如何修复错误的精度和 R 方？

Question

我尝试通过分析数据文件 Google Apps Store 来预测评级来练习线性回归，文件 csv 在 Kaggle 上。

清理并尝试应用KNeighborsRegressor运行模型后，结果，准确率和r-squared太低，我不知道为什么。
然而，预测和 y 检验之间的差异并不大，而且 MSE 非常低。

我觉得这里有一些错误，希望你能帮我改正。我希望准确率达到 90% 左右。

import re
import sys

import time
import datetime

import numpy as np
import pandas as pd

import seaborn as sns
import matplotlib.pyplot as plt

from sklearn import metrics
from sklearn import preprocessing
from sklearn.neighbors import KNeighborsRegressor
from sklearn.ensemble import RandomForestRegressor
from sklearn.model_selection import train_test_split
df = pd.read_csv('googleplaystore.csv')

df['Rating'] = df['Rating'].fillna(df['Rating'].median())


replaces = [u'\u00AE', u'\u2013', u'\u00C3', u'\u00E3', u'\u00B3', '[', ']', "'"]
for i in replaces:
    df['Current Ver'] = df['Current Ver'].astype(str).apply(lambda x : x.replace(i, ''))

regex = [r'[-+|/:/;(_)@]', r'\s+', r'[A-Za-z]+']
for j in regex:
    df['Current Ver'] = df['Current Ver'].astype(str).apply(lambda x : re.sub(j, '0', x))

df['Current Ver'] = df['Current Ver'].astype(str).apply(lambda x : x.replace('.', ',',1).replace('.', '').replace(',', '.',1)).astype(float)
df['Current Ver'] = df['Current Ver'].fillna(df['Current Ver'].median())
 df.drop([10472], axis = 0, inplace = True)
le = preprocessing.LabelEncoder()
df['App'] = le.fit_transform(df['App'])
category_list = df['Category'].unique().tolist() 
category_list = ['cat_' + word for word in category_list]
df = pd.concat([df, pd.get_dummies(df['Category'], prefix='cat')], axis=1)
df['Genres'] = df['Genres'].str.split(';').str[0]
df['Genres'].replace('Music & Audio', 'Music', inplace =True)
le = preprocessing.LabelEncoder()
df['Genres'] = le.fit_transform(df['Genres'])
le = preprocessing.LabelEncoder()
df['Content Rating'] = le.fit_transform(df['Content Rating'])
df['Price'] = df['Price'].apply(lambda x : x.strip('$'))
df['Installs'] = df['Installs'].apply(lambda x : x.strip('+').replace(',', ''))
df['Type'] = pd.get_dummies(df['Type'])

def change_size(size):
    if 'M' in size:
        x = size[:-1]
        x = float(x)*1000000
        return(x)
    elif 'k' == size[-1:]:
        x = size[:-1]
        x = float(x)*1000
        return(x)
    else:
        return None

df['Size'] = df['Size'].apply(change_size)
df['Size'] = df['Size'].fillna(value=df['Size'].median(), axis = 0)
df['new'] = pd.to_datetime(df['Last Updated'])
df['lastupdate'] = (df['new'] - df['new'].max()).dt.days
features = ['App', 'Reviews', 'Size', 'Installs', 'Type', 'Price', 'lastupdate','Content Rating', 'Genres', 'Current Ver']
features.extend(category_list)
X = df[features]
y = df['Rating']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.3, random_state = 101)
from sklearn.preprocessing import StandardScaler
sc_X = StandardScaler()
X_train = sc_X.fit_transform(X_train)
X_test = sc_X.transform(X_test)
model = KNeighborsRegressor(n_neighbors=28)
predictions = model.predict(X_test)
model.fit(X_train, y_train)

accuracy = model.score(X_test,y_test)
'Accuracy: ' + str(np.round(accuracy*100, 2)) + '%'


from sklearn import metrics

print('MAE:', metrics.mean_absolute_error(y_test, predictions))
print('MSE:', metrics.mean_squared_error(y_test, predictions))
print('RMSE:', np.sqrt(metrics.mean_squared_error(y_test, predictions)))

result = pd.DataFrame({'Actual': y_test, 'Predicted': predictions}) 
result

Answer 1

此处测量准确性的方式是与评级完全匹配（如在分类器中）。例如，如果实际评分为 4.3，预测为 4.3001，则将其计为错误。但是，这里所做的是回归，在这种情况下，MSE 和 MAE 是更好的指标。

您可以尝试将y_test 和predictions 分桶以了解“准确性”，如下所示：

>>> metrics.accuracy_score(np.around(y_test), np.around(predictions))
0.7666051660516605