remove the last element of a vector

问题:

I need to remove the last number in a groups of vectors, i.e.:

v <- 1:3
v1 <- 4:8

should become:

v <- 1:2
v1 <- 4:7

回答1:

You can use negative offsets in head (or tail), so head(x, -1) removes the last element:

R> head( 1:4, -1)
[1] 1 2 3
R> 

This also saves an additional call to length().

Edit: As pointed out by Jason, this approach is actually not faster. Can't argue with empirics. On my machine:

R> x <- rnorm(1000)
R> microbenchmark( y <- head(x, -1), y <- x[-length(x)], times=10000)
Unit: microseconds
                expr    min      lq median     uq     max
1   y <- head(x, -1) 29.412 31.0385 31.713 32.578 872.168
2 y <- x[-length(x)] 14.703 15.1150 15.565 15.955 706.880
R> 

回答2:

Use length to get the length of the object and - to remove the last one.

v[-length(v)]

A negative index in R extracts everything but the given indices.

回答3:

Dirk and Iselzer have already provided the answers. Dirk's is certainly the most straightforward, but on my system at least it is marginally slower, probably because vector subsetting with [ and length checking is cheap (and according to the source, head does use length, twice actually):

> x <- rnorm(1000)
> system.time(replicate(50000, y <- head(x, -1)))
   user  system elapsed 
   3.69    0.56    4.25 
> system.time(replicate(50000, y <- x[-length(x)]))
   user  system elapsed 
  3.504   0.552   4.058

This pattern held up for larger vector lengths and more replications. YMMV. The legibility of head certainly out-weights the marginal performance improvement of [ in most cases.

回答4:

This is another option, which has not been suggested before. NROW treats your vector as a 1-column matrix.

v[-max(NROW(v))]#1 2
v1[-max(NROW(v1))]#4 5 6 7

Based on the discussion above, this is (slightly) faster then all the other methods suggested:

x <- rnorm(1000)
system.time(replicate(50000, y <- head(x, -1)))
user  system elapsed 
3.446   0.292   3.762
system.time(replicate(50000, y <- x[-length(x)]))
user  system elapsed 
2.131   0.326   2.472
system.time(replicate(50000, y <- x[-max(NROW(x))]))
user  system elapsed 
2.076   0.262   2.342