获取跟踪以将值存储在R中 [英] Getting trace to store a value in R
问题描述
我正在R中运行nloptr软件包，这是用于非线性优化的软件包.在nloptr包中，我正在使用cobyla()命令执行优化.现在，我可以使用maxeval参数指定算法的最大迭代次数，但是执行cobyla()命令时，它仅允许我查看最终评估的输出.但是，我希望能够在该方法的每次迭代中看到所有输出.
有人建议我使用trace()命令来查看所有中间输出，但是我不知道如何获取R来存储此信息.任何人都可以向我建议如何使用trace()(或任何其他方式)完成此操作.
这是我正在使用的代码:
library(nloptr)
#########################################################################################
###
#########################################################################################
f = function(x){
ans = cos(pi*x/5)+.2*sin(4*pi*x/5)
return(ans)
}
const = function(x){
ans = numeric(1)
ans[1] = -(sin(pi*x/5)+.2*cos(4*pi*x/5))
return(ans)
}
#########################################################################################
###
#########################################################################################
x0 = runif(1,0,10)
COB = cobyla(x0,f,hin=const,lower=0,upper=10,nl.info = TRUE,
control = list(xtol_rel = 1e-16, maxeval = 2000))
所以我希望能够存储输出1,2，...，maxeval
的运行cobyla()
所以我认为我需要做的就是使用
trace(f)
或
trace(cobyla)
在执行代码之前，但我不确定如何存储运行trace()命令的值
您最好的选择是将print_level选项设置为3，并将输出发送到sink，而无需调整nloptr代码并抓取它来检索您需要的内容.
请注意，要使用print_level，您不能使用包装器功能cobyla-而是必须下拉至nloptr函数.
最后，请注意cobyla和nloptr函数与algorithm = NLOPT_LN_COBYLA的行为不同-前者将不等式约束设为> = 0，而后者将不等式约束设为< = 0.
在下面的代码中，我要做的第一件事是使用核心nloptr函数，并证明它导致的结果与使用cobyla包装器代码相同.
然后，我设置print_level = 3并将其写出到文件中.优化完成后，我将读取该文件，并使用正则表达式从输出中提取解决方案路径.
这里是这样的:
library(tidyr)
library(nloptr)
library(assertthat)
library(dplyr)
library(ggplot2)
# starting parameter values
x0.hs100 = c(1, 2, 0, 4, 0, 1, 1)
# objective function
fn.hs100 = function(x) {
(x[1]-10)^2 + 5*(x[2]-12)^2 + x[3]^4 + 3*(x[4]-11)^2 + 10*x[5]^6 +
7*x[6]^2 + x[7]^4 - 4*x[6]*x[7] - 10*x[6] - 8*x[7]
}
# inequality constraints
# NOTE: nloptr with algorithm = NLOPT_LN_COBYLA takes g(x) <= 0
hin.hs100 = function(x) {
h = numeric(4)
h[1] = 127 - 2*x[1]^2 - 3*x[2]^4 - x[3] - 4*x[4]^2 - 5*x[5]
h[2] = 282 - 7*x[1] - 3*x[2] - 10*x[3]^2 - x[4] + x[5]
h[3] = 196 - 23*x[1] - x[2]^2 - 6*x[6]^2 + 8*x[7]
h[4] = -4*x[1]^2 - x[2]^2 + 3*x[1]*x[2] -2*x[3]^2 - 5*x[6] +11*x[7]
return(-h)
}
# compute the solution
sink(file = "Output/cobyla_trace.txt", type = "output")
S1 = nloptr(x0 = x0.hs100,
eval_f = fn.hs100,
eval_g_ineq = hin.hs100,
opts = list(algorithm = "NLOPT_LN_COBYLA",
xtol_rel = 1e-8, maxeval = 2000, print_level = 3))
sink()
# inequality constraints
# NOTE: cobyla takes g(x) >= 0
hin.hs100_minus = function(x) {
h = numeric(4)
h[1] = 127 - 2*x[1]^2 - 3*x[2]^4 - x[3] - 4*x[4]^2 - 5*x[5]
h[2] = 282 - 7*x[1] - 3*x[2] - 10*x[3]^2 - x[4] + x[5]
h[3] = 196 - 23*x[1] - x[2]^2 - 6*x[6]^2 + 8*x[7]
h[4] = -4*x[1]^2 - x[2]^2 + 3*x[1]*x[2] -2*x[3]^2 - 5*x[6] +11*x[7]
return(h)
}
# compute the solution
S2 = cobyla(x0.hs100, fn.hs100, hin = hin.hs100_minus,
nl.info = TRUE, control = list(xtol_rel = 1e-8, maxeval = 2000))
# check that both return the same solution
assertthat::are_equal(S1$solution, S2$par)
# get the solutions from the output file
solutionPath = readLines(con = file("Output/cobyla_trace.txt"))
# extract the solution path data out of the raw output
solutionPathParamRaw = solutionPath[grepl("^\tx", solutionPath)]
solutionPathParamMatch = gregexpr("(-)?[0-9]+(\\.[0-9]+)?", solutionPathParamRaw, perl = TRUE)
solutionPathParam = as.data.frame(
t(
sapply(
regmatches(
solutionPathParamRaw, solutionPathParamMatch
),
as.numeric, simplify = TRUE
)
)
)
# give the columns some names
names(solutionPathParam) = paste("x", seq(1, 7), sep = "")
solutionPathParam$IterationNum = seq(1, dim(solutionPathParam)[1])
# plot the solutions
solutionPathParam %>%
gather(Parameter, Solution, -IterationNum) %>%
ggplot(aes(x = IterationNum, y = Solution, group = Parameter, color = Parameter)) +
geom_line() +
theme_bw()
ggsave("Output/SolutionPath.png")
I am running the nloptr package in R which is a package for nonlinear optimization. Within the nloptr package I am using the cobyla() command to perform my optimization. Now, I can specify the maximim number of iterations of the algorithm using the maxeval parameter, but when cobyla() command executes it only allows me to see the output for the final evaluation. However, I would like to be able to see all of the outputs at every iteration of the method.
Someone recommended to me that I use the trace() command to be able to see all of the intermediary output but I cannot figure out how to get R to store this information. Can anyone suggest to me how to accomplish this using trace() (or any other way).
Here is the code I am working with:
library(nloptr)
#########################################################################################
###
#########################################################################################
f = function(x){
ans = cos(pi*x/5)+.2*sin(4*pi*x/5)
return(ans)
}
const = function(x){
ans = numeric(1)
ans[1] = -(sin(pi*x/5)+.2*cos(4*pi*x/5))
return(ans)
}
#########################################################################################
###
#########################################################################################
x0 = runif(1,0,10)
COB = cobyla(x0,f,hin=const,lower=0,upper=10,nl.info = TRUE,
control = list(xtol_rel = 1e-16, maxeval = 2000))
So I would like to be able to store the output for running cobyla() for iterations 1,2,...,maxeval
And so what I would think I would need to do is use
trace(f)
or
trace(cobyla)
before executing the code but I am not sure how to store the values of running the trace() command
Your best bet, without tweaking the nloptr code, is to set the print_level option to 3, send the output to a sink and scrape it to retrieve what you need.
Note that to use print_level you cannot use the wrapper function cobyla -- instead you have to drop down to the nloptr function.
Lastly, note that the cobyla and the nloptr function with algorithm = NLOPT_LN_COBYLA, behave differently -- the former takes inequality constraints to be >= 0 and the latter takes them to be <=0.
In the code below, the first thing I do is to use the core nloptr function and show that it leads to the same answer as using the cobyla wrapper code.
Then, I set print_level = 3 and write that out to a file. Once the optimization is done, I read that file in, and use regular expressions to extract the solution path from the output.
Here is what that looks like:
library(tidyr)
library(nloptr)
library(assertthat)
library(dplyr)
library(ggplot2)
# starting parameter values
x0.hs100 = c(1, 2, 0, 4, 0, 1, 1)
# objective function
fn.hs100 = function(x) {
(x[1]-10)^2 + 5*(x[2]-12)^2 + x[3]^4 + 3*(x[4]-11)^2 + 10*x[5]^6 +
7*x[6]^2 + x[7]^4 - 4*x[6]*x[7] - 10*x[6] - 8*x[7]
}
# inequality constraints
# NOTE: nloptr with algorithm = NLOPT_LN_COBYLA takes g(x) <= 0
hin.hs100 = function(x) {
h = numeric(4)
h[1] = 127 - 2*x[1]^2 - 3*x[2]^4 - x[3] - 4*x[4]^2 - 5*x[5]
h[2] = 282 - 7*x[1] - 3*x[2] - 10*x[3]^2 - x[4] + x[5]
h[3] = 196 - 23*x[1] - x[2]^2 - 6*x[6]^2 + 8*x[7]
h[4] = -4*x[1]^2 - x[2]^2 + 3*x[1]*x[2] -2*x[3]^2 - 5*x[6] +11*x[7]
return(-h)
}
# compute the solution
sink(file = "Output/cobyla_trace.txt", type = "output")
S1 = nloptr(x0 = x0.hs100,
eval_f = fn.hs100,
eval_g_ineq = hin.hs100,
opts = list(algorithm = "NLOPT_LN_COBYLA",
xtol_rel = 1e-8, maxeval = 2000, print_level = 3))
sink()
# inequality constraints
# NOTE: cobyla takes g(x) >= 0
hin.hs100_minus = function(x) {
h = numeric(4)
h[1] = 127 - 2*x[1]^2 - 3*x[2]^4 - x[3] - 4*x[4]^2 - 5*x[5]
h[2] = 282 - 7*x[1] - 3*x[2] - 10*x[3]^2 - x[4] + x[5]
h[3] = 196 - 23*x[1] - x[2]^2 - 6*x[6]^2 + 8*x[7]
h[4] = -4*x[1]^2 - x[2]^2 + 3*x[1]*x[2] -2*x[3]^2 - 5*x[6] +11*x[7]
return(h)
}
# compute the solution
S2 = cobyla(x0.hs100, fn.hs100, hin = hin.hs100_minus,
nl.info = TRUE, control = list(xtol_rel = 1e-8, maxeval = 2000))
# check that both return the same solution
assertthat::are_equal(S1$solution, S2$par)
# get the solutions from the output file
solutionPath = readLines(con = file("Output/cobyla_trace.txt"))
# extract the solution path data out of the raw output
solutionPathParamRaw = solutionPath[grepl("^\tx", solutionPath)]
solutionPathParamMatch = gregexpr("(-)?[0-9]+(\\.[0-9]+)?", solutionPathParamRaw, perl = TRUE)
solutionPathParam = as.data.frame(
t(
sapply(
regmatches(
solutionPathParamRaw, solutionPathParamMatch
),
as.numeric, simplify = TRUE
)
)
)
# give the columns some names
names(solutionPathParam) = paste("x", seq(1, 7), sep = "")
solutionPathParam$IterationNum = seq(1, dim(solutionPathParam)[1])
# plot the solutions
solutionPathParam %>%
gather(Parameter, Solution, -IterationNum) %>%
ggplot(aes(x = IterationNum, y = Solution, group = Parameter, color = Parameter)) +
geom_line() +
theme_bw()
ggsave("Output/SolutionPath.png")
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