We can write a short function to return all elements of vec that match val or that are within n elements (either direction):
newfun <- function(vec, val, n) {
rows <- sapply(which(vec==val), function(x) seq(x-n, x+n, 1))
rows <- unique(sort(rows[rows>0 & rows<length(vec)]))
return(vec[rows])
}
For example:
newfun(mtcars$vs, 1, 2)
Before adding the desired range of indices to your focal index (the result from which), you need to repeat each value to the length of your range.
# set the number of values to select, before and after each focal index
n <- 1
# create a range of (relative) indices
i <- -n:n
# repeat focal indices
# add range of n prior and following indices
ix <- rep(which(mtcars$vs == 1), each = length(i)) + i
# select unique indices, truncated to the relevant range of rows,...
unique(ix[ix > 0 & ix <= nrow(mtcars)])
# [1] 2 3 4 5 6 7 8 9 10 11 12 17 18 19 20 21 22 25 26 27 28 29 31 32
# ...which then can be used to subset data
mtcars[unique(ix[ix > 0 & ix <= nrow(mtcars)]), ]
This seems to be a simple question but is not as trivial as presumably expected.
The issue is that which(mtcars$vs == 1) returns a vector rather than a single value:
[1] 3 4 6 8 9 10 11 18 19 20 21 26 28 32
If another vector -1:1 (which is c(-1L, 0L, 1L)) is added to it, the normal R rules for operations on vectors of unequal lengths apply: The recycling rule says
Any short vector operands are extended by recycling their values until they match the size of any other operands.
Therefore the shorter vector -1:1 will be recycled to the length of which(mtcars$vs == 1), i.e.,
rep(-1:1, length.out = length(which(mtcars$vs == 1)))
[1] -1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0
Therefore, the result of
which(mtcars$vs == 1) + -1:1
is the element-wise sum of the elements of both vectors where the shorter vector has been recycled to match the length of the longer vector.
[1] 2 4 7 7 9 11 10 18 20 19 21 27 27 32
which is propably not what the OP has expected.
In addition, we get the
Warning message:
In which(mtcars$vs == 1) + -1:1 :
longer object length is not a multiple of shorter object length
because which(mtcars$vs == 1) has length 14 and -1:1 has length 3.
outer()In order to select the N rows above and below each matching row, we need to add -N:N to each row number returned by which(mtcars$vs == 1):
outer(which(mtcars$vs == 1), -1:1, `+`)
[,1] [,2] [,3]
[1,] 2 3 4
[2,] 3 4 5
[3,] 5 6 7
[4,] 7 8 9
[5,] 8 9 10
[6,] 9 10 11
[7,] 10 11 12
[8,] 17 18 19
[9,] 18 19 20
[10,] 19 20 21
[11,] 20 21 22
[12,] 25 26 27
[13,] 27 28 29
[14,] 31 32 33
Now, we have an array of all row numbers. Unfortunately, it cannot be used directly for subsetting because it contains duplicates and there are row numbers which do not exist in mtcars. So the the result has to be "post-processed" before it can be used for subsetting.
library(magrittr) # piping used for clarity
rn <- outer(which(mtcars$vs == 1), -1:1, `+`) %>%
as.vector() %>%
unique() %>%
Filter(function(x) x[1 <= x & x <= nrow(mtcars)], .)
rn
[1] 2 3 4 5 6 7 8 9 10 11 12 17 18 19 20 21 22 25 26 27 28 29 31 32
mtcars[rn, ]
mpg cyl disp hp drat wt qsec vs am gear carb Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
mtcars[which(mtcars$vs == 1)[1] + c(-1:1), ]. Becausewhichwill return a series of matches (not the first one). By adding[1]you keep only the first match. You can also usemtcars[min(which(mtcars$vs == 1)) + c(-1:1), ]if you prefer. Similar logic... You can also input the number of rows you want like thisN = 1; mtcars[min(which(mtcars$vs == 1)) + c(-N:N), ]- AntoniosKmtcars[min(which(mtcars$vs == 1)) + c(-1:1), ]provided by @AntoniosK , you can also usemtcars[first(which(mtcars$vs == 1)) + c(-1:1), ]. - tmfmnkdplyrforfirst, right? - AntoniosKn <- 1i <- head(tail(filter(c(rep(FALSE, n), mtcars$vs == 1, rep(FALSE, n)), rep(1, 1+n*2)), -n), -n) > 0mtcars[i,]- GKi