sapply() with with vector of values instead of 1:n

Question

I would like to have an sapply() statement of the following form:

a <- c(4, 9, 20, 3, 10, 30)
sapply(c(a), function(x) b[x,2] - b[x,1])

Normally instead of c(a) I would have 1:x, however now I would like to go through only the values specified in a. Is this possible?

Update: The point is not what goes after function(x) - that is only intended as an example.

and what is wrong with what you have? It works perfectly, as intended. — flodel, Oct 16 '13 at 11:20
`sapply( a , print )` for example. You will see that it works, exactly as you expect. What is the problem exactly? — Simon O'Hanlon, Oct 16 '13 at 11:21
the question is not about how to solve this toy subtraction problem, the question is how to use sapply to iterate through a vector of arbitrary numbers instead of a 1:n sequence — upabove, Oct 16 '13 at 11:21
I agree with @flodel. `a` works, unless your `a` is a _position_ of certain numbers in another vector (say `1:n`). If that's the case, I see how we were thrown off of our game. — Roman Luštrik, Oct 16 '13 at 11:26

score 2 · Accepted Answer · edited Oct 16 '13 at 11:34

According to the documentation (see ?sapply), the first argument (X) to sapply can be a vector. There is no restriction on the vector saying it has to be of the form 1:x. So your current code should loop correctly over the values of a:

sapply(c(a), function(x) b[x,2]-b[x,1])

Note, that you don't need c(a) - just a is fine:

sapply(a, function(x) b[x,2]-b[x,1])

If for some reason you did not communicate to us this does not work, it is not a problem with sapply but with the function you are passing to it.

score 2 · Answer 2 · answered Oct 16 '13 at 11:01

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It isn't very clear what you are trying to achieve, but I don't think that you need sapply.

a <- c(4, 9, 20, 3, 10, 30)
b <- data.frame(x = 1:30, y = sqrt(1:30))
b[a, 2] - b[a, 1]
## [1]  -2.000000  -6.000000 -15.527864  -1.267949  -6.837722 -24.522774

answered Oct 16 '13 at 11:01

Richie Cotton

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all I want is sapply to use as 'x' the values specified in 'a' instead of iterating through a 1:n interval – upabove Oct 16 '13 at 11:02

Jonas Tundo · Answer 3 · 2013-10-16T11:13:16.420

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Do you want to perform the function only on the values specified in a but keep the original values otherwise? Because you don't provide b I show another example here.

v <- 1:40
a<-c(4,9,20,3,10,30)
sapply(v, function(x) if(x %in% a){x+1} else {x})

edited Oct 16 '13 at 11:13

answered Oct 16 '13 at 11:07

Jonas Tundo

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Yeah, I think you are right (that they are after values and not indices). However, they coincide in your example...`identical(v,seq_along(v))` – Frank Oct 16 '13 at 15:53
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Yes, but that wouldn't matter for the function. You can replace `v` with another vector. – Jonas Tundo Oct 17 '13 at 06:22

sapply() with with vector of values instead of 1:n

3 Answers3