Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How to get next number in sequence in R

I need to automate the process of getting the next number(s) in the given sequence.

Can we make a function which takes two inputs

  1. a vector of numbers(3,7,13,21 e.g.)
  2. how many next numbers

    seqNext <- function(sequ, next) {
    ..
    }
    
    seqNext( c(3,7,13,21), 3) 
    # 31 43 57
    seqNext( c(37,26,17,10), 1)
    # 5 
    
like image 380
Navin Manaswi Avatar asked Jan 07 '23 03:01

Navin Manaswi


2 Answers

By the power of maths!

x1 <- c(3,7,13,21)
dat <- data.frame(x=seq_along(x1), y=x1)

predict(lm(y ~ poly(x, 2), data=dat), newdata=list(x=5:15))
#  1   2   3   4   5   6   7   8   9  10  11 
# 31  43  57  73  91 111 133 157 183 211 241 

When dealing with successive differences that change their sign, the pattern of output values ends up switching from decreasing to increasing:

x2 <- c(37,26,17,10)

dat <- data.frame(x=seq_along(x2), y=x2)
predict(lm(y ~ poly(x,2), data=dat), newdata=list(x=1:10))

# 1      2      3      4      5      6      7      8      9     10 
#37     26     17     10      5      2      1      2      5     10
   -(11)   -(9)   -(7)    -(5)   -(3)   -(1)  -(-1)  -(-3) -(-5)
        -2     -2      -2     -2     -2    -2     -2     -2 

As a function:

seqNext <- function(x,n) {
  L <- length(x)
  dat <- data.frame(x=seq_along(x), y=x)
  unname(
    predict(lm(y ~ poly(x, 2), data=dat), newdata=list(x=seq(L+1,L+n)))
  )
}

seqNext(x1,5)
#[1] 31 43 57 73 91
seqNext(x2,5)
#[1] 5 2 1 2 5

This is also easily extensible to circumstances where the pattern might be n orders deep, e.g.:

x3 <- c(100, 75, 45, 5, -50)
diff(x3)
#[1] -25 -30 -40 -55
diff(diff(x3))
#[1]  -5 -10 -15
diff(diff(diff(x3)))
#[1] -5 -5

seqNext <- function(x,n,degree=2) {
  L <- length(x)
  dat <- data.frame(x=seq_along(x), y=x)
  unname(
    predict(lm(y ~ poly(x, degree), data=dat), newdata=list(x=seq(L+1,L+n)))
  )
}

seqNext(x3,n=5,deg=3)
#[1] -125 -225 -355 -520 -725
like image 175
thelatemail Avatar answered Jan 09 '23 18:01

thelatemail


seqNext <- function(x, n) {
  k <- length(x);  d <- diff(x[(k - 2):k])
  x[k] + 1:n * d[2] + cumsum(1:n) * diff(d[1:2])
}

seqNext(c(3,7,13,21),3) 
# [1] 31 43 57
seqNext(c(37,26,17,10),1)
# [1] 5
seqNext(c(137,126,117,110),10)
# [1] 105 102 101 102 105 110 117 126 137 150
seqNext(c(105,110,113,114),5)
# [1] 113 110 105  98  89
like image 26
Julius Vainora Avatar answered Jan 09 '23 18:01

Julius Vainora