Using multiple criteria in subset function and logical operators

For your example, I believe the following should work:

myNewDataFrame <- subset(bigfive, subset = bf11 == 1 | bf11 == 2 | bf11 == 3)

See the examples in ?subset for more. Just to demonstrate, a more complicated logical subset would be:

data(airquality)
dat <- subset(airquality, subset = (Temp > 80 & Month > 5) | Ozone < 40)

And as Chase points out, %in% would be more efficient in your example:

myNewDataFrame <- subset(bigfive, subset = bf11 %in% c(1, 2, 3))

As Chase also points out, make sure you understand the difference between | and ||. To see help pages for operators, use ?'||', where the operator is quoted.

The correct operator is %in% here. Here is an example with dummy data:

set.seed(1)
dat <- data.frame(bf11 = sample(4, 10, replace = TRUE),
                  foo = runif(10))

giving:

> head(dat)
  bf11       foo
1    2 0.2059746
2    2 0.1765568
3    3 0.6870228
4    4 0.3841037
5    1 0.7698414
6    4 0.4976992

The subset of dat where bf11 equals any of the set 1,2,3 is taken as follows using %in%:

> subset(dat, subset = bf11 %in% c(1,2,3))
   bf11       foo
1     2 0.2059746
2     2 0.1765568
3     3 0.6870228
5     1 0.7698414
8     3 0.9919061
9     3 0.3800352
10    1 0.7774452

As to why your original didn't work, break it down to see the problem. Look at what 1||2||3 evaluates to:

> 1 || 2 || 3
[1] TRUE

and you'd get the same using | instead. As a result, the subset() call would only return rows where bf11 was TRUE (or something that evaluated to TRUE).

What you could have written would have been something like:

subset(dat, subset = bf11 == 1 | bf11 == 2 | bf11 == 3)

Which gives the same result as my earlier subset() call. The point is that you need a series of single comparisons, not a comparison of a series of options. But as you can see, %in% is far more useful and less verbose in such circumstances. Notice also that I have to use | as I want to compare each element of bf11 against 1, 2, and 3, in turn. Compare:

> with(dat, bf11 == 1 || bf11 == 2)
[1] TRUE
> with(dat, bf11 == 1 | bf11 == 2)
 [1]  TRUE  TRUE FALSE FALSE  TRUE FALSE FALSE FALSE FALSE  TRUE