我有一个矩阵（初始化为零）和一组索引。如果索引中的第i个值是j，那么我想将矩阵的第（j，i）个条目设置为1.
例如：
> m = matrix(0, 10, 7)
> indices
[1] 2 9 3 4 5 1 10
结果应该是
> result
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 0 0 0 0 0 1 0
[2,] 1 0 0 0 0 0 0
[3,] 0 0 1 0 0 0 0
[4,] 0 0 0 1 0 0 0
[5,] 0 0 0 0 1 0 0
[6,] 0 0 0 0 0 0 0
[7,] 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0
[9,] 0 1 0 0 0 0 0
[10,] 0 0 0 0 0 0 1
我稍微询问了一点related question，它使用的是矢量而不是矩阵。这个问题是否有类似的简单解决方案？
答案 0 :(得分：1)
## OP's example data
m = matrix(0, 10, 7)
j <- c(2, 9, 3, 4, 5, 1, 10)
## Construct a two column matrix of indices (1st column w. rows & 2nd w. columns)
ij <- cbind(j, seq_along(j))
## Use it to subassign into the matrix
m[ij] <- 1
m
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,] 0 0 0 0 0 1 0
# [2,] 1 0 0 0 0 0 0
# [3,] 0 0 1 0 0 0 0
# [4,] 0 0 0 1 0 0 0
# [5,] 0 0 0 0 1 0 0
# [6,] 0 0 0 0 0 0 0
# [7,] 0 0 0 0 0 0 0
# [8,] 0 0 0 0 0 0 0
# [9,] 0 1 0 0 0 0 0
# [10,] 0 0 0 0 0 0 1
答案 1 :(得分：1)
对于记录，使用sapply：
indices <- c(2, 9, 3, 4, 5, 1, 10)
sapply(indices, tabulate, nbins = 10)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,] 0 0 0 0 0 1 0
# [2,] 1 0 0 0 0 0 0
# [3,] 0 0 1 0 0 0 0
# [4,] 0 0 0 1 0 0 0
# [5,] 0 0 0 0 1 0 0
# [6,] 0 0 0 0 0 0 0
# [7,] 0 0 0 0 0 0 0
# [8,] 0 0 0 0 0 0 0
# [9,] 0 1 0 0 0 0 0
# [10,] 0 0 0 0 0 0 1
对于小型数据集，您可能没有注意到性能差异，但Josh的答案（使用矩阵索引）肯定会更快 ，即使您在此处更改我的答案以使用vapply而不是sapply。