How do you create vectors with specific intervals in R?

46
votes

I have a question about creating vectors. If I do a <- 1:10, "a" has the values 1,2,3,4,5,6,7,8,9,10.

My question is how do you create a vector with specific intervals between its elements. For example, I would like to create a vector that has the values from 1 to 100 but only count in intervals of 5 so that I get a vector that has the values 5,10,15,20,...,95,100

I think that in Matlab we can do 1:5:100, how do we do this using R?

I could try doing 5*(1:20) but is there a shorter way? (since in this case I would need to know the whole length (100) and then divide by the size of the interval (5) to get the 20)

3
rvectorintervals
I know it is sometimes hard to come up with good search terms, but if I google "r sequence", the second hit is the help page of seq. The function is probably also part of most basic introductions to R. - Roland
Yea I tried several things and couldn't find it.. didn't occur to me to write "sequence".. kept thinking about intervals - Luli

3 Answers

95
votes

In R the equivalent function is seq and you can use it with the option by:

seq(from = 5, to = 100, by = 5)
# [1]   5  10  15  20  25  30  35  40  45  50  55  60  65  70  75  80  85  90  95 100

In addition to by you can also have other options such as length.out and along.with.

length.out: If you want to get a total of 10 numbers between 0 and 1, for example:

seq(0, 1, length.out = 10)
# gives 10 equally spaced numbers from 0 to 1

along.with: It takes the length of the vector you supply as input and provides a vector from 1:length(input).

seq(along.with=c(10,20,30))
# [1] 1 2 3

Although, instead of using the along.with option, it is recommended to use seq_along in this case. From the documentation for ?seq

seq is generic, and only the default method is described here. Note that it dispatches on the class of the first argument irrespective of argument names. This can have unintended consequences if it is called with just one argument intending this to be taken as along.with: it is much better to use seq_along in that case.

seq_along: Instead of seq(along.with(.))

seq_along(c(10,20,30))
# [1] 1 2 3

Hope this helps.

4
votes

Use the code

x = seq(0,100,5) #this means (starting number, ending number, interval)

the output will be 

[1]   0   5  10  15  20  25  30  35  40  45  50  55  60  65  70  75
[17]  80  85  90  95 100
0
votes

Usually, we want to divide our vector into a number of intervals. In this case, you can use a function where (a) is a vector and (b) is the number of intervals. (Let's suppose you want 4 intervals)

a <- 1:10
b <- 4

FunctionIntervalM <- function(a,b) {
  seq(from=min(a), to = max(a), by = (max(a)-min(a))/b)
}

FunctionIntervalM(a,b)
# 1.00  3.25  5.50  7.75 10.00

Therefore you have 4 intervals:

1.00 - 3.25 
3.25 - 5.50
5.50 - 7.75
7.75 - 10.00

You can also use a cut function 

  cut(a, 4)

# (0.991,3.25] (0.991,3.25] (0.991,3.25] (3.25,5.5]   (3.25,5.5]   (5.5,7.75]  
# (5.5,7.75]   (7.75,10]    (7.75,10]    (7.75,10]   
#Levels: (0.991,3.25] (3.25,5.5] (5.5,7.75] (7.75,10]