coursera林轩田的《机器学习基石》很有意思，我把一些编程作业总结在这里，参考了mac Jiang的答案：https://blog.csdn.net/a1015553840/article/details/51085129：
作业1
15-17是naive pla(perceptron learning algorithm), 算法如下:
- 初始化w
repeat {
1.寻找w(t)的下一个错误分类点（x,y）（即sign（w(t)’*x）!=y）；
2.纠正错误：w(t+1) = w(t) + y * x；
} until(每个样本都无错) - 返回w
image.png
def naive_PLA():
updates = 0
w = np.zeros(5)
while True:
halt = True
with open('hw1_15_train.dat') as csvfile:
reader = csv.reader(csvfile, delimiter='\t')
for line in reader:
x = line[0].split()
x = np.asarray(x)
x = x.astype(np.float)
x = np.insert(x, 0, 1)
y = np.array(line[1], dtype='int')
if sign(w.dot(x)) != y:
updates += 1
w += y * x
halt = False
if halt:
break
return updates
最终计算结果是：45次
image.png
为了方便处理，定义了DataSet类和PLA类
class DataSet:
def __init__(self, filename):
self.input = []
self.output = []
self.load_data(filename)
def load_data(self, filename):
with open(filename) as csvfile:
reader = csv.reader(csvfile, delimiter='\t')
for line in reader:
x = line[0].split()
x.insert(0, '1')
y = line[1]
self.input.append(x)
self.output.append(y)
class PLA:
def __init__(self, train_name='hw1_15_train.dat', test_name=None):
self.train_set = DataSet(train_name)
if test_name:
self.test_set = DataSet(test_name)
def random_cycle_pla(self, times=2000, eta=1, print_out=False):
total_updates = 0
data_set = list(zip(self.train_set.input, self.train_set.output))
for idx, _ in enumerate(range(times)):
shuffle(data_set)
current_updates = self.naive_pla(data_set, eta, print_out)
total_updates += current_updates
return total_updates / times
def naive_pla(self, data_set=None, eta=1, print_out=False):
"""naive perceptron learning algorithm"""
current_updates = 0
w = np.zeros(5)
if not data_set:
data_set = list(zip(self.train_set.input, self.train_set.output))
while True:
halt = True
for item in data_set:
x = np.array(item[0], dtype=float)
y = np.array(item[1], dtype=int)
if sign(w.dot(x)) != y:
current_updates += 1
w += eta * y * x
halt = False
if halt:
break
if print_out:
print(f'第{idx}次终止次数：{current_updates}')
return current_updates
200次平均，次数为38.145次
image.png
对w做更新时乘以eta=0.5即可：w += eta * y * x
所得的结果是：200次平均，次数为40.245
18-20题是非线性可分的问题，用pocket PLA算法，算法如下：
-
初始化w，pocket_w
{1.寻找分类错误点（x,y）
2.修正错误：w(t+1) = w(t) + y*x
3.如果w(t+1)对训练样本的错误率比pocket_w更小，则用w(t+1)替代pocket_w
} until(达到足够的迭代次数)
- 返回pocket_w
该算法每次更新后都需要计算w的所有样本的错误率，因此计算量比naive pla大，好处是对线性不可分问题也有解
image.png
只要添加计算错误率的函数errors_count, 以及计算pocket_w的函数pocket_algorithm，最后计算出错误率即可。
def errors_count(self, w, data_set):
""""统计errors发生次数"""
count = 0
for x, y in data_set:
x = np.array(x, dtype=float)
y = np.array(y, dtype=int)
if sign(w.dot(x)) != y:
count += 1
return count
def pocket_algorithm(self, update_times=50, pocket=True):
"""pocket=True： 返回pocketWeight
否则返回w"""
data_set = list(zip(self.train_set.input, self.train_set.output))
updates = 0
w = np.zeros(5)
pocket_weight = np.zeros(45)
min_errors = sys.maxsize
halt = False
while not halt:
shuffle(data_set)
for item in data_set:
x = np.array(item[0], dtype=float)
y = np.array(item[1], dtype=int)
if sign(w.dot(x)) != y:
w = w + y * x # w每次都更新
updates += 1
# print(f'updates: {updates}')
errors_count = self.errors_count(w, data_set)
if errors_count < min_errors:
min_errors = errors_count
pocket_weight = w # pocket_weight只有遇到更好的w才更新
if updates >= update_times or min_errors == 0:
halt = True
break
return pocket_weight if pocket else w
def cal_test_error_rate(self, update_times=50, times=2000, pocket=True):
train_set = list(zip(self.train_set.input, self.train_set.output))
test_set = list(zip(self.test_set.input, self.test_set.output))
train_avg_rate, test_avg_rate = 0, 0
for _ in range(times):
w = self.pocket_algorithm(update_times=update_times, pocket=pocket)
train_error_counts = self.errors_count(w, train_set)
test_error_counts = self.errors_count(w, test_set)
train_avg_rate += train_error_counts / len(train_set)
test_avg_rate += test_error_counts / len(test_set)
return train_avg_rate / times, test_avg_rate / times
2000次的平均值是：0.13131999999999994
image.png
pocket_algorithm返回w，而非最有权重w_pocket，200次的平均错误率是：0.3798
image.png
将pocket algorithm的更新次数从50改为100即可，错误率有略微的下降，计算200次的平均值为：0.11630000000000004
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