108

I need a sequence of repeated numbers, i.e. 1 1 ... 1 2 2 ... 2 3 3 ... 3 etc. The way I implemented this was:

  nyear <- 20
  names <- c(rep(1,nyear),rep(2,nyear),rep(3,nyear),rep(4,nyear),
             rep(5,nyear),rep(6,nyear),rep(7,nyear),rep(8,nyear))

which works, but is clumsy, and obviously doesn't scale well.

How do I repeat the N integers M times each in sequence?

  • I tried nesting seq() and rep() but that didn't quite do what I wanted.
  • I can obviously write a for-loop to do this, but there should be an intrinsic way to do this!
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3 Answers 3

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197

You missed the each= argument to rep():

R> n <- 3
R> rep(1:5, each=n)
 [1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
R>

so your example can be done with a simple

R> rep(1:8, each=20)
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5

Another base R option could be gl():

gl(5, 3)

Where the output is a factor:

 [1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
Levels: 1 2 3 4 5

If integers are needed, you can convert it:

as.numeric(gl(5, 3))

 [1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
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2

For your example, Dirk's answer is perfect. If you instead had a data frame and wanted to add that sort of sequence as a column, you could also use group from groupdata2 (disclaimer: my package) to greedily divide the datapoints into groups. 

# Attach groupdata2
library(groupdata2)
# Create a random data frame
df <- data.frame("x" = rnorm(27))
# Create groups with 5 members each (except last group)
group(df, n = 5, method = "greedy")
         x .groups
     <dbl> <fct>  
 1  0.891  1      
 2 -1.13   1      
 3 -0.500  1      
 4 -1.12   1      
 5 -0.0187 1      
 6  0.420  2      
 7 -0.449  2      
 8  0.365  2      
 9  0.526  2      
10  0.466  2      
# … with 17 more rows

There's a whole range of methods for creating this kind of grouping factor. E.g. by number of groups, a list of group sizes, or by having groups start when the value in some column differs from the value in the previous row (e.g. if a column is c("x","x","y","z","z") the grouping factor would be c(1,1,2,3,3).

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