CODEWITHSUNDEEP
r
user008
user008

Reputation: 23

return the values in a vector after passing in a function

Basically, I have

x<-rnorm(5).

I write a vector which takes integers 1 <= n <= 5 then returns in a vector the result from the series (1/n) * sum_{i=1}^n (1/x_i)

so 

 n=1 -> 1   * 1/x_1
 n=2 -> (1/2) * ( 1/(x_1 + x_2) )
 n=3 -> (1/3) * ( 1/(x_1 + x_2 + x_3) )
 n=4 -> (1/4) * ( 1/(x_1 + x_2 + x_3 + x_4) )

I wrote this function:

 series <-  function(n){
       n=seq(1,5,1)
       x<-rnorm(length(n))
       print(x)
  return  (  (1/n)* (1/sum(x[1:length(x[n])]))   )
}

But the result is not true, for example

 > series(5)
 [1]  1.17810059  0.85472777 -0.55077392 -0.03856963 -0.19404827
 [1] 0.8003608 0.4001804 0.2667869 0.2000902 0.1600722

for n=2 -> 1/2 * 1/x_1 + 1/x_2 = (1/2) * (1/(1.17810059+ 0.85472777)) but unfortunately, the result according to my code is 0.4001804!

P.S: I want to write the code without loops and without any function needs calling a library! just to define a simple function using the basic known functions in R and then I can save the result, if needed using Vectorize() or outer()

Upvotes: 2

Views: 67

Answers (2)

tmfmnk
tmfmnk

Reputation: 39858

Building on the basic idea from @MrFlick, you can also do:

1/seq_along(x) * 1/cumsum(x)

[1] -1.7841988 -0.6323886  0.4339966  0.2981289  0.2066433

Upvotes: 1

MrFlick
MrFlick

Reputation: 206167

The sum() function is not vectorized. It collapsing everything down to a single value. instead, you can use cumsum() to get the cumulative some of all the values in the vector thus far.

series <-  function(n){
  n <- seq(1,5,1)
  x <- rnorm(length(n))
  print(x)
  return((1/n)* (1/cumsum(x)))
}

Upvotes: 1

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