I am trying to get my head around daa.R, one of the functions in the matchingMarkets R library (links are to GitHub repositories). On lines 134-135, one finds the following if
statement
if (0 %in% (c.hist[[j]] & any(c.prefs[ ,j]==proposers[k]))){ # if no history and proposer is on preference list
c.hist[[j]][c.hist[[j]]==0][1] <- proposers[k] # then accept
}
where c.hist
and proposers
are a list and c.prefs
a matrix.
I am puzzled by the parentheses in the conditional statement. Instead of the above synthax, I would have opted for
if (0 %in% c.hist[[j]] & any(c.prefs[ ,j]==proposers[k]))
I don't understand how the original condition may work. How could R possibly check whether 0 is in (c.hist[[j]] & any(c.prefs[ ,j]==proposers[k]))
?
I am a beginner in R, so I wanted to make sure I was not missing something and tried to replicate a similar synthax with other conditions such as,
> x = list(4,3)
> y = list(5,2)
> if (3 %in% (x & any(y == 5))){z = 8}
As I expected, I got an error message
Error in x & any(y == 5) : operations are possible only for numeric, logical or complex types
whereas things go just fine when I write
if (3 %in% x & any(y == 5)){z = 8}
instead.
What am I missing? Why would the kind of conditional synthax I am puzzled by work in daa.R and not with the other conditions I tried?
When you ask R if 0 %in% x
where x
is a logical vector, R will first convert x
to a numeric vector where FALSE
becomes 0
and TRUE
becomes 1
. So essentially, asking if 0 %in% x
is like asking if x
contains any FALSE
. This is arguably pretty bad practice. A better approach would be to test if any(!x)
or !all(x)
. Worse, if x
has length 1
as it seems to be the case here, you would just test if !x
.
In light of the contorted usage, you are raising a very good question: is the code doing what it really meant to do? In R, the %in%
operator has higher precedence than &
(see ?Syntax
), thus these two statements are not the same:
0 %in% (c.hist[[j]]) & any(c.prefs[ ,j]==proposers[k])) # original code
0 %in% c.hist[[j]] & any(c.prefs[ ,j]==proposers[k]) # what you suggested
and we would need to look closely at what the code is supposed to be doing to decide if it is correct or wrong. I will just point out that you did not test your assumption properly: the error you got ("unexpected '{'") is because you forgot a closing parenthesis:
if (3 %in% (x & any(y == 5)){z = 8}
should be
if (3 %in% (x & any(y == 5))){z = 8}