【机器学习】python机器学习使用scikit-learn对模型进行评估：使用t分布及z分布评估模型误差的95%置信空间

2、当总体标准差未知，需要估计，用t检验。当n>>30，z检验和t检验结果相近，以t检验为准。但是z检验比较好计算，就在大样本时替代t。

数据准备：

import os

HOUSING_PATH = os.path.join("datasets", "housing")
import pandas as pd

def load_housing_data(housing_path=HOUSING_PATH):
    csv_path = os.path.join(housing_path, "housing.csv")
    return pd.read_csv(csv_path)
housing=load_housing_data()
housing2=housing.copy()
import numpy as np

#预处理前去掉带文字的指定列
from sklearn.preprocessing import MinMaxScaler,StandardScaler,OneHotEncoder

from sklearn.impute import SimpleImputer
housing_num = housing2.drop("ocean_proximity", axis=1)
from sklearn.preprocessing import StandardScaler
from sklearn.pipeline import Pipeline
num_pipeline = Pipeline([
        ('imputer', SimpleImputer(strategy="median")),
        ('std_scaler', StandardScaler())
    ])



from sklearn.compose import ColumnTransformer
#返回所有列名
num_attribs = list(housing_num)

cat_attribs = ["ocean_proximity"]
#找出待独热编码列的最大分类数，不然在进行测试集划分处理时，
#容易造成独热向量因测试集构成不同而列数不一致的情况
categories=housing2['ocean_proximity'].unique()
full_pipeline = ColumnTransformer([
        ("num", num_pipeline, num_attribs),
        ("cat", OneHotEncoder(categories=[categories]), cat_attribs),
    ])
#抽样后的数据，去除预测目标列，并拿出对应目标列准备数据训练
housing_labels = housing2["median_house_value"]

housing_prepared = full_pipeline.fit_transform(housing2)
from sklearn.model_selection import train_test_split
train_set, test_set,train_sety, test_sety = train_test_split(housing_prepared,housing_labels, test_size=0.1, random_state=42)
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用随机森林预测及使用t分布及z分布计算误差的95%置信区间：

from sklearn.metrics import mean_squared_error
from sklearn.ensemble import RandomForestRegressor

from sklearn.model_selection import RandomizedSearchCV
from scipy.stats import randint

forest_reg = RandomForestRegressor(random_state=200,n_estimators=10,max_features=12)
#每个迭代探索一种参数组合
forest_reg.fit(train_set, train_sety)
pre=forest_reg.predict(test_set)
print(np.sqrt(mean_squared_error(test_sety,pre)))
from scipy import stats
#用t分布计算误差95%的置信区间
def tscore(y,pre,confidence = 0.95):
    confidence = 0.95
    squared_errors = (pre - y) ** 2
    return np.sqrt(stats.t.interval(confidence, len(squared_errors) - 1,
                             loc=squared_errors.mean(),
                             scale=stats.sem(squared_errors)))
#有0.95的概率误差落在如下范围内：
print(tscore(test_sety,pre,confidence = 0.95))
#使用z分布计算
def zscore(y,pre,confidence = 0.95):
    confidence = 0.95
    squared_errors = (pre - y) ** 2
    m = len(squared_errors)
    mean = squared_errors.mean()
    zscore = stats.norm.ppf((1 + confidence) / 2)
    zmargin = zscore * squared_errors.std(ddof=1) / np.sqrt(m)
    return [np.sqrt(mean - zmargin), np.sqrt(mean + zmargin)]
print(zscore(test_sety,pre,confidence = 0.95))
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输出结果：
均方根误差：432.29251756337266
t分数置信区间：[187.94412423 581.74792449]
z分数置信区间：[188.18052438939552, 581.671498118216]

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原文地址：https://blog.csdn.net/hh1357102/article/details/126357927