I try to return the Y array from the second loop. I expect it to be (n - [k + 2]) length but I get output of n length array
a1q1 <- function (n , j , k , m, X0) {
X <- numeric(length=n)
Xn <- X0
for (i in 1:n) {
Xn <- (103515245*Xn + 12345) %% 2^32
X[i] <- Xn
}
Y <- numeric(length=n - k - 1)
for (i in k + 2:n) {
Y[i - k - 1] <- (X[i - j] + X[i - k]) %% m
}
return (Y)
}
Z <- a1q1(100, 5, 90, 11, 0)
Z
# [1] 6 5 8 3 1 10 1 1 9 6 2 2 5 7 NA NA NA NA NA NA NA NA NA NA NA NA
# [27] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
# [53] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
# [79] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
This often catches me too. You need to put parentheses around k+2
, in the second for
loop.. Try this minor adjustment below:
for (i in (k+2):n) {
Y[i-k-1] <- (X[i-j] + X[i-k]) %% m
}
@langtang has shown what you need to do.
The problem is operator precedence:
> 1+2:3
[1] 3 4
Here we are saying "add 1 to the sequence 2:3
"
> (1+2):3
[1] 3
And here we are saying "create a sequence from 1+2
to 3".
Even when you don't need to use parentheses it is often a good idea to do so. Judicious use makes the code easier to read. So for example if the intention was to actually use 1+2:3
it would be better to write 1+(2:3)
. In this way it quite obvious what is going on:
(1+2):3
vs
1+(2:3)
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