Question

我想将频率向量（即矩阵的colSums()）转换为R中原始逻辑矩阵的可能版本之一。

类似的东西：

    s <- c(1,2,3)
    # Some function of s
    # Example output:
         [,1] [,2] [,3]
    [1,]    0    0    1
    [2,]    1    0    0 
    [3,]    0    1    0
    [4,]    0    0    1
    [5,]    0    0    1
    [6,]    0    1    0

行的顺序并不重要。有人可以给我提示如何做吗？

编辑：行总和始终为1。输出可以看作是多项式数据集，其中每一行都反映一个观察值。

Answer 1

s <- c(1,2,3)
result = matrix(0, nrow = max(s), ncol = length(s))
for (i in seq_along(s)) result[1:s[i], i] = 1
result
#      [,1] [,2] [,3]
# [1,]    1    1    1
# [2,]    0    1    1
# [3,]    0    0    1

将rowums保留为1

s <- c(1,2,3)
result = matrix(0, nrow = sum(s), ncol = length(s))
result[cbind(1:sum(s), rep(seq_along(s), times = s))] = 1
result
#      [,1] [,2] [,3]
# [1,]    1    0    0
# [2,]    0    1    0
# [3,]    0    1    0
# [4,]    0    0    1
# [5,]    0    0    1
# [6,]    0    0    1

Answer 2

set.seed(523)

s <- c(1, 2, 3)
n <- 6

sapply(s, function(i) sample(c(rep(1, i), rep(0, n - i))))
#      [,1] [,2] [,3]
# [1,]    0    1    1
# [2,]    1    0    0
# [3,]    0    1    0
# [4,]    0    0    1
# [5,]    0    0    0
# [6,]    0    0    1

Answer 3

# Input:
s <- c(1,2,3)
# ...
set.seed(1) # For reproducibility
nr <- sum(s)
nc <- length(s)
mat <- matrix(0L, nrow = nr, ncol = nc)
mat[cbind(seq_len(nr), sample(rep(seq_len(nc), s)))] <- 1L

# Output:
     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    0    1
[3,]    0    1    0
[4,]    0    0    1
[5,]    0    1    0
[6,]    0    0    1

将频率向量转换为逻辑矩阵

3 个答案: