【发布时间】：2021-12-07 10:12:27
【问题描述】：
我正在努力将优化函数 optim 循环到我的整个数据帧。 下面你可以找到我的代码的一部分，当我使用标量作为 Strike09 = 1142.757 时，我的优化问题有效，但现在我想在我的整个数据集上解决这个优化问题，其中 Strike09、Strike095、...、blackscholesc11 是数据集，我想为数据集的每一行（对应于日期）优化我的函数的 5 个参数。
下面要优化的函数和使用标量作为硬参数时的工作优化问题示例
error_vector_price <- function(Strike09, Strike095, Strike1, Strike105, Strike11, Liborrate, par_x, TimetoMaturity, SPX, blackscholesc09, blackscholesc095, blackscholesc1, blackscholesc105, blackscholesc11){
MixtureCall <- function(Strike, Liborrate, par_x, TimetoMaturity, SPX){
alpha1 = log(SPX) + (par_x[2]-0.5*(par_x[4]^2)*(par_x[4]^2))*TimetoMaturity
alpha2 = log(SPX) + (par_x[3]-0.5*(par_x[5]^2)*(par_x[5]^2))*TimetoMaturity
beta1 = (par_x[4]^2)*(TimetoMaturity^0.5)
beta2 = (par_x[5]^2)*(TimetoMaturity^0.5)
theta = (par_x[1]^2)/(1+par_x[1]^2)
D1 = (-log(Strike) + alpha1 + (beta1 ^ 2)) / beta1
D2 = D1 - beta1
D3 = (-log(Strike) + alpha2 + (beta2 ^ 2)) / beta2
D4 = D3 - beta2
nD1 = sapply(D1, function(x) pnorm(x))
nD2 = sapply(D2, function(x) pnorm(x))
nD3 = sapply(D3, function(x) pnorm(x))
nD4 = sapply(D4, function(x) pnorm(x))
call1 = exp(alpha1 + beta1 * beta1 / 2) * nD1 - Strike * nD2
call2 = exp(alpha2 + beta2 * beta2 / 2) * nD3 - Strike * nD4
InstRate = (log(1 + Liborrate * TimetoMaturity)) / TimetoMaturity
Mixturecall = exp(-InstRate * TimetoMaturity) * (theta * call1 + (1 - theta) * call2)
return(Mixturecall)
}
MixturePut <- function(Strike, Liborrate, par_x, TimetoMaturity, SPX){
alpha1 = log(SPX) + (par_x[2]-0.5*(par_x[4]^2)*(par_x[4]^2))*TimetoMaturity
alpha2 = log(SPX) + (par_x[3]-0.5*(par_x[5]^2)*(par_x[5]^2))*TimetoMaturity
beta1 = (par_x[4]^2)*(TimetoMaturity^0.5)
beta2 = (par_x[5]^2)*(TimetoMaturity^0.5)
theta = (par_x[1]^2)/(1+par_x[1]^2)
D1 = (-log(Strike) + alpha1 + (beta1 ^ 2)) / beta1
D2 = D1 - beta1
D3 = (-log(Strike) + alpha2 + (beta2 ^ 2)) / beta2
D4 = D3 - beta2
nD1 = sapply(D1, function(x) pnorm(x))
nD2 = sapply(D2, function(x) pnorm(x))
nD3 = sapply(D3, function(x) pnorm(x))
nD4 = sapply(D4, function(x) pnorm(x))
call1 = exp(alpha1 + beta1 * beta1 / 2) * nD1 - Strike * nD2
call2 = exp(alpha2 + beta2 * beta2 / 2) * nD3 - Strike * nD4
InstRate = (log(1 + Liborrate * TimetoMaturity)) / TimetoMaturity
Mixturecall = exp(-InstRate * TimetoMaturity) * (theta * call1 + (1 - theta) * call2)
fwdprice = theta * exp(alpha1 + beta1 * beta1 / 2) + (1 - theta) * exp(alpha2 + beta2 * beta2 / 2)
Mixtureput = Mixturecall + (Strike - fwdprice) / (1 + Liborrate * TimetoMaturity)
return(Mixtureput)
}
model_price_vector09 <- MixturePut(Strike09, Liborrate, par_x, TimetoMaturity, SPX)
model_price_vector095 <- MixturePut(Strike095, Liborrate, par_x, TimetoMaturity, SPX)
model_price_vector1 <- MixturePut(Strike1, Liborrate, par_x, TimetoMaturity, SPX)
model_price_vector105 <- MixtureCall(Strike105, Liborrate, par_x, TimetoMaturity, SPX)
model_price_vector11 <- MixtureCall(Strike11, Liborrate, par_x, TimetoMaturity, SPX)
error_vector_price <- (((blackscholesc09 - model_price_vector09)^2+(blackscholesc095 - model_price_vector095)^2+(blackscholesc1 - model_price_vector1)^2+(blackscholesc105 - model_price_vector105)^2+(blackscholesc11 - model_price_vector11)^2))
return(error_vector_price)
}
x_seed <- c(0.64,-0.57,0.25,0.53,0.36)
results <- optim( x_seed , fn = error_vector_price, gr = NULL, Strike09 = 1142.757, Strike095 = 1206.2435, Strike1 = 1269.73, Strike105 = 1333.216, Strike11 = 1396.703, Liborrate = 0.0505750, TimetoMaturity = 0.25, SPX = 1269.73, blackscholesc09=24.995126, blackscholesc095=37.78425, blackscholesc1=57.87691, blackscholesc105=33.47135, blackscholesc11=12.671979, method = "L-BFGS-B", upper = (1))
x_star <- results$par
x_star
我现在想在我的数据集上循环这个优化问题，并为每一行获取 5 个参数来最小化我的函数。 我的数据集由 13 个变量的 152 个 obs 组成，这 13 个变量是我的优化问题的硬参数（Strike09、Strike095、Strike1、Strike105、Strike11、Liborrate、TimetoMaturity、SPX、blackscholesc09、blackscholesc095、blackscholesc1、blackscholesc105、blackscholesc11）。 这个想法如下所示，但随后的循环不起作用，我不确定循环是否是最好的方法
MIN_DATA1 =data.frame(Strike09 <- c (1142.757, 1090.971, 1111.644, 1138.833), Strike095 <- c (1206.2435, 1151.5805,1173.4020, 1202.1015), Strike1 <- c (1269.73, 1212.19, 1235.16, 1265.37),Strike105 <- c (1333.216, 1272.800, 1296.918, 1328.639), Strike11 <- c (1396.703,1333.409, 1358.676, 1391.907), Liborrate <- c (0.0505750, 0.0500500, 0.0497078, 0.0496969), TimetoMaturity <- c (0.25, 0.25, 0.25, 0.25), SPX <- c (1269.73, 1212.19, 1235.16, 1265.37), blackscholesc09 <- c(24.995126, 34.905765, 32.103535, 29.686353), blackscholesc095 <- c(37.78425, 50.31239, 45.41761, 43.50957), blackscholesc1 <- c(57.87691, 69.78892, 65.36423, 64.41497), blackscholesc105 <- c(33.47135, 44.10261, 44.12110, 41.11879), blackscholesc11<- c(12.671979, 21.055396, 21.175705, 18.883918))
for (i in 1:length(MIN_DATA)){
results[i] <- optim( x_seed , fn = error_vector_price, gr = NULL, Strike09[i], Strike095[i], Strike1[i], Strike105[i], Strike11[i], Liborrate[i], TimetoMaturity[h], SPX[i], blackscholesc09[i], blackscholesc095[i], blackscholesc1[i], blackscholesc105[i], blackscholesc11[i], method = "L-BFGS-B", upper = (0.9))
}
results
我希望我的问题足够清楚。
【问题讨论】：
-
请提供一个可重复的示例，以便您最大限度地获得帮助：How to make a great R reproducible example。
-
您可以通过嵌套来完成此操作，然后使用
purrr::map2或mapply（以一种简单的方式）。我们无能为力，因为您没有为我们提供最小的可重现示例。但是请查看tidyr::nest。 -
@Mossa，感谢您的评论。我试图添加一个可重现的示例，因为您的建议的最终实施对我来说不是 100% 清楚。再次感谢您。
标签： r loops optimization