Question

我正在尝试规范化指定维度内的数组（通过标准化，我的意思是将每个元素除以相关维度上的总和）。

我想以某种方式自动化计算，其中给出了规范化数组的维度，并且每个都使用相同的函数来生成输出。虽然我可以生成所需的金额，但我不能用一致的技术对它们进行标准化。

Mwe

a = array(1:8, c(2,2,2))

dm = 1 # dimension to sum over
ndim = length(dim(a))
margin = setdiff(seq_len(ndim), dm) # dimensions to keep in apply
a / rep(apply(a, margin, sum), each=2)

, , 1

          [,1]      [,2]
[1,] 0.3333333 0.4285714
[2,] 0.6666667 0.5714286

, , 2

          [,1]      [,2]
[1,] 0.4545455 0.4666667
[2,] 0.5454545 0.5333333


dm = 3 # dimension to sum over
ndim = length(dim(a))
margin = setdiff(seq_len(ndim), dm) 
a / rep(apply(a, margin, sum), times=2)
# but for this dimension `times`, instead of `each` was used

, , 1

          [,1]      [,2]
[1,] 0.1666667 0.3000000
[2,] 0.2500000 0.3333333

, , 2

          [,1]      [,2]
[1,] 0.8333333 0.7000000
[2,] 0.7500000 0.6666667


#for dim=2 the denominator is more difficult to generate
dm = 2 # dimension to sum over
a / c(4,6,4,6,12,14,12,14)

虽然最好不要使用循环（包括应用等），但欢迎使用任何解决方案。感谢

根据评论中的要求，关于矩阵的一个例子（如果它只是2x2阵列，这将更直接）

a = array(1:4, c(2,2))
a
#     [,1] [,2]
#[1,]    1    3
#[2,]    2    4
dm = 1 # dimension to sum over
ndim = length(dim(a))
margin = setdiff(seq_len(ndim), dm) # dimensions to keep in apply
apply(a, margin, sum)
# [1] 3 7
a / rep(apply(a, margin, sum), each=2)
#          [,1]      [,2]
#[1,] 0.3333333 0.4285714
#[2,] 0.6666667 0.5714286

dm = 2 # dimension to sum over
ndim = length(dim(a))
margin = setdiff(seq_len(ndim), dm) # dimensions to keep in apply
apply(a, margin, sum)
# [1] 4 6
a / rep(apply(a, margin, sum), times=2)
#          [,1]      [,2]
#[1,] 0.2500000 0.7500000
#[2,] 0.3333333 0.6666667

Answer 1

您可以使用sweep：

获取所寻求的一致性

normalize <- function(a,dm){
  ndim <- length(dim(a))
  margin <- setdiff(seq_len(ndim),dm)
  sweep(a,margin,apply(a, margin, sum),"/")
}

然后normalize(a,1)，normalize(a,2)和normalize(a,3)简化了您想要的数组。

规范化数组

1 个答案: