r - Most Efficient way to create a symmetric matrix

Question

I have the following matrix/ dataframe:

> e
  V1 V2 V3 V4 V5
1  0  2  3  4  5
2  0  0  6  8 10
3  0  0  0 12 15
4  0  0  0  0 20
5  0  0  0  0  0

In this case N=5 (number of rows = number of columns). I would like to fill in the missing values in this symmetric matrix (e[1,2]=e[2,1] etc.). Is there a most efficient way to fill in the missing values (N the matrix size in my case quite big)? Is there a better way than nested loops?

Answer 1

Also for speed:

2*symmpart(as.matrix(e))

Here's a benchmark:

Unit: microseconds
                                      expr      min       lq        mean    median        uq       max neval
                                  e + t(e)  572.505  597.194  655.132028  611.5420  628.4860  8424.902  1000
                             symmetrise(e) 1128.220 1154.562 1215.740071 1167.0020 1185.6585 10656.059  1000
 e[lower.tri(e)] <- e[upper.tri(e, FALSE)]  285.013  311.191  350.846885  327.1335  339.5910  8106.006  1000
                2 * symmpart(as.matrix(e))   78.392   93.953  101.330522  102.1860  107.9215   153.628  1000

It can get this speed because it creates a symmetric matrix directly.