Get rid of consecutive na in columns of a matirx - r

Question

I asked the same question here but it was closed as my post has been associated with similar questions although they are not related to my question and don't resolve it.
The dataset:
I have a huge data set saved in a matrix where the number of rows is more that one million with a dozen of columns.
The matrix looks like
data <- matrix(c(1, NA, 2, NA, 1, NA, NA, NA, 1, NA, 3, NA, 5, NA, NA, NA, 8, NA, 5, NA, 7, NA, NA, NA), ncol=3)
> data
[,1] [,2] [,3]
[1,] 1 1 8
[2,] NA NA NA
[3,] 2 3 5
[4,] NA NA NA
[5,] 1 5 7
[6,] NA NA NA
[7,] NA NA NA
[8,] NA NA NA
So if there is a missing value in certain column, then necessarily other columns will have missing values for the same row.
The question:
I would like to delete "efficiently" consecutive missing values if there are 3 or more in each column for all columns in the matrix. So I would like to delete consecutive na in a column not a row.
I already saw solutions, like this one, for my question but they were too slow for my huge data set. Do you have other suggestions which can achieve the objective efficiently? Additionally, the suggested answers (1 & 2) for my closed question are deleting if the missing values are consecutive in rows not columns.
EDIT:
Following to the comment below, the output must be like this:
[,1] [,2] [,3]
[1,] 1 1 8
[2,] NA NA NA
[3,] 2 3 5
[4,] NA NA NA
[5,] 1 5 7
EDIT:
> data
[,1] [,2] [,3] [,4]
[1,] 1 1 8 NA
[2,] NA NA NA NA
[3,] 2 3 5 NA
[4,] NA NA NA NA
[5,] 1 5 7 NA
[6,] NA NA NA NA
[7,] NA NA NA NA
[8,] NA NA NA NA
The expected output
[,1] [,2] [,3]
[1,] 1 1 8
[2,] NA NA NA
[3,] 2 3 5
[4,] NA NA NA
[5,] 1 5 7

Answer 1

If it is consecutive, then may be rle can be used
i1 <- rowSums(is.na(data)) > 0
# // or just forgot to update here
i1 <- is.na(data[,1])
data[!inverse.rle(within.list(rle(i1), {
values[values & lengths < 3] <- FALSE})),]
-output
# [,1] [,2] [,3]
#[1,] 1 1 8
#[2,] NA NA NA
#[3,] 2 3 5
#[4,] NA NA NA
#[5,] 1 5 7
Update
If we have a particular column with all NAs, then we can remove it first
data1 <- data[,colSums(!is.na(data)) != 0]
and now we apply the previous code on the selected column data
i1 <- is.na(data1[,1])
data1[!inverse.rle(within.list(rle(i1), {
values[values & lengths < 3] <- FALSE})),]
Or we may use rleid from data.table (which would be more efficient)
library(data.table)
data[as.data.table(data)[, .I[!(.N >=3 & is.na(V1))],
rleid(is.na(V1))]$V1,]

Answer 2

if there is a missing value in certain column, then necessarily other columns will have missing values for the same row.
I think this is very important information, we can take advantage of it and work only with any 1 column instead of complete dataset. Try :
vec <- data[, 1]
data[!with(rle(is.na(vec)), rep(values & lengths >= 3, lengths)), ]
# [,1] [,2] [,3]
#[1,] 1 1 8
#[2,] NA NA NA
#[3,] 2 3 5
#[4,] NA NA NA
#[5,] 1 5 7