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Efficient way to remove Sparse matrix row having large data in R

I have a large sparse Matrix. Based on the number of elements in a row, I want to remove the row of the sparse Matrix. The first step is to obtain the indexes of the row that has less that 5 elements.

number of rows = 4,000,000 number of columns = 250,000

The sparse matrix looks somewhat like this...

A header C1 C2 C3 C4 C5 ... n
First . 1 1 1 1 . .
Second . . 1 1 . . .
Third 1 . . . . . .
Fourth 1 . . . 1 . .
nth 1 . . . . . .

I could use rowSums, and remove the row based on the output.

items <- c()
for(i in 1: nrow(sparse_matrix){
  if(rowSums(sparse_matrix)[i] < 3){
    items <- append(items, i)
  }
}

However, this takes 1hr 30mins to go through around 10,000 rows which is really slow.

What would be an efficient solution to this?

First, you calculate rowSums(sparse_matrix) at every iteration of the loop, which is inefficient. Second, doing append(items, i) in a loop is also inefficient.

My solution without a loop at all:

items = which(rowSums(sparse_matrix) < 3)

Following up @AndreyShabalin's answer with an example:

example

Set up a matrix with the specified dimensions and 1 million non-zero elements:

nr <- 4e6; nc <- 2.5e4; ns <- 1e6
m <- Matrix(0, nrow = nr, ncol = nc)
set.seed(101)
i <- sample(nr, size = ns, replace = TRUE)
j <- sample(nc, size = ns, replace = TRUE)
m[cbind(i,j)] <- 1

subset

system.time(m2 <- m[rowSums(m) > 3, ]) ## 0.03 seconds
nrow(m2)  ## 543

By avoiding (1) recomputing rowSums() and (2) not growing a vector, we make this very fast.

library(microbenchmark)
microbenchmark(
  logical = m[rowSums(m) > 3, ],
  which = m[which(rowSums(m) > 3), ]
)

For reasons I don't understand, the solution with which is actually slightly faster than the logical-indexing solution (median of 30 vs 35 milliseconds...)

Unit: milliseconds
    expr      min       lq     mean   median       uq      max neval cld
 logical 33.13682 35.07974 45.65183 35.61339 37.09483 160.3382   100   b
   which 27.90955 29.90988 37.06513 30.38864 31.76437 153.8702   100  a 

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