A function to fill in a column with NA of the same type

up vote
14
down vote

favorite

I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.

for example:

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))



df[, 1:2] <- NA

yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:

df[,1] = as.numeric(NA)

df[,2] = as.character(NA)

But how do I do this collectively in a loop for all columns with all possible types of NA?

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

3

Good question +1, but why does this matter?
– Tim Biegeleisen
Dec 11 at 7:48

It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
Dec 11 at 7:49

1

But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
Dec 11 at 8:45

I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
Dec 11 at 9:27

1

Just a minor correction, you shouldn't talk about classes here but about atomic types, and it would be more idiomatic to use NA_character_ and NA_numeric_ than as.character(NA) and as.numeric(NA).
– Moody_Mudskipper
2 days ago

add a comment |

up vote
14
down vote

favorite

I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.

for example:

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))



df[, 1:2] <- NA

yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:

df[,1] = as.numeric(NA)

df[,2] = as.character(NA)

But how do I do this collectively in a loop for all columns with all possible types of NA?

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

3

Good question +1, but why does this matter?
– Tim Biegeleisen
Dec 11 at 7:48

It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
Dec 11 at 7:49

1

But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
Dec 11 at 8:45

I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
Dec 11 at 9:27

1

Just a minor correction, you shouldn't talk about classes here but about atomic types, and it would be more idiomatic to use NA_character_ and NA_numeric_ than as.character(NA) and as.numeric(NA).
– Moody_Mudskipper
2 days ago

add a comment |

up vote
14
down vote

favorite

I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.

for example:

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))



df[, 1:2] <- NA

yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:

df[,1] = as.numeric(NA)

df[,2] = as.character(NA)

But how do I do this collectively in a loop for all columns with all possible types of NA?

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

I have a data frame with many columns of different types. I would like to replace each column with NA of the corresponding class.

for example:

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"))



df[, 1:2] <- NA

yields a data frame with two logical columns, rather than numeric and character.
I know I can tell R:

df[,1] = as.numeric(NA)

df[,2] = as.character(NA)

But how do I do this collectively in a loop for all columns with all possible types of NA?

r dplyr na

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

edited Dec 11 at 8:18

zx8754

29.1k76396

edited Dec 11 at 8:18

zx8754

29.1k76396

edited Dec 11 at 8:18

zx8754

29.1k76396

asked Dec 11 at 7:36

Omry Atia

819412

asked Dec 11 at 7:36

Omry Atia

819412

asked Dec 11 at 7:36

Omry Atia

819412

3

Good question +1, but why does this matter?
– Tim Biegeleisen
Dec 11 at 7:48

It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
Dec 11 at 7:49

1

But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
Dec 11 at 8:45

I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
Dec 11 at 9:27

1

Just a minor correction, you shouldn't talk about classes here but about atomic types, and it would be more idiomatic to use NA_character_ and NA_numeric_ than as.character(NA) and as.numeric(NA).
– Moody_Mudskipper
2 days ago

add a comment |

3

Good question +1, but why does this matter?
– Tim Biegeleisen
Dec 11 at 7:48

It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
Dec 11 at 7:49

1

But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
Dec 11 at 8:45

I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
Dec 11 at 9:27

1

Just a minor correction, you shouldn't talk about classes here but about atomic types, and it would be more idiomatic to use NA_character_ and NA_numeric_ than as.character(NA) and as.numeric(NA).
– Moody_Mudskipper
2 days ago

Good question +1, but why does this matter?
– Tim Biegeleisen
Dec 11 at 7:48

It's a very weird problem, I later need to join the data frame with another frame of the original type...
– Omry Atia
Dec 11 at 7:49

But why? Please give us more context, seems like pointless (but fun) step.
– zx8754
Dec 11 at 8:45

I have a data frame created in the beginning of my program, which sometimes need to get all NA's in some columns based on a condition. This data frame needs to be joined with another data frame in the end of the program, which might not get these NA's. In order for the join to work, the two data frames need to have exactly the same types of columns.
– Omry Atia
Dec 11 at 9:27

Just a minor correction, you shouldn't talk about classes here but about atomic types, and it would be more idiomatic to use NA_character_ and NA_numeric_ than as.character(NA) and as.numeric(NA).
– Moody_Mudskipper
2 days ago

add a comment |

5 Answers
5

active

oldest

votes

up vote
9
down vote

accepted

You can use this "trick" :

df[1:nrow(df),1] <- NA

df[1:nrow(df),2] <- NA

the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.

Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :

rowIdxs <- 1:nrow(df)

df[rowIdxs ,1] <- NA

df[rowIdxs ,2] <- NA

df[rowIdxs ,3] <- NA

...

As cleverly suggested by @RonakShah, you can also use :

df[TRUE, 1] <- NA

df[TRUE, 2] <- NA

...

As pointed out by @Cath both the methods still work when you select more than one column e.g. :

df[TRUE, 1:3] <- NA

# or

df[1:nrow(df), 1:3] <- NA

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

3

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

add a comment |

up vote
8
down vote

Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.

df[!is.na(df)] <- NA

which gives,

# A tibble: 3 x 2

      x    y    

  <dbl> <chr>

1    NA <NA> 

2    NA <NA> 

3    NA <NA>

answered Dec 11 at 8:00

Sotos

27.6k51640

add a comment |

up vote
5
down vote

Another way to change all columns at once while keeping the variables' classes:

df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})



df

# A tibble: 3 x 2

#      x y    

#  <dbl> <chr>

#1    NA <NA> 

#2    NA <NA> 

#3    NA <NA>

As @digEmAll notified in comments, there is another similar but shorter way:

df <- lapply(df, function(x) as(NA,class(x)))

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

2

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

2

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

|
show 1 more comment

up vote
4
down vote

Using dplyr::na_if:

library(dplyr)



df %>% 

  mutate(x = na_if(x, x),

         y = na_if(y, y))



# # A tibble: 3 x 2

#       x y    

#   <dbl> <chr>

# 1    NA NA   

# 2    NA NA   

# 3    NA NA

If we want to mutate only subset of columns to NA, then:

# dataframe with extra column that stay unchanged

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))



df %>% 

  mutate_at(vars(x, y), funs(na_if(.,.)))



# # A tibble: 3 x 3

#       x y         z

#   <dbl> <chr> <int>

# 1    NA NA        4

# 2    NA NA        5

# 3    NA NA        6

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

add a comment |

up vote
0
down vote

Using bind_cols() from dplyr you can also do:

df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))

classes <- sapply(df, class)

df[,1:2] <- NA



bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))



     V1 V2   

  <dbl> <chr>

1    NA NA   

2    NA NA   

3    NA NA

Please note that this will change the colnames.

answered Dec 11 at 7:59

alex_555

666315

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
9
down vote

accepted

You can use this "trick" :

df[1:nrow(df),1] <- NA

df[1:nrow(df),2] <- NA

the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.

Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :

rowIdxs <- 1:nrow(df)

df[rowIdxs ,1] <- NA

df[rowIdxs ,2] <- NA

df[rowIdxs ,3] <- NA

...

As cleverly suggested by @RonakShah, you can also use :

df[TRUE, 1] <- NA

df[TRUE, 2] <- NA

...

As pointed out by @Cath both the methods still work when you select more than one column e.g. :

df[TRUE, 1:3] <- NA

# or

df[1:nrow(df), 1:3] <- NA

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

3

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

add a comment |

up vote
9
down vote

accepted

You can use this "trick" :

df[1:nrow(df),1] <- NA

df[1:nrow(df),2] <- NA

the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.

Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :

rowIdxs <- 1:nrow(df)

df[rowIdxs ,1] <- NA

df[rowIdxs ,2] <- NA

df[rowIdxs ,3] <- NA

...

As cleverly suggested by @RonakShah, you can also use :

df[TRUE, 1] <- NA

df[TRUE, 2] <- NA

...

As pointed out by @Cath both the methods still work when you select more than one column e.g. :

df[TRUE, 1:3] <- NA

# or

df[1:nrow(df), 1:3] <- NA

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

3

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

add a comment |

up vote
9
down vote

accepted

You can use this "trick" :

df[1:nrow(df),1] <- NA

df[1:nrow(df),2] <- NA

the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.

Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :

rowIdxs <- 1:nrow(df)

df[rowIdxs ,1] <- NA

df[rowIdxs ,2] <- NA

df[rowIdxs ,3] <- NA

...

As cleverly suggested by @RonakShah, you can also use :

df[TRUE, 1] <- NA

df[TRUE, 2] <- NA

...

As pointed out by @Cath both the methods still work when you select more than one column e.g. :

df[TRUE, 1:3] <- NA

# or

df[1:nrow(df), 1:3] <- NA

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

You can use this "trick" :

df[1:nrow(df),1] <- NA

df[1:nrow(df),2] <- NA

the [1:nrow(df),] basically tells R to replace all values in the column with NA and in this way the logical NA is coerced to the original type of the column before replacing the other values.

Also, if you have a lot of columns to replace and the data_frame has a lot of rows, I suggest to store the row indexes and reuse them :

rowIdxs <- 1:nrow(df)

df[rowIdxs ,1] <- NA

df[rowIdxs ,2] <- NA

df[rowIdxs ,3] <- NA

...

As cleverly suggested by @RonakShah, you can also use :

df[TRUE, 1] <- NA

df[TRUE, 2] <- NA

...

As pointed out by @Cath both the methods still work when you select more than one column e.g. :

df[TRUE, 1:3] <- NA

# or

df[1:nrow(df), 1:3] <- NA

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

edited Dec 11 at 8:47

answered Dec 11 at 7:43

digEmAll

46.2k984120

answered Dec 11 at 7:43

digEmAll

46.2k984120

answered Dec 11 at 7:43

digEmAll

46.2k984120

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

3

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

add a comment |

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

3

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

This doesn't seem to work... df is still logical :(
– Omry Atia
Dec 11 at 7:45

@OmryAtia : edited. it should work now ;)
– digEmAll
Dec 11 at 7:50

Awesome... so simple :)
– Omry Atia
Dec 11 at 7:51

why not directly df[TRUE, 1:2] <- NA?
– Cath
Dec 11 at 8:42

@Cath: sure, added in the answer, thanks !
– digEmAll
Dec 11 at 8:47

add a comment |

up vote
8
down vote

Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.

df[!is.na(df)] <- NA

which gives,

# A tibble: 3 x 2

      x    y    

  <dbl> <chr>

1    NA <NA> 

2    NA <NA> 

3    NA <NA>

answered Dec 11 at 8:00

Sotos

27.6k51640

add a comment |

up vote
8
down vote

Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.

df[!is.na(df)] <- NA

which gives,

# A tibble: 3 x 2

      x    y    

  <dbl> <chr>

1    NA <NA> 

2    NA <NA> 

3    NA <NA>

answered Dec 11 at 8:00

Sotos

27.6k51640

add a comment |

up vote
8
down vote

Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.

df[!is.na(df)] <- NA

which gives,

# A tibble: 3 x 2

      x    y    

  <dbl> <chr>

1    NA <NA> 

2    NA <NA> 

3    NA <NA>

answered Dec 11 at 8:00

Sotos

27.6k51640

Another solution that applies to all the columns can be to specify the non-NAs and replace with NA, i.e.

df[!is.na(df)] <- NA

which gives,

# A tibble: 3 x 2

      x    y    

  <dbl> <chr>

1    NA <NA> 

2    NA <NA> 

3    NA <NA>

answered Dec 11 at 8:00

Sotos

27.6k51640

answered Dec 11 at 8:00

Sotos

27.6k51640

answered Dec 11 at 8:00

Sotos

27.6k51640

answered Dec 11 at 8:00

Sotos

27.6k51640

add a comment |

up vote
5
down vote

Another way to change all columns at once while keeping the variables' classes:

df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})



df

# A tibble: 3 x 2

#      x y    

#  <dbl> <chr>

#1    NA <NA> 

#2    NA <NA> 

#3    NA <NA>

As @digEmAll notified in comments, there is another similar but shorter way:

df <- lapply(df, function(x) as(NA,class(x)))

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

2

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

2

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

|
show 1 more comment

up vote
5
down vote

Another way to change all columns at once while keeping the variables' classes:

df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})



df

# A tibble: 3 x 2

#      x y    

#  <dbl> <chr>

#1    NA <NA> 

#2    NA <NA> 

#3    NA <NA>

As @digEmAll notified in comments, there is another similar but shorter way:

df <- lapply(df, function(x) as(NA,class(x)))

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

2

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

2

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

|
show 1 more comment

up vote
5
down vote

Another way to change all columns at once while keeping the variables' classes:

df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})



df

# A tibble: 3 x 2

#      x y    

#  <dbl> <chr>

#1    NA <NA> 

#2    NA <NA> 

#3    NA <NA>

As @digEmAll notified in comments, there is another similar but shorter way:

df <- lapply(df, function(x) as(NA,class(x)))

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

Another way to change all columns at once while keeping the variables' classes:

df <- lapply(df, function(x) {type <- class(x); x <- NA; class(x) <- type; x})



df

# A tibble: 3 x 2

#      x y    

#  <dbl> <chr>

#1    NA <NA> 

#2    NA <NA> 

#3    NA <NA>

As @digEmAll notified in comments, there is another similar but shorter way:

df <- lapply(df, function(x) as(NA,class(x)))

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

edited Dec 11 at 8:37

answered Dec 11 at 8:32

Cath

19.6k43464

answered Dec 11 at 8:32

Cath

19.6k43464

answered Dec 11 at 8:32

Cath

19.6k43464

2

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

2

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

|
show 1 more comment

2

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

2

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

Also lapply(df, function(x)as(NA,class(x))) should work
– digEmAll
Dec 11 at 8:34

@digEmAll indeed and much shorter thanks!
– Cath
Dec 11 at 8:35

another base option df <- lapply(df, replace, TRUE, NA)
– docendo discimus
Dec 11 at 10:05

This works in many cases, but not always. The problem is that some classes don't have methods that automatically convert the underlying typeof, and sometimes as doesn't know how to handle classes. Try it with a POSIXct: as throws an error, and manually setting the class to c("POSIXt", "POSIXct") seems to work, but does not convert the underlying NA, the result is different then as.POSIXct(NA)
– Emil Bode
Dec 11 at 17:33

It would be better to use typeof instead of class, here it works "by chance" but will fail in the general case (e.g. factors).
– Moody_Mudskipper
2 days ago

|
show 1 more comment

up vote
4
down vote

Using dplyr::na_if:

library(dplyr)



df %>% 

  mutate(x = na_if(x, x),

         y = na_if(y, y))



# # A tibble: 3 x 2

#       x y    

#   <dbl> <chr>

# 1    NA NA   

# 2    NA NA   

# 3    NA NA

If we want to mutate only subset of columns to NA, then:

# dataframe with extra column that stay unchanged

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))



df %>% 

  mutate_at(vars(x, y), funs(na_if(.,.)))



# # A tibble: 3 x 3

#       x y         z

#   <dbl> <chr> <int>

# 1    NA NA        4

# 2    NA NA        5

# 3    NA NA        6

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

add a comment |

up vote
4
down vote

Using dplyr::na_if:

library(dplyr)



df %>% 

  mutate(x = na_if(x, x),

         y = na_if(y, y))



# # A tibble: 3 x 2

#       x y    

#   <dbl> <chr>

# 1    NA NA   

# 2    NA NA   

# 3    NA NA

If we want to mutate only subset of columns to NA, then:

# dataframe with extra column that stay unchanged

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))



df %>% 

  mutate_at(vars(x, y), funs(na_if(.,.)))



# # A tibble: 3 x 3

#       x y         z

#   <dbl> <chr> <int>

# 1    NA NA        4

# 2    NA NA        5

# 3    NA NA        6

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

add a comment |

up vote
4
down vote

Using dplyr::na_if:

library(dplyr)



df %>% 

  mutate(x = na_if(x, x),

         y = na_if(y, y))



# # A tibble: 3 x 2

#       x y    

#   <dbl> <chr>

# 1    NA NA   

# 2    NA NA   

# 3    NA NA

If we want to mutate only subset of columns to NA, then:

# dataframe with extra column that stay unchanged

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))



df %>% 

  mutate_at(vars(x, y), funs(na_if(.,.)))



# # A tibble: 3 x 3

#       x y         z

#   <dbl> <chr> <int>

# 1    NA NA        4

# 2    NA NA        5

# 3    NA NA        6

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

Using dplyr::na_if:

library(dplyr)



df %>% 

  mutate(x = na_if(x, x),

         y = na_if(y, y))



# # A tibble: 3 x 2

#       x y    

#   <dbl> <chr>

# 1    NA NA   

# 2    NA NA   

# 3    NA NA

If we want to mutate only subset of columns to NA, then:

# dataframe with extra column that stay unchanged

df = data_frame(x = c(1,2,3), y = c("a", "b", "c"), z = c(4:6))



df %>% 

  mutate_at(vars(x, y), funs(na_if(.,.)))



# # A tibble: 3 x 3

#       x y         z

#   <dbl> <chr> <int>

# 1    NA NA        4

# 2    NA NA        5

# 3    NA NA        6

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

edited Dec 11 at 8:21

answered Dec 11 at 8:11

zx8754

29.1k76396

answered Dec 11 at 8:11

zx8754

29.1k76396

answered Dec 11 at 8:11

zx8754

29.1k76396

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

add a comment |

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

Or df <- mutate_all(df,~na_if(.,.)) (or modify(df,~na_if(.,.))) while you're there :)
– Moody_Mudskipper
2 days ago

@Moody_Mudskipper I am using mutate_at as OP might want to do this on subset of columns. If they want to apply this to all columns, then why not just create an empty dataframe with 0 rows...
– zx8754
2 days ago

I don't know... OP's use case is obscure, but he mentions replacing each column by NAs.
– Moody_Mudskipper
2 days ago

add a comment |

up vote
0
down vote

Using bind_cols() from dplyr you can also do:

df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))

classes <- sapply(df, class)

df[,1:2] <- NA



bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))



     V1 V2   

  <dbl> <chr>

1    NA NA   

2    NA NA   

3    NA NA

Please note that this will change the colnames.

answered Dec 11 at 7:59

alex_555

666315

add a comment |

up vote
0
down vote

Using bind_cols() from dplyr you can also do:

df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))

classes <- sapply(df, class)

df[,1:2] <- NA



bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))



     V1 V2   

  <dbl> <chr>

1    NA NA   

2    NA NA   

3    NA NA

Please note that this will change the colnames.

answered Dec 11 at 7:59

alex_555

666315

add a comment |

up vote
0
down vote

Using bind_cols() from dplyr you can also do:

df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))

classes <- sapply(df, class)

df[,1:2] <- NA



bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))



     V1 V2   

  <dbl> <chr>

1    NA NA   

2    NA NA   

3    NA NA

Please note that this will change the colnames.

answered Dec 11 at 7:59

alex_555

666315

Using bind_cols() from dplyr you can also do:

df <- data_frame(x = c(1,2,3), y = c("a", "b", "c"))

classes <- sapply(df, class)

df[,1:2] <- NA



bind_cols(lapply(colnames(x), function(x){eval(parse(text=paste0("as.", classes[names(classes[x])], "(", df[,x],")")))}))



     V1 V2   

  <dbl> <chr>

1    NA NA   

2    NA NA   

3    NA NA

Please note that this will change the colnames.

answered Dec 11 at 7:59

alex_555

666315

answered Dec 11 at 7:59

alex_555

666315

answered Dec 11 at 7:59

alex_555

666315

answered Dec 11 at 7:59

alex_555

666315

add a comment |

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