R conditional replace with '[' and without subset()

Question

I want to conditionally replace some values in a data.frame.

Suppose I have:

a <- c(1, 4, 5, 7, 9, 8, 3, 90)
b <- c(21, 24, 25, NA, 9, 23, NA, 3)
c <- c(214, 5, NA, NA, 59, NA, 32, 12)
d <- rep(0, 8)
test.df <- data.frame(a, b, c, d)
test.df
   a  b   c d
1  1 21 214 0
2  4 24   5 0
3  5 25  NA 0
4  7 NA  NA 0
5  9  9  59 0
6  8 23  NA 0
7  3 NA  32 0
8 90  3  12 0

My first question is why the below commands do not return the same? Why the second one returns lines with NAs? What is my mistake for the second?

subset(test.df, test.df$a >=4 & !is.na(test.df$b) & test.df$c > 4)
   a  b  c d
2  4 24  5 0
5  9  9 59 0
8 90  3 12 0

test.df[test.df$a >=4 & !is.na(test.df$b) & test.df$c > 4, ]
      a  b  c  d
2     4 24  5  0
NA   NA NA NA NA
5     9  9 59  0
NA.1 NA NA NA NA
8    90  3 12  0

My second question is, based on the above criteria, how do I replace column's d values with 10 in order to get:

 test.df
   a  b   c  d
1  1 21 214  0
2  4 24   5 10
3  5 25  NA  0
4  7 NA  NA  0
5  9  9  59 10
6  8 23  NA  0
7  3 NA  32  0
8 90  3  12 10

?

Thanks!

Answer 1

1) Your criterion test.df$a >=4 & !is.na(test.df$b) & test.df$c > 4 evals to:

[1] FALSE  TRUE    NA FALSE  TRUE    NA FALSE  TRUE

As documented, subset will filters out the rows (3 and 6) where the criterion evals to NA. On the other hand, [ gives you a row of NAs for these as it is unsure if they should be included (TRUE) or excluded (FALSE).

2) I would use transform and an improved criterion:

test.df <- transform(test.df, d = ifelse(!is.na(a) &
                                         !is.na(b) &
                                         !is.na(c) &
                                         a >= 4    &
                                         c >  4, 10, d))

Answer 2

If you're interested in data.table, then this might interest you:

require(data.table) ## 1.9.2
setDT(test.df)[a >= 4 & !is.na(b) & c > 4, d := 10]
#     a  b   c  d
# 1:  1 21 214  0
# 2:  4 24   5 10
# 3:  5 25  NA  0
# 4:  7 NA  NA  0
# 5:  9  9  59 10
# 6:  8 23  NA  0
# 7:  3 NA  32  0
# 8: 90  3  12 10

setDT converts the data.frame to a data.table by reference. The condition is then evaluated and just those rows for column d where it evaluates to TRUE are replaced in-place to 10.