CODEWITHSUNDEEP
r
RalfB
RalfB

Reputation: 583

R: Matrix row operations

I have a matrix A that has a large number of rows and columns (below one example of such a matrix) that occasionally has a full row of 0 values (as in row 4 at this particular example).

I want to have a function that checks all rows of A and allows me to perform an operation on each element of these rows. Is there an easy way to do that?

I also wonder if matrix is the right data structure for this. It feels not quite right, perhaps data frames are better for that?

A = matrix(
  c(0, 0, 1, 0, 0, 0, 0,
    1, 0, 1, 1, 0, 0, 0,
    0, 0, 0, 1, 1, 0, 0,
    0, 0, 0, 0, 0, 0, 0,
    0, 0, 0, 0, 0, 1, 1,
    0, 0, 0, 1, 1, 0, 1), nrow=7,ncol=7,byrow = TRUE)

For every row of that matrix I want to determine if there are only 0's in it. If so, I want to set (for each element) the value 1/N (where N is the ncol(A)).

Sudo code:

If (sum(row of A) == 0) then row_of_A = 1/ncol(A)

Upvotes: 1

Views: 1303

Answers (1)

Roland
Roland

Reputation: 132576

Apparently you want this:

A[rowSums(A != 0) == 0,] <- 1/ncol(A)
#          [,1]      [,2]      [,3]      [,4]      [,5]      [,6]      [,7]
#[1,] 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[2,] 1.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 0.0000000
#[3,] 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000
#[4,] 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571 0.1428571
#[5,] 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000
#[6,] 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 1.0000000
#[7,] 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 0.0000000

Explanation:

  • A != 0 checks all matrix elements and returns a logical matrix with TRUE for non-zero elements.
  • We then sum the rows of that logical matrix, whereby FALSE/TRUE is coerced to 0/1.
  • We check if these rowsums are 0 and use the resulting logical vector to subset the rows.
  • We assign 1/ncol to this subset.

Benchmarks to show that apply is slower:

set.seed(42); A = matrix(sample(0:1, 5e4, TRUE), nrow=1e4)

library(microbenchmark)
microbenchmark(A[rowSums(A != 0) == 0,],
               A[!apply(A != 0, 1, any),],
               A[apply(A == 0, 1, all),])
#Unit: microseconds
#                        expr       min        lq       mean    median        uq       max neval cld
#   A[rowSums(A != 0) == 0, ]   572.202   593.298   620.7931   624.248   629.638   780.387   100  a 
# A[!apply(A != 0, 1, any), ] 14978.248 16124.652 17261.9530 17441.054 18129.975 22469.219   100   b
#  A[apply(A == 0, 1, all), ] 15182.122 16149.751 17616.8010 16561.657 17997.703 75148.079   100   b

Upvotes: 4

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