How to access elements in vectors in Lists more efficiently? - c++

Question

I have a problem where I want to combine a list of vectors, all of the same type, in a particular fashion. I want the first element of my resultant vector to be the first element of the first vector in my list, the second element should be the first element of the second vector, the third, the first of the third and so on until n where n is length of my list and then element n+1 should be the second element of the first vector. This repeats until finished.
Currently, I am doing it like this:
CharacterVector measure(nrows * expansion);
CharacterVector temp(nrows);
for(int i=0; i < measure.size(); i++){
temp = values[i % expansion];
measure[i] = temp[i / expansion];
}
return(measure);
Where values is the List of CharacterVectors. This seems incredibly inefficient, overwriting temp every single time but I don't know of a better way to access the elements in values. I don't know a lot of C++ but I assume there must be a better way.
Any and all help is greatly appreciate!
EDIT:
All vectors in 'values are of the same length nrows and values has expansion elements in it.

Answer 1

What you need is the ListOf<CharacterVector> class. As the name implies, it represents an R list which only contains CharacterVector.
The code below uses it to extract the second element of each character vector from the list. Should not be hard to adapt it to your expansion algorithm, but your example was not reproducible without a bit more context.
#include <Rcpp.h>
using namespace Rcpp ;
// [[Rcpp::export]]
CharacterVector second( ListOf<CharacterVector> values ){
int n = values.size() ;
CharacterVector res(n);
for(int i=0; i<n; i++){
res[i] = values[i][1] ;
}
return res ;
}
Then, you sourceCpp this and try it on some sample data:
> data <- list(letters, letters, LETTERS)
> second(data)
[1] "b" "b" "B"
Now about your assumption:
This seems incredibly inefficient, overwriting temp every single time
Creating a CharacterVector is pretty fast, there is no deep copy of data, so this should not have been an issue in the first place.

Answer 2

You can preconstruct the vector and can easily know at which positions the elements of the first vector should go.. i.e. measure[0], measure[n], measure[n*2] etc.. where n = mylist.size(). Here i assume of course that each vector in the list has equal size. Untested code:
CharacterVector measure(nrows * expansion);
for(int i=0; i < values.size(); ++i)
{
CharacterVector& temp = values[i];
int newPosition = i;
for( int j=0; j < temp.size(); ++j)
{
measure[newPosition ] = temp[j];
newPosition += expansion;
}
}
return(measure);