remove columns with NA values from list of matrices
-1
I have a list of matrices, and need to remove columns containing NA values, within each matrix.
I tried to use lapply with !is.na to apply this to all the matrices in the entire list, but what it returns is a list of vectors with NAs excluded, whereas I still want a list of matrices (just without the columns containing NA).
> my_list
$mat1
V1 V2 V3
[1,] 1 5 NA
[1,] 2 6 NA
[1,] 3 7 NA
[1,] 4 8 NA
$mat2
V1 V2 V3
[1,] 1 NA 9
[1,] 2 NA 10
[1,] 3 NA 11
[1,] 4 NA 12
> lapply(my_list,function(x) x[!is.na(x)])
$mat1
[1] 1 2 3 4 5 6 7 8
$mat2
[1] 1 2 3 4 9 10 11 12
The output I'm trying to get is this:
$mat1
V1 V2
[1,] 1 5
[1,] 2 6
[1,] 3 7
[1,] 4 8
$mat2
V1 V3
[1,] 1 9
[1,] 2 10
[1,] 3 11
[1,] 4 12
r matrix na
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
|
show 2 more comments
-1
I have a list of matrices, and need to remove columns containing NA values, within each matrix.
I tried to use lapply with !is.na to apply this to all the matrices in the entire list, but what it returns is a list of vectors with NAs excluded, whereas I still want a list of matrices (just without the columns containing NA).
> my_list
$mat1
V1 V2 V3
[1,] 1 5 NA
[1,] 2 6 NA
[1,] 3 7 NA
[1,] 4 8 NA
$mat2
V1 V2 V3
[1,] 1 NA 9
[1,] 2 NA 10
[1,] 3 NA 11
[1,] 4 NA 12
> lapply(my_list,function(x) x[!is.na(x)])
$mat1
[1] 1 2 3 4 5 6 7 8
$mat2
[1] 1 2 3 4 9 10 11 12
The output I'm trying to get is this:
$mat1
V1 V2
[1,] 1 5
[1,] 2 6
[1,] 3 7
[1,] 4 8
$mat2
V1 V3
[1,] 1 9
[1,] 2 10
[1,] 3 11
[1,] 4 12
r matrix na
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
4
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
1
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])
– Andre Elrico
Nov 21 '18 at 9:28
1
@RLave complete.cases looks row-wise forNA. So not practical here
– Andre Elrico
Nov 21 '18 at 9:38
3
For one matrixapply(is.na(A), 2, any)gives you the logical index for the columns with NA. To remove the columns do:A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do:lapply(L, function(A) A[, !apply(is.na(A), 2, any)])
– jogo
Nov 21 '18 at 10:02
1
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11
|
show 2 more comments
-1
-1
-1
I have a list of matrices, and need to remove columns containing NA values, within each matrix.
I tried to use lapply with !is.na to apply this to all the matrices in the entire list, but what it returns is a list of vectors with NAs excluded, whereas I still want a list of matrices (just without the columns containing NA).
> my_list
$mat1
V1 V2 V3
[1,] 1 5 NA
[1,] 2 6 NA
[1,] 3 7 NA
[1,] 4 8 NA
$mat2
V1 V2 V3
[1,] 1 NA 9
[1,] 2 NA 10
[1,] 3 NA 11
[1,] 4 NA 12
> lapply(my_list,function(x) x[!is.na(x)])
$mat1
[1] 1 2 3 4 5 6 7 8
$mat2
[1] 1 2 3 4 9 10 11 12
The output I'm trying to get is this:
$mat1
V1 V2
[1,] 1 5
[1,] 2 6
[1,] 3 7
[1,] 4 8
$mat2
V1 V3
[1,] 1 9
[1,] 2 10
[1,] 3 11
[1,] 4 12
r matrix na
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
I have a list of matrices, and need to remove columns containing NA values, within each matrix.
I tried to use lapply with !is.na to apply this to all the matrices in the entire list, but what it returns is a list of vectors with NAs excluded, whereas I still want a list of matrices (just without the columns containing NA).
> my_list
$mat1
V1 V2 V3
[1,] 1 5 NA
[1,] 2 6 NA
[1,] 3 7 NA
[1,] 4 8 NA
$mat2
V1 V2 V3
[1,] 1 NA 9
[1,] 2 NA 10
[1,] 3 NA 11
[1,] 4 NA 12
> lapply(my_list,function(x) x[!is.na(x)])
$mat1
[1] 1 2 3 4 5 6 7 8
$mat2
[1] 1 2 3 4 9 10 11 12
The output I'm trying to get is this:
$mat1
V1 V2
[1,] 1 5
[1,] 2 6
[1,] 3 7
[1,] 4 8
$mat2
V1 V3
[1,] 1 9
[1,] 2 10
[1,] 3 11
[1,] 4 12
r matrix na
r matrix na
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
edited Nov 21 '18 at 10:07
Neuroguy
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
asked Nov 21 '18 at 9:26
NeuroguyNeuroguy
447
447
4
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
1
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])
– Andre Elrico
Nov 21 '18 at 9:28
1
@RLave complete.cases looks row-wise forNA. So not practical here
– Andre Elrico
Nov 21 '18 at 9:38
3
For one matrixapply(is.na(A), 2, any)gives you the logical index for the columns with NA. To remove the columns do:A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do:lapply(L, function(A) A[, !apply(is.na(A), 2, any)])
– jogo
Nov 21 '18 at 10:02
1
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11
|
show 2 more comments
4
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
1
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])
– Andre Elrico
Nov 21 '18 at 9:28
1
@RLave complete.cases looks row-wise forNA. So not practical here
– Andre Elrico
Nov 21 '18 at 9:38
3
For one matrixapply(is.na(A), 2, any)gives you the logical index for the columns with NA. To remove the columns do:A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do:lapply(L, function(A) A[, !apply(is.na(A), 2, any)])
– jogo
Nov 21 '18 at 10:02
1
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11
4
4
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
1
1
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])– Andre Elrico
Nov 21 '18 at 9:28
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])– Andre Elrico
Nov 21 '18 at 9:28
1
1
@RLave complete.cases looks row-wise for
NA. So not practical here– Andre Elrico
Nov 21 '18 at 9:38
@RLave complete.cases looks row-wise for
NA. So not practical here– Andre Elrico
Nov 21 '18 at 9:38
3
3
For one matrix
apply(is.na(A), 2, any) gives you the logical index for the columns with NA. To remove the columns do: A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do: lapply(L, function(A) A[, !apply(is.na(A), 2, any)])– jogo
Nov 21 '18 at 10:02
For one matrix
apply(is.na(A), 2, any) gives you the logical index for the columns with NA. To remove the columns do: A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do: lapply(L, function(A) A[, !apply(is.na(A), 2, any)])– jogo
Nov 21 '18 at 10:02
1
1
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11
|
show 2 more comments
2 Answers
2
active
oldest
votes
1
For one matrix apply(is.na(mat), 2, any) gives you the logical index for the columns with NA. To remove the columns do:
mat[, !apply(is.na(mat), 2, any)]
To do this with a list of matrices do:
lapply(my_list, function(mat) mat[, !apply(is.na(mat), 2, any)])
Data:
> dput(my_list)
list(structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, NA, NA, NA,
NA), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2", "V3"
))), structure(c(1L, 2L, 3L, 4L, NA, NA, NA, NA, 9L, 10L, 11L,
12L), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2",
"V3"))))
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
add a comment |
0
Or another option is tidyverse with select_if after converting to data.frame
library(tidyverse)
map(lst1, ~ .x %>%
as.data.frame %>%
select_if(~ all(!is.na(.))))
#[[1]]
# V1 V2
#1 1 5
#2 2 6
#3 3 7
#4 4 8
#[[2]]
# V1 V3
#1 1 9
#2 2 10
#3 3 11
#4 4 12
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
For one matrix apply(is.na(mat), 2, any) gives you the logical index for the columns with NA. To remove the columns do:
mat[, !apply(is.na(mat), 2, any)]
To do this with a list of matrices do:
lapply(my_list, function(mat) mat[, !apply(is.na(mat), 2, any)])
Data:
> dput(my_list)
list(structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, NA, NA, NA,
NA), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2", "V3"
))), structure(c(1L, 2L, 3L, 4L, NA, NA, NA, NA, 9L, 10L, 11L,
12L), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2",
"V3"))))
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
add a comment |
1
For one matrix apply(is.na(mat), 2, any) gives you the logical index for the columns with NA. To remove the columns do:
mat[, !apply(is.na(mat), 2, any)]
To do this with a list of matrices do:
lapply(my_list, function(mat) mat[, !apply(is.na(mat), 2, any)])
Data:
> dput(my_list)
list(structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, NA, NA, NA,
NA), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2", "V3"
))), structure(c(1L, 2L, 3L, 4L, NA, NA, NA, NA, 9L, 10L, 11L,
12L), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2",
"V3"))))
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
add a comment |
1
1
1
For one matrix apply(is.na(mat), 2, any) gives you the logical index for the columns with NA. To remove the columns do:
mat[, !apply(is.na(mat), 2, any)]
To do this with a list of matrices do:
lapply(my_list, function(mat) mat[, !apply(is.na(mat), 2, any)])
Data:
> dput(my_list)
list(structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, NA, NA, NA,
NA), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2", "V3"
))), structure(c(1L, 2L, 3L, 4L, NA, NA, NA, NA, 9L, 10L, 11L,
12L), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2",
"V3"))))
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
For one matrix apply(is.na(mat), 2, any) gives you the logical index for the columns with NA. To remove the columns do:
mat[, !apply(is.na(mat), 2, any)]
To do this with a list of matrices do:
lapply(my_list, function(mat) mat[, !apply(is.na(mat), 2, any)])
Data:
> dput(my_list)
list(structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, NA, NA, NA,
NA), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2", "V3"
))), structure(c(1L, 2L, 3L, 4L, NA, NA, NA, NA, 9L, 10L, 11L,
12L), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("V1", "V2",
"V3"))))
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
edited Nov 21 '18 at 10:39
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
answered Nov 21 '18 at 10:29
jogojogo
9,85692135
9,85692135
add a comment |
add a comment |
0
Or another option is tidyverse with select_if after converting to data.frame
library(tidyverse)
map(lst1, ~ .x %>%
as.data.frame %>%
select_if(~ all(!is.na(.))))
#[[1]]
# V1 V2
#1 1 5
#2 2 6
#3 3 7
#4 4 8
#[[2]]
# V1 V3
#1 1 9
#2 2 10
#3 3 11
#4 4 12
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
add a comment |
0
Or another option is tidyverse with select_if after converting to data.frame
library(tidyverse)
map(lst1, ~ .x %>%
as.data.frame %>%
select_if(~ all(!is.na(.))))
#[[1]]
# V1 V2
#1 1 5
#2 2 6
#3 3 7
#4 4 8
#[[2]]
# V1 V3
#1 1 9
#2 2 10
#3 3 11
#4 4 12
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
add a comment |
0
0
0
Or another option is tidyverse with select_if after converting to data.frame
library(tidyverse)
map(lst1, ~ .x %>%
as.data.frame %>%
select_if(~ all(!is.na(.))))
#[[1]]
# V1 V2
#1 1 5
#2 2 6
#3 3 7
#4 4 8
#[[2]]
# V1 V3
#1 1 9
#2 2 10
#3 3 11
#4 4 12
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
Or another option is tidyverse with select_if after converting to data.frame
library(tidyverse)
map(lst1, ~ .x %>%
as.data.frame %>%
select_if(~ all(!is.na(.))))
#[[1]]
# V1 V2
#1 1 5
#2 2 6
#3 3 7
#4 4 8
#[[2]]
# V1 V3
#1 1 9
#2 2 10
#3 3 11
#4 4 12
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
answered Nov 21 '18 at 18:03
akrunakrun
400k13189264
400k13189264
add a comment |
add a comment |
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4
Please give a Minimal, Complete, and Verifiable example in your question!
– jogo
Nov 21 '18 at 9:27
1
lapply(my_list,function(x) x[,colSums(is.na(x)) < 1, drop = FALSE])– Andre Elrico
Nov 21 '18 at 9:28
1
@RLave complete.cases looks row-wise for
NA. So not practical here– Andre Elrico
Nov 21 '18 at 9:38
3
For one matrix
apply(is.na(A), 2, any)gives you the logical index for the columns with NA. To remove the columns do:A[, !apply(is.na(A), 2, any)]. To do this with a list of matrices do:lapply(L, function(A) A[, !apply(is.na(A), 2, any)])– jogo
Nov 21 '18 at 10:02
1
Thank you, jogo, this solved it. I added an verifiable example to the question now for clarity
– Neuroguy
Nov 21 '18 at 10:11