CODEWITHSUNDEEP
rvector
arrrrRgh
arrrrRgh

Reputation: 307

Create a vector of sequences with varying "by" arguments

I'm interested in creating a vector of sequences, all from 0 to 1, but with varying "by" arguments to seq( ).

I start with a vector I'm hoping to use for the "by" argument:

x <- c(3,6,3,9,10)

I'm wanting this to be the input to the seq( ) function:

seq(0, 1, by = 1/(x-1))

So for the first few elements of the vector it would be:

seq(0, 1, by = 1 / (3 - 1))
seq(0, 1, by = 1 / (6 -1))

Yielding vectors like:

0.0 0.5 1.0
0.0 0.2 0.4 0.6 0.8 1.0

Ultimately, I'd like the output to be a long vector that includes all of these sequences of varying lengths. Maybe a job for a loop? Or sapply / lapply?

Upvotes: 0

Views: 147

Answers (3)

rawr
rawr

Reputation: 20811

Another way: I use seq.default since you need to vectorize the by argument

x <- c(3,6,3,9,10)

vseq <- Vectorize(seq.default)

vseq(0, 1, by = 1/(x-1))

# [[1]]
# [1] 0.0 0.5 1.0
# 
# [[2]]
# [1] 0.0 0.2 0.4 0.6 0.8 1.0
# 
# [[3]]
# [1] 0.0 0.5 1.0
# 
# [[4]]
# [1] 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
# 
# [[5]]
# [1] 0.0000000 0.1111111 0.2222222 0.3333333 0.4444444 0.5555556 0.6666667 0.7777778
# [9] 0.8888889 1.0000000

Upvotes: 2

maloneypatr
maloneypatr

Reputation: 3622

Here's what the code would be for an example x = 1000

vector_x <- c()
for(i in 1:1000){

  x <- seq(0, 1, by = 1/(i-1))
  vector_x <- c(vector_x, x)

}

vector_x

Upvotes: 0

Matthew Plourde
Matthew Plourde

Reputation: 44614

sapply(1/(x-1), seq, from=0, to=1)

[[1]]
[1] 0.0 0.5 1.0

[[2]]
[1] 0.0 0.2 0.4 0.6 0.8 1.0

[[3]]
[1] 0.0 0.5 1.0

[[4]]
[1] 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000

[[5]]
 [1] 0.0000 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000

Upvotes: 4

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