我在R做了一些优化,并与我需要编写一个函数,返回一个雅可比。这是一个非常简单的雅可比 - 只是零和一个 - 但我想快速,干净地填充它。我目前的代码工作,但很sl。。我有一个四维概率数组。通过i, j, k, l
索引尺寸。我的约束是,对于每个i, j, k
,概率超过指数l
之和必须等于1干净的方式来计算雅可比阵的总和
我计算我的约束向量是这样的:
get_prob_array_from_vector <- function(prob_vector, array_dim) {
return(array(prob_vector, array_dim))
}
constraint_function <- function(prob_vector, array_dim) {
prob_array <- get_prob_array_from_vector(prob_vector, array_dim)
prob_array_sums <- apply(prob_array, MARGIN=c(1, 2, 3), FUN=sum)
return(as.vector(prob_array_sums) - 1) # Should equal zero
}
我的问题是:什么是清洁,快速计算方式雅可比的as.vector(apply(array(my_input_vector, array_dim), MARGIN=c(1, 2, 3), FUN=sum))
- 即我的constraint_function
在上面的代码 - 相对于my_input_vector
?
这是我的马虎溶液(我检查针对雅可比功能正确性从numDeriv包):
library(numDeriv)
array_dim <- c(5, 4, 3, 3)
get_prob_array_from_vector <- function(prob_vector, array_dim) {
return(array(prob_vector, array_dim))
}
constraint_function <- function(prob_vector, array_dim) {
prob_array <- get_prob_array_from_vector(prob_vector, array_dim)
prob_array_sums <- apply(prob_array, MARGIN=c(1, 2, 3), FUN=sum)
return(as.vector(prob_array_sums) - 1)
}
constraint_function_jacobian <- function(prob_vector, array_dim) {
prob_array <- get_prob_array_from_vector(prob_vector, array_dim)
jacobian <- matrix(0, Reduce("*", dim(prob_array)[1:3]), length(prob_vector))
## Must be a faster, clearner way of populating jacobian
for(i in seq_along(prob_vector)) {
dummy_vector <- rep(0, length(prob_vector))
dummy_vector[i] <- 1
dummy_array <- get_prob_array_from_vector(dummy_vector, array_dim)
dummy_array_sums <- apply(dummy_array, MARGIN=c(1, 2, 3), FUN=sum)
jacobian_row_idx <- which(dummy_array_sums != 0, arr.ind=FALSE)
stopifnot(length(jacobian_row_idx) == 1)
jacobian[jacobian_row_idx, i] <- 1
} # Is there a fast, readable one-liner that does the same as this for loop?
stopifnot(sum(jacobian) == length(prob_vector))
stopifnot(all(jacobian == 0 | jacobian == 1))
return(jacobian)
}
## Example of a probability array satisfying my constraint
my_prob_array <- array(0, array_dim)
for(i in seq_len(array_dim[1])) {
for(j in seq_len(array_dim[2])) {
my_prob_array[i, j, , ] <- diag(array_dim[3])
}
}
my_prob_array[1, 1, , ] <- 1/array_dim[3]
my_prob_array[2, 1, , ] <- 0.25 * (1/array_dim[3]) + 0.75 * diag(array_dim[3])
my_prob_vector <- as.vector(my_prob_array) # Flattened representation of my_prob_array
should_be_zero_vector <- constraint_function(my_prob_vector, array_dim)
is.vector(should_be_zero_vector)
all(should_be_zero_vector == 0) # Constraint is satistied
## Check constraint_function_jacobian for correctness using numDeriv
jacobian_analytical <- constraint_function_jacobian(my_prob_vector, array_dim)
jacobian_numerical <- jacobian(constraint_function, my_prob_vector, array_dim=array_dim)
max(abs(jacobian_analytical - jacobian_numerical)) # Very small
我的功能采取prob_vector
作为输入 - 即,我的概率阵列的扁平表示 - - 因为优化函数需要向量参数。
谢谢你,这是一个坚实的改进! – Adrian