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我正在尝试使用 R 中的 optim() 函数来解决一个简单的问题，但我在如何实现它时遇到了一些问题：

 e=tot_obs/(sum(Var1)+sum(Var2)+sum(Var3)+sum(Var4))
 output=(Var1+Var2+Var3+Var4)*e

我知道观察结果和所有变量的总数。

# Fake datasets   
# Considering that this are the observations c(1000,250,78,0,0,90)

#Known data
total_observations=1418
var1=c(1,0.3,0.5,0.01,0.05,0.6)
var2=c(500,40,40,0,0,100)
var3=c(1,0.1,0.2,0,0.1,0)
var4=c(2,0.04,0.003,0.003,0,0.05)

#Function
e=total_observations/(sum(var1)+sum(var2)+sum(var3)+sum(var4))
output=(var1+var2+var3+var4)*e

我可以在观察结果和输出之间做一个简单的关联，结果很好（~0.90）。这个给我0.97。

但现在我想测试为每个变量分配不同权重的效果。

e=tot_obs/(sum(w1*Var1)+sum(w2*Var2)+sum(w3*Var3)+sum(w4*Var4))
output=(w1*Var1+w2*Var2+w3*Var3+w4*Var4)*e
where w1+w2+w3+w4=1
and cor(observations,output)~1

我试图使用 optim() 函数，但是我完全迷路了。如果有人可以帮助我或为我指出如何做到这一点的一些好的参考资料，我将不胜感激。

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1 回答 1

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您需要solnp在包中使用函数，Rsolnp因为它允许基于相等的约束。

这个想法是建立一个函数来最小化和你的约束相等函数。

Fun <- function(param){
    e <- total_observations/(sum(param[1]*var1)+sum(param[2]*var2)+sum(param[3]*var3)+sum(param[4]*var4))
    output <- (param[1]*var1 + param[2]*var2 + param[3]*var3 + param[4]*var4)/e
    -cor(output, observations) #We want to maximize cor and therefore minimize -cor
    }

eqn <- function(param){sum(param)}

使用您的示例数据：

observations <- c(1000,250,78,0,0,90)
total_observations=1418
var1=c(1,0.3,0.5,0.01,0.05,0.6)
var2=c(500,40,40,0,0,100)
var3=c(1,0.1,0.2,0,0.1,0)
var4=c(2,0.04,0.003,0.003,0,0.05)

您的优化：

solnp(c(.1,.2,.3,.4),fun=Fun, eqfun=eqn, eqB=1)

Iter: 1 fn: -0.9793  Pars:  0.1395748 0.0008403 0.3881053 0.4714796
Iter: 2 fn: -0.9793  Pars:  0.1395531 0.0008406 0.3881409 0.4714653
solnp--> Completed in 2 iterations
$pars
[1] 0.1395530843 0.0008406453 0.3881409239 0.4714653466

$convergence
[1] 0

$values
[1] -0.9729894 -0.9793458 -0.9793458

$lagrange
             [,1]
[1,] 2.521018e-06

$hessian
           [,1]        [,2]        [,3]        [,4]
[1,]  0.4843670   5.0498894 -0.08329380  0.39560040
[2,]  5.0498894 699.5317385 -2.38763807 -0.65610831
[3,] -0.0832938  -2.3876381  0.91837245 -0.09486495
[4,]  0.3956004  -0.6561083 -0.09486495  0.43979850

$ineqx0
NULL

$nfuneval
[1] 709

$outer.iter
[1] 2

$elapsed
Time difference of 0.2371149 secs

如果将其保存到变量res中，则您要查找的内容存储在res$pars.

于 2013-09-12T09:30:20.007 回答