Question

As fast as possible, I would like to replace the first zeros in some rows of a matrix with values stored in another vector.

There is a numeric matrix where each row is a vector with some zeros.
I also have two vectors, one containing the rows, in what to be replaced, and another the new values: replace.in.these.rows and new.values. Also, I can generate the vector of first zeroes with sapply

mat <- matrix(1,5,5)
mat[c(1,8,10,14,16,22,14)] <- 0
replace.in.these.rows <- c(1,2,3)
new.values <- c(91,92,93)

corresponding.poz.of.1st.zero <- sapply(replace.in.these.rows, 
                                        function(x) which(mat [x,] == 0)[1] )

Now I would like something that iterates over the index vectors, but without a for loop possibly:

matrix[replace.in.these.rows, corresponding.poz.of.the.1st.zero ] <- new.values 

Is there a trick with indexing more than simple vectors? It could not use list or array(e.g.-by-column) as index.

By default R matrices are a set of column vectors. Do I gain anything if I store the data in a transposed form? It would mean to work on columns instead of rows.

---

Context:

This matrix stores contact ID-s of a network. This is not an adjacency matrix n x n, rather n x max.number.of.partners (or n*=30) matrix.

The network uses edgelist by default, but I wanted to store the “all links from X” together.

I assumed, but not sure if this is more efficient than always extract the information from the edgelist (multiple times each round in a simulation)

I also assumed that this linearly growing matrix form is faster than storing the same information in a same formatted list.

Some comments on these contextual assumptions are also welcome.

Answer 1

Edit: If only the first zeros are to be replace then this approach works:

first0s <-apply(mat[replace.in.these.rows, ] , 1, function(x) which(x==0)[1])
mat[cbind(replace.in.these.rows, first0s)] <- new.values
> mat
     [,1] [,2] [,3] [,4] [,5]
[1,]   91    1    1    0    1
[2,]    1    1    1    1   92
[3,]    1   93    1    1    1
[4,]    1    1    0    1    1
[5,]    1    0    1    1    1

Edit: I thought that the goal was to replace all zeros in the chosen rows and this was the approach. A completely vectorized approach:

 idxs <- which(mat==0, arr.ind=TRUE)
# This returns that rows and columns that identify the zero elements
# idxs[,"row"] %in% replace.in.these.rows
#  [1]  TRUE  TRUE FALSE FALSE  TRUE  TRUE
# That isolates the ones you want.
# idxs[ idxs[,"row"] %in% replace.in.these.rows , ]
# that shows what you will supply as the two column argument to "["
#     row col
#[1,]   1   1
#[2,]   3   2
#[3,]   1   4
#[4,]   2   5
 chosen.ones <- idxs[ idxs[,"row"] %in% replace.in.these.rows , ]
 mat[chosen.ones] <- new.values[chosen.ones[,"row"]]
# Replace the zeros with the values chosen (and duplicated if necessary) by "row".
 mat
 #---------    
 [,1] [,2] [,3] [,4] [,5]
[1,]   91    1    1   91    1
[2,]    1    1    1    1   92
[3,]    1   93    1    1    1
[4,]    1    1    0    1    1
[5,]    1    0    1    1    1