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I am trying to obtain the least common multiplier of the prime numbers in between 1 and 20:

x <- expand.grid(c(2,3,7,11,13,17,19),c(2,3,7,11,13,17,19))
l <- c()
# loop
for(i in length(x[,1])){
l[i] <- pracma::Lcm(x[i, 1], x[i, 2])
}
# gives
tail(l)
[1] NA NA NA NA NA 19

Interestingly, the loop returns only NA's besides the first and last value, this doesn't make sense since one can obtain the values by hand:

pracma::Lcm(x[4, 1], x[4, 2])
[1] 22

Can anyone spot what I am missing?

Thank you in advance!

Programming Rhino
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1 Answers1

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We can initiate l with instead of c(). Also, length is a single value. We need seq_len(nrow(x))

l <- numeric(nrow(x))

for(i in seq_len(nrow(x))) {
     l[i] <- pracma::Lcm(x[i, 1], x[i, 2])
 }

l
#[1]   2   6  14  22  26  34  38   6   3  21  33  39  51  57  14  21   7  77  91 119 133  22  33  77  11 143 187 209  26  39  91 143  13
#[34] 221 247  34  51 119 187 221  17 323  38  57 133 209 247 323  19

If we initiate as c(), then inside the loop, it would be

 l <- c(l, pracma::Lcm(x[i, 1], x[i, 2]))

Also, the pracma::Lcm is vectorized, so no need to loop over each rows

pracma::Lcm(x[,1], x[,2])
#[1]   2   6  14  22  26  34  38   6   3  21  33  39  51  57  14  21   7  77  91 119 133  22  33  77  11 143 187 209  26  39  91 143  13
#[34] 221 247  34  51 119 187 221  17 323  38  57 133 209 247 323  19

In the OP's code, the 

length(x[,1])
#[1] 49

is a single number i.e. the last row, here, i.e. only looping through that last row

Second issue is during the assignment, where we assign with a length of 0 and when we use index, it is creating a NULL

l <- c()
length(l)
#[1] 0
l[1]
#NULL
akrun
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  • Can u elaborate on why you think that would solve my problem ? – Programming Rhino Jan 02 '20 at 19:31
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    @BeginneR. There are two issues in the code, can you check `length(x[,1])` it is a single number. So, looping means it is only looping at the last row, 2) You initiated `l <- c()` Now, check `l[1]` – akrun Jan 02 '20 at 19:34