问题
I am trying to find the count of diagonal 1s in each 3x3 tile e.g.
0 0 1 1 0 0
0 1 0 0 1 0
1 0 0 or 0 0 1
from the below 15x15 matrix.
set.seed(99)
mat <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow=15)
print(mat)
[,1][,2][,3][,4][,5][,6][,7][,8][,9][,10][,11][,12][,13][,14][,15]
[1,] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 0 1 0 1 0 0 1 0 0 0 1 0 0 0 1
[3,] 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0
[4,] 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1
[5,] 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0
[6,] 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0
[7,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0
[9,] 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1
[10,] 0 0 0 0 0 0 0 0 1 0 1 1 0 1 0
[11,] 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0
[12,] 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0
[13,] 0 0 0 0 0 1 0 1 0 0 1 0 1 0 0
[14,] 1 1 0 1 1 0 0 0 0 1 0 0 0 0 1
[15,] 1 0 1 0 1 1 0 0 0 1 0 1 0 0 0
I expect the output to be 2 for the above matrix. Is there a way to do this with a for loop and if statements?
回答1:
Here's a nested for loop (using sapply()
). Note I did not have the same dataset as you so there's a different seed.
set.seed(123)
mat <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow=15)
n_by_n <- 3L
reg_diag <- diag(n_by_n)
rev_diag <- reg_diag[nrow(reg_diag):1, ]
sum(
sapply(seq_len(ncol(mat)- n_by_n + 1),
function(col) {
sapply(seq_len(nrow(mat) - n_by_n + 1),
function(row) {
tmp <- mat[row:(row + n_by_n - 1), col:(col + n_by_n - 1)]
all(tmp == reg_diag) | all(tmp == rev_diag)
})
})
)
#[1] 1
If you are only interested in diagonals and do not care about the other values in a submatrix, this splits the matrix by each diagonal and then calculated a rolling sum to see if they sum up to 3:
library(RcppRoll)
set.seed(99)
mat <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow=15)
n_by_n <- 3
diags <- row(mat)- col(mat)
cross_diags <- row(mat) + col(mat)
#could use data.table::frollsum instead of RcppRoll::roll_sumr)
sum(unlist(lapply(split(mat, diags), RcppRoll::roll_sumr, n_by_n), use.names = F) == n_by_n, na.rm = T)
#[1] 1
sum(unlist(lapply(split(mat, cross_diags), RcppRoll::roll_sumr, n_by_n), use.names = F) == n_by_n, na.rm = T)
# [1] 3
A complete base approach would be:
base_rollr <- function(x, roll) {
#from user @flodel
if (length(x) >= roll) tail(cumsum(x) - cumsum(c(rep(0, roll), head(x, -roll))), -roll + 1)
}
sum(unlist(lapply(split(mat, cross_diags), base_rollr, n_by_n), use.names = F) == n_by_n, na.rm = T)
See also: Get all diagonal vectors from matrix
And: Consecutive/Rolling sums in a vector in R
回答2:
We could use outer()
. For this we write two small vectorized functions, that count the elements of the diagonal of a 3x3 slice of our matrix; if the sum
is 3
we have a valid diagonal.
For the counterdiagonal we borrow code from this solution.
counterdiag <- function(M) M[(n<-nrow(M))^2-(1:n)*(n-1)]
Now all we need is some coordinates.
m <- n <- mapply(function(i) i:(i+2), 1:13)
And our counting functions.
fun1 <- Vectorize(function(x, y) sum(diag(mat[m[,x], n[,y]])) == 3, SIMPLIFY=FALSE)
fun2 <- Vectorize(function(x, y) sum(counterdiag(mat[m[,x], n[,y]])) == 3, SIMPLIFY=FALSE)
Usage
sum(unlist(outer(1:13, 1:13, fun1))) # diagonals
# [1] 1
sum(unlist(outer(1:13, 1:13, fun2))) # counterdiagonals
# [1] 3
来源:https://stackoverflow.com/questions/58580834/trying-to-find-the-number-of-diagonals-of-1-in-each-3x3-tile-in-a-15x15-binary-m