试图在15x15二进制矩阵中找到每个3x3瓦片中对角线1的数量? [英] Trying to find the number of diagonals of 1 in each 3x3 tile in a 15x15 binary matrix?

问题描述

我正在尝试查找每个3x3磁贴中对角线1的计数.

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

从下面的15x15矩阵中获取.

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

我希望上述矩阵的输出为2.有没有办法用for循环和if语句来做到这一点?

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?

推荐答案

这是一个嵌套的for循环(使用sapply()).请注意，我没有与您相同的数据集，所以有不同的种子.

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

如果您只对角线感兴趣，而不关心子矩阵中的其他值，则会按每个对角线拆分矩阵，然后计算滚动总和以查看它们的总和是否为3:

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

一个完整的基本方法是:

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)

另请参阅:从矩阵中获取所有对角矢量

并且: R/a向量中的连续/滚动和

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