Name
df %>% mutate_at(vars(-c(Date, thatCol)), ~replace(., is.na(.), 0))
Remove them by position
df %>% mutate_at(-c(1,4), ~replace(., is.na(.), 0))
Select them by name
df %>% mutate_at(vars(col1, thisCol, col999), ~replace(., is.na(.), 0))
Select them by position
df %>% mutate_at(c(2, 3, 5), ~replace(., is.na(.), 0))
If you want to use replace_na
df %>% mutate_at(vars(-c(Date, thatCol)), tidyr::replace_na, 0)
Note that mutate_at is soon going to be replaced by across in dplyr 1.0.0.