A pdf of this example is available here.

**SAS**

In SAS, we use the

`lag`function (section 1.4.17, p. 22) to retrieve the last value.

data fibo;

do i = 1 to 10;

fib = sum(fib, lag(fib));

if i eq 1 then fib = 1;

output;

end;

run;

proc print data=fibo;

run;

This generates the following output:

Obs i fib

1 1 1

2 2 1

3 3 2

4 4 3

5 5 5

6 6 8

7 7 13

8 8 21

9 9 34

10 10 55

**R**

In R we can loop over an array to perform the same job.

len = 10

fibvals = numeric(len)

fibvals[1] = 1

fibvals[2] = 1

for (i in 3:len) {

fibvals[i] = fibvals[i-1] + fibvals[i-2]

}

This generates the following output:

> fibvals

[1] 1 1 2 3 5 8 13 21 34 55

## 5 comments:

You, sir, have just saved me 2 hours of merrily flopping about in the wondrous world of R, not having a single clue what was going on. So, thanks.

1. Explain what each part of the code means or does. [The part: for (i in 3:len) { fibvals[i] = fibvals[I 1] + fibvals[i-2]} is called a loop and helps in generation of outputs for the recurrence equations]

2. Why is there a '3' in the loop?

Not sure what you mean to ask in #1. Everything before the for{} loop is just setting thing up.

The 3 in 3:len just means do this loop starting with the third item in the vector. So the first time we execute the loop, i = 3, and we sum fibvals[1] and fibvals[2].

Said another way, we initialize the sequence with the two 1s, and the remainder of the sequence we fill in with the loop.

this is the function for fibonacci in r.

fibo<-function(x) {

len = x

x = numeric(len)

x[1] = 1

x[2] = 1

for (i in 3:len) {

x[i] = x[i-1] + x[i-2]

}

return(max(x))

}

Would you please explain why you coded different solutions in SAS and in R? If you use array in R, please use array also in SAS, so the solutions are more comparable.

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