三个臭皮匠——集成学习AdaBoost简单代码实现示例

转载内容(引子)

第一个学习器

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import numpy as np
import pandas as pddata = {'x':[i for i in range(10)], 'y':[1, 1, 1, -1, -1, -1, 1, 1, 1, -1]}
df = pd.DataFrame(data, columns=['x', 'y'])
df['w'] = 0.1
df

封装函数(python)

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正函数（小于分界点为1，大于分界点为-1，reverse = True）
负函数（小于分界点为-1，大于分界点为1，reverse = False）

def find_min(df, reverse=True):# 正函数g_p = lambda x,s:int(x < s) * 2 - 1  # 满足时1 * 2 - 1 = 1；不满足时0 * 2 - 1 = -1# 负函数g_n = lambda x,s:int(x > s) * 2 - 1g = g_p if reverse else g_nx = df['x']e_idx = 1  # 误差权重初值x_idx = 0  # 误差权重初值对应分界点   for i in x:df_temp = df.copy(deep=True)df_temp['e'] = 0     # 将['e']属性先统一赋值0df_temp['G'] = df_temp['x'].apply(lambda x: g(x, i - 0.5))df_temp.loc[df_temp['y'] != df_temp['G'], ['e']] = df_temp['w']  # 当['y']属性与['G']属性不一致时，['e']为其权重e = df_temp['e'].sum()  # 误差(['y']与['G']不一致)权重求和if e < e_idx:           # 找最小误差权重及其对应分界点e_idx, x_idx = e, (i - 0.5)return g, round(e_idx, 3), round(x_idx, 1)

def find_func(df):g_p, e_p, s_p = find_min(df)  # 正函数返回误差及对应分界点g_n, e_n, s_n = find_min(df, reverse = False)  # 负函数返回误差及对应分界点if e_p < e_n:  # 选择误差较小的正函数(或负函数)return g_p, e_p, s_preturn g_n, e_n, s_n

def alpha(df):g, e, split = find_func(df)alpha = round((1.0 / 2) * (np.log(1 - e) - np.log(e)), 4)df['G'] = df['x'].apply(lambda x: g(x, split))df['w'] = round(df['w']* np.exp(-alpha * df['y'] * df['G']), 4)df['w'] = round(df['w'] / (df['w'].sum()), 4)return g, split, alpha

g1, split1, alpha1 = alpha(df)
alpha1

0.4236

func =lambda x:np.sign(alpha1 * g1(x, split1))
df['G'] = df['x'].apply(lambda x: int(func(x)))
df

第一轮误差

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df.loc[df['y'] != df['G'], ['x']].count()

x    3
dtype: int64

第二个学习器

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g2, split2, alpha2 = alpha(df)
alpha2

0.6505

func =lambda x:np.sign(alpha1 * g1(x, split1) + alpha2 * g2(x, split2))
df['G'] = df['x'].apply(lambda x: int(func(x)))
df

第二轮误差

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df.loc[df['y'] != df['G'], ['x']].count()

x    3
dtype: int64

第三个学习器

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g3, split3, alpha3 = alpha(df)
alpha3

0.7514

func =lambda x:np.sign(alpha1 * g1(x, split1) + alpha2 * g2(x, split2) + alpha3 * g3(x, split3))
df['G'] = df['x'].apply(lambda x: int(func(x)))
df

第三轮误差

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df.loc[df['y'] != df['G'], ['x']].count()

x    0
dtype: int64

总结

“三个臭皮匠抵过一个诸葛亮”，多个弱分类器（准确率大于50%）串起来（协作）能达到很好的准确度。

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