In this post, I will show you how to optimize your Rcpp loops so that they are 2 to 3 times faster than a standard implementation.

Context

Real data example

For this post, I will use a big.matrix which represents genotypes for 15,283 individuals, corresponding to the number of mutations (0, 1 or 2) at 287,155 different loci. Here, I will use only the first 10,000 loci (columns).

What you need to know about the big.matrix format:

  • you can easily and quickly access matrice-like objects stored on disk,
  • you can use different types of storage (I use type char to store each element on only 1 byte),
  • it is column-major ordered as standard R matrices,
  • you can access elements of a big.matrix using X[i, j] in R,
  • you can access elements of a big.matrix using X[j][i] in Rcpp,
  • you can get a RcppEigen or RcppArmadillo view of a big.matrix (see Appendix).
  • for more details, go to the GitHub repo.

Peek at the data:

print(dim(X))
## [1] 15283 10000
print(X[1:10, 1:12])
##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
##  [1,]    2    0    2    0    2    2    2    1    2     2     2     2
##  [2,]    2    0    1    2    1    1    1    1    2     1     2     2
##  [3,]    2    0    2    2    2    2    1    1    2     1     2     2
##  [4,]    2    2    0    2    0    0    0    2    2     2     0     2
##  [5,]    2    1    2    2    2    2    2    1    2     2     2     2
##  [6,]    2    1    2    1    2    2    1    1    2     2     2     2
##  [7,]    2    0    2    0    2    2    2    0    2     1     2     2
##  [8,]    2    1    1    2    1    1    1    1    2     1     2     2
##  [9,]    2    1    2    2    2    2    2    2    2     2     2     2
## [10,]    2    0    2    1    2    2    2    0    2     1     2     1

What I needed

I needed a fast matrix-vector multiplication between a big.matrix and a vector. Moreover, I could not use any RcppEigen or RcppArmadillo multiplication because I needed some options of efficiently subsetting columns or rows in my matrix (see Appendix).

Writing this multiplication in Rcpp is no more than two loops:

// [[Rcpp::depends(RcppEigen, bigmemory, BH)]]
#include 
#include 

using namespace Rcpp;

// [[Rcpp::export]]
NumericVector prod1(XPtr bMPtr, const NumericVector& x) {
  
  MatrixAccessor<char> macc(*bMPtr);

  int n = bMPtr->nrow();
  int m = bMPtr->ncol();

  NumericVector res(n);
  int i, j;
  
  for (j = 0; j < m; j++) {
    for (i = 0; i < n; i++) {
      res[i] += macc[j][i] * x[j];
    }
  }

  return res;
}

One test:

y <- rnorm(ncol(X))

print(system.time(
  test <- prod1([email protected], y)
))
##    user  system elapsed 
##   0.664   0.004   0.668

What comes next should be transposable to other applications and other types of data.

Unrolling optimization

While searching for optimizing my multiplication, I came across this Stack Overflow answer.

Unrolling in action:

// [[Rcpp::depends(RcppEigen, bigmemory, BH)]]
#include 
#include 

using namespace Rcpp;

// [[Rcpp::export]]
NumericVector prod4(XPtr bMPtr, const NumericVector& x) {
  
  MatrixAccessor<char> macc(*bMPtr);
  
  int n = bMPtr->nrow();
  int m = bMPtr->ncol();
  
  NumericVector res(n);
  int i, j;
  
  for (j = 0; j <= m - 4; j += 4) {
    for (i = 0; i < n; i++) { // unrolling optimization
      res[i] += (x[j] * macc[j][i] + x[j+1] * macc[j+1][i]) +
        (x[j+2] * macc[j+2][i] + x[j+3] * macc[j+3][i]);
    } // The parentheses are somehow important. Try without.
  }
  for (; j < m; j++) {
    for (i = 0; i < n; i++) {
      res[i] += x[j] * macc[j][i];
    }
  }
  
  return res;
}
require(microbenchmark)

print(microbenchmark(
  PROD1 = test1 <- prod1([email protected], y),
  PROD4 = test2 <- prod4([email protected], y),
  times = 5
))
## Unit: milliseconds
##   expr      min       lq     mean   median       uq      max neval
##  PROD1 609.0916 612.6428 613.7418 613.3740 616.4907 617.1096     5
##  PROD4 262.2658 267.7352 267.0268 268.0026 268.0785 269.0521     5
print(all.equal(test1, test2))
## [1] TRUE

Nice! Let’s try more. Why not using 8 or 16 rather than 4?

Rcpp::sourceCpp('https://privefl.github.io/blog/code/prods.cpp')

print(bench <- microbenchmark(
  PROD1 = prod1([email protected], y),
  PROD2 = prod2([email protected], y),
  PROD4 = prod4([email protected], y),
  PROD8 = prod8([email protected], y),
  PROD16 = prod16([email protected], y),
  times = 5
))
## Unit: milliseconds
##    expr      min       lq     mean   median       uq      max neval
##   PROD1 620.9375 627.9209 640.6087 631.1818 659.4236 663.5798     5
##   PROD2 407.6275 418.1752 417.1746 418.4589 419.0665 422.5451     5
##   PROD4 267.1687 271.4726 283.1928 271.9553 279.6698 325.6979     5
##   PROD8 241.5542 242.9120 255.4974 246.5218 267.7683 278.7307     5
##  PROD16 212.4335 213.5228 217.4781 217.1801 221.5119 222.7423     5
time <- summary(bench)[, "median"]
step <- 2^(0:4)
plot(step, time, type = "b", xaxt = "n", yaxt = "n", 
     xlab = "size of each step")
axis(side = 1, at = step)
axis(side = 2, at = round(time))

Conclusion

We have seen that unrolling can dramatically improve performances on loops. Steps of size 8 or 16 are of relatively little extra gain compared to 2 or 4.

As pointed out in the SO answer, it can behave rather differently between systems. So, if it is for your personal use, use the maximum gain (try 32!), but as I want my function to be used by others in a package, I think it’s safer to choose a step of 4.

Appendix

You can do a big.matrix-vector multiplication easily with RcppEigen or RcppArmadillo (see this code) but it lacks of efficient subsetting option.

Indeed, you still can’t use subsetting in Eigen, but this will come as said in this feature request. For Armadillo, you can but it is rather slow:

Rcpp::sourceCpp('https://privefl.github.io/blog/code/prods2.cpp')

n <- nrow(X)
ind <- sort(sample(n, size = n/2))

print(microbenchmark(
  EIGEN = test3 <- prodEigen([email protected], y),
  ARMA = test4 <- prodArma([email protected]s, y),
  ARMA_SUB = test5 <- prodArmaSub([email protected], y, ind - 1),
  times = 5
))
## Unit: milliseconds
##      expr       min        lq      mean    median        uq       max neval
##     EIGEN  567.5607  570.1843  717.2433  572.9402  576.2028 1299.3285     5
##      ARMA 1242.3581 1263.8803 1329.1212 1264.7070 1284.5612 1590.0993     5
##  ARMA_SUB  455.1174  457.5862  466.3982  461.5883  465.9056  491.7935     5
print(all(
  all.equal(test3, test), 
  all.equal(as.numeric(test4), test),
  all.equal(as.numeric(test5), test[ind])
))
## [1] TRUE