Difference between dplyr:mutate and transform when using pmin and pmax?

Question

While trying to answer this question, I encountered a difference between mutate and transform in what I expected to be equivalent operations.

# data
x <- data.frame(a=c(rep(0,10),rep(1,10),3),b=c(1:10,0,11:19,0))

#transform
transform(x,a=pmin(a,b), b=pmax(a,b))
   a  b
1  0  1
2  0  2
3  0  3
4  0  4
5  0  5
6  0  6
7  0  7
8  0  8
9  0  9
10 0 10
11 0  1
12 1 11
13 1 12
14 1 13
15 1 14
16 1 15
17 1 16
18 1 17
19 1 18
20 1 19
21 0  3

#mutate
libarary(dplyr)
x %>% mutate(a=pmin(a,b), b=pmax(a,b))
   a  b
1  0  1
2  0  2
3  0  3
4  0  4
5  0  5
6  0  6
7  0  7
8  0  8
9  0  9
10 0 10
11 0  0
12 1 11
13 1 12
14 1 13
15 1 14
16 1 15
17 1 16
18 1 17
19 1 18
20 1 19
21 0  0

Note the differences in lines 11 and 21. I suspect that mutate is mutating the data as it goes and therefore, pmax is not seeing the original data. Is this correct? Is it a bug, or by design?

Answer 1

It appears my suspicions are correct, and that it is by design to allow the use of computed variables immediately afterwards, eg:

data.frame(a=1:4,b=5:8) %>% mutate(sum=a+b, letter=letters[sum])
  a b sum letter
1 1 5   6      f
2 2 6   8      h
3 3 7  10      j
4 4 8  12      l

In order to replicate the expected behaviour from transform one needs to simply reference the variable directly:

x %>% mutate(a=pmin(x$a,x$b), b=pmax(x$a,x$b))
   a  b
1  0  1
2  0  2
3  0  3
4  0  4
5  0  5
6  0  6
7  0  7
8  0  8
9  0  9
10 0 10
11 0  1
12 1 11
13 1 12
14 1 13
15 1 14
16 1 15
17 1 16
18 1 17
19 1 18
20 1 19
21 0  3