How to avoid a loop to calculate competition index

Question

I\'ve to calculate so called competition index for a couple of the experiments. I have known position of the object and its size. I\'d like to calculate the sum of the sizes

Answer 1

I use dplyr and a join on exp. Then summarise for each (generated) id.

res <- df %>% mutate(id = row_number()) %>%
  merge(df, by='exp') %>% 
  mutate(dist = sqrt((x.x - x.y)^2 + (y.x - y.y)^2)) %>% 
  filter(dist < 2 ) %>%
  group_by(id,x.x,y.x,di.x) %>%
  summarise(comp1 = sum(di.y),
                      dist = sum(dist))

results in :

Source: local data frame [2,000 x 6]
Groups: id, x.x, y.x [?]

      id       x.x       y.x       di.x      comp1     dist
   <int>     <dbl>     <dbl>      <dbl>      <dbl>    <dbl>
1      1 127.36166  89.64637 -0.2508979 -0.2508979 0.000000
2      2  90.98491 153.17911  1.4561061  1.4561061 0.000000
3      3  58.96620 144.72710  2.7909274  2.7909274 0.000000
4      4 162.44443 132.35379  3.0175213  3.0175213 0.000000
5      5 184.52673  47.12997  1.1127618  1.1127618 0.000000
6      6  57.07334 126.03554 -0.2508979 -0.2508979 0.000000
7      7  22.28946 110.69319  1.4561061  2.5688679 1.267998
8      8  40.54007 123.32645  2.7909274  2.7909274 0.000000
9      9 179.37667  61.45213  3.0175213  3.0175213 0.000000
10    10  73.82714  67.86194  1.1127618  1.1127618 0.000000
# ... with 1,990 more rows

PS: looking at the criterium if(dist < 2 & x$exp[i] == x$exp[j]) means only a few rows match the criterium of dist < 2.

Answer 2

Loops like this are a perfect candidate for speeding up with Rcpp. The logic translates across unchanged:

library(Rcpp)

cppFunction('
List
computeIndex(const NumericVector x,
             const NumericVector y, 
             const NumericVector di,
             const CharacterVector ex)
{
    int n = x.size();
    NumericVector comp1(n), dist(n);

    for(int i = 0; i < n; ++i)
    {
        for(int j = 0; j < n; ++j)
        {
            double dx = x[j] - x[i], dy = y[j] - y[i];
            double d = std::sqrt(dx*dx + dy*dy);

            if((d < 2) && (ex[i] == ex[j]))
            {
                comp1[i] += di[j];
                dist[i] +=  d;
            }
        }
    }

    return List::create(Named("comp1") = comp1,
                        Named("dist") = dist);
}
')

res <- data.frame(computeIndex(df$x, df$y, df$di, df$exp))

Not only is this faster than the equivalent R-only code, but it avoids having to allocate any O(N^2) objects. You can also combine this with dplyr to avoid needless comparisons between rows with different exp values:

df %>%
    group_by(exp) %>%
    do({
        res <- computeIndex(.$x, .$y, .$di, .$exp)
        data.frame(., res)
    })