在R中插入符号中调整模型时如何保持参数恒定？

Question

The following code:

require(caret)
require(plyr)

portuguese_scores = read.table("https://raw.githubusercontent.com/JimGorman17/Datasets/master/student-por.csv",sep=";",header=TRUE, stringsAsFactors = FALSE)
portuguese_scores <- portuguese_scores[,!names(portuguese_scores) %in% c("school", "age", "G1", "G2")]
median_score <- summary(portuguese_scores$G3)['Median']
portuguese_scores$score_gte_than_median <- as.factor(median_score<=portuguese_scores$G3)
portuguese_scores <- portuguese_scores[,!names(portuguese_scores) %in% c("G3")]

portuguese_scores$sex <- as.numeric(mapvalues(portuguese_scores$sex, from = c("M", "F"), to = c(0, 1)))
portuguese_scores$address <- as.numeric(mapvalues(portuguese_scores$address, from = c("U", "R"), to = c(0, 1)))
portuguese_scores$famsize <- as.numeric(mapvalues(portuguese_scores$famsize, from = c("LE3", "GT3"), to = c(0, 1)))
portuguese_scores$Pstatus <- as.numeric(mapvalues(portuguese_scores$Pstatus, from = c("T", "A"), to = c(0, 1)))
portuguese_scores$Mjob <- as.numeric(mapvalues(portuguese_scores$Mjob, from = c("at_home","health","other","services","teacher"), to = c(0, 1,2,3,4)))
portuguese_scores$Fjob <- as.numeric(mapvalues(portuguese_scores$Fjob, from = c("at_home","health","other","services","teacher"), to = c(0, 1,2,3,4)))
portuguese_scores$reason <- as.numeric(mapvalues(portuguese_scores$reason, from = c("course","home","other","reputation"), to = c(0, 1,2,3)))
portuguese_scores$guardian <- as.numeric(mapvalues(portuguese_scores$guardian, from = c("father","mother","other"), to = c(0, 1,2)))
portuguese_scores$schoolsup <- as.numeric(mapvalues(portuguese_scores$schoolsup, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$famsup <- as.numeric(mapvalues(portuguese_scores$famsup, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$paid <- as.numeric(mapvalues(portuguese_scores$paid, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$activities <- as.numeric(mapvalues(portuguese_scores$activities, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$nursery <- as.numeric(mapvalues(portuguese_scores$nursery, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$higher <- as.numeric(mapvalues(portuguese_scores$higher, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$internet <- as.numeric(mapvalues(portuguese_scores$internet, from = c("no","yes"), to = c(0, 1)))
portuguese_scores$romantic <- as.numeric(mapvalues(portuguese_scores$romantic, from = c("no","yes"), to = c(0, 1)))

normalize <- function(x){ return( (x - min(x) )/( max(x) - min(x) ) )}
port_n <- data.frame(lapply(portuguese_scores[1:28], normalize), portuguese_scores[29])

set.seed(123)

train_sample <- sample(nrow(port_n), .9 * nrow(port_n))
port_train <- port_n[train_sample,]
port_test <- port_n[-train_sample,]

out1 <- train(port_train[,1:28], port_train[,29], method = "svmRadial")
out1

Generates the following output:

Support Vector Machines with Radial Basis Function Kernel 

584 samples
 28 predictor
  2 classes: 'FALSE', 'TRUE' 

No pre-processing
Resampling: Bootstrapped (25 reps) 
Summary of sample sizes: 584, 584, 584, 584, 584, 584, ... 
Resampling results across tuning parameters:

  C     Accuracy   Kappa      Accuracy SD  Kappa SD  
  0.25  0.7383930  0.4633478  0.02782725   0.05484469
  0.50  0.7382364  0.4637857  0.02883617   0.05763094
  1.00  0.7290191  0.4456935  0.02570423   0.05180727

Tuning parameter 'sigma' was held constant at a value of 0.02166535
Accuracy was used to select the optimal model using  the largest value.
The final values used for the model were sigma = 0.02166535 and C = 0.25. 

My Question:

• How can I hold c constant (at 0.25) and find the optimal sigma?

UPDATE (to all close voters):

• Do you see how caret kept sigma constant and optimized for C?
• I simply want to do the reverse. I want to keep C constant and optimize for sigma.
• And I believe that there are tuning parameters to help me do this, I just don't know the syntax.

Answer 1

In order to do this you need to use the tuneGrid argument. You need to create your own pairs for the parameters and then test them.

For example, since you want to test for C=0.25 on all occasions, you need to create a data.frame that looks like this:

svmGrid <- data.frame(C=rep(0.25,10), sigma=1:10/100)

This has the same value for C (0.25) and different values for sigma to optimize over. You need to provide these values for sigma yourself (this is only an example - use as many as you want).

In other words, according to the above data.frame, your svm model will be tested 10 times. Each time C will be constant and equal to 0.25 and sigma will take values from 0.01 to 0.1 with a step of 0.01. 10 tests will occur and the best combination will be chosen.

And then you run the model like this:

#adding the tuneGrid argument
out1 <- train(port_train[,1:28], port_train[,29], method = "svmRadial", tuneGrid=svmGrid)

Output:

> out1
Support Vector Machines with Radial Basis Function Kernel 

584 samples
 28 predictor
  2 classes: 'FALSE', 'TRUE' 

No pre-processing
Resampling: Bootstrapped (25 reps) 

Summary of sample sizes: 584, 584, 584, 584, 584, 584, ... 

Resampling results across tuning parameters:

  sigma  Accuracy   Kappa      Accuracy SD  Kappa SD  
  0.01   0.7297315  0.4417768  0.03082764   0.06044173
  0.02   0.7312643  0.4474754  0.03289345   0.06567919
  0.03   0.7301472  0.4468033  0.03618417   0.07187019
  0.04   0.7288286  0.4463212  0.03609275   0.07200966
  0.05   0.7281374  0.4466735  0.03569426   0.07055105
  0.06   0.7238098  0.4400315  0.03348371   0.06666725
  0.07   0.7213752  0.4364012  0.03467845   0.06849882
  0.08   0.7175949  0.4286502  0.04013475   0.08014780
  0.09   0.7042396  0.3981745  0.04346037   0.08864786
  0.10   0.6651296  0.3061489  0.06450228   0.14079631

Tuning parameter 'C' was held constant at a value of 0.25
Accuracy was used to select the optimal model using  the largest value.
The final values used for the model were sigma = 0.02 and C = 0.25. 

And you have your optimized sigma!