例如,我试图在每个3x3的图块中找到对角线1的计数
0 0 1 1 0 0 1 1 1 0 1 1
0 1 0 0 1 0 1 1 1 0 1 0
1 0 0 or 0 0 1 or 1 1 1 or 1 0 0
例如,仅与对角线无关,3x3区块中的其他值无关紧要。对角线是指主要对角线以及反向主要对角线。例如,以下矩阵将输出3个对角线,
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
我正在阅读矩阵列表...
set.seed(1234)
mat1 <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow = 15)
set.seed(99)
mat2 <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow = 15)
set.seed(123)
mat3 <- matrix(sample(c(0,1), 225, prob=c(0.8,0.2), replace=TRUE), nrow = 15)
mat_list <- list(mat1, mat2, mat3)
关于如何做到这一点的任何建议?
假设m
如结尾处的注解所示(尤其是条目是双精度而不是整数),位置是固定的差异,因此我们可以将rollapply
与函数identical
一起使用。
library(zoo)
nr <- nrow(m)
sum(rollapply(c(m), list(c(0, nr+1, 2*nr+2)), identical, c(1, 1, 1), align = "left")) +
sum(rollapply(c(m), list(c(0, nr-1, 2*nr-2)), identical, c(1, 1, 1), align = "left"))
## 3
m <- structure(c(1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), .Dim = c(8L,
15L), .Dimnames = list(NULL, NULL))