在R中创建多个计数器矩阵

Question

So, my goal is to take an input vector and to make an output matrix of different counters. So every time a value appears in my inputs, I want to find that counter and iterate it by 1. I understand that I'm not good at explaining this, so I illustrated a simple version below. However, I want to make 2 changes which I will enumerate after the example so that it makes sense.

nums = c(1,2,3,4,5,1,2,4,3,5)
unis = unique(nums)
counter = matrix(NA, nrow = length(nums), ncol = length(unis))
colnames(counter) = unis
for (i in 1:length(nums)){
  temp = nums[i]
  if (i == 1){
    counter[1,] = 0
    counter[1,temp] = 1
  } else {
    counter[i,] = counter[i-1,]
    counter[i,temp] = counter[i-1,temp]+1
  }
}
counter

which outputs

 > counter
      1 2 3 4 5
 [1,] 1 0 0 0 0
 [2,] 1 1 0 0 0
 [3,] 1 1 1 0 0
 [4,] 1 1 1 1 0
 [5,] 1 1 1 1 1
 [6,] 2 1 1 1 1
 [7,] 2 2 1 1 1
 [8,] 2 2 1 2 1
 [9,] 2 2 2 2 1
[10,] 2 2 2 2 2

The 2 modifications. 1) Since the real data is much larger, I would want to do this using apply or however people who know R better than me says it should be done. 2) Whereas the input is a vector where each element is only an element, how could this be generalized if an element of a vector was a tuple? For example (if nums was a tuple of 4 and 5, then it would iterate both in that step and the last line of the output would then be 2,2,2,3,2)

Thanks and if you don't understand please ask questions and I'll try to clarify

Answer 1

Using the Matrix package (which ships with a standard installation of R)

nums <- c(1,2,3,4,5,1,2,4,3,5)
apply(Matrix::sparseMatrix(i=seq_along(nums), j=nums), 2, cumsum)
#      [,1] [,2] [,3] [,4] [,5]
#  [1,]    1    0    0    0    0
#  [2,]    1    1    0    0    0
#  [3,]    1    1    1    0    0
#  [4,]    1    1    1    1    0
#  [5,]    1    1    1    1    1
#  [7,]    2    2    1    1    1
#  [8,]    2    2    1    2    1
#  [9,]    2    2    2    2    1
# [10,]    2    2    2    2    2

Note that this behaves a bit differently in a couple of ways from thelatemail's suggested solution. Which behavior you prefer will depend on what you are using this for.

Here's a small example that illustrates the differences:

nums <- c(5,2,1,1)

# My suggestion
apply(Matrix::sparseMatrix(i=seq_along(nums), j=nums), 2, cumsum)
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    0    0    0    0    1
# [2,]    0    1    0    0    1
# [3,]    1    1    0    0    1
# [4,]    2    1    0    0    1

# @thelatemail's suggestion
sapply(unique(nums), function(x) cumsum(nums==x) )
#      [,1] [,2] [,3]
# [1,]    1    0    0
# [2,]    1    1    0
# [3,]    1    1    1
# [4,]    1    1    2

---

For your second question, you could do something like this:

nums <- list(1,2,3,4,5,1,2,4,3,c(4,5))

ii <- rep(seq_along(nums), times=lengths(nums)) ## lengths() is in R>=3.2.0
jj <- unlist(nums)
apply(Matrix::sparseMatrix(i=ii, j=jj), 2, cumsum)
#       [,1] [,2] [,3] [,4] [,5]
#  [1,]    1    0    0    0    0
#  [2,]    1    1    0    0    0
#  [3,]    1    1    1    0    0
#  [4,]    1    1    1    1    0
#  [5,]    1    1    1    1    1
#  [6,]    2    1    1    1    1
#  [7,]    2    2    1    1    1
#  [8,]    2    2    1    2    1
#  [9,]    2    2    2    2    1
# [10,]    2    2    2    3    2

Answer 2

For your first query, you can get there with something like:

sapply(unique(nums), function(x) cumsum(nums==x) )

 #      [,1] [,2] [,3] [,4] [,5]
 # [1,]    1    0    0    0    0
 # [2,]    1    1    0    0    0
 # [3,]    1    1    1    0    0
 # [4,]    1    1    1    1    0
 # [5,]    1    1    1    1    1
 # [6,]    2    1    1    1    1
 # [7,]    2    2    1    1    1
 # [8,]    2    2    1    2    1
 # [9,]    2    2    2    2    1
 #[10,]    2    2    2    2    2

Answer 3

Another idea:

do.call(rbind, Reduce("+", lapply(nums, tabulate, max(unlist(nums))), accumulate = TRUE))
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    1    0    0    0    0
# [2,]    1    1    0    0    0
# [3,]    1    1    1    0    0
# [4,]    1    1    1    1    0
# [5,]    1    1    1    1    1
# [6,]    2    1    1    1    1
# [7,]    2    2    1    1    1
# [8,]    2    2    1    2    1
# [9,]    2    2    2    2    1
#[10,]    2    2    2    2    2

And generally:

x = list(1, 3, 6, c(6, 3), 2, c(4, 6, 1), c(1, 2), 3)
do.call(rbind, Reduce("+", lapply(x, tabulate, max(unlist(x))), accumulate = TRUE))
#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]    1    0    0    0    0    0
#[2,]    1    0    1    0    0    0
#[3,]    1    0    1    0    0    1
#[4,]    1    0    2    0    0    2
#[5,]    1    1    2    0    0    2
#[6,]    2    1    2    1    0    3
#[7,]    3    2    2    1    0    3
#[8,]    3    2    3    1    0    3