Using Filter(Negate(is.na), x) on a list results in unexpected behaviour
Consider
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Why does Filter(Negate(is.na), l1) not work, while Filter(Negate(is.null), l2) does work as expected (it returns all elements of l2 which are not NULL)?
The helpfile for ?Filter says
...Note that possible NA values are currently always taken as false; control over NA handling may be added in the future.
I still dont really understand the behaviour.
r
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
add a comment |
Consider
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Why does Filter(Negate(is.na), l1) not work, while Filter(Negate(is.null), l2) does work as expected (it returns all elements of l2 which are not NULL)?
The helpfile for ?Filter says
...Note that possible NA values are currently always taken as false; control over NA handling may be added in the future.
I still dont really understand the behaviour.
r
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
add a comment |
Consider
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Why does Filter(Negate(is.na), l1) not work, while Filter(Negate(is.null), l2) does work as expected (it returns all elements of l2 which are not NULL)?
The helpfile for ?Filter says
...Note that possible NA values are currently always taken as false; control over NA handling may be added in the future.
I still dont really understand the behaviour.
r
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
Consider
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Why does Filter(Negate(is.na), l1) not work, while Filter(Negate(is.null), l2) does work as expected (it returns all elements of l2 which are not NULL)?
The helpfile for ?Filter says
...Note that possible NA values are currently always taken as false; control over NA handling may be added in the future.
I still dont really understand the behaviour.
r
r
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
asked Nov 13 '18 at 8:46
Manuel RManuel R
1,44711230
1,44711230
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
is.na returns a vector for each element of the list; you want anyNA (or perhaps exactlyNA as defined below):
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Filter(Negate(is.na), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
#>
#> [[3]]
#> [1] NA
#>
#> [[4]]
#> NULL
Filter(Negate(anyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
exactlyNA <- function(x) identical(x, NA)
Filter(Negate(exactlyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
Created on 2018-11-13 by the reprex package (v0.2.1)
Your first example effectively tries to select the 1st, 3rd, 4th, and 5th elements of your list. Nothing to do with NA.
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
is.na returns a vector for each element of the list; you want anyNA (or perhaps exactlyNA as defined below):
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Filter(Negate(is.na), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
#>
#> [[3]]
#> [1] NA
#>
#> [[4]]
#> NULL
Filter(Negate(anyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
exactlyNA <- function(x) identical(x, NA)
Filter(Negate(exactlyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
Created on 2018-11-13 by the reprex package (v0.2.1)
Your first example effectively tries to select the 1st, 3rd, 4th, and 5th elements of your list. Nothing to do with NA.
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
add a comment |
is.na returns a vector for each element of the list; you want anyNA (or perhaps exactlyNA as defined below):
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Filter(Negate(is.na), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
#>
#> [[3]]
#> [1] NA
#>
#> [[4]]
#> NULL
Filter(Negate(anyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
exactlyNA <- function(x) identical(x, NA)
Filter(Negate(exactlyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
Created on 2018-11-13 by the reprex package (v0.2.1)
Your first example effectively tries to select the 1st, 3rd, 4th, and 5th elements of your list. Nothing to do with NA.
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
add a comment |
is.na returns a vector for each element of the list; you want anyNA (or perhaps exactlyNA as defined below):
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Filter(Negate(is.na), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
#>
#> [[3]]
#> [1] NA
#>
#> [[4]]
#> NULL
Filter(Negate(anyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
exactlyNA <- function(x) identical(x, NA)
Filter(Negate(exactlyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
Created on 2018-11-13 by the reprex package (v0.2.1)
Your first example effectively tries to select the 1st, 3rd, 4th, and 5th elements of your list. Nothing to do with NA.
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
is.na returns a vector for each element of the list; you want anyNA (or perhaps exactlyNA as defined below):
l1 <- list("a", NA, 1:3, NA)
l2 <- list("a", NULL, 1:3, NULL)
Filter(Negate(is.na), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
#>
#> [[3]]
#> [1] NA
#>
#> [[4]]
#> NULL
Filter(Negate(anyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
exactlyNA <- function(x) identical(x, NA)
Filter(Negate(exactlyNA), l1)
#> [[1]]
#> [1] "a"
#>
#> [[2]]
#> [1] 1 2 3
Created on 2018-11-13 by the reprex package (v0.2.1)
Your first example effectively tries to select the 1st, 3rd, 4th, and 5th elements of your list. Nothing to do with NA.
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
edited Nov 13 '18 at 9:00
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
answered Nov 13 '18 at 8:56
HughHugh
7,51343569
7,51343569
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
add a comment |
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
Nice thank you!
– Manuel R
Nov 13 '18 at 9:00
add a comment |
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