R中矩阵的每行滞后答案

【问题标题】：Lag per rows of a matrix in RR中矩阵的每行滞后
【发布时间】：2019-09-09 23:14:56
【问题描述】：

我想构造一个如下所示的矩阵：

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
[1,]    1    0    0   -1    0    0    0    0    0     0     0     0     0     0     0     0
[2,]    0    0    0    1    0    0   -1    0    0     0     0     0     0     0     0     0
[3,]    0    0    0    0    0    0    1    0    0    -1     0     0     0     0     0     0
[4,]    0    0    0    0    0    0    0    0    0     1     0     0    -1     0     0     0
[5,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0    -1

我正在考虑的方式，虽然有点不太好，是创建一个包含一个模式的向量，该模式将重复填充，逐行，矩阵

pattern <- c(1,rep(0,2),-1,rep(0,15)) #creating the pattern 
matrix(rep(pattern, 5), 5, 16, byrow = TRUE) #filling a matrix per row with repetitions of the pattern
#Returns:
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
[1,]    1    0    0   -1    0    0    0    0    0     0     0     0     0     0     0     0
[2,]    0    0    0    1    0    0   -1    0    0     0     0     0     0     0     0     0
[3,]    0    0    0    0    0    0    1    0    0    -1     0     0     0     0     0     0
[4,]    0    0    0    0    0    0    0    0    0     1     0     0    -1     0     0     0
[5,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0    -1
Warning message:
In matrix(rep(pattern, 5), 5, 16, byrow = TRUE) :
  data length [95] is not a sub-multiple or multiple of the number of columns [16]

这给了我我想要的东西，但当然有一个警告，因为我没有使用结果模式重复的所有元素。

我在想的另一种方式是，使用kronecker 产品并有类似的东西：

kronecker(diag(5),t(c(1,rep(0,2),-1)))
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
[1,]    1    0    0   -1    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[2,]    0    0    0    0    1    0    0   -1    0     0     0     0     0     0     0     0     0     0     0     0
[3,]    0    0    0    0    0    0    0    0    1     0     0    -1     0     0     0     0     0     0     0     0
[4,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0    -1     0     0     0     0
[5,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     1     0     0    -1

然后根据每行（从第二行开始）向左移动/滞后，直到获得我正在寻找的结果。在不使用lag 的情况下，有没有办法实现这种转变？或任何建议/替代解决方法？谢谢

【问题讨论】：

标签： r matrix lag shift

【解决方案1】：

1) 设置一维元素

m1 <- matrix(0, 5, 16)
m1[seq(1, length(m1), 16)] <- 1
m1[seq(16, length(m1), 16)] <- -1

2) 设置二维元素

m2 <- matrix(0, 5, 16)
m2[cbind(1:5, seq(from = 1, length = 5, by = 16 %/% 5))] <- 1
m2[cbind(1:5, seq(to = 16, length = 5, by = 16 %/% 5))] <- -1

3) 从诊断设置

len <- 5 * 16
d <- c(diag(16 - 1))
m3 <- matrix(head(d, len) - tail(d, len), ncol = 16)

【讨论】：

【解决方案2】：

这是一个使用 for 循环的解决方案，但我认为应该可以实现更简单、更快速的解决方案 -

m <- matrix(0, nrow = 5, ncol = 16)

colnum <- seq(1, by = 3, length.out = nrow(m))

for(i in 1:nrow(m)) {
  m[i, colnum[i]:(colnum[i]+3)] <- c(1, 0, 0, -1)
}

m
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
[1,]    1    0    0   -1    0    0    0    0    0     0     0     0     0     0     0     0
[2,]    0    0    0    1    0    0   -1    0    0     0     0     0     0     0     0     0
[3,]    0    0    0    0    0    0    1    0    0    -1     0     0     0     0     0     0
[4,]    0    0    0    0    0    0    0    0    0     1     0     0    -1     0     0     0
[5,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0    -1

【讨论】：