Two equivalent commands give different returns in r

Question

Why do the following two commands not return the same output?

x <- sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5))
  sum(x==1)

sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5)) %>%
  sum(.==1)

The first of the 2 command always gives me the right answer (something around 25) and the 2nd command returns a number that is way too high like 52. What did I understand wrong about the pipe operator here?

score 9 · Accepted Answer · answered Nov 29 '21 at 17:21

9

Just wrap it with {}

sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5)) %>%
   {sum(.==1)}

The issue is that .==1 is considered as a second argument. It can be matched if we do

sum(x, x == 1)

As we are doing sample, make sure to specify the set.seed as well

set.seed(24)
x <- sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5))
  sum(x==1)

set.seed(24)
sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5)) %>%
   {sum(.==1)}

answered Nov 29 '21 at 17:21

akrun

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Thank you it works with the extra brackets. I am aware of the seeds but I tried it more often and therefore knew it wasn't just a coincidence but I coded something wrong. – Dave Twickenham Nov 29 '21 at 17:33
3

@DaveTwickenham i understand, that part i added just for other users in case they want to replicate – akrun Nov 29 '21 at 17:34

score 4 · Answer 2 · answered Nov 29 '21 at 20:49

4

you can also use a lambda function to avoid the mistake:

set.seed(24)
sample(0:1, 50, replace = TRUE, prob = c(0.5, 0.5)) |>
  (\(x) sum(x))()

answered Nov 29 '21 at 20:49

denis

5,580
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rg255 · Answer 3 · 2021-11-30T09:29:36.323

Another approach could be to move your logical operation to the sample() function

sample(c(0:1) == TRUE, 
       size = 50, 
       replace = TRUE, 
       prob = c(0.5, 0.5)) %>%
  sum

You do ask why — the sum function will sum all the values it receives, not just the first vector. When you are piping, it takes the output of the last step to the first argument in sum(...) so by also providing x==1 you are providing two vectors and essentially saying sum(x, x==1). You can see this with sum(c(1,1,1), c(1,1,1)) for example (which returns six, summing the all values in both vectors), or in your code:

set.seed(1)
x <- sample(0:1, 
            size = 50, 
            replace = TRUE, 
            prob = c(0.5, 0.5))
sum(x, x==1)

set.seed(1)
sample(0:1, 
       size = 50, 
       replace = TRUE, 
       prob = c(0.5, 0.5)) %>%
  sum(x==1)

Further, if x <- c(1, 0, 1, 0, 1, 1), then sum(x, x==1) or x %>% sum(x==1) is effectively

sum(c(1, 0, 1, 0, 1, 1), as.numeric(c(TRUE, FALSE, TRUE, FALSE, TRUE, TRUE)))

Two equivalent commands give different returns in r

3 Answers3