我正在通过视频讲座和作业问题自学CalTech的“从数据中学习”课程。我遇到了一个家庭作业问题：我编写的代码似乎可以正常工作，但是我获得的答案与解决方案密钥中提供的答案不匹配。问题如下。
在以下问题中，我们将PLA与SVM进行比较，硬边界为1 线性可分离的数据集。对于每次运行，您将创建自己的 目标函数f和数据集D。取d = 2并选择一条随机线 在飞机上作为目标函数f（通过随机抽取两个， [-1上的均匀分布点1]×[-1； 1]然后走线 通过它们），其中线的一侧映射到+1，而 其他映射到-1。选择数据集的输入xn作为随机数 X中的点= [-1; 1]×[-1； 1]，然后评估目标函数 每个xn以获得对应的输出yn。如果所有数据点都打开 在生产线的一侧，放弃运行并开始新的运行。启动PLA 与全零向量并为每个PLA选择错误分类的点 随机迭代。运行PLA查找最终假设gPLA和 衡量f和gPLA之间的差异，因为P [f（x）6 = gPLA（x）]（您 可以准确地计算出该值，也可以通过生成一个 足够大的独立点集进行评估）。现在开始 通过对SVM进行相同数据求解来找到最终假设gSVM min_w; b 1 / 2w ^ Tw s.t.使用二次编程yn *（w ^ T * xn + b）≥1 关于原始或双重问题。衡量f之间的分歧 和gSVM作为P [f（x）！= gSVM（x）]，并计算支持数量 每次运行中得到的向量。
这些问题需要运行1000次迭代 针对样本中的N = 10和N = 100个数据点进行实验， 计算支持向量机方法表现更好的时间比例 而不是PLA方法，以及支持向量的数量。
我的尝试：我尝试使用quadprog软件包对R中的问题进行二次编程，以解决给定优化问题的对偶。 对于N = 100（3），我得到了支持向量平均数的正确答案，但是我对SVM性能优于PLA的比例的回答（对于N = 100，约为90％，对于N = 10，约为80％ ）超出了解决方案键的答案（N = 100时为70％，N = 10时为60％）。我的代码/概念中有毛刺吗？我在实施Solve.QP时非常谨慎。我在每一步都添加了注释，以表明我打算做什么。
N= 10 (resp 100)
runs = 1 # Initializing the number of runs
count = 0 # Count of number of times Eout for SVM < Eout for PLA (SVM performs better)
sv = matrix(nrow = 1, ncol = 1000) # number of support vectors in each iteration
sv[,] = 0
while(runs<1001){
x <- runif(2, min = -1, max = 1) # Picking a random point in (-1, 1)
y <- runif(2, min = -1, max = 1)
fit = (lm(y~x))
t = summary(fit)$coefficients[,1]
f <- function(x){ # A random line in the plane
t[2]*x + t[1]
}
A = matrix(ncol=N, nrow=2) # Training data set, generated randomly, as required by question
b = matrix(ncol=N, nrow=1)
for(i in 1:N){
A[, i] <- c(runif(2, min = -1, max = 1))
b[1, i] <- sign(A[2, i] - f(A[1, i]))
}
if(length(range(b)) ==1){next} # Move to the next iteration if all training points lie on the same side of the line
w <- matrix(ncol=1, nrow=3) # Weight vector, to be calculated using PLA
w[, 1] = 0
g <- function(z){
t(w) %*% z
}
i = 1
while(i < N+1){ # Running the Perceptron Learning Algorithm
j = sample(1:N, 1)
if((sign(g(c(1, A[, j]))) == b[1, j]) == 0){
w = w + b[1, j]*c(1, A[, j])
}
i = i + 1
}
S = matrix(ncol=10000, nrow=2) # Testing data set, random points in XY plane
for(v in 1:10000){
S[, v] <- c(runif(2, min = -10000, max = 10000))
}
m = 0 # counter of Eout(PLA)
v = 1
while(v < 10001){
if(sign(g(c(1, S[, v]))) != sign(S[, v][2] - f(S[, v][1]))){
m = m + 1
}
v = v + 1
}
# SVM
require(quadprog)
# Implementing classification using the quadprog package, solve.QP command
Dmat <- matrix(nrow = N, ncol = N)
dvec <- matrix(nrow = N, ncol = 1)
Amat <- matrix(nrow= N+1, ncol = N)
Amat[,] = 0
bvec= matrix(nrow = N+1, ncol =1)
bvec[,1]=0
for(i in 1:N){
for(j in 1:N){
Dmat[i, j] = b[i]*b[j]*t(A[,i])%*%A[,j]
dvec[i,1] = 1
if(i>1 && i==j+1) {
Amat[i,j] <- 1
}
else Amat[i,j] = 0
bvec[i] = 0
}
}
Dmat = Dmat + 10^(-8)*diag(N) # Done to solve the problem of non-positive definite matrix
Amat[1,] = t(b)
Amat[N+1,N]=1 # Amat has its first row as the vector b (as one constraint is b^T*alpha = 0), and the rest as identity because we need all alpha_i's greater than or equal to 0
Amt = t(Amat)
sol <- solve.QP(Dmat, dvec, Amt, bvec, meq=1) #meq = 1 because the first inequality must be treated as an equality
l = as.vector(sol["solution"])
alpha=0
for(i in 1:N){
alpha[i]=as.numeric(rapply(l,c)[i]) # alpha records the obtained parameters
}
w1 = 0 # 2-column weight vector (excludes intercept term)
for(n in 1:N){
w1 = w1 + alpha[n]*b[n]*A[,n] # Formula for obtaining weights using alpha's
}
for(n0 in 1:N){
if(abs(alpha[n0]) > 10^(-1)){
break # Picking an index corresponding to an actual support vector
}
}
for(n in 1:N){
if(abs(alpha[n]) > 10^(-1) & alpha[n]>0){
sv[cntr] = sv[cntr]+1 # Support vector counter
}
}
w0 = (1-t(w1)%*%A[,n0]*b[n0])/b[n0] # Formula for first entry of final weight vector, corresponding to intercept term
w2 = c(w0,w1) # final weight vector
m1 = 0 # Eout counter for SVM
v1 = 1
g1 <- function(z){ #function: multiplication by the weight vector
t(w2) %*% z
}
while(v1 < 10001){ # Counting the number of errors
if(sign(g1(c(1, S[, v1]))) != sign(S[, v1][2] - f(S[, v1][1]))){
m1= m1 + 1
}
v1 = v1 + 1
}
if(m>m1){count = count + 1/1000} # If out-of-sample error m in PLA is higher than out-of-sample error in SVM, add to the count
runs = runs+1/1000
}
count*100 # corresponds to the percentage of times SVM performed better:
code yields about 90% for N =100 and about 80% for N = 10, but solution key says 70% for N = 100 and 60% for N = 10