When trying to replace values, “missing values are not allowed in subscripted assignments of data frames”

Question

I have a table that has two columns: whether you were sick (H01) and the number of days sick (H03). However, the number of days sick is NA if H01 == false, and I would like to s

Answer 1

Following works. Watch out there is no comma in sub setting:

x <- data.frame(a=c(NA,2:5), b=c(1:5))

x$a[x$a==2] <- 99

Answer 2

I realise the question is very old, but I think the most elegant solution is by using the which() function:

 pe94.person[which(pe94.person$H01 == 12),]$H03 <- 0

should do what the original poster asked for. Because which() drops the NAs and keeps the (positions of the) TRUE results only.

Answer 3

There might be an NA somewhere in the column that is causing the error. Run the index on a specific column instead of the entire data frame.

movies[movies$Actors == "N/A",] = NA #ERROR
movies$Actors[movies$Actors == "N/A"] = NA #Works

Answer 4

Simply use the subset() function to exclude all NA from the string.

It works as x[subset & !is.na(subset)]. Look at this data:

> x <- data.frame(a = c(T,F,T,F,NA,F,T, F, NA,NA,T,T,F),
>                 b = c(F,T,T,F,T, T,NA,NA,F, T, T,F,F))

Subsetting with [ operator returns this:

> x[x$b == T & x$a == F, ]

         a    b
2    FALSE TRUE
NA      NA   NA
6    FALSE TRUE
NA.1    NA   NA
NA.2    NA   NA

And subset() does what we want:

> subset(x, b == T & a == F)

      a    b
2 FALSE TRUE
6 FALSE TRUE

To change the values of subsetted variables:

> ss <- subset(x, b == T & a == F)
> x[rownames(ss), 'a'] <- T

> x[c(2,6), ]

     a    b
2 TRUE TRUE
6 TRUE TRUE

Answer 5

It is due to missingness in H01 variable.

> x <- data.frame(a=c(NA,2:5), b=c(1:5))
> x
   a b
1 NA 1
2  2 2
3  3 3
4  4 4
5  5 5
> x[x$a==2,]$b <- 99
Error in `[<-.data.frame`(`*tmp*`, x$a == 1, , value = list(a = NA_integer_,  : 
  missing values are not allowed in subscripted assignments of data frames

The assignment won't work because x$a has a missing value.

Subsetting first works:

> z <- x[x$a==2,]
> z$b <- 99
> z <- x[x$a==2,]
> z
    a  b
NA NA NA
2   2  2

But that's because the [<- function apparently can't handle missing values in its extraction indices, even though [ can:

> `[<-`(x,x$a==2,,99)
Error in `[<-.data.frame`(x, x$a == 2, , 99) : 
  missing values are not allowed in subscripted assignments of data frames

So instead, trying specifying your !is.na(x$a) part when you're doing the assignment:

> `[<-`(x,!is.na(x$a) & x$a==2,'b',99)
   a  b
1 NA  1
2  2 99
3  3  3
4  4  4
5  5  5

Or, more commonly:

> x[!is.na(x$a) & x$a==2,]$b <- 99
> x
   a  b
1 NA  1
2  2 99
3  3  3
4  4  4
5  5  5

Note that this behavior is described in the documentation:

The replacement methods can be used to add whole column(s) by specifying non-existent column(s), in which case the column(s) are added at the right-hand edge of the data frame and numerical indices must be contiguous to existing indices. On the other hand, rows can be added at any row after the current last row, and the columns will be in-filled with missing values. Missing values in the indices are not allowed for replacement.

Answer 6

You can use ifelse, like so

pe94.person$foo <- ifelse(!is.na(pe94.person$H01) & pe94.person$H01 == 12, 0, pe94.person$H03)

check if foo meets your criteria and then go ahead and assign it to pe94.person$H03 directly. I find it safer to assign it a new variable and usually use that in subsequent analysis.