如何绘制三种判别函数的决策边界？

问题描述

这是我的代码，无需使用机器学习库即可使用LDA和QDA。

STrainmatrix = SynTraining.values
STestmatrix = SynTest.values
# read in datasets
Xtrain = STrainmatrix[:,:-1]
ytrain = STrainmatrix[:,-1].astype(int)
Xtest = STestmatrix[:,:-1]
ytest = STestmatrix[:,-1].astype(int)
nte,nf = Xtest.shape        # get the dimension of testing samples
                            # nte is the number of testing samples
                            # nf is the number of features/dimensions
Pw0 = Pw1 = 0.5

# training process -- derive the model
# also measure the runtime including both training and testing time
t0 = time.time()           # start time
arr = Xtrain[ytrain == 0]       # get training samples for class 0
covs0 = np.cov(np.transpose(arr))    # calculate Sigma
means0 = np.mean(arr,axis=0)        # mean along rows
arr = Xtrain[ytrain == 1]       # get training samples for class 1
covs1 = np.cov(np.transpose(arr))    # calculate Sigma
means1 = np.mean(arr,axis=0)        # mean along rows
covavg = (covs0 + covs1) / 2               # average of Sigmas
varavg = np.sum(np.diagonal(covavg)) / nf  # average of variances
meanavg = (means0 + means1) /2
# testing process -- predict
y = []                # save the classification label of testing samples
for i in range(nte):
    g0 = -(np.dot(Xtest[i]-means0,Xtest[i]-means0)) / (2 * varavg) + np.log(Pw0)
    g1 = -(np.dot(Xtest[i]-means1,Xtest[i]-means1)) / (2 * varavg) + np.log(Pw1)
    if g0 >= g1:
        y.append(0)
    else:
        y.append(1)
t1 = time.time()           # ending time
# calculate overall classification accuracy
acc1 = np.count_nonzero(y == ytest) / len(y)
run1 = t1-t0
print(f'The overall accuracy is: {acc1},finished in {run1} seconds') 


# training process -- derive the model
# also measure the runtime including both training and testing time
t2 = time.time()           # start time
arr = Xtrain[ytrain == 0]       # get training samples for class 0
covs0 = np.cov(np.transpose(arr))    # calculate Sigma
means0 = np.mean(arr,axis=0)        # mean along rows
covavg = (covs0 + covs1) / 2               # average of Sigmas
varavg = np.sum(np.diagonal(covavg)) / nf  # average of variances 
meanavg = (means0 + means1) /2

# testing process -- predict
y1 = []                # save the classification label of testing samples
for i in range(nte):
    g0 = -((np.dot(np.dot(Xtest[i]-means0,inv(covs0)),(Xtest[i]-means0).T)) /(2)) + np.log(Pw0)
    g1 = -((np.dot(np.dot(Xtest[i]-means1,inv(covs1)),(Xtest[i]-means1).T)) /(2)) + np.log(Pw1)
    if g0 >= g1:
        y1.append(0)
    else:
        y1.append(1)
t3 = time.time()           # ending time
# calculate overall classification accuracy
acc2 = np.count_nonzero(y1 == ytest) / len(y1)
run2 = t3-t2
print(f'The overall accuracy is: {acc2},finished in {run2} seconds') 

# training process -- derive the model
# also measure the runtime including both training and testing time
t4 = time.time()           # start time
arr = Xtrain[ytrain == 0]       # get training samples for class 0
covs0 = np.cov(np.transpose(arr))    # calculate Sigma
means0 = np.mean(arr,axis=0)        # mean along rows
covavg = (covs0 + covs1) / 2               # average of Sigmas
varavg = np.sum(np.diagonal(covavg)) / nf  # average of variances   
meanavg = (means0 + means1) /2

# testing process -- predict
y2 = []                # save the classification label of testing samples
for i in range(nte):
    g0 = -((np.dot(np.dot(Xtest[i]- means0,(Xtest[i]- means0).T))/(2)) - (np.log(det(inv(covs0))))/2 + np.log(Pw0)
    g1 = -((np.dot(np.dot(Xtest[i]- means1,(Xtest[i]- means1).T))/(2)) - (np.log(det(inv(covs1))))/2 + np.log(Pw1)
    if g0 >= g1:
        y2.append(0)
    else:
        y2.append(1)
t5 = time.time()           # ending time
# calculate overall classification accuracy
acc3 = np.count_nonzero(y2 == ytest) / len(y2)
run3 = t5-t4
print(f'The overall accuracy is: {acc3},finished in {run3} seconds')

这是执行算法并为我提供准确性和运行时间的代码。如何绘制这三种情况的决策边界？我也试图找到类的准确性，但是我不明白这意味着什么以及我将如何做到这一点。下面是我的尝试。

解决方法

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