r语言 - nloptr在达到maxeval后没有返回到目前为止的最低解

Question

我正在寻找具有非线性约束的非线性函数的良好(不一定是最好的)解决方案。我已经搜索了最合适的方法，我认为nlopt是GN_ISRES算法的最佳选择。

函数是在以下约束条件下最小化两组点之间的距离

• 所有点必须为整数
• 点的坐标必须是偶数或奇数
• 没有一个点可以与另一个
相同

下面是我的可复制示例代码，我的主要问题是，即使从一个坏的解决方案开始，当达到maxeval时，存储的解决方案仍然是开始的解决方案，而不是迭代的最佳解决方案。

为什么会出现这种情况，是否有一种方法可以使它返回到目前为止的最佳解决方案?

library(nloptr)
coords = data.frame(
name = c(1,2,3,4,5),
lat = c(45,46,48,42,36),
lon = c(1,3,2,5,4)
)
hexbin_coords = data.frame(
name = c(1,2,3,4,5),
x = c(1,2,3,4,5),
y = c(41,42,43,44,45)
)
eval_f <- function(z, lat, lon) {
y = z[((length(z)/2)+1):length(z)]
x = z[1:(length(z)/2)]
distance = sum(sqrt((x - lat)^2 + (y - lon)^2))
return(distance)
}
## constraints
eval_g <- function(z, lat, lon) {
y = z[((length(z)/2)+1):length(z)]
x = z[1:(length(z)/2)]
sum1 = sum((x - y) %% 2)
sum2 = sum(z %% 1)
#sum2 = 0
sum3 = any(duplicated(apply(data.frame(x,rep("a",length(x)),y), 1, paste, collapse="")))
return(sum1 + sum2 + sum3)
}
opts <- list( "algorithm"= "NLOPT_GN_ISRES",
"xtol_rel"= 1.0e-7,
"maxeval"= 10,
"print_level" = 3)
sol <- nloptr(x0 = c(hexbin_coords$x, hexbin_coords$y),
eval_f = eval_f,
eval_g_eq = eval_g,
lb = rep(0,10),
ub = rep(50,10),
opts = opts,
lat = coords$lat,
lon = coords$lon
)
print(sol)

Answer 1

不幸的是，nloptr不直接提供此功能，因此唯一的选择是通过print_level=3获取值的输出:

library(stringr) # For string matching and extracting
sol.output <- capture.output(sol <- nloptr(x0 = c(hexbin_coords$x, hexbin_coords$y),
eval_f = eval_f,
eval_g_eq = eval_g,
lb = rep(0,10),
ub = rep(50,10),
opts = opts,
lat = coords$lat,
lon = coords$lon)) # capture the "print" output to a variable
sol.hist <- sol.output[grepl("tx",sol.output)] #get all lines starting with a tab and x (Those are the lines with the variables)
sol.hist <- sapply(stringr::str_extract_all(sol.hist,"[0-9]+\.[0-9]+"),as.numeric) # Extract all numbers and convert to numeric
rownames(sol.hist) <- paste0("Iteration_",seq_len(nrow(sol.hist))) #set some descriptive names
colnames(sol.hist) <- paste0("Variable_",seq_len(ncol(sol.hist)))
sol.hist
Iteration_1 Iteration_2 Iteration_3 Iteration_4 Iteration_5 Iteration_6 Iteration_7 Iteration_8
Variable_1            1   41.870830   12.911006    2.279217    2.857534   24.370609   21.847667   11.074417
Variable_2            2   28.649218   20.234235    9.663595   37.479727    7.707112   17.469032   48.587274
Variable_3            3   28.547109   48.541501   29.354267   24.670474   16.181966    4.559475   21.171942
Variable_4            4    4.529856    0.221904   15.666075   13.072797   24.905793   39.300537   29.968914
Variable_5            5   13.938569   25.595069   43.422727   15.680567   27.393603    8.527784   47.716471
Variable_6           41   17.493988   31.714159   16.504085   48.342804   34.445824   32.355841    2.347884
Variable_7           42   20.647009   42.209615   23.079238   22.102841    7.842979   27.584622    7.954957
Variable_8           43   33.884236   24.857512    4.001447   39.029162   39.050927   38.151746   19.158068
Variable_9           44   39.405297    4.173269   48.558713   48.976114   48.358011   30.155023    6.904914
Variable_10          45   20.769793   29.435024    0.124530   21.440616   24.843518   28.395304    6.854763

使用这种方法，您不仅可以获得"最好的";价值是所有价值的完整历史。因此，您还可以绘制优化图形等